# 12. Processing Data in Groups#

The open-access textbook Minimalist Data Wrangling with Python by Marek Gagolewski is, and will remain, freely available for everyone’s enjoyment (also in PDF; a printed version can be ordered from Amazon: AU CA DE ES FR IT JP NL PL SE UK US). It is a non-profit project. Although available online, it is a whole course; it should be read from the beginning to the end. Refer to the Preface for general introductory remarks. Any bug/typo reports/fixes are appreciated. Also, make sure to check out my other book, Deep R Programming [34].

Let us consider another subset of the US Centres for Disease Control and Prevention National Health and Nutrition Examination Survey, this time carrying some body measures (P_BMX) together with demographics (P_DEMO).

"teaching-data/master/marek/nhanes_p_demo_bmx_2020.csv",
comment="#")
nhanes = (
nhanes
.loc[
(nhanes.DMDBORN4 <= 2) & (nhanes.RIDAGEYR >= 18),
["RIDAGEYR", "BMXWT", "BMXHT", "BMXBMI", "RIAGENDR", "DMDBORN4"]
]  # age >= 18 and only US and non-US born
.rename({
"RIDAGEYR": "age",
"BMXWT": "weight",
"BMXHT": "height",
"BMXBMI": "bmival",
"RIAGENDR": "gender",
"DMDBORN4": "usborn"
}, axis=1)  # rename columns
.dropna()   # remove missing values
.reset_index(drop=True)
)

We consider only the adult (at least 18 years old) participants, whose country of birth (the US or not) is well-defined. Let us recode the usborn and gender variables (for readability) and introduce the BMI categories:

nhanes.loc[:, "usborn"] = (
nhanes.usborn.astype("category")
.cat.rename_categories(["yes", "no"]).astype("str")       # recode usborn
)
nhanes.loc[:, "gender"] = (
nhanes.gender.astype("category")
.cat.rename_categories(["male", "female"]).astype("str")  # recode gender
)
nhanes.loc[:, "bmicat"] = pd.cut(                             # new column
nhanes.bmival,
bins=  [ 0,             18.5,        25,            30,       np.inf ],
labels=[   "underweight",    "normal",  "overweight",  "obese"       ]
)

Here is a preview of this data frame:

##    age  weight  height  bmival  gender usborn      bmicat
## 0   29    97.1   160.2    37.8  female     no       obese
## 1   49    98.8   182.3    29.7    male    yes  overweight
## 2   36    74.3   184.2    21.9    male    yes      normal
## 3   68   103.7   185.3    30.2    male    yes       obese
## 4   76    83.3   177.1    26.6    male    yes  overweight

We have a mix of categorical (gender, US born-ness, BMI category) and numerical (age, weight, height, BMI) variables. Unless we had encoded qualitative variables as integers, this would not be possible with plain matrices, at least not with a single one.

In this section, we will treat the categorical columns as grouping variables, so that we can, e.g., summarise or visualise the data in each group separately, because it is likely that data distributions vary across different factor levels. This is much like having many data frames stored in one object, e.g., the heights of women and men separately.

nhanes is thus an example of heterogeneous data at their best.

## 12.1. Basic Methods#

DataFrame and Series objects are equipped with the groupby methods, which assist in performing a wide range of operations in data groups defined by one or more data frame columns (compare [92]).

They return objects of class DataFrameGroupBy and SeriesGroupby:

type(nhanes.groupby("gender"))
## <class 'pandas.core.groupby.generic.DataFrameGroupBy'>
type(nhanes.groupby("gender").height)  # or (...)["height"]
## <class 'pandas.core.groupby.generic.SeriesGroupBy'>

Important

When we wish to browse the list of available attributes in the pandas manual, it is worth knowing that DataFrameGroupBy and SeriesGroupBy are separate types. Still, they have many methods and slots in common, because they both inherit from (extend) the GroupBy class.

Exercise 12.1

Skim through the documentation of the said classes.

For example, the pandas.DataFrameGroupBy.size method determines the number of observations in each group:

nhanes.groupby("gender").size()
## gender
## female    4514
## male      4271
## dtype: int64

It returns an object of type Series. We can also perform the grouping with respect to a combination of levels in two qualitative columns:

nhanes.groupby(["gender", "bmicat"]).size()
## gender  bmicat
## female  underweight      93
##         normal         1161
##         overweight     1245
##         obese          2015
## male    underweight      65
##         normal         1074
##         overweight     1513
##         obese          1619
## dtype: int64

This yields a Series with a hierarchical index (as discussed in Section 10.1.3). Nevertheless, we can always call reset_index to convert it to standalone columns:

nhanes.groupby(["gender", "bmicat"]).size().rename("counts").reset_index()
##    gender       bmicat  counts
## 0  female  underweight      93
## 1  female       normal    1161
## 2  female   overweight    1245
## 3  female        obese    2015
## 4    male  underweight      65
## 5    male       normal    1074
## 6    male   overweight    1513
## 7    male        obese    1619

Take note of the rename part. It gave us some readable column names.

Furthermore, it is possible to group rows in a data frame using a list of any Series objects, i.e., not just column names in a given data frame; see Section 16.2.3 for an example.

Exercise 12.2

(*) Note the difference between pandas.GroupBy.count and pandas.GroupBy.size methods (by reading their documentation).

### 12.1.1. Aggregating Data in Groups#

The DataFrameGroupBy and SeriesGroupBy classes are equipped with several well-known aggregation functions. For example:

nhanes.groupby("gender").mean(numeric_only=True).reset_index()
##    gender        age     weight      height     bmival
## 0  female  48.956580  78.351839  160.089189  30.489189
## 1    male  49.653477  88.589932  173.759541  29.243620

The arithmetic mean was computed only on numeric columns[1]. Further, a few common aggregates are generated by describe:

nhanes.groupby("gender").height.describe().reset_index()
##    gender   count        mean       std  ...    25%    50%    75%    max
## 0  female  4514.0  160.089189  7.035483  ...  155.3  160.0  164.8  189.3
## 1    male  4271.0  173.759541  7.702224  ...  168.5  173.8  178.9  199.6
##
## [2 rows x 9 columns]

But we can always apply a custom list of functions by using aggregate:

(nhanes.
loc[:, ["gender", "height", "weight"]].
groupby("gender").
aggregate([np.mean, np.median, len, lambda x: (np.max(x)+np.min(x))/2]).
reset_index()
)
##    gender      height               ...     weight
##                  mean median   len  ...       mean median   len <lambda_0>
## 0  female  160.089189  160.0  4514  ...  78.351839   74.1  4514     143.45
## 1    male  173.759541  173.8  4271  ...  88.589932   85.0  4271     139.70
##
## [2 rows x 9 columns]

The result’s columns slot features a hierarchical index.

Note

The column names in the output object are generated by reading the applied functions’ __name__ slots, see, e.g., print(np.mean.__name__).

mr = lambda x: (np.max(x)+np.min(x))/2
mr.__name__ = "midrange"
(nhanes.
loc[:, ["gender", "height", "weight"]].
groupby("gender").
aggregate([np.mean, mr]).
reset_index()
)
##    gender      height              weight
##                  mean midrange       mean midrange
## 0  female  160.089189    160.2  78.351839   143.45
## 1    male  173.759541    172.1  88.589932   139.70

### 12.1.2. Transforming Data in Groups#

We can easily transform individual columns relative to different data groups by means of the transform method for GroupBy objects.

def std0(x, axis=None):
return np.std(x, axis=axis, ddof=0)
std0.__name__ = "std0"

def standardise(x):
return (x-np.mean(x, axis=0))/std0(x, axis=0)

nhanes["height_std"] = (
nhanes.
loc[:, ["height", "gender"]].
groupby("gender").
transform(standardise)
)

##    age  weight  height  bmival  gender usborn      bmicat  height_std
## 0   29    97.1   160.2    37.8  female     no       obese    0.015752
## 1   49    98.8   182.3    29.7    male    yes  overweight    1.108960
## 2   36    74.3   184.2    21.9    male    yes      normal    1.355671
## 3   68   103.7   185.3    30.2    male    yes       obese    1.498504
## 4   76    83.3   177.1    26.6    male    yes  overweight    0.433751

The new column gives the relative z-scores: a woman with a relative z-score of 0 has height of 160.1 cm, whereas a man with the same z-score has height of 173.8 cm.

We can check that the means and standard deviations in both groups are equal to 0 and 1:

(nhanes.
loc[:, ["gender", "height", "height_std"]].
groupby("gender").
aggregate([np.mean, std0])
)
##             height              height_std
##               mean      std0          mean std0
## gender
## female  160.089189  7.034703 -1.351747e-15  1.0
## male    173.759541  7.701323  3.145329e-16  1.0

Interestingly, it is likely a bug in pandas that groupby("gender").aggregate([np.std]) somewhat passes ddof=1 to numpy.std, hence our using a custom function.

Exercise 12.3

Create a data frame comprised of the five tallest men and the five tallest women.

### 12.1.3. Manual Splitting into Subgroups (*)#

It turns out that GroupBy objects and their derivatives are iterable; compare Section 3.4. As a consequence, the grouped data frames and series can be easily processed manually in case where the built-in methods are insufficient (i.e., not so rarely).

Let us consider a small sample of our data frame.

.loc[:, ["gender", "weight", "height"]].groupby("gender")
)
list(grouped)
## [('female',    gender  weight  height
## 0  female    97.1   160.2), ('male',   gender  weight  height
## 1   male    98.8   182.3
## 2   male    74.3   184.2
## 3   male   103.7   185.3
## 4   male    83.3   177.1)]

The way Python formatted the above output is imperfect, so we need to contemplate it for a tick. We see that when iterating through a GroupBy object, we get access to pairs giving all the levels of the grouping variable and the subsets of the input data frame corresponding to these categories.

Here is a simple example where we make use of the above fact:

for level, df in grouped:
# level is a string label
# df is a data frame - we can do whatever we want
print(f"There are {df.shape[0]} subject(s) with gender={level}.")
## There are 1 subject(s) with gender=female.
## There are 4 subject(s) with gender=male.

We see that splitting followed by manual processing of the chunks in a loop is quite tedious in the case where we would merely like to compute some basic aggregates. These scenarios are extremely common. No wonder why the pandas developers introduced a convenient interface in the form of the pandas.DataFrame.groupby and pandas.Series.groupby methods and the DataFrameGroupBy and SeriesGroupby classes. Still, for more ambitious tasks, the low-level way to perform the splitting will come in handy.

Exercise 12.4

(**) Using the manual splitting and matplotlib.pyplot.boxplot, draw a box-and-whisker plot of heights grouped by BMI category (four boxes side by side).

Exercise 12.5

(**) Using the manual splitting, compute the relative z-scores of the height column separately for each BMI category.

Example 12.6

Let us also demonstrate that the splitting can be done manually without the use of pandas. Namely, calling numpy.split(a, ind) returns a list with a (being an array-like object, e.g., a vector, a matrix, or a data frame) partitioned rowwisely into len(ind)+1 chunks at indexes given by ind. For example:

a = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]
for e in np.split(a, [2, 6]):
print(repr(e))
## array(['one', 'two'], dtype='<U5')
## array(['three', 'four', 'five', 'six'], dtype='<U5')
## array(['seven', 'eight', 'nine'], dtype='<U5')

To split a data frame into groups defined by a categorical column, we can first sort it with respect to the criterion of interest, for instance, the gender data:

nhanes_srt = nhanes.sort_values("gender", kind="stable")

Then, we can use numpy.unique to fetch the indexes of first occurrences of each series of identical labels:

levels, where = np.unique(nhanes_srt.gender, return_index=True)
levels, where
## (array(['female', 'male'], dtype=object), array([   0, 4514]))

This can now be used for dividing the sorted data frame into chunks:

nhanes_grp = np.split(nhanes_srt, where[1:])  # where[0] is not interesting

We obtained a list of data frames split at rows specified by where[1:]. Here is a preview of the first and the last row in each chunk:

for i in range(len(levels)):
# process (levels[i], nhanes_grp[i])
print(f"level='{levels[i]}'; preview:")
print(nhanes_grp[i].iloc[ [0, -1], : ], end="\n\n")
## level='female'; preview:
##       age  weight  height  bmival  gender usborn bmicat  height_std
## 0      29    97.1   160.2    37.8  female     no  obese    0.015752
## 8781   67    82.8   147.8    37.9  female     no  obese   -1.746938
##
## level='male'; preview:
##       age  weight  height  bmival gender usborn      bmicat  height_std
## 1      49    98.8   182.3    29.7   male    yes  overweight    1.108960
## 8784   74    59.7   167.5    21.3   male     no      normal   -0.812788

Within each subgroup, we can apply any operation we have learnt so far: our imagination is the only major limiting factor. For instance, we can aggregate some columns:

nhanes_agg = [
dict(
level=t.gender.iloc[0],  # they are all the same here – take first
height_mean=np.round(np.mean(t.height), 2),
weight_mean=np.round(np.mean(t.weight), 2)
)
for t in nhanes_grp
]
print(nhanes_agg[0])
## {'level': 'female', 'height_mean': 160.09, 'weight_mean': 78.35}
print(nhanes_agg[1])
## {'level': 'male', 'height_mean': 173.76, 'weight_mean': 88.59}

The resulting list of dictionaries can be combined to form a data frame:

pd.DataFrame(nhanes_agg)
##     level  height_mean  weight_mean
## 0  female       160.09        78.35
## 1    male       173.76        88.59

Furthermore, a simple trick to allow grouping with respect to more than one column is to apply numpy.unique on a string vector that combines the levels of the grouping variables, e.g., by concatenating them like nhanes_srt.gender + "___" + nhanes_srt.bmicat (assuming that nhanes_srt is ordered with respect to these two criteria).

## 12.2. Plotting Data in Groups#

The seaborn package is particularly convenient for plotting grouped data – it is highly interoperable with pandas.

### 12.2.1. Series of Box Plots#

Figure 12.1 depicts a box plot with four boxes side by side:

sns.boxplot(x="bmival", y="gender", hue="usborn",
data=nhanes, palette="Paired")
plt.show()

Let us contemplate for a while how easy it is now to compare the BMI distribution in different groups. Here, we have two grouping variables, as specified by the y and hue arguments.

Exercise 12.7

Create a similar series of violin plots.

Exercise 12.8

(*) Add the average BMIs in each group to the above box plot using matplotlib.pyplot.plot. Check ylim to determine the range on the y-axis.

### 12.2.2. Series of Bar Plots#

In Figure 12.2, on the other hand, we have a bar plot representing a contingency table (obtained in a different way than in Chapter 11):

sns.barplot(
y="counts", x="gender", hue="bmicat", palette="Paired",
data=(
nhanes.
groupby(["gender", "bmicat"]).
size().
rename("counts").
reset_index()
)
)
plt.show()
Exercise 12.9

Draw a similar bar plot where the bar heights sum to 100% for each gender.

Exercise 12.10

Using the two-sample chi-squared test, verify whether the BMI category distributions for men and women differ significantly from each other.

### 12.2.3. Semitransparent Histograms#

Figure 12.3 illustrates that playing with semitransparent objects can make comparisons easy. By passing common_norm=False, we scaled each density histogram separately, which is the behaviour we desire if samples are of different lengths.

sns.histplot(data=nhanes, x="weight", hue="usborn",
element="step", stat="density", common_norm=False)
plt.show()

### 12.2.4. Scatterplots with Group Information#

Scatterplots for grouped data can display category information using points of different colours and/or styles, compare Figure 12.4.

sns.scatterplot(x="height", y="weight", hue="gender", style="gender",
data=nhanes, alpha=0.2, markers=["o", "v"])
plt.show()

### 12.2.5. Grid (Trellis) Plots#

Grid plot (also known as trellis, panel, or lattice plots) are a way to visualise data separately for each factor level. All the plots share the same coordinate ranges which makes them easily comparable. For instance, Figure 12.5 depicts a series of histograms of weights grouped by a combination of two categorical variables.

grid = sns.FacetGrid(nhanes, col="gender", row="usborn")
grid.map(sns.histplot, "weight", stat="density", color="lightgray")
# plt.show()  # not required...
Exercise 12.11

Important

Grid plots can bear any kind of data visualisation we have discussed so far (e.g., histograms, bar plots, scatterplots).

Exercise 12.12

Draw a trellis plot with scatterplots of weight vs height grouped by BMI category and gender.

### 12.2.6. Comparing ECDFs with the Kolmogorov–Smirnov Test (*)#

Figure 12.6 compares the empirical cumulative distribution functions of the weight distributions for US and non-US born participants.

for usborn, weight in nhanes.groupby("usborn").weight:
sns.ecdfplot(data=weight, legend=False, label=usborn)
plt.legend(title="usborn")
plt.show()

We have used manual splitting of the weight column into subgroups and then plotted the two ECDFs separately, because a call to seaborn.ecdfplot(data=nhanes, x="weight", hue="usborn") does not honour our wish to use alternating lines styles (most likely due to a bug).

A two-sample Kolmogorov–Smirnov test can be used to check whether two ECDFs $$\hat{F}_n'$$ (e.g., the weight of the US-born participants) and $$\hat{F}_m''$$ (e.g., the weight of non-US-born persons) are significantly different from each other:

$\begin{split} \left\{ \begin{array}{rll} H_0: & \hat{F}_n' = \hat{F}_n'' & \text{(null hypothesis)}\\ H_1: & \hat{F}_n' \neq \hat{F}_n'' & \text{(two-sided alternative)} \\ \end{array} \right. \end{split}$

The test statistic will be a variation of the one-sample setting discussed in Section 6.2.3. Namely, let:

$\hat{D}_{n,m} = \sup_{t\in\mathbb{R}} | \hat{F}_n'(t) - \hat{F}_m''(t) |.$

Computing the above is slightly trickier than in the previous case[2], but luckily an appropriate procedure is already implemented in scipy.stats:

x12 = nhanes.set_index("usborn").weight
x1 = x12.loc["yes"]  # first sample
x2 = x12.loc["no"]   # second sample
Dnm = scipy.stats.ks_2samp(x1, x2)[0]
Dnm
## 0.22068075889911914

Assuming significance level $$\alpha=0.001$$, the critical value is approximately (for larger $$n$$ and $$m$$) equal to:

$K_{n,m} = \sqrt{ -\frac{ \log(\alpha/2) (n+m) }{ 2nm } }.$
alpha = 0.001
np.sqrt(-np.log(alpha/2) * (len(x1)+len(x2)) / (2*len(x1)*len(x2)))
## 0.04607410479813944

As usual, we reject the null hypothesis when $$\hat{D}_{n,m}\ge K_{n,m}$$, which is exactly the case here (at significance level $$0.1\%$$). In other words, weights of US- and non-US-born participants differ significantly.

Important

Frequentist hypothesis testing only takes into account the deviation between distributions that is explainable due to sampling effects (the assumed randomness of the data generation process). For large sample sizes, even very small deviations[3] will be deemed statistically significant, but it does not mean that we should consider them as practically significant. For instance, if a very costly, environmentally unfriendly, and generally inconvenient for everyone upgrade leads to a process’ improvement such that we reject the null hypothesis stating that two distributions are equal, but it turns out that the gains are ca. 0.5%, the good old common sense should be applied.

Exercise 12.13

Compare between the ECDFs of weights of men and women who are between 18 and 25 years old. Determine whether they are significantly different.

Important

Some statistical textbooks and many research papers in the social sciences (amongst many others) employ the significance level of $$\alpha=5\%$$, which is often criticised as too high[4]. Many stakeholders aggressively push towards constant improvements in terms of inventing bigger, better, faster, more efficient things. In this context, larger $$\alpha$$ allows for generating more sensational discoveries. This is because it considers smaller differences as already significant. This all adds to what we call the reproducibility crisis in the empirical sciences.

We, on the other hand, claim that it is better to err on the side of being cautious. This, in the long run, is more sustainable.

### 12.2.7. Comparing Quantiles#

Plotting quantiles in two samples against each other can also give us some further (informal) insight with regard to the possible distributional differences. Figure 12.7 depicts an example Q-Q plot (see also the one-sample version in Section 6.2.2), where we see that the distributions have similar shapes (points more or less lie on a straight line), but they are shifted and/or scaled (if they were, they would be on the identity line).

x = nhanes.weight.loc[nhanes.usborn == "yes"]
y = nhanes.weight.loc[nhanes.usborn == "no"]
xd = np.sort(x)
yd = np.sort(y)
if len(xd) > len(yd):  # interpolate between quantiles in a longer sample
xd = np.quantile(xd, np.arange(1, len(yd)+1)/(len(yd)+1))
else:
yd = np.quantile(yd, np.arange(1, len(xd)+1)/(len(xd)+1))
plt.plot(xd, yd, "o")
plt.axline((xd[len(xd)//2], xd[len(xd)//2]), slope=1,
linestyle=":", color="gray")  # identity line
plt.xlabel(f"Sample quantiles (weight; usborn=yes)")
plt.ylabel(f"Sample quantiles (weight; usborn=no)")
plt.show()

Notice that we interpolated between the quantiles in a larger sample to match the length of the shorter vector.

Let us consider a small sample of white, rather sweet wines from a much larger wine quality dataset.

"teaching-data/master/other/sweetwhitewine_train2.csv",
comment="#")
## 0  10.625271  10.340159    0
## 1   9.066111  18.593274    1
## 2  10.806395   6.206685    0
## 3  13.432876   2.739529    0
## 4   9.578162   3.053025    0

We are given each wine’s alcohol and residual sugar content, as well as a binary categorical variable stating whether a group of sommeliers deem a given beverage quite bad (1) or not (0). Figure 12.8 reveals that subpar wines are rather low in… alcohol and, to some extent, sugar.

sns.scatterplot(x="alcohol", y="sugar", data=wine_train,
plt.xlabel("alcohol")
plt.ylabel("sugar")
plt.show()

Someone answer the door! We have a delivery of a few new wine bottles. Interestingly, their alcohol and sugar contents have been given on their respective labels.

"teaching-data/master/other/sweetwhitewine_test2.csv",
comment="#").iloc[:, :-1]
##      alcohol      sugar
## 0   9.315523  10.041971
## 1  12.909232   6.814249
## 2   9.051020  12.818683
## 3   9.567601  11.091827
## 4   9.494031  12.053790

We would like to determine which of the wines from the test set might be not-bad without asking an expert for their opinion. In other words, we would like to exercise a classification task (see, e.g., [8, 47]). More formally:

Important

Assume we are given a set of training points $$\mathbf{X}\in\mathbb{R}^{n\times m}$$ and the corresponding reference outputs $$\boldsymbol{y}\in\{L_1,L_2,\dots,L_l\}^n$$ in the form of a categorical variable with l distinct levels. The aim of a classification algorithm is to predict what the outputs for each point from a possibly different dataset $$\mathbf{X}'\in\mathbb{R}^{n'\times m}$$, i.e., $$\hat{\boldsymbol{y}}'\in\{L_1,L_2,\dots,L_l\}^{n'}$$, might be.

In other words, we are asked to fill the gaps in a categorical variable. Recall that in a regression problem (Section 9.2), the reference outputs were numerical.

Exercise 12.14

Which of the following are instances of classification problems and which are regression tasks?

• Detect email spam.

• Predict a market stock price (good luck with that).

• Assess credit risk.

• Detect tumour tissues in medical images.

• Predict time-to-recovery of cancer patients.

• Recognise smiling faces on photographs (kind of creepy).

• Detect unattended luggage in airport security camera footage.

What kind of data should you gather to tackle them?

### 12.3.1. K-Nearest Neighbour Classification#

One of the simplest approaches to classification – good enough for such an introductory course – is based on the information about a test point’s nearest neighbours living in the training sample; compare Section 8.4.4.

Fix $$k\ge 1$$. Namely, to classify some $$\boldsymbol{x}'\in\mathbb{R}^m$$:

1. Find the indexes $$N_k(\boldsymbol{x}')=\{i_1,\dots,i_k\}$$ of the $$k$$ points from $$\mathbf{X}$$ closest to $$\boldsymbol{x}'$$, i.e., ones that fulfil for all $$j\not\in\{i_1,\dots,i_k\}$$:

$\|\mathbf{x}_{i_1,\cdot}-\boldsymbol{x}'\| \le\dots\le \| \mathbf{x}_{i_k,\cdot} -\boldsymbol{x}' \| \le \| \mathbf{x}_{j,\cdot} -\boldsymbol{x}' \|.$
2. Classify $$\boldsymbol{x}'$$ as $$\hat{y}'=\mathrm{mode}(y_{i_1},\dots,y_{i_k})$$, i.e., assign it the label that most frequently occurs amongst its $$k$$ nearest neighbours. If a mode is nonunique, resolve the ties, for example, at random.

It is thus a similar algorithm to k-nearest neighbour regression (Section 9.2.1). We only replaced the quantitative mean with the qualitative mode.

This is a variation on the theme: if you don’t know what to do in a given situation, try to mimic what the majority of people around you are doing. Or, if you don’t know what to think about a particular wine, but amongst the five most similar ones (in terms of alcohol and sugar content) three were said to be awful, say that you don’t like it because it’s not sweet enough. Thanks to this, others will take you for a very refined wine taster.

Let us apply a 5-nearest neighbour classifier on the standardised version of the dataset: as we are about to use a technique based on pairwise distances, it would be best if the variables were on the same scale. Thus, we first compute the z-scores for the training set:

X_train = np.array(wine_train.loc[:, ["alcohol", "sugar"]])
means = np.mean(X_train, axis=0)
sds = np.std(X_train, axis=0)
Z_train = (X_train-means)/sds

Then, we determine the z-scores for the test set:

Z_test = (np.array(wine_test.loc[:, ["alcohol", "sugar"]])-means)/sds

Let us stress that we referred to the aggregates computed for the training set. This is a good example of a situation where we cannot simply use a built-in method from pandas. Instead, we apply what we have learned about numpy.

To make the predictions, we will use the following function:

def knn_class(X_test, X_train, y_train, k):
nnis = scipy.spatial.KDTree(X_train).query(X_test, k)[1]
nnls = y_train[nnis]  # same as: y_train[nnis.reshape(-1)].reshape(-1, k)
return scipy.stats.mode(nnls.reshape(-1, k), axis=1, keepdims=False)[0]

First, we fetched the indexes of each test point’s nearest neighbours (amongst the points in the training set). Then, we read their corresponding labels; they are stored in a matrix with $$k$$ columns. Finally, we computed the modes in each row. As a consequence, we have each point in the test set classified.

And now:

k = 5
y_pred = knn_class(Z_test, Z_train, y_train, k)
y_pred[:10]  # preview
## array([1, 0, 0, 1, 1, 0, 1, 0, 0, 1])

Note

Unfortunately, scipy.stats.mode does not resolve possible ties at random: e.g., the mode of $$(1, 1, 1, 2, 2, 2)$$ is always 1. Nevertheless, in our case, $$k$$ is odd and the number of possible classes is $$l=2$$, so the mode is always unique.

Figure 12.9 shows how nearest neighbour classification categorises different regions of a section of the two-dimensional plane. The greater the $$k$$, the smoother the decision boundaries. Naturally, in regions corresponding to few training points, we do not expect the classification accuracy to be good enough[5].

x1 = np.linspace(Z_train[:, 0].min(), Z_train[:, 0].max(), 100)
x2 = np.linspace(Z_train[:, 1].min(), Z_train[:, 1].max(), 100)
xg1, xg2 = np.meshgrid(x1, x2)
Xg12 = np.column_stack((xg1.reshape(-1), xg2.reshape(-1)))
ks = [5, 25]
for i in range(len(ks)):
plt.subplot(1, len(ks), i+1)
yg12 = knn_class(Xg12, Z_train, y_train, ks[i])
plt.scatter(Z_train[y_train == 0, 0], Z_train[y_train == 0, 1],
c="black", marker="o", alpha=0.5)
plt.scatter(Z_train[y_train == 1, 0], Z_train[y_train == 1, 1],
c="#DF536B", marker="v", alpha=0.5)
plt.contourf(x1, x2, yg12.reshape(len(x2), len(x1)),
cmap="gist_heat", alpha=0.5)
plt.title(f"$k={ks[i]}$", fontdict=dict(fontsize=10))
plt.xlabel("alcohol")
if i == 0: plt.ylabel("sugar")
plt.show()
Example 12.15

(*) The same with the scikit-learn package:

import sklearn.neighbors
knn = sklearn.neighbors.KNeighborsClassifier(k)
knn.fit(Z_train, y_train)
y_pred2 = knn.predict(Z_test)

We can verify that the results are identical to the ones above by calling:

np.all(y_pred2 == y_pred)
## True

### 12.3.2. Assessing the Quality of Predictions#

It is time to reveal the truth: our test wines, it turns out, have already been assessed by some experts.

"teaching-data/master/other/sweetwhitewine_test2.csv",
comment="#")
y_test[:10]  # preview
## array([1, 0, 0, 1, 0, 0, 1, 0, 1, 1])

The accuracy score is the most straightforward measure of the similarity between these true labels (denoted $$\boldsymbol{y}'=(y_1',\dots,y_{n'}')$$) and the ones predicted by the classifier (denoted $$\hat{\boldsymbol{y}}'=(\hat{y}_1',\dots,\hat{y}_{n'}')$$). It is defined as a ratio between the correctly classified instances and all the instances:

$\text{Accuracy}(\boldsymbol{y}', \hat{\boldsymbol{y}}') = \frac{\sum_{i=1}^{n'} \mathbf{1}(y_i' = \hat{y}_i')}{n'},$

where the indicator function $$\mathbf{1}(y_i' = \hat{y}_i')=1$$ if and only if $$y_i' = \hat{y}_i'$$ and $$0$$ otherwise.

Computing the above for our test sample gives:

np.mean(y_test == y_pred)
## 0.706

Thus, 71% of the wines were correctly classified with regard to their true quality. Before we get too enthusiastic, let us note that our dataset is slightly imbalanced in terms of the distribution of label counts:

pd.Series(y_test).value_counts()  # contingency table
## 0    330
## 1    170
## dtype: int64

It turns out that the majority of the wines (330 out of 500) in our sample are truly good. Notice that a dummy classifier which labels all the wines as great would have accuracy of 66%. Our k-nearest neighbour approach to wine quality assessment is not that usable after all.

It is therefore always beneficial to analyse the corresponding confusion matrix, which is a two-way contingency table summarising the correct decisions and errors we make.

C = pd.DataFrame(
dict(y_pred=y_pred, y_test=y_test)
).value_counts().unstack(fill_value=0)
C
## y_test    0   1
## y_pred
## 0       272  89
## 1        58  81

In the binary classification case ($$l=2$$) such as this one, its entries are usually referred to as (see also the table below):

• TN – the number of cases where the true $$y_i'=0$$ and the predicted $$\hat{y}_i'=0$$ (true negative),

• TP – the number of instances such that the true $$y_i'=1$$ and the predicted $$\hat{y}_i'=1$$ (true positive),

• FN – how many times the true $$y_i'=1$$ but the predicted $$\hat{y}_i'=0$$ (false negative),

• FN – how many times the true $$y_i'=0$$ but the predicted $$\hat{y}_i'=1$$ (false positive).

The terms positive and negative refer to the output predicted by a classifier, i.e., they indicate whether some $$\hat{y}_i'$$ is equal to 1 and 0, respectively.

Table 12.1 The different cases of true vs predicted labels in a binary classification task $$(l=2)$$#

$$y_i'=0$$

$$y_i'=1$$

$$\hat{y}_i'=0$$

True Negative

False Negative (Type II error)

$$\hat{y}_i'=1$$

False Positive (Type I error)

True Positive

Ideally, the number of false positives and false negatives should be as low as possible. The accuracy score only takes the raw number of true negatives (TN) and true positives (TP) into account:

$\text{Accuracy}(\boldsymbol{y}', \hat{\boldsymbol{y}}') = \frac{\text{TN}+\text{TP}}{\text{TN}+\text{TP}+\text{FN}+\text{FP}}.$

Consequently, it might not be a good metric in imbalanced classification problems.

There are, fortunately, some more meaningful measures in the case where class 1 is less prevalent and where mispredicting it is considered more hazardous than making an inaccurate prediction with respect to class 0. This is because most will agree that it is better to be surprised by a vino mislabelled as bad, than be disappointed with a highly recommended product where we have already built some expectations around it. Further, not getting diagnosed as having COVID-19 where we are genuinely sick can be more dangerous for the people around us than being asked to stay at home with nothing but a headache.

Precision answers the question: If the classifier outputs 1, what is the probability that this is indeed true?

$\text{Precision}(\boldsymbol{y}', \hat{\boldsymbol{y}}') = \frac{\text{TP}}{\text{TP}+\text{FP}} = \frac{\sum_{i=1}^{n'} y_i' \hat{y}_i'}{\sum_{i=1}^{n'} \hat{y}_i'} .$
C = np.array(C)  # convert to matrix
C[1,1]/(C[1,1]+C[1,0])  # precision
## 0.5827338129496403
np.sum(y_test*y_pred)/np.sum(y_pred)  # equivalently
## 0.5827338129496403

When a classifier labels a vino as bad, in 58% of cases it is veritably undrinkable.

Recall (sensitivity, hit rate, or true positive rate) addresses the question: If the true class is 1, what is the probability that the classifier will detect it?

$\text{Recall}(\boldsymbol{y}', \hat{\boldsymbol{y}}') = \frac{\text{TP}}{\text{TP}+\text{FN}} = \frac{\sum_{i=1}^{n'} y_i' \hat{y}_i'}{\sum_{i=1}^{n'} {y}_i'}.$
C[1,1]/(C[1,1]+C[0,1])  # recall
## 0.4764705882352941
np.sum(y_test*y_pred)/np.sum(y_test)  # equivalently
## 0.4764705882352941

Only 48% of the really bad wines will be filtered out by the classifier.

F-measure (or $$F_1$$-measure), is the harmonic[6] mean of precision and recall in the case where we would rather have them aggregated into a single number:

$\text{F}(\boldsymbol{y}', \hat{\boldsymbol{y}}') = \frac{1}{ \frac{ \frac{1}{\text{Precision}}+\frac{1}{\text{Recall}} }{2} } = \left( \frac{1}{2} \left( \text{Precision}^{-1}+\text{Recall}^{-1} \right) \right)^{-1} = \frac{\text{TP}}{\text{TP} + \frac{\text{FP} + \text{FN}}{2}}.$
C[1,1]/(C[1,1]+0.5*C[0,1]+0.5*C[1,0])  # F
## 0.5242718446601942

Overall, we can conclude that our classifier is rather weak.

Exercise 12.16

Would you use precision or recall in each of the following settings?

• Medical diagnosis,

• medical screening,

• suggestions of potential matches in a dating app,

• plagiarism detection,

• wine recommendation.

### 12.3.3. Splitting into Training and Test Sets#

The training set was used as a source of knowledge about our problem domain. The k-nearest neighbour classifier is technically model-free. As a consequence, to generate a new prediction, we need to be able to query all the points in the database every time.

Nonetheless, most statistical/machine learning algorithms, by construction, generalise the patterns discovered in the dataset in the form of mathematical functions (oftentimes, very complicated ones), that are fitted by minimising some error metric. Linear regression analysis by means of the least squares approximation uses exactly this kind of approach. Logistic regression for a binary response variable would be a conceptually similar classifier, but it is beyond our introductory course.

Either way, we used a separate test set to verify the quality of our classifier on so-far unobserved data, i.e., its predictive capabilities. We do not want our model to fit to the training data too closely. This could lead to its being completely useless when filling the gaps between the points it was exposed to. This is like being a student who can only repeat what the teacher says, and when faced with a slightly different real-world problem, they panic and say complete gibberish.

In the above example, the training and test sets were created by yours truly. Still, normally, it is the data scientist who splits a single data frame into two parts themself; see Section 10.5.4. This way, they can mimic the situation where some test observations become available after the learning phase is complete.

### 12.3.4. Validating Many Models (Parameter Selection) (*)#

In statistical modelling, there usually are many hyperparameters that should be tweaked. For example:

• which independent variables should be used for model building,

• how they should be preprocessed; e.g., which of them should be standardised,

• if an algorithm has some tunable parameters, what is the best combination thereof; for instance, which $$k$$ should we use in the k-nearest neighbours search.

At initial stages of data analysis, we usually tune them up by trial and error. Later, but this is already beyond the scope of this introductory course, we are used to exploring all the possible combinations thereof (exhaustive grid search) or making use of some local search-based heuristics (e.g., greedy optimisers such as hill climbing).

These always involve verifying the performance of many different classifiers, for example, 1-, 3-, 9, and 15-nearest neighbours-based ones. For each of them, we need to compute separate quality metrics, e.g., F-measures. Then, the classifier which yields the highest score is picked as the best. Unfortunately, if we do it recklessly, this can lead to overfitting, this time with respect to the test set. The obtained metrics might be too optimistic and can poorly reflect the real performance of the solution on future data.

Assuming that our dataset carries a decent number of observations, to overcome this problem, we can perform a random training/validation/test split:

• training sample (e.g., 60% of randomly chosen rows) – for model construction,

• validation sample (e.g., 20%) – used to tune the hyperparameters of many classifiers and to choose the best one,

• test (hold-out) sample (e.g., the remaining 20%) – used to assess the goodness of fit of the best classifier.

This common-sense approach is not limited to classification. We can validate different regression models in the same way.

Important

We would like to obtain a good estimate of a classifier’s performance on previously unobserved data. For this reason, the test (hold-out) sample must neither be used in the training nor the validation phase.

Exercise 12.17

Determine the best parameter setting for the k-nearest neighbour classification of the color variable based on standardised versions of some physicochemical features (chosen columns) of wines in the wine_quality_all dataset. Create a 60/20/20% dataset split. For each $$k=1, 3, 5, 7, 9$$, compute the corresponding F-measure on the validation test. Evaluate the quality of the best classifier on the test set.

Note

(*) Instead of a training/validation/test split, we can use various cross-validation techniques, especially on smaller datasets. For instance, in a 5-fold cross-validation, we split the original training set randomly into five disjoint parts: $$A, B, C, D, E$$ (more or less of the same size). We use each combination of four chunks as training sets and the remaining part as the validation set, for which we generate the predictions and then compute, say, the F-measure:

training set

validation set

F-measure

$$B\cup C\cup D\cup E$$

$$A$$

$$F_A$$

$$A\cup C\cup D\cup E$$

$$B$$

$$F_B$$

$$A\cup B\cup D\cup E$$

$$C$$

$$F_C$$

$$A\cup B\cup C\cup E$$

$$D$$

$$F_D$$

$$A\cup B\cup C\cup D$$

$$E$$

$$F_E$$

In the end, we can determine the average F-measure, $$(F_A+F_B+F_C+F_D+F_E)/5$$, as a basis for assessing different classifiers’ quality.

Once the best classifier is chosen, we can use the whole training sample to fit the final model and then consider the separate test sample to assess its quality.

Furthermore, for highly imbalanced labels, some form of stratified sampling might be necessary. Such problems are typically explored in more advanced courses in statistical learning.

Exercise 12.18

(**) Redo the above exercise (assessing the wine colour classifiers), but this time maximise the F-measure obtained by a 5-fold cross-validation.

So far, we have been implicitly assuming that either each dataset comes from a single homogeneous distribution, or we have a categorical variable that naturally defines the groups that we can split the dataset into. Nevertheless, it might be the case that we are given a sample coming from a distribution mixture, where some subsets behave differently, but a grouping variable has not been provided at all (e.g., we have height and weight data but no information about the subjects’ sexes).

Clustering (also known as segmentation or quantisation; see, e.g., [2, 94]) methods can be used to partition a dataset into groups based only on the spatial structure of the points’ relative densities. In the k-means method, which we discuss below, the cluster structure is determined based on the points’ proximity to $$k$$ carefully chosen group centroids; compare Section 8.4.2.

### 12.4.1. K-Means Method#

Fix $$k \ge 2$$. In the k-means method[7], we seek $$k$$ pivot points, $$\boldsymbol{c}_1, \boldsymbol{c}_2,\dots,\boldsymbol{c}_k\in\mathbb{R}^m$$, such that the sum of squared distances between the input points in $$\mathbf{X}\in\mathbb{R}^{n\times m}$$ and their closest pivots is minimised:

$\text{minimise}\ \sum_{i=1}^n \min\left\{ \| \mathbf{x}_{i,\cdot} - \boldsymbol{c}_{1} \|^2, \| \mathbf{x}_{i,\cdot} - \boldsymbol{c}_{2} \|^2, \dots, \| \mathbf{x}_{i,\cdot} - \boldsymbol{c}_{k} \|^2 \right\} \qquad\text{w.r.t. }{\boldsymbol{c}_1, \boldsymbol{c}_2,\dots,\boldsymbol{c}_k}.$

Let us introduce the label vector $$\boldsymbol{l}$$ such that:

$l_i = \mathrm{arg}\min_{j} \| \mathbf{x}_{i,\cdot} - \boldsymbol{c}_{j} \|^2,$

i.e., it is the index of the pivot closest to $$\mathbf{x}_{i,\cdot}$$.

We will consider all the points $$\mathbf{x}_{i,\cdot}$$ with $$i$$ such that $$l_i=j$$ as belonging to the same, $$j$$-th, cluster (point group). This way $$\boldsymbol{l}$$ defines a partition of the original dataset into $$k$$ nonempty, mutually disjoint subsets.

Now, the above optimisation task can be equivalently rewritten as:

$\text{minimise}\ \sum_{i=1}^n \| \mathbf{x}_{i,\cdot} - \boldsymbol{c}_{l_i} \|^2 \qquad\text{w.r.t. }{\boldsymbol{c}_1, \boldsymbol{c}_2,\dots,\boldsymbol{c}_k}.$

And this is why we refer to the above objective function as the (total) within-cluster sum of squares (WCSS). This problem looks easier, but let us not be tricked; $$l_i$$s depend on $$\boldsymbol{c}_j$$s. They vary together. We have just made it less explicit.

It can be shown that given a fixed label vector $$\boldsymbol{l}$$ representing a partition, $$\boldsymbol{c}_j$$ must be the centroid (Section 8.4.2) of the points assigned thereto:

$\boldsymbol{c}_j = \frac{1}{n_j} \sum_{i: l_i=j} \mathbf{x}_{i,\cdot},$

where $$n_j=|\{i: l_i=j\}|$$ gives the number of $$i$$s such that $$l_i=j$$, i.e., the size of the $$j$$-th cluster.

Here is an example dataset (see below for a scatterplot):

X = np.genfromtxt("https://raw.githubusercontent.com/gagolews/" +
"teaching-data/master/marek/blobs1.txt", delimiter=",")

We can call scipy.cluster.vq.kmeans2 to find $$k=2$$ clusters:

import scipy.cluster.vq
C, l = scipy.cluster.vq.kmeans2(X, 2)

The discovered cluster centres are stored in a matrix with $$k$$ rows and $$m$$ columns, i.e., the $$j$$-th row gives $$\mathbf{c}_j$$.

C
## array([[ 0.99622971,  1.052801  ],
##        [-0.90041365, -1.08411794]])

The label vector is:

l
## array([1, 1, 1, ..., 0, 0, 0], dtype=int32)

As usual in Python, indexing starts at 0. So for $$k=2$$ we only obtain the labels 0 and 1.

Figure 12.10 depicts the two clusters together with the cluster centroids. We use l as a colour selector in my_colours[l] (this is a clever instance of the integer vector-based indexing). It seems that we correctly discovered the very natural partitioning of this dataset into two clusters.

plt.scatter(X[:, 0], X[:, 1], c=np.array(["black", "#DF536B"])[l])
plt.plot(C[:, 0], C[:, 1], "yX")
plt.axis("equal")
plt.show()

Here are the cluster sizes:

np.bincount(l)  # or, e.g., pd.Series(l).value_counts()
## array([1017, 1039])

The label vector l can be added as a new column in the dataset. Here is a preview:

Xl = pd.DataFrame(dict(X1=X[:, 0], X2=X[:, 1], l=l))
Xl.sample(5, random_state=42)  # some randomly chosen rows
##             X1        X2  l
## 184  -0.973736 -0.417269  1
## 1724  1.432034  1.392533  0
## 251  -2.407422 -0.302862  1
## 1121  2.158669 -0.000564  0
## 1486  2.060772  2.672565  0

We can now enjoy all the techniques for processing data in groups that we have discussed so far. In particular, computing the columnwise means gives nothing else than the above cluster centroids:

Xl.groupby("l").mean()
##          X1        X2
## l
## 0  0.996230  1.052801
## 1 -0.900414 -1.084118

The label vector l can be recreated by computing the distances between all the points and the centroids and then picking the indexes of the closest pivots:

l_test = np.argmin(scipy.spatial.distance.cdist(X, C), axis=1)
np.all(l_test == l)  # verify they are identical
## True

Important

By construction[8], the k-means method can only detect clusters of convex shapes (such as Gaussian blobs).

Exercise 12.19

Perform the clustering of the wut_isolation dataset and notice how nonsensical, geometrically speaking, the returned clusters are.

Exercise 12.20

Determine a clustering of the wut_twosplashes dataset and display the results on a scatterplot. Compare them with those obtained on the standardised version of the dataset. Recall what we said about the Euclidean distance and its perception being disturbed when a plot’s aspect ratio is not 1:1.

Note

(*) An even simpler classifier than the k-nearest neighbours one described above builds upon the concept of the nearest centroids. Namely, it first determines the centroids (componentwise arithmetic means) of the points in each class. Then, a new point (from the test set) is assigned to the class whose centroid is the closest thereto. The implementation of such a classifier is left as a rather straightforward exercise to the reader. As an application, we recommend using it to extrapolate the results generated by the k-means method (for different $$k$$s) to previously unobserved data, e.g., all points on a dense equidistant grid.

### 12.4.2. Solving K-means Is Hard#

Unfortunately, the k-means method – the identification of label vectors/cluster centres that minimise the total within-cluster sum of squares – relies on solving a computationally hard combinatorial optimisation problem (e.g., [58]). In other words, the search for the truly (i.e., globally) optimal solution takes, for larger $$n$$ and $$k$$, an impractically long time.

As a consequence, we must rely on some approximate algorithms which all have one drawback in common. Namely, whatever they return can be suboptimal. Hence, they can constitute a possibly meaningless solution.

The documentation of scipy.cluster.vq.kmeans2 is of course honest about it. It states that the method attempts to minimise the Euclidean distance between observations and centroids. Further, sklearn.cluster.KMeans, implementing a similar algorithm, mentions that the procedure is very fast […], but it falls in local minima. That is why it can be useful to restart it several times.

To understand what it all means, it will be very educational to study this issue in more detail. This is because the discussed approach to clustering is not the only hard problem in data science (selecting an optimal set of independent variables with respect to AIC or BIC in linear regression is another example).

### 12.4.3. Lloyd’s Algorithm#

Technically, there is no such thing as the k-means algorithm. There are many procedures, based on numerous different heuristics, that attempt to solve the k-means problem. Unfortunately, neither of them is perfect. This is not possible.

Perhaps the most widely known and easiest to understand method is traditionally attributed to Lloyd [61]. It is based on the fixed-point iteration and. For a given $$\mathbf{X}\in\mathbb{R}^{n\times m}$$ and $$k \ge 2$$:

1. Pick initial cluster centres $$\boldsymbol{c}_{1}, \dots, \boldsymbol{c}_{k}$$, for example, randomly.

2. For each point in the dataset, $$\mathbf{x}_{i,\cdot}$$, determine the index of its closest centre $$l_i$$:

$l_i = \mathrm{arg}\min_{j} \| \mathbf{x}_{i,\cdot} - \boldsymbol{c}_{j} \|^2.$
3. Compute the centroids of the clusters defined by the label vector $$\boldsymbol{l}$$, i.e., for every $$j=1,2,\dots,k$$:

$\boldsymbol{c}_j = \frac{1}{n_j} \sum_{i: l_i=j} \mathbf{x}_{i,\cdot},$

where $$n_j=|\{i: l_i=j\}|$$ gives the size of the $$j$$-th cluster.

4. If the objective function (total within-cluster sum of squares) has not changed significantly since the last iteration (say, the absolute value of the difference between the last and the current loss is less than $$10^{-9}$$), then stop and return the current $$\boldsymbol{c}_1,\dots,\boldsymbol{c}_k$$ as the result. Otherwise, go to Step 2.

Exercise 12.21

(*) Implement the Lloyd algorithm in the form of a function kmeans(X, C), where X is the data matrix (n-by-m) and where the rows in C, being a k-by-m matrix, give the initial cluster centres.

### 12.4.4. Local Minima#

The way the above algorithm is constructed implies what follows.

Important

Lloyd’s method guarantees that the centres $$\boldsymbol{c}_1,\dots,\boldsymbol{c}_k$$ it returns cannot be significantly improved any further by repeating Steps 2 and 3 of the algorithm. Still, it does not necessarily mean that they yield the globally optimal (the best possible) WCSS. We might as well get stuck in a local minimum, where there is no better positioning thereof in the neighbourhoods of the current cluster centres; compare Figure 12.11. Yet, had we looked beyond them, we could have found a superior solution.

A variant of the Lloyd method is implemented in scipy.cluster.vq.kmeans2, where the initial cluster centres are picked at random. Let us test its behaviour by analysing three chosen country-wise categories from the 2016 Sustainable Society Indices dataset.

"teaching-data/master/marek/ssi_2016_categories.csv",
comment="#")
X = ssi.set_index("Country").loc[:,
["PersonalDevelopmentAndHealth", "WellBalancedSociety", "Economy"]
].rename({
"PersonalDevelopmentAndHealth": "Health",
"WellBalancedSociety": "Balance",
"Economy": "Economy"
}, axis=1)  # rename columns
n = X.shape[0]
X.loc[["Australia", "Germany", "Poland", "United States"], :]  # preview
##                  Health   Balance   Economy
## Country
## Australia      8.590927  6.105539  7.593052
## Germany        8.629024  8.036620  5.575906
## Poland         8.265950  7.331700  5.989513
## United States  8.357395  5.069076  3.756943

It is a three-dimensional dataset, where each point (row) corresponds to a different country. Let us find a partition into $$k=3$$ clusters.

k = 3
np.random.seed(123)  # reproducibility matters
C1, l1 = scipy.cluster.vq.kmeans2(X, k)
C1
## array([[7.99945084, 6.50033648, 4.36537659],
##        [7.6370645 , 4.54396676, 6.89893746],
##        [6.24317074, 3.17968018, 3.60779268]])

The objective function (total within-cluster sum of squares) at the returned cluster centres is equal to:

import scipy.spatial.distance
def get_wcss(X, C):
D = scipy.spatial.distance.cdist(X, C)**2
return np.sum(np.min(D, axis=1))

get_wcss(X, C1)
## 446.5221283436733

Is it good or not necessarily? We are unable to tell. What we can do, however, is to run the algorithm again, this time from a different starting point:

np.random.seed(1234)  # different seed - different initial centres
C2, l2 = scipy.cluster.vq.kmeans2(X, k)
C2
## array([[7.80779013, 5.19409177, 6.97790733],
##        [6.31794579, 3.12048584, 3.84519706],
##        [7.92606993, 6.35691349, 3.91202972]])
get_wcss(X, C2)
## 437.51120966832775

It is a better solution (we are lucky; it might as well have been worse). But is it the best possible? Again, we cannot tell, alone in the dark.

Does a potential suboptimality affect the way the data points are grouped? It is indeed the case here. Let us look at the contingency table for the two label vectors:

pd.DataFrame(dict(l1=l1, l2=l2)).value_counts().unstack(fill_value=0)
## l2   0   1   2
## l1
## 0    8   0  43
## 1   39   6   0
## 2    0  57   1

Important

Clusters are essentially unordered. The label vector $$(1, 1, 2, 2, 1, 3)$$ represents the same clustering as the label vectors $$(3, 3, 2, 2, 3, 1)$$ and $$(2, 2, 3, 3, 2, 1)$$.

By looking at the contingency table, we see that clusters 0, 1, and 2 in l1 correspond, respectively, to clusters 2, 0, and 1 in l2 (via a kind of majority voting). We can relabel the elements in l1 to get a more readable result:

l1p = np.array([2, 0, 1])[l1]
pd.DataFrame(dict(l1p=l1p, l2=l2)).value_counts().unstack(fill_value=0)
## l2    0   1   2
## l1p
## 0    39   6   0
## 1     0  57   1
## 2     8   0  43

Much better. It turns out that 8+6+1 countries are categorised differently. We would definitely not want to initiate any diplomatic crisis because of our not knowing that the above algorithm might return suboptimal solutions.

Exercise 12.22

(*) Determine which countries are affected.

### 12.4.5. Random Restarts#

There will never be any guarantees, but we can increase the probability of generating a satisfactory solution by simply restarting the method multiple times from many randomly chosen points and picking the best[9] solution (the one with the smallest WCSS) identified as the result.

Let us make 1,000 such restarts:

wcss, Cs = [], []
for i in range(1000):
C, l = scipy.cluster.vq.kmeans2(X, k, seed=i)
Cs.append(C)
wcss.append(get_wcss(X, C))

The best of the local minima (no guarantee that it is the global one, again) is:

np.min(wcss)
## 437.51120966832775

It corresponds to the cluster centres:

Cs[np.argmin(wcss)]
## array([[7.80779013, 5.19409177, 6.97790733],
##        [7.92606993, 6.35691349, 3.91202972],
##        [6.31794579, 3.12048584, 3.84519706]])

They are the same as C2 above (up to a permutation of labels). We were lucky[10], after all.

It is very educational to look at the distribution of the objective function at the identified local minima to see that, proportionally, in the case of this dataset it is not rare to end up in a quite bad solution; see Figure 12.12.

plt.hist(wcss, bins=100)
plt.show()

Also, Figure 12.13 depicts all the cluster centres to which the algorithm converged. We see that we should not be trusting the results generated by a single run of a heuristic solver to the k-means problem.

Example 12.23

(*) The scikit-learn package implements an algorithm that is similar to the Lloyd’s one. The method is equipped with the n_init parameter (which defaults to 10) which automatically applies the aforementioned restarting.

import sklearn.cluster
np.random.seed(123)
km = sklearn.cluster.KMeans(k)  # KMeans(k, n_init=10)
km.fit(X)
## KMeans(n_clusters=3)
km.inertia_  # WCSS – not optimal!
## 437.54671889589287

Still, there are no guarantees: the solution is suboptimal too. As an exercise, pass n_init=100, n_init=1000, and n_init=10000 and determine the returned WCSS.

Note

It is theoretically possible that a developer from the scikit-learn team, when they see the above result, will make a tweak in the algorithm so that after an update to the package, the returned minimum will be better. This cannot be deemed a bug fix, though, as there are no bugs here. Improving the behaviour of the method in this example will lead to its degradation in others. There is no free lunch in optimisation.

Note

Some datasets are more well-behaving than others. The k-means method is overall quite usable, but we must always be cautious.

We recommend always performing at least 100 random restarts. Also, if a report from data analysis does not say anything about the number of tries performed, we should assume that the results are gibberish[11]. People will complain about our being a pain, but we know better; compare Rule#9.

Exercise 12.24

Run the k-means method, $$k=8$$, on the sipu_unbalance dataset from many random sets of cluster centres. Note the value of the total within-cluster sum of squares. Also, plot the cluster centres discovered. Do they make sense? Compare these to the case where you start the method from the following cluster centres which are close to the global minimum.

$\begin{split} \mathbf{C} = \left[ \begin{array}{cc} -15 & 5 \\ -12 & 10 \\ -10 & 5 \\ 15 & 0 \\ 15 & 10 \\ 20 & 5 \\ 25 & 0 \\ 25 & 10 \\ \end{array} \right]. \end{split}$

An overall good introduction to classification is [47] and [8]. Nevertheless, as we said earlier, we recommend going through a solid course in matrix algebra and mathematical statistics first, e.g., [20, 39] and [21, 38, 40]. For advanced theoretical (probabilistic, information-theoretic) results, see, e.g., [9, 22].

Hierarchical clustering algorithms (see, e.g., [32, 65]) are also noteworthy, because they do not require asking for a fixed number of clusters. Furthermore, density-based algorithms (DBSCAN and its variants) [12, 25, 59] utilise the notion of fixed-radius search that we have introduced in Section 8.4.4.

There are quite a few ways that aim to assess the quality of clustering results, but their meaningfulness is somewhat limited; see [36] for discussion.

## 12.6. Exercises#

Exercise 12.25

Name the data type of the objects that the DataFrame.groupby method returns.

Exercise 12.26

What is the relationship between the GroupBy, DataFrameGroupBy, and SeriesGroupBy classes?

Exercise 12.27

What are relative z-scores and how can we compute them?

Exercise 12.28

Why and when the accuracy score might not be the best way to quantify a classifier’s performance?

Exercise 12.29

What is the difference between recall and precision, both in terms of how they are defined and where they are the most useful?

Exercise 12.30

Explain how the k-nearest neighbour classification and regression algorithms work. Why do we say that they are model-free?

Exercise 12.31

In the context of k-nearest neighbour classification, why it might be important to resolve the potential ties at random when computing the mode of the neighbours’ labels?

Exercise 12.32

What is the purpose of a training/test and a training/validation/test set split?

Exercise 12.33

Give the formula for the total within-cluster sum of squares.

Exercise 12.34

Are there any cluster shapes that cannot be detected by the k-means method?

Exercise 12.35

Why do we say that solving the k-means problem is hard?

Exercise 12.36

Why restarting Lloyd’s algorithm many times is necessary? Why are reports from data analysis that do not mention the number of restarts not trustworthy?