Python Pandas匹配距离另一个Dataframe最近的索引

Question

df.index = 10,100,1000

df2.index = 1,2,11,50,101,500,1001
Just sample

I need to match closest index from df2 compare with df by these conditions

1. df2.index have to > df.index
2. only one closest value

for example output

df     |   df2
10     |   11
100    |   101
1000   |   1001

Now I can do it with for-loop and it's extremely slow

And I used new_df2 to keep index instead of df2

new_df2 = pd.DataFrame(columns = ["value"])
for col in df.index:
    for col2 in df2.index:
        if(col2 > col):
            new_df2.loc[col2] = df2.loc[col2]
            break
        else:
            df2 = df2[1:] #delete first row for index speed

How to avoid for-loop in this case Thank.

Answer 1

Not sure how robust this is, but you can sort df2 so it's index is decreasing, and use asof to find the most recent index label matching each key in df 's index:

df2.sort_index(ascending=False, inplace=True)
df['closest_df2'] = df.index.map(lambda x: df2.index.asof(x))

df
Out[19]: 
      a  closest_df2
10    1           11
100   2          101
1000  3         1001