在Pandas Dataframe上进行groupby之后如何进行条件计数? [英] How to do a conditional count after groupby on a Pandas Dataframe?

问题描述

我有以下数据框:

   key1  key2
0    a   one
1    a   two
2    b   one
3    b   two
4    a   one
5    c   two

现在，我想按key1对数据帧进行分组，并用值"one"对列key2进行计数，以获得以下结果:

Now, I want to group the dataframe by the key1 and count the column key2 with the value "one" to get this result:

   key1  
0    a   2
1    b   1
2    c   0

我只是得到通常的计数:

I just get the usual count with:

df.groupby(['key1']).size()

但是我不知道如何插入条件.

But I don't know how to insert the condition.

我尝试过这样的事情:

df.groupby(['key1']).apply(df[df['key2'] == 'one'])

但是我不能再进一步了.我该怎么办?

But I can't get any further. How can I do this?

推荐答案

我认为您需要先添加条件:

I think you need add condition first:

#if need also category c with no values of 'one'
df11=df.groupby('key1')['key2'].apply(lambda x: (x=='one').sum()).reset_index(name='count')
print (df11)
  key1  count
0    a      2
1    b      1
2    c      0

或将 categorical 与key1一起使用，然后丢失值由size添加:

Or use categorical with key1, then missing value is added by size:

df['key1'] = df['key1'].astype('category')
df1 = df[df['key2'] == 'one'].groupby(['key1']).size().reset_index(name='count') 
print (df1)
  key1  count
0    a      2
1    b      1
2    c      0

---

如果需要所有组合:

---

If need all combinations:

df2 = df.groupby(['key1', 'key2']).size().reset_index(name='count') 
print (df2)
  key1 key2  count
0    a  one      2
1    a  two      1
2    b  one      1
3    b  two      1
4    c  two      1

df3 = df.groupby(['key1', 'key2']).size().unstack(fill_value=0)
print (df3)
key2  one  two
key1          
a       2    1
b       1    1
c       0    1

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