modin.pandas.Series.diff¶
- Series.diff(periods: int = 1)[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
Series
with 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