如何在python pandas的同一列上进行分组并取唯一计数和某个值计数作为聚合? [英] How to do group by and take count of unique and count of some value as aggregate on same column in python pandas?

问题描述

我的问题与我以前的

My question is related to my previous Question but it's different. So I am asking the new question.

在上述问题中，请参阅@jezrael的答案.

In above question see the answer of @jezrael.

df = pd.DataFrame({'col1':[1,1,1],
                   'col2':[4,4,6],
                   'col3':[7,7,9],
                   'col4':[3,3,5]})

print (df)
   col1  col2  col3  col4
0     1     4     7     3
1     1     4     7     3
2     1     6     9     5

df1 = df.groupby(['col1','col2']).agg({'col3':'size','col4':'nunique'})
df1['result_col'] = df1['col3'].div(df1['col4'])
print (df1)
           col4  col3  result_col
col1 col2                        
1    4        1     2         2.0
     6        1     1         1.0

现在在这里我要对col4的特定值进行计数.假设我也想在同一查询中计算col4 == 3.

Now here I want to take count for the specific value of col4 . Say I also want to take count of col4 == 3 in the same query.

df.groupby(['col1','col2']).agg({'col3':'size','col4':'nunique'}) ... + count(col4=='3')

如何在上述相同的查询中执行此操作，但我已经尝试过但未获得解决方案.

How to do this in same above query I have tried bellow but not getting solution.

df.groupby(['col1','col2']).agg({'col3':'size','col4':'nunique','col4':'x: lambda x[x == 7].count()'})

推荐答案

通过提前将col4==3作为列进行一些预处理.然后使用aggregate

Do some preprocessing by including the col4==3 as a column ahead of time. Then use aggregate

df.assign(result_col=df.col4.eq(3).astype(int)).groupby(
    ['col1', 'col2']
).agg(dict(col3='size', col4='nunique', result_col='sum'))

           col3  result_col  col4
col1 col2                        
1    4        2           2     1
     6        1           0     1

---

旧答案

---

old answers

g = df.groupby(['col1', 'col2'])
g.agg({'col3':'size','col4': 'nunique'}).assign(
    result_col=g.col4.apply(lambda x: x.eq(3).sum()))

           col3  col4  result_col
col1 col2                        
1    4        2     1           2
     6        1     1           0

---

稍微重新排列

---

slightly rearranged

g = df.groupby(['col1', 'col2'])
final_df = g.agg({'col3':'size','col4': 'nunique'})
final_df.insert(1, 'result_col', g.col4.apply(lambda x: x.eq(3).sum()))
final_df

           col3  result_col  col4
col1 col2                        
1    4        2           2     1
     6        1           0     1

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