Handling categorical data

So far, we have only been working with numerical values. However, it is not uncommon that real-world datasets contain one or more categorical feature columns. When we are talking about categorical data, we have to further distinguish between nominal and ordinal features. Ordinal features can be understood as categorical values that can be sorted or ordered. For example, T-shirt size would be an ordinal feature, because we can define an order XL > L > M. In contrast, nominal features don't imply any order and, to continue with the previous example, we could think of T-shirt color as a nominal feature since it typically doesn't make sense to say that, for example, red is larger than blue.

Before we explore different techniques to handle such categorical data, let's create a new data frame to illustrate the problem:

>>> import pandas as pd
>>> df = pd.DataFrame([
...            ['green', 'M', 10.1, 'class1'], 
...            ['red', 'L', 13.5, 'class2'], 
...            ['blue', 'XL', 15.3, 'class1']])
>>> df.columns = ['color', 'size', 'price', 'classlabel']
>>> df
   color size  price classlabel
0  green    M   10.1     class1
1    red    L   13.5     class2
2   blue   XL   15.3     class1

As we can see in the preceding output, the newly created DataFrame contains a nominal feature (color), an ordinal feature (size), and a numerical feature (price) column. The class labels (assuming that we created a dataset for a supervised learning task) are stored in the last column. The learning algorithms for classification that we discuss in this book do not use ordinal information in class labels.

Mapping ordinal features

To make sure that the learning algorithm interprets the ordinal features correctly, we need to convert the categorical string values into integers. Unfortunately, there is no convenient function that can automatically derive the correct order of the labels of our size feature. Thus, we have to define the mapping manually. In the following simple example, let's assume that we know the difference between features, for example, Mapping ordinal features.

>>> size_mapping = {
...                 'XL': 3,
...                 'L': 2,
...                 'M': 1}
>>> df['size'] = df['size'].map(size_mapping)
>>> df
   color  size  price classlabel
0  green     1   10.1     class1
1    red     2   13.5     class2
2   blue     3   15.3     class1

If we want to transform the integer values back to the original string representation at a later stage, we can simply define a reverse-mapping dictionary inv_size_mapping = {v: k for k, v in size_mapping.items()} that can then be used via the pandas' map method on the transformed feature column similar to the size_mapping dictionary that we used previously.

Encoding class labels

Many machine learning libraries require that class labels are encoded as integer values. Although most estimators for classification in scikit-learn convert class labels to integers internally, it is considered good practice to provide class labels as integer arrays to avoid technical glitches. To encode the class labels, we can use an approach similar to the mapping of ordinal features discussed previously. We need to remember that class labels are not ordinal, and it doesn't matter which integer number we assign to a particular string-label. Thus, we can simply enumerate the class labels starting at 0:

>>> import numpy as np
>>> class_mapping = {label:idx for idx,label in
...                  enumerate(np.unique(df['classlabel']))}
>>> class_mapping
{'class1': 0, 'class2': 1}

Next we can use the mapping dictionary to transform the class labels into integers:

>>> df['classlabel'] = df['classlabel'].map(class_mapping)
>>> df
   color  size  price  classlabel
0  green     1   10.1           0
1    red     2   13.5           1
2   blue     3   15.3           0

We can reverse the key-value pairs in the mapping dictionary as follows to map the converted class labels back to the original string representation:

>>> inv_class_mapping = {v: k for k, v in class_mapping.items()}
>>> df['classlabel'] = df['classlabel'].map(inv_class_mapping)
>>> df
   color  size  price classlabel
0  green     1   10.1     class1
1    red     2   13.5     class2
2   blue     3   15.3     class1

Alternatively, there is a convenient LabelEncoder class directly implemented in scikit-learn to achieve the same:

>>> from sklearn.preprocessing import LabelEncoder
>>> class_le = LabelEncoder()
>>> y = class_le.fit_transform(df['classlabel'].values)
>>> y
array([0, 1, 0])

Note that the fit_transform method is just a shortcut for calling fit and transform separately, and we can use the inverse_transform method to transform the integer class labels back into their original string representation:

>>> class_le.inverse_transform(y)
array(['class1', 'class2', 'class1'], dtype=object)

Performing one-hot encoding on nominal features

In the previous section, we used a simple dictionary-mapping approach to convert the ordinal size feature into integers. Since scikit-learn's estimators treat class labels without any order, we used the convenient LabelEncoder class to encode the string labels into integers. It may appear that we could use a similar approach to transform the nominal color column of our dataset, as follows:

>>> X = df[['color', 'size', 'price']].values
>>> color_le = LabelEncoder()
>>> X[:, 0] = color_le.fit_transform(X[:, 0])
>>> X
array([[1, 1, 10.1],
       [2, 2, 13.5],
       [0, 3, 15.3]], dtype=object)

After executing the preceding code, the first column of the NumPy array X now holds the new color values, which are encoded as follows:

  • blue → 0
  • green → 1
  • red → 2

If we stop at this point and feed the array to our classifier, we will make one of the most common mistakes in dealing with categorical data. Can you spot the problem? Although the color values don't come in any particular order, a learning algorithm will now assume that green is larger than blue, and red is larger than green. Although this assumption is incorrect, the algorithm could still produce useful results. However, those results would not be optimal.

A common workaround for this problem is to use a technique called one-hot encoding. The idea behind this approach is to create a new dummy feature for each unique value in the nominal feature column. Here, we would convert the color feature into three new features: blue, green, and red. Binary values can then be used to indicate the particular color of a sample; for example, a blue sample can be encoded as blue=1, green=0, red=0. To perform this transformation, we can use the OneHotEncoder that is implemented in the scikit-learn.preprocessing module:

>>> from sklearn.preprocessing import OneHotEncoder
>>> ohe = OneHotEncoder(categorical_features=[0])
>>> ohe.fit_transform(X).toarray()
array([[  0. ,   1. ,   0. ,   1. ,  10.1],
       [  0. ,   0. ,   1. ,   2. ,  13.5],
       [  1. ,   0. ,   0. ,   3. ,  15.3]])

When we initialized the OneHotEncoder, we defined the column position of the variable that we want to transform via the categorical_features parameter (note that color is the first column in the feature matrix X). By default, the OneHotEncoder returns a sparse matrix when we use the transform method, and we converted the sparse matrix representation into a regular (dense) NumPy array for the purposes of visualization via the toarray method. Sparse matrices are simply a more efficient way of storing large datasets, and one that is supported by many scikit-learn functions, which is especially useful if it contains a lot of zeros. To omit the toarray step, we could initialize the encoder as OneHotEncoder(…,sparse=False) to return a regular NumPy array.

An even more convenient way to create those dummy features via one-hot encoding is to use the get_dummies method implemented in pandas. Applied on a DataFrame, the get_dummies method will only convert string columns and leave all other columns unchanged:

>>> pd.get_dummies(df[['price', 'color', 'size']])
   price  size  color_blue  color_green  color_red
0   10.1     1           0            1          0
1   13.5     2           0            0          1
2   15.3     3           1            0          0
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