One way is to set b and c as the index of your frames respectively, and use join followed by reset_index:
df1.set_index('b').join(df2.set_index('c')).reset_index()
b a d
0 a 0 Alex
1 b 1 Alex
2 c 2 Alex
3 d 3 Alex
This will be faster than the merge/drop method on large dataframes, mostly because drop is slow. @Bill's method is faster than my suggestion, and @W-B & @PiRsquared easily outspeed the other suggestions:
import timeit
df1 = pd.concat((df1 for _ in range(1000)))
df2 = pd.concat((df2 for _ in range(1000)))
def index_method(df1 = df1, df2 = df2):
return df1.set_index('b').join(df2.set_index('c')).reset_index()
def merge_method(df1 = df1, df2=df2):
return df1.merge(df2, left_on='b', right_on='c').drop('c', axis='columns')
def rename_method(df1 = df1, df2 = df2):
return df1.rename({'b': 'c'}, axis=1).merge(df2)
def index_method2(df1 = df1, df2 = df2):
return df1.join(df2.set_index('c'), on='b')
def assign_method(df1 = df1, df2 = df2):
return df1.set_index('b').assign(c=df2.set_index('c').d).reset_index()
def map_method(df1 = df1, df2 = df2):
return df1.assign(d=df1.b.map(dict(df2.values)))
>>> timeit.timeit(index_method, number=10) / 10
0.7853091600998596
>>> timeit.timeit(merge_method, number=10) / 10
1.1696729859002517
>>> timeit.timeit(rename_method, number=10) / 10
0.4291436871004407
>>> timeit.timeit(index_method2, number=10) / 10
0.5037374985004135
>>> timeit.timeit(assign_method, number=10) / 10
0.0038641377999738325
>>> timeit.timeit(map_method, number=10) / 10
0.006620216699957382
