pandas.core.groupby.SeriesGroupBy.diff#

SeriesGroupBy.diff(periods=1, axis=<no_default>)[source] #

First discrete difference of element.

Calculates the difference of each element compared with another element in the group (default is element in previous row).

Parameters:

periodsint, default 1

Periods to shift for calculating difference, accepts negative values.

axisaxis to shift, default 0

Take difference over rows (0) or columns (1).

Deprecated since version 2.1.0: For axis=1, operate on the underlying object instead. Otherwise the axis keyword is not necessary.

Returns:

Series or DataFrame

First differences.

See also

Series.groupby

Apply a function groupby to a Series.

DataFrame.groupby

Apply a function groupby to each row or column of a DataFrame.

Examples

For SeriesGroupBy:

>>> lst = ['a', 'a', 'a', 'b', 'b', 'b']
>>> ser = pd.Series([7, 2, 8, 4, 3, 3], index=lst)
>>> ser
a 7
a 2
a 8
b 4
b 3
b 3
dtype: int64
>>> ser.groupby(level=0).diff()
a NaN
a -5.0
a 6.0
b NaN
b -1.0
b 0.0
dtype: float64

For DataFrameGroupBy:

>>> data = {'a': [1, 3, 5, 7, 7, 8, 3], 'b': [1, 4, 8, 4, 4, 2, 1]}
>>> df = pd.DataFrame(data, index=['dog', 'dog', 'dog',
...  'mouse', 'mouse', 'mouse', 'mouse'])
>>> df
 a b
 dog 1 1
 dog 3 4
 dog 5 8
mouse 7 4
mouse 7 4
mouse 8 2
mouse 3 1
>>> df.groupby(level=0).diff()
 a b
 dog NaN NaN
 dog 2.0 3.0
 dog 2.0 4.0
mouse NaN NaN
mouse 0.0 0.0
mouse 1.0 -2.0
mouse -5.0 -1.0