There are several ways to do this. Which one I'd use would depend more upon the context, because I'd probably write the one which generalized most naturally.
First, explicitly (useful if we're sticking with 2D or if I might be interested in other transformations):
>>> L = [(2,3), (4,5), (6,7)]
>>> M = [(y,x) for x,y in L]
>>> M
[(3, 2), (5, 4), (7, 6)]
or maybe using slice notation (seq[start:stop:step]):
>>> M = [v[::-1] for v in L]
>>> M
[(3, 2), (5, 4), (7, 6)]
Or, using words instead:
>>> M = [tuple(reversed(v)) for v in L]
>>> M
[(3, 2), (5, 4), (7, 6)]
>>> M = list(tuple(reversed(v)) for v in L)
>>> M
[(3, 2), (5, 4), (7, 6)]
And if the objects are really vectors, then maybe I'd use a rotation matrix. All depends, but I have to admit if there were only two things to swap I'd reach for the first method.
Umm, wouldn't you try to use the *fastest* one, all else being equal? Using `L = [(randint(0,10),randint(0,10)) for i in range(10000)]`, the first one wins hands down. A little (numerical, not # of ops) analysis shows that the first two options are slightly superlinear in the length of the list while the last is nearly exactly O(n^2). Anyway, go with the first one if you really only have pairs!
Asked: 2012-02-08 10:41:42 +0100
Seen: 876 times
Last updated: Feb 08 '12
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