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

Question

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
Index
Val
0
1
1
2
2
3
3
4
4
0
5
1
6
-1
7
-2
8
-3
I am looking for the output:
Index
F(Val)
0
4
1
4
2
4
3
4
4
6
5
6
6
7
7
8
8
NaN
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:
Index
G(Val)
0
4
1
4
2
4
3
4
4
1
5
1
6
-1
7
-2
8
NaN
Thanks!

Answer 1

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

Answer 2

Using 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