根据 Pandas 中现有列的函数创建新列的动态方法 [英] Dynamic way to create new columns as a function of existing columns in pandas

问题描述

我正在寻找一种更具编程性的方法来创建多个新列作为 Pandas DataFrame 中现有列的函数.

I'm looking for a more programmatic way of creating multiple new columns as a function of existing columns in a Pandas DataFrame.

我有 14 列 Level_2 - Level_15.我想迭代地创建 14 个新列，将列 2-15、3-15、4-15 等相加.

I have 14 columns Level_2 - Level_15. I want to iteratively create 14 new columns that sum columns 2-15, then 3-15, then 4-15 and so on.

现在我的代码看起来像这样

Right now my code looks something like this

cols['2_sum'] = cols.Level_2 + cols.Level_3 + cols.Level_4 + cols.Level_5 + cols.Level_6 + cols.Level_7 + cols.Level_8 + cols.Level_9 + cols.Level_10 + cols.Level_11 + cols.Level_12 + cols.Level_13 + cols.Level_14 + cols.Level_15
cols['3_sum'] = cols.Level_3 + cols.Level_4 + cols.Level_5 + cols.Level_6 + cols.Level_7 + cols.Level_8 + cols.Level_9 + cols.Level_10 + cols.Level_11 + cols.Level_12 + cols.Level_13 + cols.Level_14 + cols.Level_15
cols['4_sum'] = cols.Level_4 + cols.Level_5 + cols.Level_6 + cols.Level_7 + cols.Level_8 + cols.Level_9 + cols.Level_10 + cols.Level_11 + cols.Level_12 + cols.Level_13 + cols.Level_14 + cols.Level_15 

是否有更多的 Pandas 或 Pythonic 方法可以做到这一点?

Is there a more pandas or pythonic way to do this?

谢谢！

推荐答案

这是一个例子:

示例数据:

In [147]: df = pd.DataFrame(np.random.rand(3, 15),
     ...:                   columns=['ID'] + ['Level_{}'.format(x) for x in range(2, 16)])
     ...:

In [148]: df
Out[148]:
         ID   Level_2   Level_3   Level_4   Level_5   Level_6   Level_7   Level_8   Level_9  Level_10  Level_11  \
0  0.851407  0.957810  0.204217  0.848265  0.168324  0.010265  0.191499  0.787552  0.648678  0.424462  0.038888
1  0.354270  0.442843  0.631624  0.081120  0.357300  0.211621  0.177321  0.316312  0.836935  0.445603  0.267165
2  0.998240  0.341875  0.590768  0.475935  0.071915  0.720590  0.041327  0.926167  0.671880  0.516845  0.450720

   Level_12  Level_13  Level_14  Level_15
0  0.465109  0.508491  0.282262  0.848373
1  0.205415  0.399493  0.537186  0.774417
2  0.131734  0.554596  0.253658  0.104193

解决方案:

In [149]: for n in range(15, 1, -1):
     ...:     df['{}_sum'.format(15-n+2)] = df.filter(regex=r'Level_\d+').iloc[:, :n].sum(1)
     ...:

结果:

In [150]: df
Out[150]:
         ID   Level_2   Level_3   Level_4   Level_5   Level_6   Level_7   Level_8   Level_9  Level_10    ...     \
0  0.851407  0.957810  0.204217  0.848265  0.168324  0.010265  0.191499  0.787552  0.648678  0.424462    ...
1  0.354270  0.442843  0.631624  0.081120  0.357300  0.211621  0.177321  0.316312  0.836935  0.445603    ...
2  0.998240  0.341875  0.590768  0.475935  0.071915  0.720590  0.041327  0.926167  0.671880  0.516845    ...

      6_sum     7_sum     8_sum     9_sum    10_sum    11_sum    12_sum    13_sum    14_sum    15_sum
0  4.745067  4.279958  4.241070  3.816608  3.167931  2.380379  2.188880  2.178615  2.010292  1.162027
1  3.973259  3.767844  3.500679  3.055076  2.218140  1.901828  1.724508  1.512887  1.155587  1.074468
2  4.939755  4.808021  4.357301  3.840456  3.168576  2.242409  2.201082  1.480492  1.408577  0.932643

[3 rows x 29 columns]

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