[英]How to multiply different columns in different dataframes using Pandas
我有 2 个要相乘的数据帧。 我想将 dataframe 1 中的多列与 dataframe 2 中的一列相乘
raw_material_LCI = dataframe1[["climate change","ozone depletion",
"ionising radiation, hh","photochemical ozone formation, hh",
"particulate matter","human toxicity, non-cancer",
"human toxicity, cancer","acidification",
"eutrophication, freshwater","eutrophication, marine",
"eutrophication, terrestrial","ecotoxicity, freshwater",
"land use", "resource use, fossils","resource use, minerals and metals",
"water scarcity"]] * dataframe2["mass_frac"]
上面的代码返回一个 dataframe,其中所有值都是 NaN。 列的名称都是其中包含数值的字段。
我决定尝试将 dataframe1 与一个值相乘,看看它是否有效,例如下面的示例
raw_material_LCI = dataframe1[["climate change","ozone depletion",
"ionising radiation, hh","photochemical ozone formation, hh",
"particulate matter","human toxicity, non-cancer",
"human toxicity, cancer","acidification",
"eutrophication, freshwater","eutrophication, marine",
"eutrophication, terrestrial","ecotoxicity, freshwater",
"land use", "resource use, fossils","resource use, minerals and metals",
"water scarcity"]] * 0.7
具有单个值的示例返回带有数字的 dataframe,因此它可以工作。 有谁知道为什么一开始的乘法不起作用? 我看过 Python 中多篇关于在不同数据框中乘以列的文章,但找不到解决方案。
当您将两个数据帧相乘时,您必须同时对齐行和列索引;当您将 DataFrame 乘以一个系列时,您必须对齐行索引:
>>> df
A B C D E
0 0.787081 0.350508 0.058542 0.492340 0.489379
1 0.512436 0.501375 0.108115 0.960808 0.841969
2 0.055247 0.305830 0.976043 0.016188 0.006424
3 0.303570 0.914876 0.157100 0.767454 0.340381
4 0.446077 0.595001 0.307799 0.115410 0.568281
5 0.226516 0.636902 0.086790 0.079260 0.402414
6 0.451920 0.526025 0.012470 0.931610 0.267155
7 0.472778 0.137005 0.227569 0.941355 0.584782
8 0.944396 0.769115 0.497214 0.531419 0.570797
9 0.788023 0.310288 0.336480 0.585466 0.432246
>>> sr
0 0.920878
1 0.445332
2 0.894407
3 0.613317
4 0.242270
5 0.299121
6 0.843052
7 0.279014
8 0.526778
9 0.249538
dtype: float64
所以,这会产生nan
值:
>>> df * sr
A B C D E
0 0.724805 0.322775 0.053910 0.453385 0.450658
1 0.228204 0.223279 0.048147 0.427878 0.374956
2 0.049413 0.273536 0.872980 0.014479 0.005745
3 0.186185 0.561109 0.096352 0.470693 0.208762
4 0.108071 0.144151 0.074571 0.027961 0.137678
5 0.067756 0.190511 0.025961 0.023708 0.120371
6 0.380992 0.443466 0.010513 0.785396 0.225226
7 0.131912 0.038226 0.063495 0.262651 0.163162
8 0.497487 0.405153 0.261921 0.279940 0.300683
9 0.196642 0.077429 0.083965 0.146096 0.107862
但沿索引轴使用mul
按预期工作:
>>> df.mul(sr, axis=0) # but not df.mul(sr) (same as df*sr)
A B C D E
0 0.724805 0.322775 0.053910 0.453385 0.450658
1 0.228204 0.223279 0.048147 0.427878 0.374956
2 0.049413 0.273536 0.872980 0.014479 0.005745
3 0.186185 0.561109 0.096352 0.470693 0.208762
4 0.108071 0.144151 0.074571 0.027961 0.137678
5 0.067756 0.190511 0.025961 0.023708 0.120371
6 0.380992 0.443466 0.010513 0.785396 0.225226
7 0.131912 0.038226 0.063495 0.262651 0.163162
8 0.497487 0.405153 0.261921 0.279940 0.300683
9 0.196642 0.077429 0.083965 0.146096 0.107862
如果您的系列和 dataframe 的长度不同,您会得到部分结果:
>>> df.mul(sr.iloc[:5], axis=0)
A B C D E
0 0.724805 0.322775 0.053910 0.453385 0.450658
1 0.228204 0.223279 0.048147 0.427878 0.374956
2 0.049413 0.273536 0.872980 0.014479 0.005745
3 0.186185 0.561109 0.096352 0.470693 0.208762
4 0.108071 0.144151 0.074571 0.027961 0.137678
5 NaN NaN NaN NaN NaN
6 NaN NaN NaN NaN NaN
7 NaN NaN NaN NaN NaN
8 NaN NaN NaN NaN NaN
9 NaN NaN NaN NaN NaN
>>> df.mul(sr.iloc[5:], axis=0)
A B C D E
0 NaN NaN NaN NaN NaN
1 NaN NaN NaN NaN NaN
2 NaN NaN NaN NaN NaN
3 NaN NaN NaN NaN NaN
4 NaN NaN NaN NaN NaN
5 0.067756 0.190511 0.025961 0.023708 0.120371
6 0.380992 0.443466 0.010513 0.785396 0.225226
7 0.131912 0.038226 0.063495 0.262651 0.163162
8 0.497487 0.405153 0.261921 0.279940 0.300683
9 0.196642 0.077429 0.083965 0.146096 0.107862
注意在实例之间具有相同的索引。
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