迭代数据框中的每个列，将每个值与另一个数据框中的另一列中的值匹配

Question

I basically have two data frames. Let's say aa and bb. I want to look all the values in the first column of bb that are in the first column of aa and if they are I have to get column 2 of aa and add it to a new column in bb (if there is not much I'll put a 0). Let's see if looking at some code it makes more sense. I've done it using apply and a function:

aa=pd.DataFrame({'a':[1,2,3,4,5],'b':[6,7,8,9,0]})
bb=pd.DataFrame({'c':[11,2,13,4,15],'d':['f','h','j','k','l']})

   a  b
0  1  6
1  2  7
2  3  8
3  4  9
4  5  0

    c  d
0  11  f
1   2  h
2  13  j
3   4  k
4  15  l


def set_time_session (row):
    element = row['c']
    if element in aa['a'].unique():
        return aa['b'][aa['a']==element]
    else:
        return 0

column = bb.apply(set_time_session,axis=1)
bb['newcolumn']=column

       c  d  newcolumn
0  11  f          0
1   2  h          7
2  13  j          0
3   4  k          9
4  15  l          0

This actually works, but when done in a dataframe with 200000 rows it takes forever to complete. I'm sure the is a better and faster way to do it. Thanks!

Answer 1

Try this:

res = pd.merge(aa, bb, left_on='a', right_on='c', how='inner', left_index=True)
bb['newcolumn']= res.reindex(range(len(aa))).fillna(0)['b']
print(bb)