# 11. Handling Categorical Data#

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].

So far, we have been mostly dealing with quantitative (numeric) data, on which we were able to apply various mathematical operations, such as computing the arithmetic mean or taking the square thereof. Naturally, not every transformation must always make sense in every context (e.g., multiplying temperatures – what does it mean when we say that it is twice as hot today as compared to yesterday?), but still, the possibilities were plenty.

Qualitative data (also known as categorical data, factors, or enumerated types) such as eye colour, blood type, or a flag whether a patient is ill, on the other hand, take a small number of unique values. They support an extremely limited set of admissible operations. Namely, we can only determine whether two entities are equal or not.

In datasets involving many features, which we shall cover in Chapter 12, categorical variables are often used for observation grouping (e.g., so that we can compute the best and average time for marathoners in each age category or draw box plots for finish times of men and women separately). Also, they may serve as target variables in statistical classification tasks (e.g., so that we can determine if an email is “spam” or “not spam”).

## 11.1. Representing and Generating Categorical Data#

Common ways to represent a categorical variable with $$l$$ distinct levels $$\{L_1,L_2,\dots,L_l\}$$ is by storing it as:

• a vector of strings,

• a vector of integers between 0 (inclusive) and $$l$$ (exclusive[1]).

These two are easily interchangeable.

For $$l=2$$ (binary data), another convenient representation is by means of logical vectors. This can be extended to a so-called one-hot encoded representation using a logical vector of length $$l$$.

Let us consider the data on the original whereabouts of the top 16 marathoners (the 37th PZU native Marathon dataset):

marathon = pd.read_csv("https://raw.githubusercontent.com/gagolews/" +
"teaching-data/master/marek/37_pzu_warsaw_marathon_simplified.csv",
comment="#")
cntrs = np.array(marathon.country, dtype="str")
cntrs16 = cntrs[:16]
cntrs16
## array(['KE', 'KE', 'KE', 'ET', 'KE', 'KE', 'ET', 'MA', 'PL', 'PL', 'IL',
##        'PL', 'KE', 'KE', 'PL', 'PL'], dtype='<U2')


These are two-letter ISO 3166 country codes, encoded of course as strings (notice the dtype="str" argument).

Calling pandas.unique allows us to determine the set of distinct categories:

cat_cntrs16 = pd.unique(cntrs16)
cat_cntrs16
## array(['KE', 'ET', 'MA', 'PL', 'IL'], dtype='<U2')


Hence, cntrs16 is a categorical vector of length $$n=16$$ (len(cntrs16)) with data assuming one of $$l=5$$ different levels (len(cat_cntrs16)).

Note

We could have also used numpy.unique discussed in Section 5.5.3, but it would sort the distinct values lexicographically. In other words, they would not be listed in the order of appearance.

### 11.1.1. Encoding and Decoding Factors#

To encode a label vector using a set of consecutive nonnegative integers, we can call pandas.factorize:

codes_cntrs16, cat_cntrs16 = pd.factorize(cntrs16)  # sort=False
cat_cntrs16
## array(['KE', 'ET', 'MA', 'PL', 'IL'], dtype='<U2')
codes_cntrs16
## array([0, 0, 0, 1, 0, 0, 1, 2, 3, 3, 4, 3, 0, 0, 3, 3])


The code sequence 0, 0, 0, 1, … corresponds to the 1st, 1st, 1st, 2nd, … level in cat_cntrs16, i.e., Kenya, Kenya, Kenya, Ethiopia, ….

Important

When we represent categorical data using numeric codes, it is possible to introduce non-occurring levels. Such information can be useful, e.g., we could explicitly indicate that there were no runners from Australia in the top 16.

Even though we can represent categorical variables using a set of integers, it does not mean that they become instances of a quantitative type. Arithmetic operations thereon do not really make sense.

The values between $$0$$ (inclusive) and $$5$$ (exclusive) can be used to index a given array of length $$l=5$$. As a consequence, to decode our factor, we can simply write:

cat_cntrs16[codes_cntrs16]
## array(['KE', 'KE', 'KE', 'ET', 'KE', 'KE', 'ET', 'MA', 'PL', 'PL', 'IL',
##        'PL', 'KE', 'KE', 'PL', 'PL'], dtype='<U2')


We can use any other set of labels now:

np.array(["Kenya", "Ethiopia", "Morocco", "Poland", "Israel"])[codes_cntrs16]
## array(['Kenya', 'Kenya', 'Kenya', 'Ethiopia', 'Kenya', 'Kenya',
##        'Ethiopia', 'Morocco', 'Poland', 'Poland', 'Israel', 'Poland',
##        'Kenya', 'Kenya', 'Poland', 'Poland'], dtype='<U8')


This is an instance of the relabelling of a categorical variable.

Exercise 11.1

(**) Here is a way of recoding a variable, i.e., changing the order of labels and permuting the numeric codes:

new_codes = np.array([3, 0, 2, 4, 1])  # an example permutation of labels
new_cat_cntrs16 = cat_cntrs16[new_codes]
new_cat_cntrs16
## array(['PL', 'KE', 'MA', 'IL', 'ET'], dtype='<U2')


Then we make use of the fact that numpy.argsort applied on a vector representing a permutation, determines its very inverse:

new_codes_cntrs16 = np.argsort(new_codes)[codes_cntrs16]
new_codes_cntrs16
## array([1, 1, 1, 4, 1, 1, 4, 2, 0, 0, 3, 0, 1, 1, 0, 0])


Verification:

np.all(cntrs16 == new_cat_cntrs16[new_codes_cntrs16])
## True

Exercise 11.2

(**) Determine the set of unique values in cntrs16 in the order of appearance (and not sorted lexicographically), but without using pandas.unique nor pandas.factorize. Then, encode cntrs16 using this level set.

Hint: check out the return_index argument to numpy.unique and numpy.searchsorted.

Furthermore, pandas includes a special dtype for storing categorical data. Namely, we can write:

cntrs16_series = pd.Series(cntrs16, dtype="category")


or, equivalently:

cntrs16_series = pd.Series(cntrs16).astype("category")


These two yield a Series object displayed as if it was represented using string labels:

cntrs16_series.head()  # preview
## 0    KE
## 1    KE
## 2    KE
## 3    ET
## 4    KE
## dtype: category
## Categories (5, object): ['ET', 'IL', 'KE', 'MA', 'PL']


Instead, it is encoded using the aforementioned numeric representation:

np.array(cntrs16_series.cat.codes)
## array([2, 2, 2, 0, 2, 2, 0, 3, 4, 4, 1, 4, 2, 2, 4, 4], dtype=int8)
cntrs16_series.cat.categories
## Index(['ET', 'IL', 'KE', 'MA', 'PL'], dtype='object')


This time the labels are sorted lexicographically.

Most often we will be storing categorical data in data frames as ordinary strings, unless a relabelling on-the-fly is required:

(marathon.iloc[:16, :].country.astype("category")
.cat.reorder_categories(
["KE",    "IL",     "MA",      "ET",       "PL"]
)
.cat.rename_categories(
["Kenya", "Israel", "Morocco", "Ethiopia", "Poland"]
).astype("str")
## 0       Kenya
## 1       Kenya
## 2       Kenya
## 3    Ethiopia
## 4       Kenya
## Name: country, dtype: object


### 11.1.2. Binary Data as Logical and Probability Vectors#

Binary data is a special case of the qualitative setting, where we only have $$l=2$$ categories. For example:

• 0, e.g., healthy/fail/off/non-spam/absent/…),

• 1, e.g., ill/success/on/spam/present/…).

Usually, the interesting or noteworthy category is denoted with 1.

Important

When converting logical to numeric, False becomes 0 and True becomes 1. Conversely, 0 is converted to False and anything else (including -0.326) to True.

Hence, instead of working with vectors of 0s and 1s, we might equivalently be playing with logical arrays. For example:

np.array([True, False, True, True, False]).astype(int)
## array([1, 0, 1, 1, 0])


The other way around:

np.array([-2, -0.326, -0.000001, 0.0, 0.1, 1, 7643]).astype(bool)
## array([ True,  True,  True, False,  True,  True,  True])


or, equivalently:

np.array([-2, -0.326, -0.000001, 0.0, 0.1, 1, 7643]) != 0
## array([ True,  True,  True, False,  True,  True,  True])


Important

It is not rare to work with vectors of probabilities, where the i-th element therein, say p[i], denotes the likelihood of an observation’s belonging to class 1. Consequently, the probability of being a member of class 0 is 1-p[i]. In the case where we would rather work with crisp classes, we can simply apply the conversion (p>=0.5) to get a logical vector.

Exercise 11.3

Given a numeric vector x, create a vector of the same length as x whose i-th element is equal to "yes" if x[i] is in the unit interval and to "no" otherwise. Use numpy.where, which can act as a vectorised version of the if statement.

### 11.1.3. One-Hot Encoding (*)#

Let $$\boldsymbol{x}$$ be a vector of $$n$$ integer labels in $$\{0,...,l-1\}$$. Its one-hot encoded version is a 0/1 (or, equivalently, logical) matrix $$\mathbf{R}$$ of shape $$n$$-by-$$l$$ such that $$r_{i,j}=1$$ if and only if $$x_i = j$$.

For example, if $$\boldsymbol{x}=(0, 1, 2, 1)$$ and $$l=4$$, then:

$\begin{split} \mathbf{R} = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]. \end{split}$

One can easily verify that each row consists of one and only one 1 (the number of 1s per one row is 1). Such a representation is adequate when solving a multiclass classification problem by means of l binary classifiers. For example, if spam, bacon, and hot dogs are on the menu, then spam is encoded as $$(1, 0, 0)$$, i.e., yeah-spam, nah-bacon, and nah-hot dog. We can build three binary classifiers, each narrowly specialising in sniffing one particular type of food.

Example 11.4

Write a function to one-hot encode a given categorical vector represented using character strings.

Example 11.5

Write a function to decode a one-hot encoded matrix.

Example 11.6

(*) We can also work with matrices like $$\mathbf{P}\in[0,1]^{n\times l}$$, where $$p_{i,j}$$ denotes the probability of the $$i$$-th object’s belonging to the $$j$$-th class. Given an example matrix of this kind, verify that in each row the probabilities sum to 1 (up to a small numeric error). Then, decode such a matrix by choosing the greatest element in each row.

### 11.1.4. Binning Numeric Data (Revisited)#

Numerical data can be converted to categorical via binning (quantisation). Even though this results in information loss, it may open some new possibilities. In fact, we needed binning to draw all the histograms in Chapter 4. Also, reporting observation counts in each bin instead of raw data enables us to include them in printed reports (in the form of tables).

Note

As strong proponents of openness and transparency, we always encourage all entities (governments, universities, non-profits, corporations, etc.) to share raw, unabridged versions of their datasets under the terms of some open data license. This is to enable public scrutiny and to get the most out of the possibilities they can bring for benefit of the community.

Of course, sometimes the sharing of unprocessed information can violate the privacy of the subjects. In such a case, it might be a good idea to communicate them in a binned form.

Note

Rounding is a kind of binning. In particular, numpy.round allows for rounding to the nearest tenths, hundredths, …, as well as tens, hundreds, …. It is useful if data are naturally imprecise, and we do not want to give the impression that it is otherwise. Nonetheless, rounding can easily introduce tied observations, which are problematic on their own; see Section 5.5.3.

Consider the 16 best marathon finish times (in minutes):

mins = np.array(marathon.mins)
mins16 = mins[:16]
mins16
## array([129.32, 130.75, 130.97, 134.17, 134.68, 135.97, 139.88, 143.2 ,
##        145.22, 145.92, 146.83, 147.8 , 149.65, 149.88, 152.65, 152.88])


numpy.searchsorted can be used to determine the interval where each value in mins falls.

bins = [130, 140, 150]
codes_mins16 = np.searchsorted(bins, mins16)
codes_mins16
## array([0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3])


By default, the intervals are of the form $$(a, b]$$ (not including $$a$$, including $$b$$). Code 0 corresponds to values less than the first bin bound, whereas code 3 – greater than or equal to the last bound:

pandas.cut gives us another interface to the same binning method. It returns a vector-like object with dtype="category", with very readable labels generated automatically (and ordered; see Section 11.4.7):

cut_mins16 = pd.Series(pd.cut(mins16, [-np.inf, 130, 140, 150, np.inf]))
cut_mins16.iloc[ [0, 1, 6, 7, 13, 14, 15] ].astype("str")  # preview
## 0      (-inf, 130.0]
## 1     (130.0, 140.0]
## 6     (130.0, 140.0]
## 7     (140.0, 150.0]
## 13    (140.0, 150.0]
## 14      (150.0, inf]
## 15      (150.0, inf]
## dtype: object
cut_mins16.cat.categories.astype("str")
## Index(['(-inf, 130.0]', '(130.0, 140.0]', '(140.0, 150.0]',
##        '(150.0, inf]'],
##       dtype='object')

Example 11.7

(*) We can create a set of the corresponding categories manually, for example, as follows:

bins2 = np.r_[-np.inf, bins, np.inf]
np.array(
[f"({bins2[i]}, {bins2[i+1]}]" for i in range(len(bins2)-1)]
)
## array(['(-inf, 130.0]', '(130.0, 140.0]', '(140.0, 150.0]',
##        '(150.0, inf]'], dtype='<U14')

Exercise 11.8

(*) Check out the numpy.histogram_bin_edges function which tries to determine some informative interval bounds based on a few simple heuristics. Recall that numpy.linspace and numpy.geomspace can be used for generating equidistant bounds on linear and logarithmic scales, respectively.

### 11.1.5. Generating Pseudorandom Labels#

numpy.random.choice allows us to create a pseudorandom sample with categories picked with given probabilities:

np.random.seed(123)
np.random.choice(
a=["spam", "bacon", "eggs", "tempeh"],
p=[   0.7,     0.1,   0.15,     0.05],
replace=True,
size=16
)
## array(['spam', 'spam', 'spam', 'spam', 'bacon', 'spam', 'tempeh', 'spam',
##        'spam', 'spam', 'spam', 'bacon', 'spam', 'spam', 'spam', 'bacon'],
##       dtype='<U6')


If we generate a sufficiently long vector, we will expect "spam" to occur ca. 70% times, and "tempeh" to be drawn in 5% of the cases, etc.

## 11.2. Frequency Distributions#

### 11.2.1. Counting#

pandas.Series.value_counts creates a frequency table in the form of a Series object equipped with a readable index (element labels):

pd.Series(cntrs16).value_counts()  # sort=True, ascending=False
## KE    7
## PL    5
## ET    2
## MA    1
## IL    1
## dtype: int64


By default, data are ordered with respect to the counts, decreasingly.

If we already have an array of integer codes between $$0$$ and $$l-1$$, numpy.bincount will return the number of times each code appears therein.

counts_cntrs16 = np.bincount(codes_cntrs16)
counts_cntrs16
## array([7, 2, 1, 5, 1])


A vector of counts can easily be turned into a vector of proportions (fractions) or percentages (if we multiply them by 100):

counts_cntrs16 / np.sum(counts_cntrs16) * 100
## array([43.75, 12.5 ,  6.25, 31.25,  6.25])


Almost 31.25% of the top runners were from Poland (this marathon is held in Warsaw after all…).

Exercise 11.9

Using numpy.argsort, sort counts_cntrs16 increasingly together with the corresponding items in cat_cntrs16.

### 11.2.2. Two-Way Contingency Tables: Factor Combinations#

Some datasets may feature many categorical columns, each having possibly different levels. Let us now consider all the runners in the marathon dataset:

marathon.loc[:, "age"] = marathon.category.str.slice(1)  # first two chars
marathon.loc[marathon.age >= "60", "age"] = "60+"  # too few runners aged 70+
marathon = marathon.loc[:, ["sex", "age", "country"]]
##   sex age country
## 0   M  20      KE
## 1   M  20      KE
## 2   M  20      KE
## 3   M  20      ET
## 4   M  30      KE


The three columns are: sex, age (in 10-year brackets), and country. We can of course analyse the data distribution in each column individually, but this we leave as an exercise. Instead, we note that some interesting patterns might also arise when we study the combinations of levels of different variables.

Here are the levels of the sex and age variables:

pd.unique(marathon.sex)
## array(['M', 'F'], dtype=object)
pd.unique(marathon.age)
## array(['20', '30', '50', '40', '60+'], dtype=object)


We have $$2\cdot 5=10$$ combinations thereof. We can use pandas.DataFrame.value_counts to determine the number of observations at each two-dimensional level:

counts2 = marathon.loc[:, ["sex", "age"]].value_counts()
counts2
## sex  age
## M    30     2200
##      40     1708
##      20      879
##      50      541
## F    30      449
##      40      262
##      20      240
## M    60+     170
## F    50       43
##      60+      19
## dtype: int64


These can be converted to a two-way contingency table, which is a matrix that gives the number of occurrences of each pair of values from the two factors:

V = counts2.unstack(fill_value=0)
V
## age   20    30    40   50  60+
## sex
## F    240   449   262   43   19
## M    879  2200  1708  541  170


For example, there were 19 women aged at least 60 amongst the marathoners. Jolly good.

The marginal (one-dimensional) frequency distributions can be recreated by computing the rowwise and columnwise sums of V:

np.sum(V, axis=1)
## sex
## F    1013
## M    5498
## dtype: int64
np.sum(V, axis=0)
## age
## 20     1119
## 30     2649
## 40     1970
## 50      584
## 60+     189
## dtype: int64


### 11.2.3. Combinations of Even More Factors#

pandas.DataFrame.value_counts can also be used with a combination of more than two categorical variables:

counts3 = (marathon
.loc[
marathon.country.isin(["PL", "UA", "SK"]),
["country", "sex", "age"]
]
.value_counts()
.rename("count")
.reset_index()
)
counts3
##    country sex  age  count
## 0       PL   M   30   2081
## 1       PL   M   40   1593
## 2       PL   M   20    824
## 3       PL   M   50    475
## 4       PL   F   30    422
## 5       PL   F   40    248
## 6       PL   F   20    222
## 7       PL   M  60+    134
## 8       PL   F   50     26
## 9       PL   F  60+      8
## 10      UA   M   20      8
## 11      UA   M   30      8
## 12      UA   M   50      3
## 13      UA   F   30      2
## 14      UA   M   40      2
## 15      SK   F   50      1
## 16      SK   M   40      1
## 17      SK   M  60+      1


The display is in the long format (compare Section 10.6.2), because we cannot nicely print a three-dimensional array. Yet, we can always partially unstack the dataset, for aesthetic reasons:

counts3.set_index(["country", "sex", "age"]).unstack()
##              count
## age             20      30      40     50    60+
## country sex
## PL      F    222.0   422.0   248.0   26.0    8.0
##         M    824.0  2081.0  1593.0  475.0  134.0
## SK      F      NaN     NaN     NaN    1.0    NaN
##         M      NaN     NaN     1.0    NaN    1.0
## UA      F      NaN     2.0     NaN    NaN    NaN
##         M      8.0     8.0     2.0    3.0    NaN


Let us again appreciate how versatile is the concept of data frames. Not only can we represent data to be investigated (one row per observation, variables possibly of mixed types), but also we can store the results of such analyses (neatly formatted tables).

## 11.3. Visualising Factors#

Methods for visualising categorical data are by no means fascinating (unless we use them as grouping variables in more complex datasets, but this is a topic that we cover in Chapter 12).

### 11.3.1. Bar Plots#

Example data:

x = (marathon.age.astype("category")
.cat.reorder_categories(["20", "30", "40", "50", "60+"])
.value_counts(sort=False)
)
x
## 20     1119
## 30     2649
## 40     1970
## 50      584
## 60+     189
## Name: age, dtype: int64


Bar plots are self-explanatory and hence will do the trick most of the time; see Figure 11.1.

ind = np.arange(len(x))  # 0, 1, 2, 3, 4
plt.bar(ind, height=x, color="lightgray", edgecolor="black", alpha=0.8)
plt.xticks(ind, x.index)
plt.show()


The ind vector gives the x-coordinates of the bars; here: consecutive integers. By calling matplotlib.pyplot.xticks we assign them readable labels.

Exercise 11.10

Draw a bar plot for the five most prevalent foreign (i.e., excluding Polish) marathoners’ original whereabouts which features an additional bar that represents “all other” countries. Depict percentages instead of counts, so that the total bar height is 100%. Assign a different colour to each bar.

A bar plot is a versatile tool for visualising the counts also in the two-variable case; see Figure 11.2. Let us now use a pleasant wrapper around matplotlib.pyplot.bar offered by the statistical data visualisation package called seaborn [91] (written by Michael Waskom).

v = (marathon.loc[:, ["sex", "age"]].value_counts(sort=False)
.rename("count").reset_index()
)
sns.barplot(x="age", hue="sex", y="count", data=v)
plt.show()


Note

It is customary to call a single function from seaborn and then perform a series of additional calls to matplotlib to tweak the display details. We should remember that the former uses the latter to achieve its goals, not the other way around. seaborn is particularly convenient for plotting grouped data.

Exercise 11.11

(*) Draw a similar chart using matplotlib.pyplot.bar.

Exercise 11.12

(**) Create a stacked bar plot similar to the one in Figure 11.3, where we have horizontal bars for data that have been normalised so that, for each sex, their sum is 100%.

### 11.3.2. Political Marketing and Statistics#

Even such a simple chart as bar plot can be manipulated. In the second round of the presidential elections that were held in Poland in 2020, Andrzej Duda won against Rafał Trzaskowski. In Figure 11.4, we have the official results that might be presented by the pro-government media outlets:

plt.bar([1, 2], height=[51.03, 48.97], width=0.25,
color="lightgray", edgecolor="black", alpha=0.8)
plt.xticks([1, 2], ["Duda", "Trzaskowski"])
plt.ylabel("%")
plt.xlim(0, 3)
plt.ylim(48.9, 51.1)
plt.show()


Another media conglomerate could have reported it like in Figure 11.5:

plt.bar([1, 2], height=[51.03, 48.97], width=0.25,
color="lightgray", edgecolor="black", alpha=0.8)
plt.xticks([1, 2], ["Duda", "Trzaskowski"])
plt.ylabel("%")
plt.xlim(0, 3)
plt.ylim(0, 250)
plt.yticks([0, 100])
plt.show()


Important

We should always read the axis tick marks. And when drawing our own bar plots, we must never trick the reader; this is unethical; compare Rule#9.

### 11.3.3. Pie Cha… Don’t Even Trip#

We are definitely not going to discuss the infamous pie charts, because their use in data analysis has been widely criticised for a long time (it is difficult to judge the ratios of areas of their slices). Do not draw them. Ever. Good morning.

### 11.3.4. Pareto Charts (*)#

As a general (empirical) rule, it is usually the case that most instances of something’s happening (usually 70–90%) are due to only a few causes (10–30%). This is known as the Pareto principle (with 80% vs 20% being an often cited rule of thumb).

Example 11.13

In Chapter 6, we modelled the US cities’ population dataset using the Pareto distribution (the very same Pareto, but a different, yet related mathematical object). We discovered that only ca. 14% of the settlements (those with 10,000 or more inhabitants) are home to as much as 84% of the population. Hence, we may say that this data domain follows the Pareto rule.

Here is a dataset fabricated by the Clinical Excellence Commission in New South Wales, Australia, listing the most frequent causes of medication errors:

cat_med = np.array([
"Unauthorised drug", "Wrong IV rate", "Wrong patient", "Dose missed",
"Underdose", "Wrong calculation","Wrong route", "Wrong drug",
"Wrong time", "Technique error", "Duplicated drugs", "Overdose"
])
counts_med = np.array([1, 4, 53, 92, 7, 16, 27, 76, 83, 3, 9, 59])
np.sum(counts_med)  # total number of medication errors
## 430


Let us display the dataset ordered with respect to the counts, decreasingly:

med = pd.Series(counts_med, index=cat_med).sort_values(ascending=False)
med
## Dose missed          92
## Wrong time           83
## Wrong drug           76
## Overdose             59
## Wrong patient        53
## Wrong route          27
## Wrong calculation    16
## Duplicated drugs      9
## Underdose             7
## Wrong IV rate         4
## Technique error       3
## Unauthorised drug     1
## dtype: int64


Pareto charts may aid in visualising the datasets where the Pareto principle is likely to hold, at least approximately. They include bar plots with some extras:

• bars are listed in decreasing order,

• the cumulative percentage curve is added.

The plotting of the Pareto chart is a little tricky, because it involves using two different Y axes (as usual, fine-tuning the figure and studying the manual of the matplotlib package is left as an exercise.)

x = np.arange(len(med))  # 0, 1, 2, ...
p = 100.0*med/np.sum(med)  # percentages

fig, ax1 = plt.subplots()
ax1.set_xticks(x-0.5, med.index, rotation=60)
ax1.set_ylabel("%")
ax1.bar(x, height=p, color="lightgray", edgecolor="black")
ax2 = ax1.twinx()  # creates a new coordinate system with a shared x-axis
ax2.plot(x, np.cumsum(p), "ro-")
ax2.grid(visible=False)
ax2.set_ylabel("cumulative %")

fig.tight_layout()
plt.show()


In Figure 11.6, we can read that the first five causes (less than 40%) correspond to ca. 85% of the medication errors. More precisely, the cumulative probabilities are:

med.cumsum()/np.sum(med)
## Dose missed          0.213953
## Wrong time           0.406977
## Wrong drug           0.583721
## Overdose             0.720930
## Wrong patient        0.844186
## Wrong route          0.906977
## Wrong calculation    0.944186
## Duplicated drugs     0.965116
## Underdose            0.981395
## Wrong IV rate        0.990698
## Technique error      0.997674
## Unauthorised drug    1.000000
## dtype: float64


Note that there is an explicit assumption here that a single error is only due to a single cause. Also, we presume that each medication error has a similar degree of severity.

Policymakers and quality controllers often rely on similar simplifications. They most probably are going to be addressing only the top causes. If we ever wondered why some processes (mal)function the way they do, there is a hint above. Inventing something more effective yet so simple at the same time requires much more effort.

It would be also nice to report the number of cases where no mistakes are made and the cases where errors are insignificant. Healthcare workers are doing a wonderful job for our communities, especially in the public system. Why add to their stress?

### 11.3.5. Heat Maps#

Two-way contingency tables can be depicted by means of a heatmap, where each count influences the corresponding cell’s colour intensity; see Figure 11.7.

from matplotlib import cm
V = marathon.loc[:, ["sex", "age"]].value_counts().unstack(fill_value=0)
sns.heatmap(V, annot=True, fmt="d", cmap=cm.get_cmap("copper"))
plt.show()


As an exercise, draw a similar heatmap using matplotlib.pyplot.imshow.

## 11.4. Aggregating and Comparing Factors#

### 11.4.1. A Mode#

The only operation on categorical data on which we can rely is counting.

counts = marathon.country.value_counts()
## PL    6033
## GB      71
## DE      38
## FR      33
## SE      30
## Name: country, dtype: int64


Therefore, as far as qualitative data aggregation is concerned, what we are left with is the mode, i.e., the most frequently occurring value.

counts.index[0]  # counts are sorted
## 'PL'


Important

A mode might be ambiguous.

It turns out that amongst the fastest 22 runners (a nicely round number), there is a tie between Kenya and Poland – both meet our definition of a mode:

counts = marathon.country.iloc[:22].value_counts()
counts
## KE    7
## PL    7
## ET    3
## IL    3
## MA    1
## MD    1
## Name: country, dtype: int64


To avoid any bias, it is always best to report all the potential mode candidates:

counts.loc[counts == counts.max()].index
## Index(['KE', 'PL'], dtype='object')


If one value is required, though, we can pick one at random (calling numpy.random.choice).

### 11.4.2. Binary Data as Logical Vectors#

Recall that we are used to representing binary data as logical vectors or, equivalently, vectors of 0s and 1s.

Perhaps the most useful arithmetic operation on logical vectors is the sum. For example:

np.sum(marathon.country == "PL")
## 6033


This gave the number of elements that are equal to "PL" (because the sum of 0s and 1s is equal to the number of 1s in the sequence). Note that (country == "PL") is a logical vector that represents a binary categorical variable with levels: not-Poland (False) and Poland (True).

If we divide the above result by the length of the vector, we will get the proportion:

np.mean(marathon.country == "PL")
## 0.9265857779142989


About 93% of the runners were from Poland. As this is greater than 0.5, "PL” is definitely the mode.

Exercise 11.14

What is the meaning of numpy.all, numpy.any, numpy.min, numpy.max, numpy.cumsum, and numpy.cumprod applied on logical vectors?

Note

(**) Having the 0/1 (or zero/nonzero) vs False/True correspondence allows us to perform some logical operations using integer arithmetic. In mathematics, 0 is the annihilator of multiplication and the neutral element of addition, whereas 1 is the neutral element of multiplication. In particular, assuming that p and q are logical values and a and b are numeric ones, we have, what follows:

• p+q != 0 means that at least one value is True and p+q == 0 if and only if both are False;

• more generally, p+q == 2 if both elements are True, p+q == 1 if only one is True (we call it exclusive-or, XOR), and p+q == 0 if both are False;

• p*q != 0 means that both values are True and p*q == 0 holds whenever at least one is False;

• 1-p corresponds to the negation of p (changes 1 to 0 and 0 to 1);

• p*a + (1-p)*b is equal to a if p is True and equal to b otherwise.

### 11.4.3. Pearson’s Chi-Squared Test (*)#

The Kolmogorov–Smirnov test that we described in Section 6.2.3 verifies whether a given sample differs significantly from a hypothesised continuous[2] distribution, i.e., it works for numeric data.

For binned/categorical data, we can use a classical and easy-to-understand test developed by Karl Pearson in 1900. It is supposed to judge whether the differences between the observed proportions $$\hat{p}_1,\dots,\hat{p}_l$$ and the theoretical ones $$p_1,\dots,p_l$$ are significantly large or not:

$\begin{split} \left\{ \begin{array}{rll} H_0: & \hat{p}_i=p_i\ \text{ for all }i=1,\dots,l & \text{(null hypothesis)}\\ H_1: & \hat{p}_i\neq p_i\ \text{ for some }i=1,\dots,l & \text{(alternative hypothesis)} \\ \end{array} \right. \end{split}$

Having such a test is beneficial, e.g., when the data we have at hand are based on small surveys that are supposed to serve as estimates of what might be happening in a larger population.

Going back to our political example from Section 11.3.2, it turns out that one of the pre-election polls indicated that $$c=516$$ out of $$n=1017$$ people would vote for the first candidate. We have $$\hat{p}_1=50.74\%$$ (Duda) and $$\hat{p}_2=49.26\%$$ (Trzaskowski). If we would like to test whether the observed proportions are significantly different from each other, we could test them against the theoretical distribution $$p_1=50\%$$ and $$p_2=50\%$$, which states that there is a tie between the competitors (up to a sampling error).

A natural test statistic is based on the relative squared differences:

$\hat{T} = n \sum_{i=1}^l \frac{\left(\hat{p}_i-p_i\right)^2}{p_i}.$
c, n = 516, 1017
p_observed = np.array([c, n-c]) / n
p_expected = np.array([0.5, 0.5])
T = n * np.sum( (p_observed-p_expected)**2 / p_expected )
T
## 0.2212389380530986


Similarly to the continuous case in Section 6.2.3, we should reject the null hypothesis, if:

$\hat{T} \ge K.$

The critical value $$K$$ is computed based on the fact that, if the null hypothesis is true, $$\hat{T}$$ follows the $$\chi^2$$ (chi-squared, hence the name of the test) distribution with $$l-1$$ degrees of freedom, see scipy.stats.chi2.

We thus need to query the theoretical quantile function to determine the test statistic that is not exceeded in 99.9% of the trials (under the null hypothesis):

alpha = 0.001  # significance level
scipy.stats.chi2.ppf(1-alpha, len(p_observed)-1)
## 10.827566170662733


As $$\hat{T} < K$$ (because $$0.22 < 10.83$$), we cannot deem the two proportions significantly different. In other words, this poll did not indicate (at the significance level $$0.1\%$$) any of the candidates as a clear winner.

Exercise 11.15

Assuming $$n=1017$$, determine the smallest $$c$$, i.e., the number of respondents claiming they would vote for Duda, that leads to the rejection of the null hypothesis.

### 11.4.4. Two-Sample Pearson’s Chi-Squared Test (*)#

Let us consider the data depicted in Figure 11.3 and test whether the runners’ age distributions differ significantly between men and women.

V = marathon.loc[:, ["sex", "age"]].value_counts().unstack(fill_value=0)
c1, c2 = np.array(V)  # first row, second row
c1  # women
## array([240, 449, 262,  43,  19])
c2  # men
## array([ 879, 2200, 1708,  541,  170])


There are $$l=5$$ age categories. First, denote the total number of observations in both groups with $$n'$$ and $$n''$$.

n1 = c1.sum()
n2 = c2.sum()
n1, n2
## (1013, 5498)


The observed proportions in the first group (females), denoted as $$\hat{p}_1',\dots,\hat{p}_l'$$, are, respectively:

p1 = c1/n1
p1
## array([0.23692004, 0.44323791, 0.25863771, 0.04244817, 0.01875617])


Here are the proportions in the second group (males), $$\hat{p}_1'',\dots,\hat{p}_l''$$:

p2 = c2/n2
p2
## array([0.15987632, 0.40014551, 0.31065842, 0.09839942, 0.03092033])


We would like to verify whether the corresponding proportions are equal (up to some sampling error):

$\begin{split} \left\{ \begin{array}{rll} H_0: & \hat{p}_i'=\hat{p}_i''\ \text{ for all }i=1,\dots,l & \text{(null hypothesis)}\\ H_1: & \hat{p}_i'\neq \hat{p}_i''\ \text{ for some }i=1,\dots,l & \text{(alternative hypothesis)} \\ \end{array} \right. \end{split}$

In other words, we are interested whether the categorical data in the two groups come from the same discrete probability distribution.

Taking the estimated expected proportions:

$\bar{p}_i = \frac{n_i' \hat{p}_i' + n_i'' \hat{p}_i''}{n'+n''},$

for all $$i=1,\dots,l$$, the test statistic this time is equal to:

$\hat{T} = n' \sum_{i=1}^l \frac{\left(\hat{p}_i' -\bar{p}_i\right)^2}{\bar{p}_i} + n'' \sum_{i=1}^l \frac{\left(\hat{p}_i''-\bar{p}_i\right)^2}{\bar{p}_i},$

which is a variation on the one-sample theme presented in the previous subsection.

pp = (n1*p1+n2*p2)/(n1+n2)
T = n1 * np.sum( (p1-pp)**2 / pp ) + n2 * np.sum( (p2-pp)**2 / pp )
T
## 75.31373854741857


It can be shown that, if the null hypothesis is true, the test statistic approximately follows the $$\chi^2$$ distribution with $$l-1$$ degrees of freedom[3]. The critical value $$K$$ is equal to:

alpha = 0.001  # significance level
scipy.stats.chi2.ppf(1-alpha, len(p1)-1)
## 18.46682695290317


As $$\hat{T} \ge K$$ (because $$75.31 \ge 18.47$$), we reject the null hypothesis. And so, the runners’ age distribution differs across sexes (at significance level 0.1%).

### 11.4.5. Measuring Association (*)#

Let us consider the Australian Bureau of Statistics National Health Survey 2018 data on the prevalence of certain medical conditions as a function of age. Here is the extracted contingency table:

l = [
["Arthritis", "Asthma", "Back problems", "Cancer (malignant neoplasms)",
"Chronic obstructive pulmonary disease", "Diabetes mellitus",
"Heart, stroke and vascular disease", "Kidney disease",
"Mental and behavioural conditions", "Osteoporosis"],
["15-44", "45-64", "65+"]
]
C = 1000*np.array([
[ 360.2,     1489.0,      1772.2],
[1069.7,      741.9,       433.7],
[1469.6,     1513.3,       955.3],
[  28.1,      162.7,       237.5],
[ 103.8,      207.0,       251.9],
[ 135.4,      427.3,       607.7],
[  94.0,      344.4,       716.0],
[  29.6,       67.7,       123.3],
[2218.9,     1390.6,       725.0],
[  36.1,      312.3,       564.7],
]).astype(int)
pd.DataFrame(C, index=l[0], columns=l[1])
##                                          15-44    45-64      65+
## Arthritis                               360000  1489000  1772000
## Asthma                                 1069000   741000   433000
## Back problems                          1469000  1513000   955000
## Cancer (malignant neoplasms)             28000   162000   237000
## Chronic obstructive pulmonary disease   103000   207000   251000
## Diabetes mellitus                       135000   427000   607000
## Heart, stroke and vascular disease       94000   344000   716000
## Kidney disease                           29000    67000   123000
## Mental and behavioural conditions      2218000  1390000   725000
## Osteoporosis                             36000   312000   564000


Cramér’s $$V$$ is one of a few ways to measure the degree of association between two categorical variables. It is equal to 0 (the lowest possible value) if the two variables are independent (there is no association between them) and 1 (the highest possible value) if they are tied.

Given a two-way contingency table $$C$$ with $$n$$ rows and $$m$$ columns and assuming that:

$T = \sum_{i=1}^n \sum_{j=1}^m \frac{(c_{i,j} - e_{i,j})^2}{e_{i,j}},$

where:

$e_{i,j} = \frac{ \left(\sum_{k=1}^m c_{i,k}\right) \left(\sum_{k=1}^n c_{k,j}\right) }{ \sum_{i=1}^n \sum_{j=1}^m c_{i,j} },$

the Cramér coefficient is given by:

$V = \sqrt{ \frac{T}{\min\{n-1, m-1\} \sum_{i=1}^n \sum_{j=1}^m c_{i,j}} }.$

Here, $$c_{i,j}$$ gives the actually observed counts and $$e_{i, j}$$ denotes the number that we would expect to see if the two variables were really independent.

scipy.stats.contingency.association(C)
## 0.316237999724298


The above means that there might be a small association between age and the prevalence of certain conditions. In other words, it might be the case that some conditions are more prevalent in different age groups than others.

Exercise 11.16

Compute the Cramér $$V$$ using only numpy functions.

Example 11.17

(**) We can easily verify the hypothesis whether $$V$$ does not differ significantly from $$0$$, i.e., whether the variables are independent. Looking at $$T$$, we see that this is essentially the test statistic in Pearson’s chi-squared goodness-of-fit test.

E = C.sum(axis=1).reshape(-1, 1) * C.sum(axis=0).reshape(1, -1) / C.sum()
T = np.sum((C-E)**2 / E)
T
## 3715440.465191512


If the data are really independent, $$T$$ follows the chi-squared distribution $$n + m - 1$$. As a consequence, the critical value $$K$$ is equal to:

alpha = 0.001  # significance level
scipy.stats.chi2.ppf(1-alpha, C.shape[0] + C.shape[1] - 1)
## 32.90949040736021


As $$T$$ is much greater than $$K$$, we conclude (at significance level $$0.1\%$$) that the health conditions are not independent of age.

Exercise 11.18

(*) Take a look at Table 19: Comorbidity of selected chronic conditions in the National Health Survey 2018, where we clearly see that many disorders co-occur. Visualise them on some heatmaps and bar plots (including data grouped by sex and age).

### 11.4.6. Binned Numeric Data#

Generally, modes are meaningless for continuous data, where repeated values are – at least theoretically – highly unlikely. It might make sense to compute them on binned data, though.

Looking at a histogram, e.g., in Figure 4.2, the mode is the interval corresponding to the highest bar (hopefully assuming there is only one). If we would like to obtain a single number, we can choose for example the middle of this interval as the mode.

For numeric data, the mode will heavily depend on the coarseness and type of binning; compare Figure 4.4 and Figure 6.8. Thus, the question “what is the most popular income” is overall a quite difficult one to answer.

Exercise 11.19

Compute some informative modes for the uk_income_simulated_2020 dataset. Play around with different numbers of bins on linear and logarithmic scales and see how they affect the mode.

### 11.4.7. Ordinal Data (*)#

The case where the categories can be linearly ordered, is called ordinal data. For instance, Australian university grades are: F (fail) < P (pass) < C (credit) < D (distinction) < HD (high distinction), some questionnaires use Likert-type scales such as “strongly disagree” < “disagree” < “neutral” < “agree” < “strongly agree”, etc.

A linear ordering allows us to go beyond the mode. This is because, due to the existence of order statistics and observation ranks, we can also easily define sample quantiles. Nevertheless, the standard methods for resolving ties will not work: we need to be careful.

For example, the median of a sample of student grades (P, P, C, D, HD) is C, but (P, P, C, D, HD, HD) is either C or D - we can choose one at random or just report that the solution is ambiguous (C+? D-? C/D?).

Another option, of course, is to treat ordinal data as numbers (e.g., F = 0, P = 1, …, HD = 4). In the latter example, the median would be equal to 2.5.

There are some cases, though, where the conversion of labels to consecutive integers is far from optimal – because it gives the impression that the “distance” between different levels is always equal (linear).

Exercise 11.20

(**) The grades_results dataset represents the grades (F, P, C, D, HD) of 100 students attending an imaginary course in an Australian university. You can load it in the form of an ordered categorical Series by calling:

grades = np.genfromtxt("https://raw.githubusercontent.com/gagolews/" +
categories=["F", "P", "C", "D", "HD"], ordered=True))
grades.value_counts()  # note the order of labels
## F     30
## P     29
## C     23
## HD    22
## D     19
## dtype: int64


How would you determine the average grade represented as a number between 0 and 100, taking into account that for a P you need at least 50%, C is given for ≥ 60%, D for ≥ 70%, and HD for only (!) 80% of the points. Come up with a pessimistic, optimistic, and best-shot estimate, and then compare your result to the true corresponding scores listed in the grades_scores dataset.

## 11.5. Exercises#

Exercise 11.21

Does it make sense to compute the arithmetic mean of a categorical variable?

Exercise 11.22

Name the basic use cases for categorical data.

Exercise 11.23

(*) What is a Pareto chart?

Exercise 11.24

How can we deal with the case of the mode being nonunique?

Exercise 11.25

What is the meaning of the sum and mean for binary data (logical vectors)?

Exercise 11.26

What is the meaning of numpy.mean((x > 0) & (x < 1)), where x is a numeric vector?

Exercise 11.27

List some ways to visualise multidimensional categorical data (combinations of two or more factors).

Exercise 11.28

(*) State the null hypotheses verified by the one- and two-sample chi-squared goodness-of-fit tests.

Exercise 11.29

(*) How is Cramér’s V defined and what values does it take?