pandas.core.groupby.SeriesGroupBy.fillna#

SeriesGroupBy.fillna(value=None, method=None, axis=None, inplace=False, limit=None, downcast=None)[source]#

Fill NA/NaN values using the specified method within groups.

Parameters
valuescalar, dict, Series, or DataFrame

Value to use to fill holes (e.g. 0), alternately a dict/Series/DataFrame of values specifying which value to use for each index (for a Series) or column (for a DataFrame). Values not in the dict/Series/DataFrame will not be filled. This value cannot be a list. Users wanting to use the value argument and not method should prefer Series.fillna() as this will produce the same result and be more performant.

method{{‘bfill’, ‘ffill’, None}}, default None

Method to use for filling holes. 'ffill' will propagate the last valid observation forward within a group. 'bfill' will use next valid observation to fill the gap.

axis{0 or ‘index’, 1 or ‘columns’}

Unused, only for compatibility with DataFrameGroupBy.fillna().

inplacebool, default False

Broken. Do not set to True.

limitint, default None

If method is specified, this is the maximum number of consecutive NaN values to forward/backward fill within a group. In other words, if there is a gap with more than this number of consecutive NaNs, it will only be partially filled. If method is not specified, this is the maximum number of entries along the entire axis where NaNs will be filled. Must be greater than 0 if not None.

downcastdict, default is None

A dict of item->dtype of what to downcast if possible, or the string ‘infer’ which will try to downcast to an appropriate equal type (e.g. float64 to int64 if possible).

Returns
Series

Object with missing values filled within groups.

See also

ffill

Forward fill values within a group.

bfill

Backward fill values within a group.

Examples

>>> ser = pd.Series([np.nan, np.nan, 2, 3, np.nan, np.nan])
>>> ser
0    NaN
1    NaN
2    2.0
3    3.0
4    NaN
5    NaN
dtype: float64

Propagate non-null values forward or backward within each group.

>>> ser.groupby([0, 0, 0, 1, 1, 1]).fillna(method="ffill")
0    NaN
1    NaN
2    2.0
3    3.0
4    3.0
5    3.0
dtype: float64

>>> ser.groupby([0, 0, 0, 1, 1, 1]).fillna(method="bfill")
0    2.0
1    2.0
2    2.0
3    3.0
4    NaN
5    NaN
dtype: float64

Only replace the first NaN element within a group.

>>> ser.groupby([0, 0, 0, 1, 1, 1]).fillna(method="ffill", limit=1)
0    NaN
1    NaN
2    2.0
3    3.0
4    3.0
5    NaN
dtype: float64