modin.pandas.Series.diff¶
- Series.diff(periods=1) Series[source]¶
- First discrete difference of element. - Calculates the difference of a Series element compared with another element in the Series (default is element in previous row). - Parameters:
- periods (int, default 1) – Periods to shift for calculating difference, accepts negative values. 
- Returns:
- Snowpark pandas - Serieswith the first differences of the Series.
- Return type:
- Snowpark pandas - Series
 - Notes - For boolean dtypes, this uses operator.xor() rather than operator.sub(). The result is calculated according to current dtype in Series, however dtype of the result is always float64. - Examples - Difference with previous row - >>> s = pd.Series([1, 1, 2, 3, 5, 8]) >>> s.diff() 0 NaN 1 0.0 2 1.0 3 1.0 4 2.0 5 3.0 dtype: float64 - Difference with 3rd previous row - >>> s.diff(periods=3) 0 NaN 1 NaN 2 NaN 3 2.0 4 4.0 5 6.0 dtype: float64 - Difference with following row - >>> s.diff(periods=-1) 0 0.0 1 -1.0 2 -1.0 3 -2.0 4 -3.0 5 NaN dtype: float64