Modify data frame based on the row index

Given a pandas dataframe (20, 40), I would like to modify the first 10 rows of the first 20 columns using the value index. For example, if:

df.iloc[5,6] = 0.98,

I would like to modify the value in the following way

new df.iloc[5,6] = 0.98 ** -(1/5)
where 5 is the row index.

And I should do the same for every value between the first 10 rows and the first 20 columns.

Can anyone help me? Thank you very much in advance.

CodePudding user response：

Can you explain what you want to do in a more general way? I don't understand why you chose 5 here.

The way to make columns off other columns is

df["new column"] = df["column1"] ** (-1/df["column2"])

the way you do it with the index is the same

    df["new column"] = df["column1"] ** (-1/df.index)

CodePudding user response：

You can do this operation in-place with the following snippet.

from numpy.random import default_rng
from pandas import DataFrame
from string import ascii_lowercase

rng = default_rng(0)
df = DataFrame(
    (data := rng.integers(1, 10, size=(4, 5))),
    columns=[*ascii_lowercase[:data.shape[1]]]
)

print(df)
   a  b  c  d  e
0  8  6  5  3  3
1  1  1  1  2  8
2  6  9  5  6  9
3  7  6  5  6  9

# you would replace :3, :4 with :10, :20 for your data
df.iloc[:3, :4] **= (-1 / df.index)

print(df)
   a         b         c         d  e
0  0  0.166667  0.447214  0.693361  3
1  1  1.000000  1.000000  0.793701  8
2  0  0.111111  0.447214  0.550321  9
3  7  6.000000  5.000000  6.000000  9

In the event your index is not a simple RangeIndex you can use numpy.arange to mimic this:

from numpy import arange

df.iloc[:3, :4] **= (-1 / arange(df.shape[0]))

print(df)
   a         b         c         d  e
0  0  0.166667  0.447214  0.693361  3
1  1  1.000000  1.000000  0.793701  8
2  0  0.111111  0.447214  0.550321  9
3  7  6.000000  5.000000  6.000000  9

Note: If 0 is in your index, like it is in this example, you'll encounter a RuntimeWarning of dividing by 0.