how to do a proper pandas group by, where, sum - pandas

Question

i am having a hard time using a group by + where to apply a sum to a broader range.
given this code:
from io import StringIO
import numpy as np
f = pd.read_csv(StringIO("""
fund_id,l_s,val
fund1,L,10
fund1,L,20
fund1,S,30
fund2,L,15
fund2,L,25
fund2,L,35
"""))
# fund total - works as expected
f['fund_total'] = f.groupby('fund_id')['val'].transform(np.sum)
# fund L total - applied only to L rows.
f['fund_total_l'] = f[f['l_s'] == "L"].groupby('fund_id')['val'].transform(np.sum)
f
this code gets me close:
numbers are correct, but i would like fund_total_l column to show 30 for all rows of fund1 (not just L). I want a fund level summary, but sum filtered by the l_s column
i know i can do this with multiple steps, but this needs to be a single operation. i can use a separate generic function if that helps.
playground: https://repl.it/repls/UnusualImpeccableDaemons

Answer 1

Use Series.where, to create NaN, these will be ignored in your sum:
f['val_temp'] = f['val'].where(f['l_s'] == "L")
f['fund_total_l'] = f.groupby('fund_id')['val_temp'].transform('sum')
f = f.drop(columns='val_temp')
Or in one line using assign:
df['fun_total_l'] = (
f.assign(val_temp=f['val'].where(f['l_s'] == "L"))
.groupby('fund_id')['val_temp'].transform('sum')
)
Another way would be to partly use your solution, but then use DataFrame.reindex to get the original index back and then use ffill and bfill to fill up our NaN:
f['fund_total_l'] = (
f[f['l_s'] == "L"]
.groupby('fund_id')['val']
.transform('sum')
.reindex(f.index)
.ffill()
.bfill()
)
fund_id l_s val fund_total_l
0 fund1 L 10 30.0
1 fund1 L 20 30.0
2 fund1 S 30 30.0
3 fund2 L 15 75.0
4 fund2 L 25 75.0
5 fund2 L 35 75.0

Answer 2

I think there is a more elegant solution, but I'm not able to broadcast the results back to the individual rows.
Essentially, with a boolean mask of all the "L" rows
f.groupby("fund_id").apply(lambda g:sum(g["val"]*(g["l_s"]=="L")))
you obtain
fund_id
fund1 30
fund2 75
dtype: int64
now we can just merge after using reset_index to obtain
pd.merge(f, f.groupby("fund_id").apply(lambda g:sum(g["val"]*(g["l_s"]=="L"))).reset_index(), on="fund_id")
to obtain
fund_id l_s val 0
0 fund1 L 10 30
1 fund1 L 20 30
2 fund1 S 30 30
3 fund2 L 15 75
4 fund2 L 25 75
5 fund2 L 35 75
However, I'd guess that the merging is not necessary and can be obtained directly in apply