CODEWITHSUNDEEP
pythonpandasnumpy
Suresh Raja
Suresh Raja

Reputation: 745

Divide each pair of column with each other

I am looking to divide each pair of successive column and repopulate the calculated value. For example, I have the following DataFrame below. In the data frame below, I want to divide B/A and the D/C. Then the result of B/A should be repopulated in column B and the result of D/C should be populated in column D.

Note that my actual DataFrame is very large. It has 86 columns. I would prefer to have an automated scheme that either loops through all the columns (that is 86 columns) that changes the values in 43 columns or a built-in Pandas function that does this operation.

       A            B           C           D
0   2.056494    -3.002088   0.516822    -1.338846
1   0.082295    1.387734    -0.495226   1.119553
2   0.298618    -0.130158   0.804705    -0.120110
3   0.178088    1.137238    1.331856    -0.472720
4   -0.378670   1.649041    -0.240723   2.044113
5   3.602587    1.152502    -0.170646   -0.961922
6   -0.285846   -0.154891   1.492879    0.752487
7   -0.412809   1.076796    -2.001025   -0.954021

Thanks for reading this and appreciate your help.

Upvotes: 1

Views: 87

Answers (1)

DSM
DSM

Reputation: 353059

You can use .iloc to slice every other column, and then .values to remove the indices on both axes so it'll align correctly:

>>> df.iloc[:, 1::2] /= df.iloc[:, ::2].values
>>> df
          A          B         C         D
0  2.056494  -1.459809  0.516822 -2.590536
1  0.082295  16.862920 -0.495226 -2.260691
2  0.298618  -0.435868  0.804705 -0.149260
3  0.178088   6.385820  1.331856 -0.354933
4 -0.378670  -4.354823 -0.240723 -8.491557
5  3.602587   0.319910 -0.170646  5.636944
6 -0.285846   0.541869  1.492879  0.504051
7 -0.412809  -2.608461 -2.001025  0.476766

.iloc allows us to index positionally using the standard Python slicing:

>>> df.iloc[:, 1::2]
           B         D
0  -1.459809 -2.590536
1  16.862920 -2.260691
2  -0.435868 -0.149260
3   6.385820 -0.354933
4  -4.354823 -8.491557
5   0.319910  5.636944
6   0.541869  0.504051
7  -2.608461  0.476766

And without the .values, we'd still have the column names, which would give us

>>> df.iloc[:, 1::2] / df.iloc[:, ::2] 
    A   B   C   D
0 NaN NaN NaN NaN
1 NaN NaN NaN NaN
2 NaN NaN NaN NaN
3 NaN NaN NaN NaN
4 NaN NaN NaN NaN
5 NaN NaN NaN NaN
6 NaN NaN NaN NaN
7 NaN NaN NaN NaN

Upvotes: 5

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