Call func on self producing a Series with the same axis shape as self.
Function to use for transforming the data. If a function, must either work when passed a Series or when passed to Series.apply. If func is both list-like and dict-like, dict-like behavior takes precedence.
Accepted combinations are:
function
string function name
list-like of functions and/or function names, e.g. [np.exp, 'sqrt']
dict-like of axis labels -> functions, function names or list-like of such.
Unused. Parameter needed for compatibility with DataFrame.
Positional arguments to pass to func.
Keyword arguments to pass to func.
A Series that must have the same length as self.
See also
Series.aggOnly perform aggregating type operations.
Series.applyInvoke function on a Series.
Notes
Functions that mutate the passed object can produce unexpected behavior or errors and are not supported. See Mutating with User Defined Function (UDF) methods for more details.
Examples
>>> df = pd.DataFrame({'A': range(3), 'B': range(1, 4)})
>>> df
A B
0 0 1
1 1 2
2 2 3
>>> df.transform(lambda x: x + 1)
A B
0 1 2
1 2 3
2 3 4
Even though the resulting Series must have the same length as the input Series, it is possible to provide several input functions:
>>> s = pd.Series(range(3))
>>> s
0 0
1 1
2 2
dtype: int64
>>> s.transform([np.sqrt, np.exp])
sqrt exp
0 0.000000 1.000000
1 1.000000 2.718282
2 1.414214 7.389056
You can call transform on a GroupBy object:
>>> df = pd.DataFrame({
... "Date": [
... "2015-05-08", "2015-05-07", "2015-05-06", "2015-05-05",
... "2015-05-08", "2015-05-07", "2015-05-06", "2015-05-05"],
... "Data": [5, 8, 6, 1, 50, 100, 60, 120],
... })
>>> df
Date Data
0 2015-05-08 5
1 2015-05-07 8
2 2015-05-06 6
3 2015-05-05 1
4 2015-05-08 50
5 2015-05-07 100
6 2015-05-06 60
7 2015-05-05 120
>>> df.groupby('Date')['Data'].transform('sum')
0 55
1 108
2 66
3 121
4 55
5 108
6 66
7 121
Name: Data, dtype: int64
>>> df = pd.DataFrame({
... "c": [1, 1, 1, 2, 2, 2, 2],
... "type": ["m", "n", "o", "m", "m", "n", "n"]
... })
>>> df
c type
0 1 m
1 1 n
2 1 o
3 2 m
4 2 m
5 2 n
6 2 n
>>> df['size'] = df.groupby('c')['type'].transform(len)
>>> df
c type size
0 1 m 3
1 1 n 3
2 1 o 3
3 2 m 4
4 2 m 4
5 2 n 4
6 2 n 4
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Licensed under the 3-clause BSD License.
https://pandas.pydata.org/pandas-docs/version/2.3.0/reference/api/pandas.Series.transform.html