一种方法是分别设置b和c作为帧的索引，并使用join后跟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
这将比merge/drop大数据帧上的方法快，主要是因为drop它很慢。@Bill的方法比我的建议快，而@WB和@PiRsquared轻松超越了其他建议：
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
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