用sklearn缩放的pandas数据框列-Python 实用宝典

用sklearn缩放的pandas数据框列

我有一个带有混合类型列的pandas数据框,我想将sklearn的min_max_scaler应用于某些列。理想情况下,我想就地进行这些转换,但还没有找到一种方法来进行。我编写了以下有效的代码: import pandas as pd import numpy as np from sklearn import preprocessing scaler = preprocessing.MinMaxScaler() dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']}) min_max_scaler = preprocessing.MinMaxScaler() def scaleColumns(df, cols_to_scale): for col in cols_to_scale: df[col] = pd.DataFrame(min_max_scaler.fit_transform(pd.DataFrame(dfTest[col])),columns=[col]) return df dfTest A B C 0 14.00 103.02 big 1 90.20 107.26 small 2 90.95 110.35 …

问题:用sklearn缩放的pandas数据框列

我有一个带有混合类型列的pandas数据框,我想将sklearn的min_max_scaler应用于某些列。理想情况下,我想就地进行这些转换,但还没有找到一种方法来进行。我编写了以下有效的代码:

import pandas as pd
import numpy as np
from sklearn import preprocessing

scaler = preprocessing.MinMaxScaler()

dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
min_max_scaler = preprocessing.MinMaxScaler()

def scaleColumns(df, cols_to_scale):
    for col in cols_to_scale:
        df[col] = pd.DataFrame(min_max_scaler.fit_transform(pd.DataFrame(dfTest[col])),columns=[col])
    return df

dfTest

    A   B   C
0    14.00   103.02  big
1    90.20   107.26  small
2    90.95   110.35  big
3    96.27   114.23  small
4    91.21   114.68  small

scaled_df = scaleColumns(dfTest,['A','B'])
scaled_df

A   B   C
0    0.000000    0.000000    big
1    0.926219    0.363636    small
2    0.935335    0.628645    big
3    1.000000    0.961407    small
4    0.938495    1.000000    small

我很好奇这是否是进行此转换的首选/最有效的方法。有没有办法可以使用df.apply更好呢?

我也很惊讶我无法使用以下代码:

bad_output = min_max_scaler.fit_transform(dfTest['A'])

如果我将整个数据帧传递给缩放器,则它可以工作:

dfTest2 = dfTest.drop('C', axis = 1) good_output = min_max_scaler.fit_transform(dfTest2) good_output

我很困惑为什么将系列传递给定标器会失败。在上面的完整工作代码中,我希望只将一个系列传递给缩放器,然后将dataframe column =设置为缩放的序列。我已经看到这个问题在其他几个地方问过,但找不到一个好的答案。任何帮助了解这里发生的事情将不胜感激!

I have a pandas dataframe with mixed type columns, and I'd like to apply sklearn's min_max_scaler to some of the columns. Ideally, I'd like to do these transformations in place, but haven't figured out a way to do that yet. I've written the following code that works:

import pandas as pd
import numpy as np
from sklearn import preprocessing

scaler = preprocessing.MinMaxScaler()

dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
min_max_scaler = preprocessing.MinMaxScaler()

def scaleColumns(df, cols_to_scale):
    for col in cols_to_scale:
        df[col] = pd.DataFrame(min_max_scaler.fit_transform(pd.DataFrame(dfTest[col])),columns=[col])
    return df

dfTest

    A   B   C
0    14.00   103.02  big
1    90.20   107.26  small
2    90.95   110.35  big
3    96.27   114.23  small
4    91.21   114.68  small

scaled_df = scaleColumns(dfTest,['A','B'])
scaled_df

A   B   C
0    0.000000    0.000000    big
1    0.926219    0.363636    small
2    0.935335    0.628645    big
3    1.000000    0.961407    small
4    0.938495    1.000000    small

I'm curious if this is the preferred/most efficient way to do this transformation. Is there a way I could use df.apply that would be better?

I'm also surprised I can't get the following code to work:

bad_output = min_max_scaler.fit_transform(dfTest['A'])

If I pass an entire dataframe to the scaler it works:

dfTest2 = dfTest.drop('C', axis = 1) good_output = min_max_scaler.fit_transform(dfTest2) good_output

I'm confused why passing a series to the scaler fails. In my full working code above I had hoped to just pass a series to the scaler then set the dataframe column = to the scaled series. I've seen this question asked a few other places, but haven't found a good answer. Any help understanding what's going on here would be greatly appreciated!


回答 0

我不确定以前的版本是否pandas阻止了此操作,但现在以下代码段对我来说效果很好,并且无需使用就可以产生所需的内容apply

>>> import pandas as pd
>>> from sklearn.preprocessing import MinMaxScaler


>>> scaler = MinMaxScaler()

>>> dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],
                           'B':[103.02,107.26,110.35,114.23,114.68],
                           'C':['big','small','big','small','small']})

>>> dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A', 'B']])

>>> dfTest
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

I am not sure if previous versions of pandas prevented this but now the following snippet works perfectly for me and produces exactly what you want without having to use apply

>>> import pandas as pd
>>> from sklearn.preprocessing import MinMaxScaler


>>> scaler = MinMaxScaler()

>>> dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],
                           'B':[103.02,107.26,110.35,114.23,114.68],
                           'C':['big','small','big','small','small']})

>>> dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A', 'B']])

>>> dfTest
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

回答 1

像这样?

dfTest = pd.DataFrame({
           'A':[14.00,90.20,90.95,96.27,91.21],
           'B':[103.02,107.26,110.35,114.23,114.68], 
           'C':['big','small','big','small','small']
         })
dfTest[['A','B']] = dfTest[['A','B']].apply(
                           lambda x: MinMaxScaler().fit_transform(x))
dfTest

    A           B           C
0   0.000000    0.000000    big
1   0.926219    0.363636    small
2   0.935335    0.628645    big
3   1.000000    0.961407    small
4   0.938495    1.000000    small

Like this?

dfTest = pd.DataFrame({
           'A':[14.00,90.20,90.95,96.27,91.21],
           'B':[103.02,107.26,110.35,114.23,114.68], 
           'C':['big','small','big','small','small']
         })
dfTest[['A','B']] = dfTest[['A','B']].apply(
                           lambda x: MinMaxScaler().fit_transform(x))
dfTest

    A           B           C
0   0.000000    0.000000    big
1   0.926219    0.363636    small
2   0.935335    0.628645    big
3   1.000000    0.961407    small
4   0.938495    1.000000    small

回答 2

正如pir的评论中提到的那样-该.apply(lambda el: scale.fit_transform(el))方法将产生以下警告:

DeprecationWarning:在0.17中弃用1d数组作为数据,它将在0.19中引发ValueError。如果数据具有单个功能,则使用X.reshape(-1,1)来重塑数据,如果包含单个样本,则使用X.reshape(1,-1)来重塑数据。

将您的列转换为numpy数组应该可以完成这项工作(我更喜欢StandardScaler):

from sklearn.preprocessing import StandardScaler
scale = StandardScaler()

dfTest[['A','B','C']] = scale.fit_transform(dfTest[['A','B','C']].as_matrix())

- 编辑 2018年11月(已针对熊猫0.23.4测试)-

作为罗布·默里提到的意见,大熊猫的电流(v0.23.4)版本.as_matrix()的回报FutureWarning。因此,应将其替换为.values

from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()

scaler.fit_transform(dfTest[['A','B']].values)

- 编辑 2019年5月(已针对熊猫0.24.2测试)-

正如joelostblom在评论中提到的那样:“因此0.24.0,建议使用.to_numpy()代替.values。”

更新的示例:

import pandas as pd
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
dfTest = pd.DataFrame({
               'A':[14.00,90.20,90.95,96.27,91.21],
               'B':[103.02,107.26,110.35,114.23,114.68],
               'C':['big','small','big','small','small']
             })
dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A','B']].to_numpy())
dfTest
      A         B      C
0 -1.995290 -1.571117    big
1  0.436356 -0.603995  small
2  0.460289  0.100818    big
3  0.630058  0.985826  small
4  0.468586  1.088469  small

As it is being mentioned in pir's comment - the .apply(lambda el: scale.fit_transform(el)) method will produce the following warning:

DeprecationWarning: Passing 1d arrays as data is deprecated in 0.17 and will raise ValueError in 0.19. Reshape your data either using X.reshape(-1, 1) if your data has a single feature or X.reshape(1, -1) if it contains a single sample.

Converting your columns to numpy arrays should do the job (I prefer StandardScaler):

from sklearn.preprocessing import StandardScaler
scale = StandardScaler()

dfTest[['A','B','C']] = scale.fit_transform(dfTest[['A','B','C']].as_matrix())

-- Edit Nov 2018 (Tested for pandas 0.23.4)--

As Rob Murray mentions in the comments, in the current (v0.23.4) version of pandas .as_matrix() returns FutureWarning. Therefore, it should be replaced by .values:

from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()

scaler.fit_transform(dfTest[['A','B']].values)

-- Edit May 2019 (Tested for pandas 0.24.2)--

As joelostblom mentions in the comments, "Since 0.24.0, it is recommended to use .to_numpy() instead of .values."

Updated example:

import pandas as pd
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
dfTest = pd.DataFrame({
               'A':[14.00,90.20,90.95,96.27,91.21],
               'B':[103.02,107.26,110.35,114.23,114.68],
               'C':['big','small','big','small','small']
             })
dfTest[['A', 'B']] = scaler.fit_transform(dfTest[['A','B']].to_numpy())
dfTest
      A         B      C
0 -1.995290 -1.571117    big
1  0.436356 -0.603995  small
2  0.460289  0.100818    big
3  0.630058  0.985826  small
4  0.468586  1.088469  small

回答 3

df = pd.DataFrame(scale.fit_transform(df.values), columns=df.columns, index=df.index)

这应该在没有折旧警告的情况下起作用。

df = pd.DataFrame(scale.fit_transform(df.values), columns=df.columns, index=df.index)

This should work without depreciation warnings.


回答 4

您只能使用以下方法进行操作 pandas

In [235]:
dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
df = dfTest[['A', 'B']]
df_norm = (df - df.min()) / (df.max() - df.min())
print df_norm
print pd.concat((df_norm, dfTest.C),1)

          A         B
0  0.000000  0.000000
1  0.926219  0.363636
2  0.935335  0.628645
3  1.000000  0.961407
4  0.938495  1.000000
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

You can do it using pandas only:

In [235]:
dfTest = pd.DataFrame({'A':[14.00,90.20,90.95,96.27,91.21],'B':[103.02,107.26,110.35,114.23,114.68], 'C':['big','small','big','small','small']})
df = dfTest[['A', 'B']]
df_norm = (df - df.min()) / (df.max() - df.min())
print df_norm
print pd.concat((df_norm, dfTest.C),1)

          A         B
0  0.000000  0.000000
1  0.926219  0.363636
2  0.935335  0.628645
3  1.000000  0.961407
4  0.938495  1.000000
          A         B      C
0  0.000000  0.000000    big
1  0.926219  0.363636  small
2  0.935335  0.628645    big
3  1.000000  0.961407  small
4  0.938495  1.000000  small

回答 5

我知道这是一个很老的评论,但仍然:

不要使用单括号(dfTest['A']),而应使用双括号(dfTest[['A']])

即:min_max_scaler.fit_transform(dfTest[['A']])

我相信这会取得理想的结果。

I know it's a very old comment, but still:

Instead of using single bracket (dfTest['A']), use double brackets (dfTest[['A']]).

i.e: min_max_scaler.fit_transform(dfTest[['A']]).

I believe this will give the desired result.


本文由 Python 实用宝典 作者:Python实用宝典 发表,其版权均为 Python 实用宝典 所有,文章内容系作者个人观点,不代表 Python 实用宝典 对观点赞同或支持。如需转载,请注明文章来源。
17

抱歉,评论已关闭!