pandas function to find index of the first future instance where column is less than each row's value

I'm new to Stack Overflow, and I just have a question about solving a problem in pandas. I am looking to create a function that returns the index of the first future instance where a column is less than each row's value for that column.

For example, consider the dataframe:

import numpy as np
import pandas as pd
df = pd.DataFrame({'Val': [1, 2, 3, 4, 0, 1, -1, -2, -3]}, index = np.arange(0,9))
df

I am looking for the output:

Or the series/array equivalent of F(Val).

I've been able to solve this quite easily using for loops, but obviously this is extremely slow on the large dataset I am working with an not a very elegant or optimal solution. My hope is that the solution is an efficient pandas function that employs vectorization.

Also, as a bonus question (if anyone can assist), how might the maximum value between each row's index and the F(Val) index be computed using vectorization? The output should look like:

Thanks!

You can use:

grp = df['Val'].lt(df['Val'].shift()).shift(fill_value=0).cumsum()
df['F(Val)'] = df.groupby(grp).transform(lambda x: x.index[-1]).shift(-1)
print(df)

# Output
   Val  F(Val)
0    1     4.0
1    2     4.0
2    3     4.0
3    4     4.0
4    0     6.0
5    1     6.0
6   -1     7.0
7   -2     8.0
8   -3     NaN

Using [tag:numpy] broadcasting and the lower triangle:

a = df['Val'].to_numpy()
m = np.tril(a[:,None]<=a, k=-1)
df['F(Val)'] = np.where(m.any(0), m.argmax(0), np.nan)

Same logic with expanding:

df['F(Val)'] = (df.loc[::-1, 'Val'].expanding()
                  .apply(lambda x: s.idxmax() if len(s:=(x.iloc[-2::-1]<=x.iloc[-1]))
                         else np.nan)
               )

Output (with a difference to the provided one):

   Val  F(Val)
0    1     5.0  # here the next is 5
1    2     4.0
2    3     4.0
3    4     4.0
4    2     5.0
5   -2     7.0
6   -1     7.0
7   -2     8.0
8   -3     NaN