Participer au site avec un Tip
Rechercher
 

Améliorations / Corrections

Vous avez des améliorations (ou des corrections) à proposer pour ce document : je vous remerçie par avance de m'en faire part, cela m'aide à améliorer le site.

Emplacement :

Description des améliorations :

Classe « Series »

Méthode pandas.Series.transform

Signature de la méthode transform

def transform(self, func: Union[Callable, str, List[Union[Callable, str]], Dict[Optional[Hashable], Union[Callable, str, List[Union[Callable, str]]]]], axis: Union[str, int] = 0, *args, **kwargs) -> Union[ForwardRef('DataFrame'), ForwardRef('Series')] 

Description

transform.__doc__

Call ``func`` on self producing a Series with transformed values.

Produced Series will have same axis length as self.

Parameters
----------
func : function, str, list-like or dict-like
    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.
axis : {0 or 'index'}
        Parameter needed for compatibility with DataFrame.
*args
    Positional arguments to pass to `func`.
**kwargs
    Keyword arguments to pass to `func`.

Returns
-------
Series
    A Series that must have the same length as self.

Raises
------
ValueError : If the returned Series has a different length than self.

See Also
--------
Series.agg : Only perform aggregating type operations.
Series.apply : Invoke function on a Series.

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