如何在熊猫中并排附加两个数据框列

Question

我有一个df，它包含所有数值列。 我想为每列查找cumprod并将每列结果的结果并排附加。 这个怎么做。 我希望此并列结果可以方便我进行比较。

例如：

我的输入df：

        col1      col2      col3
0   1.000000  1.000000  1.000000
1   0.998766  0.999490  0.998892
2   0.997779  0.999081  0.998005
3   0.996299  0.998469  0.996676
4   0.994573  0.997754  0.995126
5   0.993095  0.997140  0.993797
6   0.991125  0.996322  0.992027
7   0.989648  0.995708  0.990699
8   0.988171  0.995094  0.989372
9   0.986695  0.994480  0.988045
10  0.984729  0.993660  0.986276
11  0.983010  0.992943  0.984730

df的累积产品：

        col1      col2      col3
0   1.000000  1.000000  1.000000
1   0.998766  0.999490  0.998892
2   0.996547  0.998572  0.996899
3   0.992859  0.997043  0.993585
4   0.987471  0.994803  0.988742
5   0.980653  0.991958  0.982609
6   0.971949  0.988310  0.974775
7   0.961887  0.984069  0.965708
8   0.950509  0.979241  0.955444
9   0.937863  0.973836  0.944022
10  0.923541  0.967662  0.931066
11  0.907850  0.960833  0.916849

预期产量：

        col1      col1      col2      col2      col3      col3
0   1.000000  1.000000  1.000000  1.000000  1.000000  1.000000
1   0.998766  0.998766  0.999490  0.999490  0.998892  0.998892
2   0.997779  0.996547  0.999081  0.998572  0.998005  0.996899
3   0.996299  0.992859  0.998469  0.997043  0.996676  0.993585
4   0.994573  0.987471  0.997754  0.994803  0.995126  0.988742
5   0.993095  0.980653  0.997140  0.991958  0.993797  0.982609
6   0.991125  0.971949  0.996322  0.988310  0.992027  0.974775
7   0.989648  0.961887  0.995708  0.984069  0.990699  0.965708
8   0.988171  0.950509  0.995094  0.979241  0.989372  0.955444
9   0.986695  0.937863  0.994480  0.973836  0.988045  0.944022
10  0.984729  0.923541  0.993660  0.967662  0.986276  0.931066
11  0.983010  0.907850  0.992943  0.960833  0.984730  0.916849

注意：如果我更喜欢用cum_of_coln而不是coln的列名

获取我使用的cum_prod的代码，

print df
print df.cumprod()

Answer 1

计算cumprod ，然后使用cytoolz并交错列标题：

from toolz import interleave

df2 = df.cumprod().add_prefix('cum_of_')
df3 = pd.concat([df, df2], axis=1)[list(interleave([df, df2]))]

或者，您可以使用sorted ：

df2 = df.cumprod().add_prefix('cum_of_')
df3 = pd.concat([df, df2], axis=1)
df3 = df3[sorted(df3, key=lambda x: x.split('_')[-1])]

第三种选择是排序后对列标题进行更改。 应该相当有效率。

df3 = pd.concat([df,  df.cumprod()], axis=1).sort_index(axis=1)
c = df3.columns.values
c[1::2] = 'cum_of_' + c[1::2]
df3.columns = c

df3.head()
        col1  cum_of_col1      col2  cum_of_col2      col3  cum_of_col3
0   1.000000     1.000000  1.000000     1.000000  1.000000     1.000000
1   0.998766     0.998766  0.999490     0.999490  0.998892     0.998892
2   0.997779     0.996548  0.999081     0.998571  0.998005     0.996899
3   0.996299     0.992860  0.998469     0.997043  0.996676     0.993586
4   0.994573     0.987471  0.997754     0.994803  0.995126     0.988743

Answer 2

使用concat并按列表concat生成的列表重新排序：

cols = [item for x in df.columns for item in (x, 'cum_of_' + x)]
df = pd.concat([df, df.cumprod().add_prefix('cum_of_')], axis=1)[cols]

print (df)
        col1  cum_of_col1      col2  cum_of_col2      col3  cum_of_col3
0   1.000000     1.000000  1.000000     1.000000  1.000000     1.000000
1   0.998766     0.998766  0.999490     0.999490  0.998892     0.998892
2   0.997779     0.996548  0.999081     0.998571  0.998005     0.996899
3   0.996299     0.992860  0.998469     0.997043  0.996676     0.993586
4   0.994573     0.987471  0.997754     0.994803  0.995126     0.988743
5   0.993095     0.980653  0.997140     0.991958  0.993797     0.982610
6   0.991125     0.971949  0.996322     0.988310  0.992027     0.974775
7   0.989648     0.961888  0.995708     0.984068  0.990699     0.965709
8   0.988171     0.950510  0.995094     0.979240  0.989372     0.955445
9   0.986695     0.937863  0.994480     0.973835  0.988045     0.944023
10  0.984729     0.923541  0.993660     0.967661  0.986276     0.931067
11  0.983010     0.907850  0.992943     0.960832  0.984730     0.916850

Answer 3

使用pd.assign直接附加列：

df.assign(**df.cumprod().add_prefix('cumprod_'))

 col1      col2      col3  cumprod_col1  cumprod_col2  cumprod_col3
0   1.000000  1.000000  1.000000      1.000000      1.000000      1.000000
1   0.998766  0.999490  0.998892      0.998766      0.999490      0.998892
2   0.997779  0.999081  0.998005      0.996548      0.998571      0.996899
3   0.996299  0.998469  0.996676      0.992860      0.997043      0.993586
4   0.994573  0.997754  0.995126      0.987471      0.994803      0.988743
5   0.993095  0.997140  0.993797      0.980653      0.991958      0.982610
6   0.991125  0.996322  0.992027      0.971949      0.988310      0.974775
7   0.989648  0.995708  0.990699      0.961888      0.984068      0.965709
8   0.988171  0.995094  0.989372      0.950510      0.979240      0.955445
9   0.986695  0.994480  0.988045      0.937863      0.973835      0.944023
10  0.984729  0.993660  0.986276      0.923541      0.967661      0.931067
11  0.983010  0.992943  0.984730      0.907850      0.960832      0.916850

如果要将列排序为col1 - cumprod_col1... ，则可以使用reindex_axis按字母顺序对列进行排序，在这种情况下，已使用add_suffix添加了后缀

df = df.assign(**df.cumprod().add_suffix('_cumprod'))
df = df.reindex_axis(sorted(df.columns), axis=1)

      col1    col1_cumprod      col2  col2_cumprod      col3  col3_cumprod
0   1.000000      1.000000  1.000000      1.000000  1.000000      1.000000
1   0.998766      0.998766  0.999490      0.999490  0.998892      0.998892
2   0.997779      0.996548  0.999081      0.998571  0.998005      0.996899
3   0.996299      0.992860  0.998469      0.997043  0.996676      0.993586
4   0.994573      0.987471  0.997754      0.994803  0.995126      0.988743
5   0.993095      0.980653  0.997140      0.991958  0.993797      0.982610
6   0.991125      0.971949  0.996322      0.988310  0.992027      0.974775
7   0.989648      0.961888  0.995708      0.984068  0.990699      0.965709
8   0.988171      0.950510  0.995094      0.979240  0.989372      0.955445
9   0.986695      0.937863  0.994480      0.973835  0.988045      0.944023
10  0.984729      0.923541  0.993660      0.967661  0.986276      0.931067
11  0.983010      0.907850  0.992943      0.960832  0.984730      0.916850