Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
610 views
in Technique[技术] by (71.8m points)

python - How to do a conditional count after groupby on a Pandas Dataframe?

I have the following dataframe:

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

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

I just get the usual count with:

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

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

I tried things like this:

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

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

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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

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

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...