我试图使用scikit-learn的LabelEncoder来编码字符串标签的pandas DataFrame。由于数据帧有许多(50+)列，我想避免为每一列创建一个LabelEncoder对象;我宁愿只有一个大的LabelEncoder对象，它可以跨所有数据列工作。
将整个DataFrame扔到LabelEncoder中会产生以下错误。请记住，我在这里使用的是虚拟数据;实际上，我正在处理大约50列的字符串标记数据，所以需要一个解决方案，不引用任何列的名称。
import pandas
from sklearn import preprocessing
df = pandas.DataFrame({
'pets': ['cat', 'dog', 'cat', 'monkey', 'dog', 'dog'],
'owner': ['Champ', 'Ron', 'Brick', 'Champ', 'Veronica', 'Ron'],
'location': ['San_Diego', 'New_York', 'New_York', 'San_Diego', 'San_Diego',
'New_York']
})
le = preprocessing.LabelEncoder()
le.fit(df)
回溯(最近一次调用):
文件“”，第1行，在
文件"/Users/bbalin/anaconda/lib/python2.7/site-packages/sklearn/预处理/label.py"，第103行
y = column_or_1d(y, warn=True)
文件"/Users/bbalin/anaconda/lib/python2.7/site-packages/sklearn/utils/validation.py"，第306行，在column_or_1d中
raise ValueError("错误的输入形状{0}".format(形状))
ValueError:错误的输入形状(6,3)
对于如何解决这个问题有什么想法吗?
根据对@PriceHardman解决方案提出的意见，我将提出以下版本的类:
class LabelEncodingColoumns(BaseEstimator, TransformerMixin):
def __init__(self, cols=None):
pdu._is_cols_input_valid(cols)
self.cols = cols
self.les = {col: LabelEncoder() for col in cols}
self._is_fitted = False
def transform(self, df, **transform_params):
"""
Scaling ``cols`` of ``df`` using the fitting
Parameters
----------
df : DataFrame
DataFrame to be preprocessed
"""
if not self._is_fitted:
raise NotFittedError("Fitting was not preformed")
pdu._is_cols_subset_of_df_cols(self.cols, df)
df = df.copy()
label_enc_dict = {}
for col in self.cols:
label_enc_dict[col] = self.les[col].transform(df[col])
labelenc_cols = pd.DataFrame(label_enc_dict,
# The index of the resulting DataFrame should be assigned and
# equal to the one of the original DataFrame. Otherwise, upon
# concatenation NaNs will be introduced.
index=df.index
)
for col in self.cols:
df[col] = labelenc_cols[col]
return df
def fit(self, df, y=None, **fit_params):
"""
Fitting the preprocessing
Parameters
----------
df : DataFrame
Data to use for fitting.
In many cases, should be ``X_train``.
"""
pdu._is_cols_subset_of_df_cols(self.cols, df)
for col in self.cols:
self.les[col].fit(df[col])
self._is_fitted = True
return self
这个类适合编码器的训练集，并在转换时使用适合的版本。代码的初始版本可以在这里找到。
根据对@PriceHardman解决方案提出的意见，我将提出以下版本的类:
class LabelEncodingColoumns(BaseEstimator, TransformerMixin):
def __init__(self, cols=None):
pdu._is_cols_input_valid(cols)
self.cols = cols
self.les = {col: LabelEncoder() for col in cols}
self._is_fitted = False
def transform(self, df, **transform_params):
"""
Scaling ``cols`` of ``df`` using the fitting
Parameters
----------
df : DataFrame
DataFrame to be preprocessed
"""
if not self._is_fitted:
raise NotFittedError("Fitting was not preformed")
pdu._is_cols_subset_of_df_cols(self.cols, df)
df = df.copy()
label_enc_dict = {}
for col in self.cols:
label_enc_dict[col] = self.les[col].transform(df[col])
labelenc_cols = pd.DataFrame(label_enc_dict,
# The index of the resulting DataFrame should be assigned and
# equal to the one of the original DataFrame. Otherwise, upon
# concatenation NaNs will be introduced.
index=df.index
)
for col in self.cols:
df[col] = labelenc_cols[col]
return df
def fit(self, df, y=None, **fit_params):
"""
Fitting the preprocessing
Parameters
----------
df : DataFrame
Data to use for fitting.
In many cases, should be ``X_train``.
"""
pdu._is_cols_subset_of_df_cols(self.cols, df)
for col in self.cols:
self.les[col].fit(df[col])
self._is_fitted = True
return self
这个类适合编码器的训练集，并在转换时使用适合的版本。代码的初始版本可以在这里找到。
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
