modin.pandas.DataFrame.diff

DataFrame.diff(periods=1, axis=0) → Self[source] (https://github.com/snowflakedb/snowpark-python/blob/v1.42.0/.tox/docs/lib/python3.9/site-packages/modin/pandas/base.py#L1432-L1448)

First discrete difference of element.

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

Parameters:

  • periods (int, default 1) – Periods to shift for calculating difference, accepts negative values.

  • axis ({0 or 'index', 1 or 'columns'}, default 0) – Take difference over rows (0) or columns (1).

Returns:

DataFrame with the first differences of the Series.

Return type:

DataFrame

Notes

For boolean dtypes, this uses operator.xor() rather than operator.sub(). The result is calculated according to current dtype in DataFrame, however dtype of the result is always float64.

Examples

>>> df = pd.DataFrame({'a': [1, 2, 3, 4, 5, 6],
...                    'b': [1, 1, 2, 3, 5, 8],
...                    'c': [1, 4, 9, 16, 25, 36]})
>>> df 
   a  b   c
0  1  1   1
1  2  1   4
2  3  2   9
3  4  3  16
4  5  5  25
5  6  8  36
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Difference with previous row

>>> df.diff() 
    a    b     c
0  NaN  NaN   NaN
1  1.0  0.0   3.0
2  1.0  1.0   5.0
3  1.0  1.0   7.0
4  1.0  2.0   9.0
5  1.0  3.0  11.0
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Difference with previous column

>>> df.diff(axis=1) 
    a   b   c
0 None  0   0
1 None -1   3
2 None -1   7
3 None -1  13
4 None  0  20
5 None  2  28
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Difference with 3rd previous row

>>> df.diff(periods=3) 
    a    b     c
0  NaN  NaN   NaN
1  NaN  NaN   NaN
2  NaN  NaN   NaN
3  3.0  2.0  15.0
4  3.0  4.0  21.0
5  3.0  6.0  27.0
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Difference with following row

>>> df.diff(periods=-1) 
    a    b     c
0 -1.0  0.0  -3.0
1 -1.0 -1.0  -5.0
2 -1.0 -1.0  -7.0
3 -1.0 -2.0  -9.0
4 -1.0 -3.0 -11.0
5  NaN  NaN   NaN
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