Pandas groupby get filtered sum over total sum

I believe you can use a division on groupby.transform() with sum and assign using .assign() after filtering so as to align on ythe index:

df[df['C']].assign(Ratio=df['Value']/df.groupby('Group')['Value'].transform('sum'))

If more than 1 True per group, use:

m=(df.groupby(['Group','C'],as_index=False,sort=False)['Value'].sum()
  .query('C==True').assign(Sum=df.groupby(['Group'])['Value'].transform('sum')))
m[['Group']].assign(Ratio=m['Value']/m['Sum'])

   Group     Ratio
0      1  0.333333
3      2  0.571429

You can divide by aggregate filtered rows with all rows and then convert Series to one column DataFrame:

filt = df.loc[df['C']].groupby('Group')['Value'].sum()
tot = df.groupby('Group')['Value'].sum()
df1 = filt.div(tot, fill_value=0).to_frame('ratio')
print (df1)
          ratio
Group          
1      0.333333
2      0.571429

Your solution is possible with change .agg working with all columns to GroupBy.apply for return Series, but if large data/ many unique groups it should be slow:

df = (df.groupby('Group')
        .apply(lambda x: x.loc[x.C, 'Value'].sum() / x.Value.sum())
        .to_frame('ratio'))
print (df)
          ratio
Group          
1      0.333333
2      0.571429

Solutions working nice also with only False group:

df = pd.DataFrame([[0, 2, False], [1, 2, True], [1, 4, False], 
                   [2, 6, False], [2, 8, True]], columns=["Group", "Value", "C"])


df1 = (df.groupby('Group')
        .apply(lambda x: x.loc[x.C, 'Value'].sum() / x.Value.sum())
        .to_frame('ratio'))
print (df1)
          ratio
Group          
0      0.000000
1      0.333333
2      0.571429

filt = df.loc[df['C']].groupby('Group')['Value'].sum()
tot = df.groupby('Group')['Value'].sum()

print (df1)
          ratio
Group          
0      0.000000
1      0.333333
2      0.571429

You could use apply:

result = df.groupby('Group').apply(lambda x: pd.Series({'ratio' : (x.Value * x.C).sum() / x.Value.sum()}))
print(result)

Output

          ratio
Group          
1      0.333333
2      0.571429