On Fri, Mar 13, 2009 at 2:12 PM, John Posner wrote: > >> I have 2 lists >> a = [(4, 1), (7, 3), (3, 2), (2, 4)] >> b = [2, 4, 1, 3] >> >> Now, I want to order _a_ (a[1]) based on _b_. >> i.e. the second element in tuple should be the same as >> b. >> i.e. Output would be [(3, 2), (2, 4), (4, 1), (7, 3)] > > Essentially, you're sorting a list. The Pythonic approach is to use the > sort() function, hiding the details in a "custom comparison function": > > def compare_func(first, second): > b = [2, 4, 1, 3] > return cmp(b.index(first[1]), b.index(second[1])) > > if __name__ == '__main__': > a = [(4, 1), (7, 3), (3, 2), (2, 4)] > a.sort(cmp=compare_func) > print a
A version that runs in O(n ln n) instead of O(n^2 ln n): a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [2, 4, 1, 3] bdict = dict((v,k) for (k,v) in enumerate(b)) a.sort(key=lambda k: bdict[k[1]]) If you don't want to build the intermediary dict, a less efficient version that runs in O(n^2): a.sort(key=lambda k: b.index(k[1])) Which is mostly similar to John's solution, but still more efficient because it only does a b.index call once per 'a' item instead of twice per comparison. -Miles -- http://mail.python.org/mailman/listinfo/python-list
