An ordering question
Hi, 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)] I did the same as follows: l = len(a) * [None] for (k, v) in a: ... for i, e in enumerate(b): ... if e == v: ...l[i] = (k, v) This works, but the code -for python- looks very kludgy. I thought for ~2 hours to see whether I can do it in a line or 2, but I cannot seem to find a mechanism. Can someone help me out? -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
Kottiyath wrote: Hi, 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)] I did the same as follows: l = len(a) * [None] for (k, v) in a: ... for i, e in enumerate(b): ... if e == v: ...l[i] = (k, v) This works, but the code -for python- looks very kludgy. I thought for ~2 hours to see whether I can do it in a line or 2, but I cannot seem to find a mechanism. Can someone help me out? How about: a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [2, 4, 1, 3] d = dict((v, k) for k, v in a) c = [(d[s], s) for s in b] c [(3, 2), (2, 4), (4, 1), (7, 3)] Understanding how it works is left as an exercise for the reader. :-) -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
MRAB wrote: Kottiyath wrote: Hi, 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)] I did the same as follows: l = len(a) * [None] for (k, v) in a: ... for i, e in enumerate(b): ... if e == v: ...l[i] = (k, v) This works, but the code -for python- looks very kludgy. I thought for ~2 hours to see whether I can do it in a line or 2, but I cannot seem to find a mechanism. Can someone help me out? How about: a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [2, 4, 1, 3] d = dict((v, k) for k, v in a) c = [(d[s], s) for s in b] c [(3, 2), (2, 4), (4, 1), (7, 3)] Understanding how it works is left as an exercise for the reader. :-) Actually, a more general solution is: a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [2, 4, 1, 3] d = dict((t[1], t) for t in a) c = [d[s] for s in b] c [(3, 2), (2, 4), (4, 1), (7, 3)] -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
Kottiyath n.kottiy...@gmail.com writes: Hi, 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)] [...] whether I can do it in a line or 2, a.sort(key=lambda (x, y): b[y - 1], reverse=True) -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
Hrvoje Niksic wrote: Kottiyath n.kottiy...@gmail.com writes: Hi, 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)] [...] whether I can do it in a line or 2, a.sort(key=lambda (x, y): b[y - 1], reverse=True) that's hilarious! i have no idea how it works, but unfortunately it's not general: a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [4, 2, 1, 3] a.sort(key=lambda xy: b[xy[1] - 1], reverse=True) a [(4, 1), (2, 4), (3, 2), (7, 3)] (in 3.0 there's no unpacking in function args) did you just play around until you got the right answer? how many different tries did you need? the answer i think the original person was looking for is: a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [2, 4, 1, 3] a.sort(key=lambda xy: b.index(xy[1])) a [(3, 2), (2, 4), (4, 1), (7, 3)] a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [4, 2, 1, 3] a.sort(key=lambda xy: b.index(xy[1])) a [(2, 4), (3, 2), (4, 1), (7, 3)] however, it's not very efficient. it would be better to do: a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [4, 2, 1, 3] idx = dict((b[i], i) for i in range(len(b))) a.sort(key=lambda xy: idx[xy[1]]) a [(2, 4), (3, 2), (4, 1), (7, 3)] for large lists andrew -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
MRAB wrote: a = [(4, 1), (7, 3), (3, 2), (2, 4)] b = [2, 4, 1, 3] d = dict((v, k) for k, v in a) c = [(d[s], s) for s in b] c [(3, 2), (2, 4), (4, 1), (7, 3)] ah, that is more efficient than the suggestions i posted. andrew -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
andrew cooke and...@acooke.org writes: Hrvoje Niksic wrote: Kottiyath n.kottiy...@gmail.com writes: Hi, 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)] [...] whether I can do it in a line or 2, a.sort(key=lambda (x, y): b[y - 1], reverse=True) that's hilarious! i have no idea how it works [...] did you just play around until you got the right answer? how many different tries did you need? Glad to have made someone laugh, but it in fact wasn't meant to. :-) It was a combination of misunderstanding the question *and* trying out the answer. I understood that the second element of each tuple should be sorted according to the corresponding element in b. This part is partly correct, but I took corresponding to mean that it should be sorted by the value of b[x], x being the second element of each tuple. Since those elements appear to be 1-based, I added -1. This produced a list in exactly the reverse order, so I added reverse=True. Your solution are probably the two lines the OP was looking for. -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
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. Very nice! This solution is pretty much a direct transliteration of the original problem, as the OP *might* have articulated it: List A consists of ordered pairs. Sort list A according to the position in list B of the pair's second member. A solution is truly Pythonic when it possesses this directness. -John -- http://mail.python.org/mailman/listinfo/python-list
Parameter sublists [was: An ordering question]
On Fri, 13 Mar 2009 18:56:30 +0100, Hrvoje Niksic hnik...@xemacs.org wrote: [snip] a.sort(key=lambda (x, y): b[y - 1], reverse=True) Huh? I had no idea one could do this: def g( ( ( x, y ), z ) ): ... return y ... g( ((1,2),3) ) 2 What should I have read to learn that trick? -- To email me, substitute nowhere-spamcop, invalid-net. -- http://mail.python.org/mailman/listinfo/python-list
Re: Parameter sublists [was: An ordering question]
On Fri, Mar 13, 2009 at 5:30 PM, Peter Pearson ppear...@nowhere.invalid wrote: On Fri, 13 Mar 2009 18:56:30 +0100, Hrvoje Niksic hnik...@xemacs.org wrote: [snip] a.sort(key=lambda (x, y): b[y - 1], reverse=True) Huh? I had no idea one could do this: def g( ( ( x, y ), z ) ): ... return y ... g( ((1,2),3) ) 2 What should I have read to learn that trick? Don't bother. It's been excised in Python 3.0. Cheers, Chris -- I have a blog: http://blog.rebertia.com -- http://mail.python.org/mailman/listinfo/python-list
Re: Parameter sublists [was: An ordering question]
On Mar 14, 5:39 am, Chris Rebert c...@rebertia.com wrote: On Fri, Mar 13, 2009 at 5:30 PM, Peter Pearson ppear...@nowhere.invalid wrote: On Fri, 13 Mar 2009 18:56:30 +0100, Hrvoje Niksic hnik...@xemacs.org wrote: [snip] a.sort(key=lambda (x, y): b[y - 1], reverse=True) Huh? I had no idea one could do this: def g( ( ( x, y ), z ) ): ... return y ... g( ((1,2),3) ) 2 What should I have read to learn that trick? Don't bother. It's been excised in Python 3.0. Cheers, Chris -- I have a blog:http://blog.rebertia.com Thank you very much. These many solutions ? I think programming is not my forte :-) :-) -- http://mail.python.org/mailman/listinfo/python-list
Re: An ordering question
En Fri, 13 Mar 2009 17:33:51 -0200, Hrvoje Niksic hnik...@xemacs.org escribió: andrew cooke and...@acooke.org writes: Hrvoje Niksic wrote: Kottiyath n.kottiy...@gmail.com writes: 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)] a.sort(key=lambda (x, y): b[y - 1], reverse=True) that's hilarious! i have no idea how it works Glad to have made someone laugh, but it in fact wasn't meant to. :-) It was a combination of misunderstanding the question *and* trying out the answer. I understood that the second element of each tuple should be sorted according to the corresponding element in b. This part is partly correct, but I took corresponding to mean that it should be sorted by the value of b[x], x being the second element of each tuple. Since those elements appear to be 1-based, I added -1. This produced a list in exactly the reverse order, so I added reverse=True. This shows how important is to provide a good, unambiguous example in any specification. Not just a correct example, but one that also shows how things are not to be done. In case of doubt, people always look at the examples for clarification. -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
