根据另一个数据框中的列填充数据框中的空值

Question

I have a dataframe df1 like this.

I want to fill the nan and the number 0 in column score with mutiple values in another dataframe df2 according to the different names.

How could I do this?

Answer 1

Option 1
Short version

df1.score = df1.score.mask(df1.score.eq(0)).fillna(
    df1.name.map(df2.set_index('name').score)
)
df1

  name  score
0    A   10.0
1    B   32.0
2    A   10.0
3    C   30.0
4    B   20.0
5    A   45.0
6    A   10.0
7    A   10.0

---

Option 2
Interesting version using searchsorted . df2 must be sorted by 'name' .

i = np.where(np.isnan(df1.score.mask(df1.score.values == 0).values))[0]
j = df2.name.values.searchsorted(df1.name.values[i])
df1.score.values[i] = df2.score.values[j]
df1

  name  score
0    A   10.0
1    B   32.0
2    A   10.0
3    C   30.0
4    B   20.0
5    A   45.0
6    A   10.0
7    A   10.0

Answer 2

If df1 and df2 are your dataframes, you can create a mapping and then call pd.Series.replace :

df1 = pd.DataFrame({'name' : ['A', 'B', 'A', 'C', 'B', 'A', 'A', 'A'], 
                     'score': [0, 32, 0, np.nan, np.nan, 45, np.nan, np.nan]})
df2 = pd.DataFrame({'name' : ['A', 'B', 'C'], 'score' : [10, 20, 30]})

print(df1)

  name  score
0    A    0.0
1    B   32.0
2    A    0.0
3    C    NaN
4    B    NaN
5    A   45.0
6    A    NaN
7    A    NaN

print(df2) 

  name  score
0    A     10
1    B     20
2    C     30

mapping = dict(df2.values)

df1.loc[(df1.score.isnull()) | (df1.score == 0), 'score'] =\
               df1[(df1.score.isnull()) | (df1.score == 0)].name.replace(mapping)

print(df1)

  name  score
0    A   10.0
1    B   32.0
2    A   10.0
3    C   30.0
4    B   20.0
5    A   45.0
6    A   10.0
7    A   10.0

Answer 3

Or using merge , fillna

import pandas as pd
import numpy as np

df1.loc[df.score==0,'score']=np.nan
df1.merge(df2,on='name',how='left').fillna(method='bfill',axis=1)[['name','score_x']]\
    .rename(columns={'score_x':'score'})

Answer 4

This method changes the order (the result will be sorted by name ).

df1.set_index('name').replace(0, np.nan).combine_first(df2.set_index('name')).reset_index()

  name  score
0    A     10
1    A     10
2    A     45
3    A     10
4    A     10
5    B     32
6    B     20
7    C     30