pandas - how to create multiple columns in groupby with conditional?
3
I need to group a dataframe, but I need to create two columns, one that is a simple count and another that is a count with conditional, as in the example:
The qtd_ok column counts only those that have 'OK'
I tried this, but I do not know how to add the total count in the same groupby:
df.groupby(['column1', 'column2', 'column3']).apply(lambda x : x['status'].sum() == 'OK')
python pandas dataframe pandas-groupby
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
add a comment |
3
I need to group a dataframe, but I need to create two columns, one that is a simple count and another that is a count with conditional, as in the example:
The qtd_ok column counts only those that have 'OK'
I tried this, but I do not know how to add the total count in the same groupby:
df.groupby(['column1', 'column2', 'column3']).apply(lambda x : x['status'].sum() == 'OK')
python pandas dataframe pandas-groupby
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
add a comment |
3
3
3
I need to group a dataframe, but I need to create two columns, one that is a simple count and another that is a count with conditional, as in the example:
The qtd_ok column counts only those that have 'OK'
I tried this, but I do not know how to add the total count in the same groupby:
df.groupby(['column1', 'column2', 'column3']).apply(lambda x : x['status'].sum() == 'OK')
python pandas dataframe pandas-groupby
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
I need to group a dataframe, but I need to create two columns, one that is a simple count and another that is a count with conditional, as in the example:
The qtd_ok column counts only those that have 'OK'
I tried this, but I do not know how to add the total count in the same groupby:
df.groupby(['column1', 'column2', 'column3']).apply(lambda x : x['status'].sum() == 'OK')
python pandas dataframe pandas-groupby
python pandas dataframe pandas-groupby
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
edited Nov 21 '18 at 14:43
jpp
99.5k2161110
99.5k2161110
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
asked Nov 21 '18 at 14:15
Hiago BonamelliHiago Bonamelli
617
617
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
1
First create helper column A with assign and then aggregate by agg functions sum for count only OK values and size for count all values per groups:
df = (df.assign(A=(df['status']== 'OK'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
Sample:
df = pd.DataFrame({
'column1':['a'] * 9,
'column2':['a'] * 4 + ['b'] * 5,
'column3':list('aaabaabbb'),
'status':list('aabaaabba'),
})
print (df)
column1 column2 column3 status
0 a a a a
1 a a a a
2 a a a b
3 a a b a
4 a b a a
5 a b a a
6 a b b b
7 a b b b
8 a b b a
df = (df.assign(A=(df['status']== 'a'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
print (df)
column1 column2 column3 qtd_ok qtd
0 a a a 2 3
1 a a b 1 1
2 a b a 2 2
3 a b b 1 3
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
1
Thank you!! It worked
– Hiago Bonamelli
Nov 21 '18 at 15:33
1
Good one to know, +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
pd.crosstab
You can use pd.crosstab with margins=True:
# data from @jezrael
list_of_lists = df.iloc[:, :-1].values.T.tolist()
condition = df['status'].eq('a')
res = pd.crosstab(list_of_lists, condition, margins=True)
.drop('All', level=0).reset_index()
print(res)
status column1 column2 column3 False True All
0 a a a 1 2 3
1 a a b 0 1 1
2 a b a 0 2 2
3 a b b 2 1 3
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
1
Another nice solution +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
Just an idea to count with groupby with lambda which can further be enhanced ..
>>> df
colum1 colum2 colum3 status
0 unit1 section1 content1 OK
1 unit1 section1 content1 OK
2 unit1 section1 content1 error
3 unit1 section1 content2 OK
4 unit1 section2 content1 OK
5 unit1 section2 content1 OK
6 unit1 section2 content2 error
7 unit1 section2 content2 error
8 unit1 section2 content2 OK
using groupby with lambda..
>>> df.groupby(['colum1','colum2', 'colum3'])['status'].apply(lambda x: x[x.str.contains('OK', case=False)].count()).reset_index()
colum1 colum2 colum3 status
0 unit1 section1 content1 2
1 unit1 section1 content2 1
2 unit1 section2 content1 2
3 unit1 section2 content2 1
Also can use case=False for ignorecase for ok.
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
First create helper column A with assign and then aggregate by agg functions sum for count only OK values and size for count all values per groups:
df = (df.assign(A=(df['status']== 'OK'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
Sample:
df = pd.DataFrame({
'column1':['a'] * 9,
'column2':['a'] * 4 + ['b'] * 5,
'column3':list('aaabaabbb'),
'status':list('aabaaabba'),
})
print (df)
column1 column2 column3 status
0 a a a a
1 a a a a
2 a a a b
3 a a b a
4 a b a a
5 a b a a
6 a b b b
7 a b b b
8 a b b a
df = (df.assign(A=(df['status']== 'a'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
print (df)
column1 column2 column3 qtd_ok qtd
0 a a a 2 3
1 a a b 1 1
2 a b a 2 2
3 a b b 1 3
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
1
Thank you!! It worked
– Hiago Bonamelli
Nov 21 '18 at 15:33
1
Good one to know, +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
First create helper column A with assign and then aggregate by agg functions sum for count only OK values and size for count all values per groups:
df = (df.assign(A=(df['status']== 'OK'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
Sample:
df = pd.DataFrame({
'column1':['a'] * 9,
'column2':['a'] * 4 + ['b'] * 5,
'column3':list('aaabaabbb'),
'status':list('aabaaabba'),
})
print (df)
column1 column2 column3 status
0 a a a a
1 a a a a
2 a a a b
3 a a b a
4 a b a a
5 a b a a
6 a b b b
7 a b b b
8 a b b a
df = (df.assign(A=(df['status']== 'a'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
print (df)
column1 column2 column3 qtd_ok qtd
0 a a a 2 3
1 a a b 1 1
2 a b a 2 2
3 a b b 1 3
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
1
Thank you!! It worked
– Hiago Bonamelli
Nov 21 '18 at 15:33
1
Good one to know, +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
1
1
First create helper column A with assign and then aggregate by agg functions sum for count only OK values and size for count all values per groups:
df = (df.assign(A=(df['status']== 'OK'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
Sample:
df = pd.DataFrame({
'column1':['a'] * 9,
'column2':['a'] * 4 + ['b'] * 5,
'column3':list('aaabaabbb'),
'status':list('aabaaabba'),
})
print (df)
column1 column2 column3 status
0 a a a a
1 a a a a
2 a a a b
3 a a b a
4 a b a a
5 a b a a
6 a b b b
7 a b b b
8 a b b a
df = (df.assign(A=(df['status']== 'a'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
print (df)
column1 column2 column3 qtd_ok qtd
0 a a a 2 3
1 a a b 1 1
2 a b a 2 2
3 a b b 1 3
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
First create helper column A with assign and then aggregate by agg functions sum for count only OK values and size for count all values per groups:
df = (df.assign(A=(df['status']== 'OK'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
Sample:
df = pd.DataFrame({
'column1':['a'] * 9,
'column2':['a'] * 4 + ['b'] * 5,
'column3':list('aaabaabbb'),
'status':list('aabaaabba'),
})
print (df)
column1 column2 column3 status
0 a a a a
1 a a a a
2 a a a b
3 a a b a
4 a b a a
5 a b a a
6 a b b b
7 a b b b
8 a b b a
df = (df.assign(A=(df['status']== 'a'))
.groupby(['column1', 'column2', 'column3'])['A']
.agg([('qtd_ok','sum'),('qtd','size')])
.astype(int)
.reset_index())
print (df)
column1 column2 column3 qtd_ok qtd
0 a a a 2 3
1 a a b 1 1
2 a b a 2 2
3 a b b 1 3
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
answered Nov 21 '18 at 14:21
jezraeljezrael
331k24273351
331k24273351
1
Thank you!! It worked
– Hiago Bonamelli
Nov 21 '18 at 15:33
1
Good one to know, +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
Thank you!! It worked
– Hiago Bonamelli
Nov 21 '18 at 15:33
1
Good one to know, +1
– pygo
Nov 21 '18 at 15:54
1
1
Thank you!! It worked
– Hiago Bonamelli
Nov 21 '18 at 15:33
Thank you!! It worked
– Hiago Bonamelli
Nov 21 '18 at 15:33
1
1
Good one to know, +1
– pygo
Nov 21 '18 at 15:54
Good one to know, +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
pd.crosstab
You can use pd.crosstab with margins=True:
# data from @jezrael
list_of_lists = df.iloc[:, :-1].values.T.tolist()
condition = df['status'].eq('a')
res = pd.crosstab(list_of_lists, condition, margins=True)
.drop('All', level=0).reset_index()
print(res)
status column1 column2 column3 False True All
0 a a a 1 2 3
1 a a b 0 1 1
2 a b a 0 2 2
3 a b b 2 1 3
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
1
Another nice solution +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
pd.crosstab
You can use pd.crosstab with margins=True:
# data from @jezrael
list_of_lists = df.iloc[:, :-1].values.T.tolist()
condition = df['status'].eq('a')
res = pd.crosstab(list_of_lists, condition, margins=True)
.drop('All', level=0).reset_index()
print(res)
status column1 column2 column3 False True All
0 a a a 1 2 3
1 a a b 0 1 1
2 a b a 0 2 2
3 a b b 2 1 3
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
1
Another nice solution +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
1
1
pd.crosstab
You can use pd.crosstab with margins=True:
# data from @jezrael
list_of_lists = df.iloc[:, :-1].values.T.tolist()
condition = df['status'].eq('a')
res = pd.crosstab(list_of_lists, condition, margins=True)
.drop('All', level=0).reset_index()
print(res)
status column1 column2 column3 False True All
0 a a a 1 2 3
1 a a b 0 1 1
2 a b a 0 2 2
3 a b b 2 1 3
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
pd.crosstab
You can use pd.crosstab with margins=True:
# data from @jezrael
list_of_lists = df.iloc[:, :-1].values.T.tolist()
condition = df['status'].eq('a')
res = pd.crosstab(list_of_lists, condition, margins=True)
.drop('All', level=0).reset_index()
print(res)
status column1 column2 column3 False True All
0 a a a 1 2 3
1 a a b 0 1 1
2 a b a 0 2 2
3 a b b 2 1 3
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
edited Nov 21 '18 at 15:36
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
answered Nov 21 '18 at 14:41
jppjpp
99.5k2161110
99.5k2161110
1
Another nice solution +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
Another nice solution +1
– pygo
Nov 21 '18 at 15:54
1
1
Another nice solution +1
– pygo
Nov 21 '18 at 15:54
Another nice solution +1
– pygo
Nov 21 '18 at 15:54
add a comment |
1
Just an idea to count with groupby with lambda which can further be enhanced ..
>>> df
colum1 colum2 colum3 status
0 unit1 section1 content1 OK
1 unit1 section1 content1 OK
2 unit1 section1 content1 error
3 unit1 section1 content2 OK
4 unit1 section2 content1 OK
5 unit1 section2 content1 OK
6 unit1 section2 content2 error
7 unit1 section2 content2 error
8 unit1 section2 content2 OK
using groupby with lambda..
>>> df.groupby(['colum1','colum2', 'colum3'])['status'].apply(lambda x: x[x.str.contains('OK', case=False)].count()).reset_index()
colum1 colum2 colum3 status
0 unit1 section1 content1 2
1 unit1 section1 content2 1
2 unit1 section2 content1 2
3 unit1 section2 content2 1
Also can use case=False for ignorecase for ok.
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
add a comment |
1
Just an idea to count with groupby with lambda which can further be enhanced ..
>>> df
colum1 colum2 colum3 status
0 unit1 section1 content1 OK
1 unit1 section1 content1 OK
2 unit1 section1 content1 error
3 unit1 section1 content2 OK
4 unit1 section2 content1 OK
5 unit1 section2 content1 OK
6 unit1 section2 content2 error
7 unit1 section2 content2 error
8 unit1 section2 content2 OK
using groupby with lambda..
>>> df.groupby(['colum1','colum2', 'colum3'])['status'].apply(lambda x: x[x.str.contains('OK', case=False)].count()).reset_index()
colum1 colum2 colum3 status
0 unit1 section1 content1 2
1 unit1 section1 content2 1
2 unit1 section2 content1 2
3 unit1 section2 content2 1
Also can use case=False for ignorecase for ok.
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
add a comment |
1
1
1
Just an idea to count with groupby with lambda which can further be enhanced ..
>>> df
colum1 colum2 colum3 status
0 unit1 section1 content1 OK
1 unit1 section1 content1 OK
2 unit1 section1 content1 error
3 unit1 section1 content2 OK
4 unit1 section2 content1 OK
5 unit1 section2 content1 OK
6 unit1 section2 content2 error
7 unit1 section2 content2 error
8 unit1 section2 content2 OK
using groupby with lambda..
>>> df.groupby(['colum1','colum2', 'colum3'])['status'].apply(lambda x: x[x.str.contains('OK', case=False)].count()).reset_index()
colum1 colum2 colum3 status
0 unit1 section1 content1 2
1 unit1 section1 content2 1
2 unit1 section2 content1 2
3 unit1 section2 content2 1
Also can use case=False for ignorecase for ok.
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
Just an idea to count with groupby with lambda which can further be enhanced ..
>>> df
colum1 colum2 colum3 status
0 unit1 section1 content1 OK
1 unit1 section1 content1 OK
2 unit1 section1 content1 error
3 unit1 section1 content2 OK
4 unit1 section2 content1 OK
5 unit1 section2 content1 OK
6 unit1 section2 content2 error
7 unit1 section2 content2 error
8 unit1 section2 content2 OK
using groupby with lambda..
>>> df.groupby(['colum1','colum2', 'colum3'])['status'].apply(lambda x: x[x.str.contains('OK', case=False)].count()).reset_index()
colum1 colum2 colum3 status
0 unit1 section1 content1 2
1 unit1 section1 content2 1
2 unit1 section2 content1 2
3 unit1 section2 content2 1
Also can use case=False for ignorecase for ok.
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
edited Nov 21 '18 at 15:48
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
answered Nov 21 '18 at 15:32
pygopygo
3,0551619
3,0551619
add a comment |
add a comment |
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