Yves Dorfsman wrote: > Peter Otten <[EMAIL PROTECTED]> wrote: > >> > A slightly similar problem: If I want to "merge," say, list1=[1,2,3] >> > with list2=[4,5,6] to obtain [1,4,2,5,3,6], is there some clever way >> > with "zip" to do so? > >> >>> items = [None] * 6 >> >>> items[::2] = 1,2,3 >> >>> items[1::2] = 4,5,6 >> >>> items >> [1, 4, 2, 5, 3, 6] > > My problem with this solution is that you depend on knowing how many > elements you have in each list ahead of time.
$ python -m timeit -s"a = [1,2,3]*100; b = [4,5,6]*100; from operator import add" "list(reduce(add, zip(a, b)))" 1000 loops, best of 3: 637 usec per loop $ python -m timeit -s"a = [1,2,3]*100; b = [4,5,6]*100" "items=[None]*(2*len(a)); items[::2] = a; items[1::2] = b" 10000 loops, best of 3: 23.4 usec per loop The speed gain is significant. Why should I throw away useful information if I have it? I'd even be willing to convert one arbitrary iterable to a list to get the length information. $ python -m timeit -s"a = [1,2,3]*100; b = [4,5,6]*100" "aa = list(a); items=[None]*(2*len(aa)); items[::2] = aa; items[1::2] = b" 10000 loops, best of 3: 29.5 usec per loop > Assuming that both list > are of the same length, then, I find the following more elegant: > > list1=[1,2,3] > list2=[4,5,6] > > reduce(lambda x, y: x+y, zip(list1, list2)) > > of course, zip creates tuples, so you end up with a tuple, therefore if > you need for your solution to be a list: > list(reduce(lambda x, y: x+y, zip(list1, list2))) I'd rather use a plain old for-loop: >>> from itertools import izip >>> a = [1,2,3] >>> b = [4,5,6] >>> items = [] >>> for chunk in izip(a, b): ... items.extend(chunk) ... >>> items [1, 4, 2, 5, 3, 6] $ python -m timeit -s"a = [1,2,3]*100; b = [4,5,6]*100; from itertools import izip" "items = [] > for chunk in izip(a, b): items.extend(chunk)" 1000 loops, best of 3: 242 usec per loop $ python -m timeit -s"a = [1,2,3]*100; b = [4,5,6]*100; from itertools import izip" "items = []; extend = items.extend for chunk in izip(a, b): extend(chunk)" 10000 loops, best of 3: 70.9 usec per loop > reduce(lambda x, y: x+y, list(zip(list1, list2)) ) list(zip(...)) has no effect here. Possible fixes >>> reduce(lambda x, y: x + list(y), zip(list1, list2), []) [1, 4, 2, 5, 3, 6] >>> reduce(lambda x, y: x.extend(y) or x, zip(list1, list2), []) [1, 4, 2, 5, 3, 6] $ python -m timeit -s"a = [1,2,3]*100; b = [4,5,6]*100" "reduce(lambda x, y: x.extend(y) or x, zip(a, b), [])" 1000 loops, best of 3: 368 usec per loop are more complicated than elegant. Not recommended. Peter -- http://mail.python.org/mailman/listinfo/python-list
