What does print(df.groupby('k')['v'].agg(lambda s: (s ** 2).sum()).to_dict()) print? pandas GroupBy and Aggregation

Answer, C. {'a': 5, 'b': 9}

A. {'a': 5, 'b': 3}

Applies the squaring and summing correctly to group a but reports group b's own plain value instead of its square.

B. {'a': 3, 'b': 3}

Sums the plain values of each group instead of squaring them first.

C. {'a': 5, 'b': 9}

Correct. The lambda squares every value in a group before summing them, so group a becomes 1 squared plus 2 squared, 5, and group b becomes 3 squared alone, 9.

D. {'a': 9, 'b': 9}

Squares the already summed total of each group instead of squaring the individual values before adding them.

A custom function passed to agg() receives the whole group as a series and can do arbitrary computation on it before returning a single summary value. Whatever that function computes, sum of squares here, runs independently inside each group, exactly like a built in aggregation would.