pandas.core.groupby.DataFrameGroupBy.diff#

DataFrameGroupBy.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