给出df2 DF，其中
df2[car_colname]
car_30 car_90 ...
2005-02-21 0.117660 0.004845 ...
2005-02-22 0.145713 0.000901 ...
2005-02-23 0.074074 0.002139 ...
2005-02-24 0.076934 -0.004596 ...
2005-02-25 0.085200 -0.016227 ...
2005-04-08 0.097728 -0.005937 ...
2005-04-11 0.222366 0.007553 ...
和
df2[sd_colname]
std_30 std_90 ...
2005-02-21 0.052266 0.004150 ...
2005-02-22 0.052266 0.004150 ...
2005-02-23 0.052266 0.004150 ...
2005-02-24 0.052266 0.004150 ...
2005-02-25 0.052266 0.004150 ...
2005-04-08 0.061682 0.006731 ...
2005-04-11 0.061682 0.006731 ...
我想按行df2[car_colname]除以df2[sd_colname]，以便第一行的结果来自：0.117660 / 0.052266 = 2.25117667317185和0.004845 / 0.004150 = 1.16746987951807，依此类推...
2005-02-21 2.25117667317185 1.16746987951807 ...
2005-02-22 2.78791183561015 0.21710843373494 ...
2005-02-23 1.41725022002832 0.515421686746988 ...
2005-02-24 1.4719703057437 -1.10746987951807 ...
2005-02-25 1.6301228331994 -3.91012048192771 ...
2005-04-08 1.58438442333258 -0.882038330114396 ...
2005-04-11 3.60503874712234 1.12212152726192 ...
我尝试df2[car_colname] / df2[sd_colname]或df2[car_colname].div(df2[sd_colname])无济于事。
答案 0 :(得分：2)
您可能需要将columns更改为两个dfs
df1.columns=df1.columns.str.split('_').str[-1]
df2.columns=df2.columns.str.split('_').str[-1]
df1.div(df2)
Out[786]:
30 90
2005-02-21 2.251177 1.167470
2005-02-22 2.787912 0.217108
2005-02-23 1.417250 0.515422
2005-02-24 1.471970 -1.107470
2005-02-25 1.630123 -3.910120
2005-04-08 1.584384 -0.882038
2005-04-11 3.605039 1.122122
答案 1 :(得分：0)
只需替换列名称，
df1.div(df2.rename(columns = (dict(zip(df2.columns, df1.columns)))))
car_30 car_90
2005-02-21 2.251177 1.167470
2005-02-22 2.787912 0.217108
2005-02-23 1.417250 0.515422
2005-02-24 1.471970 -1.107470
2005-02-25 1.630123 -3.910120
2005-04-08 1.584384 -0.882038
2005-04-11 3.605039 1.122122