python - 按 2 个类别分组然后求和

Question

以上输出来自:

df.groupby('croho subonderdeel').sum()

我计算了每个类别的毕业生总数，但我也想按列计算。例如，仅接收第一列“2011 MAN”的输出。

我尝试了以下方法:

df.groupby('croho subonderdeel','2011 MAN').sum() 

然后我收到以下错误:

ValueError: No axis named 2011 MAN for object type <class 'pandas.core.frame.DataFrame'>

然后我想也许我需要对“2011 MAN”求和，而不是分组两次。所以我尝试了:

df.groupby('croho subonderdeel').sum('2011 MAN')

然后我收到此错误:

TypeError: f() takes 1 positional argument but 2 were given

有人可以向我解释一下，为什么我尝试的两种方法都不可行？也许我可以自己解决这个问题。

Answer 1

您需要在[]中指定列，例如:

df.groupby('croho subonderdeel')['2011 MAN'].sum() 

您还可以指定多列:

df.groupby('croho subonderdeel')['2011 MAN', '2012 MAN'].sum() 

此外，如果需要2列输出添加参数as_index=False:

df.groupby('croho subonderdeel', as_index=False)['2011 MAN'].sum() 

或者:

df.groupby('croho subonderdeel')['2011 MAN'].sum().reset_index()

<小时/>

但是如果想要按 2 个类别(2 列)进行分组，请将 [] 添加到 groupby:

df.groupby(['croho subonderdeel', 'another col'])['2011 MAN'].sum()

<小时/>

示例:

df = pd.DataFrame({'another col':list('efefef'),
                   '2011 MAN':[4,5,4,5,5,4],
                   '2011 WROUW':[7,8,9,4,2,3],
                   '2012 MAN':[1,3,5,7,1,0],
                   '2012 WROUW':[5,3,6,9,2,4],
                   'croho subonderdeel':list('aaabbb')})

print (df)
   2011 MAN  2011 WROUW  2012 MAN  2012 WROUW another col croho subonderdeel
0         4           7         1           5           e                  a
1         5           8         3           3           f                  a
2         4           9         5           6           e                  a
3         5           4         7           9           f                  b
4         5           2         1           2           e                  b
5         4           3         0           4           f                  b

print(df.groupby('croho subonderdeel')['2011 MAN'].sum())
croho subonderdeel
a    13
b    14
Name: 2011 MAN, dtype: int64

print(df.groupby('croho subonderdeel', as_index=False)['2011 MAN'].sum())
  croho subonderdeel  2011 MAN
0                  a        13
1                  b        14

print(df.groupby('croho subonderdeel')['2011 MAN'].sum().reset_index())
  croho subonderdeel  2011 MAN
0                  a        13
1                  b        14

<小时/>

print(df.groupby('croho subonderdeel')['2011 MAN', '2012 WROUW'].sum())
                    2011 MAN  2012 WROUW
croho subonderdeel                      
a                         13          14
b                         14          15

print(df.groupby('croho subonderdeel', as_index=False)['2011 MAN', '2012 WROUW'].sum())
  croho subonderdeel  2011 MAN  2012 WROUW
0                  a        13          14
1                  b        14          15

<小时/>

print (df.groupby(['croho subonderdeel', 'another col'])['2011 MAN'].sum())
croho subonderdeel  another col
a                   e              8
                    f              5
b                   e              5
                    f              9
Name: 2011 MAN, dtype: int64

print (df.groupby(['croho subonderdeel', 'another col'], as_index=False)['2011 MAN'].sum())
  croho subonderdeel another col  2011 MAN
0                  a           e         8
1                  a           f         5
2                  b           e         5
3                  b           f         9

<小时/>

print (df.groupby(['croho subonderdeel', 'another col']).sum())
                                2011 MAN  2011 WROUW  2012 MAN  2012 WROUW
croho subonderdeel another col                                            
a                  e                   8          16         6          11
                   f                   5           8         3           3
b                  e                   5           2         1           2
                   f                   9           7         7          13

print (df.groupby(['croho subonderdeel', 'another col'], as_index=False).sum())
 croho subonderdeel another col  2011 MAN  2011 WROUW  2012 MAN  2012 WROUW
0                  a           e         8          16         6          11
1                  a           f         5           8         3           3
2                  b           e         5           2         1           2
3                  b           f         9           7         7          13