熊猫中嵌套的ifelse替代品

Question

Suppose we've got a test dataset:

value   group
123 1
120 1
NA  1
130 1
23  2
22  2
24  2
NA  2

Now we want to replace missing values with group -wise median values. In R we can do it using a nested ifelse call.

first.med <- median(test[test$group == 1, ]$value, na.rm = T)
second.med <- median(test[test$group == 2, ]$value, na.rm = T)

test$value <- ifelse(is.na(test$value) & test$group == 1, first.med
                     ifelse(is.na(test$value) & test$group == 2, second.med,
                            test$value))

I though about applying the numpy.where function or the pandas.DataFrame.Set.map method as showcased here , but both techniques do not support nesting. I can think of a list comprehension to do this, but I wish to know if there is an alternative in the realm of NumPy/pandas. Thank you in advance.

Answer 1

In this case, you can use a groupby to fill by the group median:

In [16]: df.groupby('group')['value'].apply(lambda x: x.fillna(x.median()))
Out[16]: 
0    123
1    120
2    123
3    130
4     23
5     22
6     24
7     23
dtype: float64

Although in general, both of those methods can be nested just fine. Eg, you could do:

In [23]: medians = df.groupby('group')['value'].median()

In [24]: np.where(pd.isnull(df['value']), 
           np.where(df['group'] == 1, medians.loc[1], medians.loc[2]),    
           df['value'])
Out[24]: array([ 123.,  120.,  123.,  130.,   23.,   22.,   24.,   23.])

Answer 2

df = pd.DataFrame({'value' : [123,120,np.nan ,130,23 ,22 ,24 ,np.nan] , 'group' : [1 , 1 ,1 , 1 , 2 , 2 , 2 , 2] })

def replace_with_median(df):
    df['value'][pd.isnull(df['value'])] = df['value'].median()
    return df

df.groupby('group').apply(replace_with_median)