Groupby A:
In [0]: grp = df.groupby('A')
Within each group, sum over B and broadcast the values using transform. Then sort by B:
In [1]: grp[['B']].transform(sum).sort('B')
Out[1]:
B
2 -2.829710
5 -2.829710
1 0.253651
4 0.253651
0 0.551377
3 0.551377
Index the original df by passing the index from above. This will re-order the A values by the aggregate sum of the B values:
In [2]: sort1 = df.ix[grp[['B']].transform(sum).sort('B').index]
In [3]: sort1
Out[3]:
A B C
2 baz -0.528172 False
5 baz -2.301539 True
1 bar -0.611756 True
4 bar 0.865408 False
0 foo 1.624345 False
3 foo -1.072969 True
Finally, sort the 'C' values within groups of 'A' using the sort=False option to preserve the A sort order from step 1:
In [4]: f = lambda x: x.sort('C', ascending=False)
In [5]: sort2 = sort1.groupby('A', sort=False).apply(f)
In [6]: sort2
Out[6]:
A B C
A
baz 5 baz -2.301539 True
2 baz -0.528172 False
bar 1 bar -0.611756 True
4 bar 0.865408 False
foo 3 foo -1.072969 True
0 foo 1.624345 False
Clean up the df index by using reset_index with drop=True:
In [7]: sort2.reset_index(0, drop=True)
Out[7]:
A B C
5 baz -2.301539 True
2 baz -0.528172 False
1 bar -0.611756 True
4 bar 0.865408 False
3 foo -1.072969 True
0 foo 1.624345 False