在Pandas数据框中转换分类数据

I have a dataframe with this type of data (too many columns):

col1        int64
col2        int64
col3        category
col4        category
col5        category

Columns seems like this:

Name: col3, dtype: category
Categories (8, object): [B, C, E, G, H, N, S, W]

I want to convert all value in columns to integer like this:

[1, 2, 3, 4, 5, 6, 7, 8]

I solved this for one column by this:

dataframe['c'] = pandas.Categorical.from_array(dataframe.col3).codes

Now I have two columns in my dataframe – old col3 and new c and need to drop old columns.

That’s bad practice. It’s work but in my dataframe many columns and I don’t want do it manually.

How do this pythonic and just cleverly?

First, to convert a Categorical column to its numerical codes, you can do this easier with: dataframe['c'].cat.codes.
Further, it is possible to select automatically all columns with a certain dtype in a dataframe using select_dtypes. This way, you can apply above operation on multiple and automatically selected columns.

First making an example dataframe:

In [75]: df = pd.DataFrame({'col1':[1,2,3,4,5], 'col2':list('abcab'),  'col3':list('ababb')})

In [76]: df['col2'] = df['col2'].astype('category')

In [77]: df['col3'] = df['col3'].astype('category')

In [78]: df.dtypes
Out[78]:
col1       int64
col2    category
col3    category
dtype: object

Then by using select_dtypes to select the columns, and then applying .cat.codes on each of these columns, you can get the following result:

In [80]: cat_columns = df.select_dtypes(['category']).columns

In [81]: cat_columns
Out[81]: Index([u'col2', u'col3'], dtype='object')

In [83]: df[cat_columns] = df[cat_columns].apply(lambda x: x.cat.codes)

In [84]: df
Out[84]:
   col1  col2  col3
0     1     0     0
1     2     1     1
2     3     2     0
3     4     0     1
4     5     1     1

This works for me:

pandas.factorize( ['B', 'C', 'D', 'B'] )[0]

Output:

[0, 1, 2, 0]

If your concern was only that you making a extra column and deleting it later, just dun use a new column at the first place.

dataframe = pd.DataFrame({'col1':[1,2,3,4,5], 'col2':list('abcab'),  'col3':list('ababb')})
dataframe.col3 = pd.Categorical.from_array(dataframe.col3).codes

You are done. Now as Categorical.from_array is deprecated, use Categorical directly

dataframe.col3 = pd.Categorical(dataframe.col3).codes

If you also need the mapping back from index to label, there is even better way for the same

dataframe.col3, mapping_index = pd.Series(dataframe.col3).factorize()

check below

print(dataframe)
print(mapping_index.get_loc("c"))

Here multiple columns need to be converted. So, one approach i used is ..

for col_name in df.columns:
    if(df[col_name].dtype == 'object'):
        df[col_name]= df[col_name].astype('category')
        df[col_name] = df[col_name].cat.codes

This converts all string / object type columns to categorical. Then applies codes to each type of category.

For converting categorical data in column C of dataset data, we need to do the following:

from sklearn.preprocessing import LabelEncoder 
labelencoder= LabelEncoder() #initializing an object of class LabelEncoder
data['C'] = labelencoder.fit_transform(data['C']) #fitting and transforming the desired categorical column.

@Quickbeam2k1 ,see below –

dataset=pd.read_csv('Data2.csv')
np.set_printoptions(threshold=np.nan)
X = dataset.iloc[:,:].values

Using sklearn

from sklearn.preprocessing import LabelEncoder
labelencoder_X=LabelEncoder()
X[:,0] = labelencoder_X.fit_transform(X[:,0])

What I do is, I replace values.

Like this-

df['col'].replace(to_replace=['category_1', 'category_2', 'category_3'], value=[1, 2, 3], inplace=True)

In this way, if the col column has categorical values, they get replaced by the numerical values.

For a certain column, if you don’t care about the ordering, use this

df['col1_num'] = df['col1'].apply(lambda x: np.where(df['col1'].unique()==x)[0][0])

If you care about the ordering, specify them as a list and use this

df['col1_num'] = df['col1'].apply(lambda x: ['first', 'second', 'third'].index(x))