如果我们有单列来做标签编码和它的逆变换，当python中有多列时，很容易做到这一点

def stringtocategory(dataset):
    '''
    @author puja.sharma
    @see The function label encodes the object type columns and gives label      encoded and inverse tranform of the label encoded data
    @param dataset dataframe on whoes column the label encoding has to be done
    @return label encoded and inverse tranform of the label encoded data.
   ''' 
   data_original = dataset[:]
   data_tranformed = dataset[:]
   for y in dataset.columns:
       #check the dtype of the column object type contains strings or chars
       if (dataset[y].dtype == object):
          print("The string type features are  : " + y)
          le = preprocessing.LabelEncoder()
          le.fit(dataset[y].unique())
          #label encoded data
          data_tranformed[y] = le.transform(dataset[y])
          #inverse label transform  data
          data_original[y] = le.inverse_transform(data_tranformed[y])
   return data_tranformed,data_original

在这里和其他地方进行了大量的搜索和实验后，我认为你的答案是:

pd.DataFrame(列= df.columns, data = LabelEncoder () .fit_transform (df.values.flatten ()) .reshape (df.shape))

这将跨列保留类别名称:

import pandas as pd
from sklearn.preprocessing import LabelEncoder

df = pd.DataFrame([['A','B','C','D','E','F','G','I','K','H'],
                   ['A','E','H','F','G','I','K','','',''],
                   ['A','C','I','F','H','G','','','','']], 
                  columns=['A1', 'A2', 'A3','A4', 'A5', 'A6', 'A7', 'A8', 'A9', 'A10'])

pd.DataFrame(columns=df.columns, data=LabelEncoder().fit_transform(df.values.flatten()).reshape(df.shape))

    A1  A2  A3  A4  A5  A6  A7  A8  A9  A10
0   1   2   3   4   5   6   7   9   10  8
1   1   5   8   6   7   9   10  0   0   0
2   1   3   9   6   8   7   0   0   0   0

下面是我一次性转换多列的解决方案，以及精确的inverse_transform

from sklearn import preprocessing
columns = ['buying','maint','lug_boot','safety','cls']  # columns names where transform is required
for X in columns:
  exec(f'le_{X} = preprocessing.LabelEncoder()')  #create label encoder with name "le_X", where X is column name
  exec(f'df.{X} = le_{X}.fit_transform(df.{X})')  #execute fit transform for column X with respective lable encoder "le_X", where X is column name
df.head()  # to display transformed results

for X in columns:
  exec(f'df.{X} = le_{X}.inverse_transform(df.{X})')  #execute inverse_transform for column X with respective lable encoder "le_X", where X is column name
df.head() # to display Inverse transformed results of df

这并没有直接回答你的问题(Naputipulu Jon和PriceHardman对此有精彩的回答)

但是，对于一些分类任务等，您可以使用

pandas.get_dummies(input_df) 

这可以输入带有分类数据的数据框架，并返回带有二进制值的数据框架。变量值被编码到结果数据框架中的列名中。更多的

主要使用@Alexander回答，但必须做一些更改-

cols_need_mapped = ['col1', 'col2']

mapper = {col: {cat: n for n, cat in enumerate(df[col].astype('category').cat.categories)} 
     for col in df[cols_need_mapped]}

for c in cols_need_mapped :
    df[c] = df[c].map(mapper[c])

然后，为了将来重用，你可以将输出保存到json文档中，当你需要它时，你可以读入并使用.map()函数，就像我上面所做的那样。

从scikit-learn 0.20开始，你可以使用sklearn.compose.ColumnTransformer和sklearn.预处理. onehotencoder:

如果你只有分类变量，OneHotEncoder直接:

from sklearn.preprocessing import OneHotEncoder

OneHotEncoder(handle_unknown='ignore').fit_transform(df)

如果你有异构类型的特性:

from sklearn.compose import make_column_transformer
from sklearn.preprocessing import RobustScaler
from sklearn.preprocessing import OneHotEncoder

categorical_columns = ['pets', 'owner', 'location']
numerical_columns = ['age', 'weigth', 'height']
column_trans = make_column_transformer(
    (categorical_columns, OneHotEncoder(handle_unknown='ignore'),
    (numerical_columns, RobustScaler())
column_trans.fit_transform(df)

文档中有更多选项:http://scikit-learn.org/stable/modules/compose.html#columntransformer-for-heterogeneous-data