根据value_counts()更改 pandas 数据框中的值 [英] Change values in pandas dataframe according to value_counts()

问题描述

我有以下熊猫数据框:

import pandas as pd 
from pandas import Series, DataFrame

data = DataFrame({'Qu1': ['apple', 'potato', 'cheese', 'banana', 'cheese', 'banana', 'cheese', 'potato', 'egg'],
              'Qu2': ['sausage', 'banana', 'apple', 'apple', 'apple', 'sausage', 'banana', 'banana', 'banana'],
              'Qu3': ['apple', 'potato', 'sausage', 'cheese', 'cheese', 'potato', 'cheese', 'potato', 'egg']})

当值计数大于或等于某个数字时，我想根据value_counts()更改Qu1，Qu2，Qu3列中的值

I'd like to change values in columns Qu1,Qu2,Qu3 according to value_counts() when value count great or equal some number

例如Qu1列

>>> pd.value_counts(data.Qu1) >= 2
cheese     True
potato     True
banana     True
apple     False
egg       False

我想保留值cheese，potato，banana，因为每个值至少出现两次.

I'd like to keep values cheese,potato,banana, because each value has at least two appearances.

根据值apple和egg，我想创建值others

From values apple and egg I'd like to create valueothers

对于列Qu2不变:

>>> pd.value_counts(data.Qu2) >= 2
banana     True
apple      True
sausage    True

最终结果如附件test_data

test_data = DataFrame({'Qu1': ['other', 'potato', 'cheese', 'banana', 'cheese', 'banana', 'cheese', 'potato', 'other'],
                  'Qu2': ['sausage', 'banana', 'apple', 'apple', 'apple', 'sausage', 'banana', 'banana', 'banana'],
                  'Qu3': ['other', 'potato', 'other', 'cheese', 'cheese', 'potato', 'cheese', 'potato', 'other']})

谢谢！

推荐答案

我将创建一个形状相同的数据框，其中对应的条目为值计数:

I would create a dataframe of same shape where the corresponding entry is the value count:

data.apply(lambda x: x.map(x.value_counts()))
Out[229]: 
   Qu1  Qu2  Qu3
0    1    2    1
1    2    4    3
2    3    3    1
3    2    3    3
4    3    3    3
5    2    2    3
6    3    4    3
7    2    4    3
8    1    4    1

然后，使用 df.where 中的结果返回相应条目小于2的其他":

And, use the results in df.where to return "other" where the corresponding entry is smaller than 2:

data.where(data.apply(lambda x: x.map(x.value_counts()))>=2, "other")

      Qu1      Qu2     Qu3
0   other  sausage   other
1  potato   banana  potato
2  cheese    apple   other
3  banana    apple  cheese
4  cheese    apple  cheese
5  banana  sausage  potato
6  cheese   banana  cheese
7  potato   banana  potato
8   other   banana   other

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