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.
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)
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.
