-
GroupBy.cumcount(ascending=True)
[source] -
Number each item in each group from 0 to the length of that group - 1.
Essentially this is equivalent to
1>>>
self
.
apply
(
lambda
x: Series(np.arange(
len
(x)), x.index))
Parameters: ascending : bool, default True
If False, number in reverse, from length of group - 1 to 0.
Examples
1234567891011121314151617181920212223242526>>> df
=
pd.DataFrame([[
'a'
], [
'a'
], [
'a'
], [
'b'
], [
'b'
], [
'a'
]],
... columns
=
[
'A'
])
>>> df
A
0
a
1
a
2
a
3
b
4
b
5
a
>>> df.groupby(
'A'
).cumcount()
0
0
1
1
2
2
3
0
4
1
5
3
dtype: int64
>>> df.groupby(
'A'
).cumcount(ascending
=
False
)
0
3
1
2
2
1
3
1
4
0
5
0
dtype: int64
GroupBy.cumcount()

2025-01-10 15:47:30
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