Solution to Question 27 is no longer supported.

[tomleary]

I was able to get the correct result with the following:

df.groupby('grps')['vals'].nlargest(3).groupby('grps').sum()

but I'm sure there's a more elegant way to do it than by using the groupby method twice in one line.

[Akshitlohani]

we can also use lambda function
df.groupby('grps')['vals'].apply(lambda x: x.nlargest(3,'last').sum())