pandas 在groupby函数中计算空值 [英] Pandas count null values in a groupby function

问题描述

df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar', 'foo', 'bar', 'foo', 'foo'],
               'B' : ['one', 'one', 'two', 'three', 'two', 'two', 'one', 'three'],
               'C' : [np.nan, 'bla2', np.nan, 'bla3', np.nan, np.nan, np.nan, np.nan]})

输出:

     A      B     C
0  foo    one   NaN
1  bar    one  bla2
2  foo    two   NaN
3  bar  three  bla3
4  foo    two   NaN
5  bar    two   NaN
6  foo    one   NaN
7  foo  three   NaN

我想使用groupby来计算foo的不同组合的NaN数.

I would like to use groupby in order to count the number of NaN's for the different combinations of foo.

预期输出(EDIT):

Expected Output (EDIT):

     A      B     C    D
0  foo    one   NaN    2
1  bar    one  bla2    0
2  foo    two   NaN    2
3  bar  three  bla3    0
4  foo    two   NaN    2
5  bar    two   NaN    1
6  foo    one   NaN    2
7  foo  three   NaN    1

当前我正在尝试:

df['count']=df.groupby(['A'])['B'].isnull().transform('sum')

但这不起作用...

谢谢

推荐答案

我认为您需要如果需要过滤器，请先添加 boolean indexing :

If need filter first add boolean indexing:

df = df[df['A'] == 'foo']
df2 = df.C.isnull().groupby([df['A'],df['B']]).sum().astype(int)
print (df2)
A    B    
foo  one      2
     three    1
     two      2

或更简单:

df = df[df['A'] == 'foo']
df2 = df['B'].value_counts()
print (df2)
one      2
two      2
three    1
Name: B, dtype: int64

解决方案非常相似，仅添加

Solution is very similar, only add transform:

df['D'] = df.C.isnull().groupby([df['A'],df['B']]).transform('sum').astype(int)
print (df)
     A      B     C  D
0  foo    one   NaN  2
1  bar    one  bla2  0
2  foo    two   NaN  2
3  bar  three  bla3  0
4  foo    two   NaN  2
5  bar    two   NaN  1
6  foo    one   NaN  2
7  foo  three   NaN  1

类似的解决方案:

df['D'] = df.C.isnull()
df['D'] = df.groupby(['A','B'])['D'].transform('sum').astype(int)
print (df)
     A      B     C  D
0  foo    one   NaN  2
1  bar    one  bla2  0
2  foo    two   NaN  2
3  bar  three  bla3  0
4  foo    two   NaN  2
5  bar    two   NaN  1
6  foo    one   NaN  2
7  foo  three   NaN  1

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