How to do a conditional count after groupby on a Pandas Dataframe?
Question:
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?
Answers:
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
You can count the occurence of ‘one’ for the groupby dataframe, in the column ‘key2’ like this:
df.groupby('key1')['key2'].apply(lambda x: x[x == 'one'].count())
yield
key1
a 2
b 1
c 0
Name: key2, dtype: int64
Option 1
df.set_index('key1').key2.eq('one').sum(level=0).astype(int).reset_index()
key1 key2
0 a 2
1 b 1
2 c 0
Option 2
df.key2.eq('one').groupby(df.key1).sum().astype(int).reset_index()
key1 key2
0 a 2
1 b 1
2 c 0
Option 3
f, u = df.key1.factorize()
pd.DataFrame(dict(key1=u, key2=np.bincount(f, df.key2.eq('one')).astype(int)))
key1 key2
0 a 2
1 b 1
2 c 0
Option 4
pd.crosstab(df.key1, df.key2.eq('one'))[True].rename('key2').reset_index()
key1 key2
0 a 2
1 b 1
2 c 0
Option 5
pd.get_dummies(df.key1).mul(
df.key2.eq('one'), 0
).sum().rename_axis('key1').reset_index(name='key2')
key1 key2
0 a 2
1 b 1
2 c 0
You can do this with applying groupby() on both keys and unstack().
df = df.groupby(['key1', 'key2']).size().unstack()
Maybe not the fastest solution, but you can create new data frame with column of ones if key2 is equal to ‘one’.
df2 = df.assign(oneCount =
lambda x: [1 if row.key2 == 'one' else 0 for index, row in x.iterrows()])
key1 key2 oneCount
0 a one 1
1 a two 0
2 b one 1
3 b two 0
4 a one 1
5 c two 0
And then aggregate it.
df3 = df2.groupby('key1').agg({"oneCount":sum}).reset_index()
key1 oneCount
0 a 2
1 b 1
2 c 0
I need count 2 columns (lambda with two arguments) as the example:
Pandas dataframe groupby func
, in the column key2
like this:
df.groupby('key1')['key2'].apply(lambda x: x[x == 'one'].count())
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?
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
You can count the occurence of ‘one’ for the groupby dataframe, in the column ‘key2’ like this:
df.groupby('key1')['key2'].apply(lambda x: x[x == 'one'].count())
yield
key1
a 2
b 1
c 0
Name: key2, dtype: int64
Option 1
df.set_index('key1').key2.eq('one').sum(level=0).astype(int).reset_index()
key1 key2
0 a 2
1 b 1
2 c 0
Option 2
df.key2.eq('one').groupby(df.key1).sum().astype(int).reset_index()
key1 key2
0 a 2
1 b 1
2 c 0
Option 3
f, u = df.key1.factorize()
pd.DataFrame(dict(key1=u, key2=np.bincount(f, df.key2.eq('one')).astype(int)))
key1 key2
0 a 2
1 b 1
2 c 0
Option 4
pd.crosstab(df.key1, df.key2.eq('one'))[True].rename('key2').reset_index()
key1 key2
0 a 2
1 b 1
2 c 0
Option 5
pd.get_dummies(df.key1).mul(
df.key2.eq('one'), 0
).sum().rename_axis('key1').reset_index(name='key2')
key1 key2
0 a 2
1 b 1
2 c 0
You can do this with applying groupby() on both keys and unstack().
df = df.groupby(['key1', 'key2']).size().unstack()
Maybe not the fastest solution, but you can create new data frame with column of ones if key2 is equal to ‘one’.
df2 = df.assign(oneCount =
lambda x: [1 if row.key2 == 'one' else 0 for index, row in x.iterrows()])
key1 key2 oneCount
0 a one 1
1 a two 0
2 b one 1
3 b two 0
4 a one 1
5 c two 0
And then aggregate it.
df3 = df2.groupby('key1').agg({"oneCount":sum}).reset_index()
key1 oneCount
0 a 2
1 b 1
2 c 0
I need count 2 columns (lambda with two arguments) as the example:
Pandas dataframe groupby func
, in the column key2
like this:
df.groupby('key1')['key2'].apply(lambda x: x[x == 'one'].count())