Set custom sort (first and last) without knowing all values

I have a pandas DataFrame with the following information

         id          desc ammount
356 89 2521 ShouldBeFirst    0.00
356 89 2521  ShouldBeLast   19.00
356 89 2521   RandomValue   39.00
356 89 2521  RandomValue2   29.00
123 45 6789  RandomValue3   29.99
123 45 6789 ShouldBeFirst    0.00
123 45 6789  ShouldBeLast   99.00
123 45 6789  RandomValue3   39.00
123 45 6789  RandomValue2   19.00

What I would like is to sort the DataFrame based on the ID (which can easily be done with df.sort_values('id', ascending=True)) and after that, set the first row of that ID to always be ShouldBeFirst, and the last row with that ID to ShouldBeLast, like this:

         id          desc ammount
123 45 6789 ShouldBeFirst    0.00
123 45 6789  RandomValue2   19.00
123 45 6789  RandomValue3   29.99
123 45 6789  RandomValue3   39.00
123 45 6789  ShouldBeLast   99.00
356 89 2521 ShouldBeFirst    0.00
356 89 2521  RandomValue2   29.00
356 89 2521   RandomValue   39.00
356 89 2521  ShouldBeLast   19.00

I've seen several threads on how to custom sort a pandas DataFrame, but they all require to list all possible values. I do not have all the values on the column desc.

Create an intermediate desc_order column by mapping known values to predefined order then sort the dataframe

df['desc_order'] = df['desc'].map({'ShouldBeFirst': 0, 'ShouldBeLast': 2}).fillna(1)
df.sort_values(['id', 'desc_order', 'ammount']).drop(columns=['desc_order'])

            id           desc  ammount
5  123 45 6789  ShouldBeFirst     0.00
8  123 45 6789   RandomValue2    19.00
4  123 45 6789   RandomValue3    29.99
7  123 45 6789   RandomValue3    39.00
6  123 45 6789   ShouldBeLast    99.00
0  356 89 2521  ShouldBeFirst     0.00
3  356 89 2521   RandomValue2    29.00
2  356 89 2521    RandomValue    39.00
1  356 89 2521   ShouldBeLast    19.00

You can use a custom dictionary mapping the first and last and fillna for the other values:

order = {'ShouldBeFirst': 0, 'ShouldBeLast': 2}

def sorter(s):
    if s.name == 'desc':
        return s.map(order).fillna(1)
    else:
        return s

out = df.sort_values(by=['id', 'desc', 'ammount'], key=sorter)

Or using numpy.lexsort:

order = {'ShouldBeFirst': 0, 'ShouldBeLast': 2}

df.iloc[np.lexsort([df['ammount'], df['desc'].map(order).fillna(1), df['id']])]

Output:

            id           desc  ammount
5  123 45 6789  ShouldBeFirst     0.00
8  123 45 6789   RandomValue2    19.00
4  123 45 6789   RandomValue3    29.99
7  123 45 6789   RandomValue3    39.00
6  123 45 6789   ShouldBeLast    99.00
0  356 89 2521  ShouldBeFirst     0.00
3  356 89 2521   RandomValue2    29.00
2  356 89 2521    RandomValue    39.00
1  356 89 2521   ShouldBeLast    19.00

Try this......
import pandas as pd

# Assuming the DataFrame is already available as 'df'
df_sorted = df.sort_values('id', ascending=True)

mask = df_sorted.duplicated('id', keep=False)
df_sorted.loc[mask, 'order'] = 1
df_sorted.loc[~mask, 'order'] = [0, 2] * (df_sorted['id'].nunique() // 2 + 1)

df_sorted = df_sorted.sort_values(['id', 'order']).drop('order', axis=1)

print(df_sorted)