对分组的熊猫数据框中的行求和并返回 NaN

Question

Example

import pandas as pd
import numpy as np
d = {'l':  ['left', 'right', 'left', 'right', 'left', 'right'],
     'r': ['right', 'left', 'right', 'left', 'right', 'left'],
     'v': [-1, 1, -1, 1, -1, np.nan]}
df = pd.DataFrame(d)

Problem

When a grouped dataframe contains a value of np.NaN I want the grouped sum to be NaN as is given by the skipna=False flag for pd.Series.sum and also pd.DataFrame.sum however, this

In [235]: df.v.sum(skipna=False)
Out[235]: nan

However, this behavior is not reflected in the pandas.DataFrame.groupby object

In [237]: df.groupby('l')['v'].sum()['right']
Out[237]: 2.0

and cannot be forced by applying the np.sum method directly

In [238]: df.groupby('l')['v'].apply(np.sum)['right']
Out[238]: 2.0

Workaround

I can workaround this by doing

check_cols = ['v']
df['flag'] = df[check_cols].isnull().any(axis=1)
df.groupby('l')['v', 'flag'].apply(np.sum).apply(
    lambda x: x if not x.flag else np.nan,
    axis=1
)

but this is ugly. Is there a better method?

Answer 1

I think it's inherent to pandas. A workaround can be :

df.groupby('l')['v'].apply(array).apply(sum)

to mimic the numpy way,

or

df.groupby('l')['v'].apply(pd.Series.sum,skipna=False) # for series, or
df.groupby('l')['v'].apply(pd.DataFrame.sum,skipna=False) # for dataframes.

to call the good function.

Answer 2

I'm not sure where this falls on the ugliness scale, but it works:

>>> series_sum = pd.core.series.Series.sum
>>> df.groupby('l')['v'].agg(series_sum, skipna=False)
l
left     -3
right   NaN
Name: v, dtype: float64

I just dug up the sum method you used when you took df.v.sum , which supports the skipna option:

>>> help(df.v.sum)
Help on method sum in module pandas.core.generic:

sum(axis=None, skipna=None, level=None, numeric_only=None, **kwargs) method 
of pandas.core.series.Series instance

Answer 3

Is that what you want?

In [24]: df.groupby('l')['v'].agg(lambda x: np.nan if x.isnull().any() else x.sum())
Out[24]:
l
left    -3.0
right    NaN
Name: v, dtype: float64

or

In [22]: df.groupby('l')['v'].agg(lambda x: x.sum() if x.notnull().all() else np.nan)
Out[22]:
l
left    -3.0
right    NaN
Name: v, dtype: float64

Answer 4

df.groupby(xxx).yyy.apply(lambda x: x.sum(skipna=False))