Opposite of melt in python pandas
Question:
I cannot figure out how to do "reverse melt" using Pandas in python.
This is my starting data
label type value
0 x a 1
1 x b 2
2 x c 3
3 y a 4
4 y b 5
5 y c 6
6 z a 7
7 z b 8
8 z c 9
This is the output I would like to have:
label a b c
x 1 2 3
y 4 5 6
z 7 8 9
I’m sure there is an easy way to do this, but I don’t know how.
Answers:
there are a few ways:
using .pivot:
>>> origin.pivot(index='label', columns='type')['value']
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[3 rows x 3 columns]
using pivot_table:
>>> origin.pivot_table(values='value', index='label', columns='type')
value
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[3 rows x 3 columns]
or .groupby followed by .unstack:
>>> origin.groupby(['label', 'type'])['value'].aggregate('mean').unstack()
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[3 rows x 3 columns]
DataFrame.set_index + DataFrame.unstack
df.set_index(['label','type'])['value'].unstack()
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
simplifying the passing of pivot arguments
df.pivot(*df)
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[*df]
#['label', 'type', 'value']
For expected output we need DataFrame.reset_index and DataFrame.rename_axis
df.pivot(*df).rename_axis(columns = None).reset_index()
label a b c
0 x 1 2 3
1 y 4 5 6
2 z 7 8 9
if there are duplicates in a,b columns we could lose information so we need GroupBy.cumcount
print(df)
label type value
0 x a 1
1 x b 2
2 x c 3
3 y a 4
4 y b 5
5 y c 6
6 z a 7
7 z b 8
8 z c 9
0 x a 1
1 x b 2
2 x c 3
3 y a 4
4 y b 5
5 y c 6
6 z a 7
7 z b 8
8 z c 9
df.pivot_table(index = ['label',
df.groupby(['label','type']).cumcount()],
columns = 'type',
values = 'value')
type a b c
label
x 0 1 2 3
1 1 2 3
y 0 4 5 6
1 4 5 6
z 0 7 8 9
1 7 8 9
Or:
(df.assign(type_2 = df.groupby(['label','type']).cumcount())
.set_index(['label','type','type_2'])['value']
.unstack('type'))
(
df.pivot(index="label",columns="type",values="value").
reset_index(level=0,drop=True)
)
Another method:
df.groupby(["label","type"])["value"].sum().unstack()
Another:
pd.crosstab(index=df.label,columns=df.type,values=df.value,aggfunc="su
I cannot figure out how to do "reverse melt" using Pandas in python.
This is my starting data
label type value
0 x a 1
1 x b 2
2 x c 3
3 y a 4
4 y b 5
5 y c 6
6 z a 7
7 z b 8
8 z c 9
This is the output I would like to have:
label a b c
x 1 2 3
y 4 5 6
z 7 8 9
I’m sure there is an easy way to do this, but I don’t know how.
there are a few ways:
using .pivot:
>>> origin.pivot(index='label', columns='type')['value']
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[3 rows x 3 columns]
using pivot_table:
>>> origin.pivot_table(values='value', index='label', columns='type')
value
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[3 rows x 3 columns]
or .groupby followed by .unstack:
>>> origin.groupby(['label', 'type'])['value'].aggregate('mean').unstack()
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[3 rows x 3 columns]
DataFrame.set_index + DataFrame.unstack
df.set_index(['label','type'])['value'].unstack()
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
simplifying the passing of pivot arguments
df.pivot(*df)
type a b c
label
x 1 2 3
y 4 5 6
z 7 8 9
[*df]
#['label', 'type', 'value']
For expected output we need DataFrame.reset_index and DataFrame.rename_axis
df.pivot(*df).rename_axis(columns = None).reset_index()
label a b c
0 x 1 2 3
1 y 4 5 6
2 z 7 8 9
if there are duplicates in a,b columns we could lose information so we need GroupBy.cumcount
print(df)
label type value
0 x a 1
1 x b 2
2 x c 3
3 y a 4
4 y b 5
5 y c 6
6 z a 7
7 z b 8
8 z c 9
0 x a 1
1 x b 2
2 x c 3
3 y a 4
4 y b 5
5 y c 6
6 z a 7
7 z b 8
8 z c 9
df.pivot_table(index = ['label',
df.groupby(['label','type']).cumcount()],
columns = 'type',
values = 'value')
type a b c
label
x 0 1 2 3
1 1 2 3
y 0 4 5 6
1 4 5 6
z 0 7 8 9
1 7 8 9
Or:
(df.assign(type_2 = df.groupby(['label','type']).cumcount())
.set_index(['label','type','type_2'])['value']
.unstack('type'))
(
df.pivot(index="label",columns="type",values="value").
reset_index(level=0,drop=True)
)
Another method:
df.groupby(["label","type"])["value"].sum().unstack()
Another:
pd.crosstab(index=df.label,columns=df.type,values=df.value,aggfunc="su