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Python Pandas:对所有列进行分组并计算不同的价值吗?

更新时间:2022-12-10 21:15:59

您可以使用 stack 系列,然后 Series.groupby SeriesGroupBy.nunique

You can use stack for Series and then Series.groupby with SeriesGroupBy.nunique:

df1 = df.set_index('column1').stack()

print (df1.groupby(level=[0,1]).nunique(dropna=False).unstack())

示例:

print (df)
  column1 column2 column3 column4
0    name    True    True     NaN
1    name     NaN    True     NaN
2   name1     NaN    True    True
3   name1    True    True    True

df1 = df.set_index('column1').stack(dropna=False)
print (df1)
column1         
name     column2    True
         column3    True
         column4     NaN
         column2     NaN
         column3    True
         column4     NaN
name1    column2     NaN
         column3    True
         column4    True
         column2    True
         column3    True
         column4    True
dtype: object

print (df1.groupby(level=[0,1]).nunique(dropna=False).unstack(fill_value=0))
         column2  column3  column4
column1                           
name           2        1        1
name1          2        1        1

print (df1.groupby(level=[0,1]).nunique().unstack(fill_value=0))
         column2  column3  column4
column1                           
name           1        1        0
name1          1        1        1






另一种应用了两次的解决方案

print (df.groupby('column1')
         .apply(lambda x: x.iloc[:,1:].apply(lambda y: y.nunique(dropna=False))))
         column2  column3  column4
column1                           
name           2        1        1
name1          2        1        1

print (df.groupby('column1').apply(lambda x: x.iloc[:,1:].apply(lambda y: y.nunique())))
         column2  column3  column4
column1                           
name           1        1        0
name1          1        1        1