Reputation: 71
df1 = pd.DataFrame({'foo': ['one', 'one', 'one', 'two', 'two', 'two'],
'bar': ['A', 'B', 'C', 'A', 'B', 'C']})
foo | bar | |
---|---|---|
0 | one | A |
1 | one | B |
2 | one | C |
3 | two | A |
4 | two | B |
5 | two | C |
I would like to convert this to
foo | val1 | val2 | val3 |
---|---|---|---|
One | A | B | C |
Two | A | B | C |
And the code I tried is:
pd.pivot_table(df1,index='foo',aggfunc=['first'])
But the above code is returning only the first value
Upvotes: 7
Views: 1663
Reputation: 28709
Taking a cue from @HenryEcker's solution, a combination of groupby and pivot can get the result:
(df1.assign(counter = lambda df: df.groupby('foo').cumcount())
.pivot('foo', 'counter', 'bar')
.rename(columns = lambda col: f"val{col}" if col!=0 else "val")
.rename_axis(columns=None)
)
val val1 val2
foo
one A B C
two A B C
Upvotes: 0
Reputation: 153510
Try:
df1.groupby([df1.groupby('foo').cumcount() + 1,
'foo']).first()['bar'].unstack(0).add_prefix('val').reset_index()
Output:
foo val1 val2 val3
0 one A B C
1 two A B C
Upvotes: 2
Reputation: 159
pd.pivot_table(df1,index='foo',columns=df1.groupby(df1.foo).cumcount()+1,
values='bar',aggfunc=(lambda x:min(x)), dropna=True).add_prefix('val')
output:
val1 val2 val 3
foo
one A B C
two A B C
Upvotes: 0
Reputation: 185
df1 = pd.DataFrame({'foo': ['one', 'one', 'one', 'two', 'two', 'two'],
'bar': ['A', 'B', 'C', 'A', 'B', 'C']})
df1['i'] = df1['bar'].apply(lambda x: "val "+x)
pd.pivot(df1,index='foo',columns='i',values='bar')
Upvotes: -1
Reputation: 35686
We can enumerate groups with groupby cumcount
and use those as the pivot columns then add_prefix
to the numerical values and reset_index
to return the 'foo' values to the columns:
new_df = (
df1.pivot_table(index='foo',
columns=df1.groupby('foo').cumcount() + 1,
values='bar',
aggfunc='first')
.add_prefix('val')
.reset_index()
)
foo val1 val2 val3
0 one A B C
1 two A B C
See how df1.groupby('foo').cumcount() + 1
makes the columns:
foo columns
0 one 1 # First instance of "one"
1 one 2 # Second instance of "one"
2 one 3 # Third instance of "one"
3 two 1
4 two 2
5 two 3
Code to generate the above DataFrame:
demo_df = pd.DataFrame({
'foo': df1['foo'],
'columns': df1.groupby('foo').cumcount() + 1
})
Upvotes: 6
Reputation: 195553
Another solution:
x = df1.pivot("foo", "bar", "bar")
x.columns = [f"var{i}" for i in range(1, len(x.columns) + 1)]
x = x.reset_index()
print(x)
Prints:
foo var1 var2 var3
0 one A B C
1 two A B C
Upvotes: 1