I hope I'm not going too off topic with this. Just some clarifications :) Is the original solution suggested to be O(n^2) because of the nested > remove(i) call within the loop?
Exactly. > Would that have a logarithmic complexity instead because the list becomes > shorter everytime it finds a 0 and removes it? If we have no zeroes, then a simple count would eliminate the iteration. But instead, we still try to remove values so we get O(n*n) If we have a list of *only* zeroes and we reduce our list to length zero, we iterate the full length, remove an element until we have none and we would have something like this: n + (n-1) + (n-2) ... +1) Famous pattern <https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF> that results in O(0.5n^2 + 0.5n) --> We can drop the constants as above, and also we can skip the linear part here since n*n will be way larger than just n if we have relatively large n. Example with n = 100 and n = 1000 we can see the quadratic growth: n=100: 5050 n=1000: 500500 Even though n increased by a factor of 10, our values increased by a factor of ~99.1 in this case. A list is not really a good choice for removing elements from random places due to reordering, that's why its O(n). You could time this and verify for yourself. python -m timeit "listerine = [0]*100;listerine.remove(0)" We have to subtract list creation from timing. python -m timeit "listerine = [0]*100" And just increase those numbers as we go on. And then your solution has two O(n) calls (the count and the list comp) so > that is O(n) yea? Yes. O(constant*n) -> O(n) is just the canonical form of writing it. It's still linear growth. Simplest example: 2*100 && 2*1000 -> we have tenfolds increase even if remove that constant. > Then you have the memory complexity of needing to allocate temporary lists > for the final nonzero and zero lists. I concur. This very example though has small ints which are cached by python and we can go into ranges of millions and we will still talk about a few megs. import sys > maxint = sys.maxint > by_nonzero = lambda i: i or maxint > l = [4,0,8,0,0,2,1,-5] > l.sort(key=by_nonzero) > print l > # [-5, 1, 2, 4, 8, 0, 0, 0] Yes, this is a good approach. That's where my method will fail. I just wrote it for this very example and because list comprehensions are super optimized. :) Also, there's the sorted method, which works not just on lists but on any iterable, but of course it does a copy of the original iterable if you don't want it altered. e.g: import string, operator, random dict_compr = {letter : random.randint(0, 10) for letter in string. ascii_lowercase} sorted_by_values = sorted(dict_compr.iteritems(), key=operator.itemgetter(1 )) On Monday, November 6, 2017 at 7:47:27 PM UTC+1, Justin Israel wrote: > > > > On Tue, Nov 7, 2017, 4:58 AM Balazs Pataki <blaise...@gmail.com > <javascript:>> wrote: > >> You could also try this, which is a bit more pythonic, also faster. >> >> def re_sort(to_sort): >>> # we need those zeroes so we check how many of them we have >>> nr_of_zeroes = to_sort.count(0) >>> # list comprehension that gets rid of zeroes >>> filtered = [item for item in to_sort if item != 0] >>> # we append together the filtered part and inplace add a list of >>> zeroes of length nr_of_zeroes >>> return filtered + [0]*nr_of_zeroes >> >> >> While this has time complexity of O(n), yours is O(n^2), so the longer >> your list gets, the worse your performance gets. >> > > I'm not great at big O notation. Is the original solution suggested to be > O(n^2) because of the nested remove(i) call within the loop? Would that > have a logarithmic complexity instead because the list becomes shorter > everytime it finds a 0 and removes it? Otherwise it would not remove at all > for nonzero. > > And then your solution has two O(n) calls (the count and the list comp) so > that is O(n) yea? Then you have the memory complexity of needing to > allocate temporary lists for the final nonzero and zero lists. > > Here is one more approach that uses a sort in place. It has both examples > depending on whether you want the nonzero numbers sorted or not. > > > by_nonzero = lambda i: 0 if i else 1 > > l = [4,0,8,0,0,2,1,-5] > l.sort(key=by_nonzero) > print l > # [4, 8, 2, 1, -5, 0, 0, 0] > > > import sys > maxint = sys.maxint > > by_nonzero = lambda i: i or maxint > > l = [4,0,8,0,0,2,1,-5] > l.sort(key=by_nonzero) > print l > # [-5, 1, 2, 4, 8, 0, 0, 0] > > > > >> >> On Sunday, November 5, 2017 at 3:07:32 PM UTC+1, Virbhadra Gupta wrote: >>> >>> >>> >>> i was trying to sort list by only 0 and other Numbers like [2,0,0,4] to >>> [2,4,0,0] >>> i come up with this code but in this code variable list is changing. i >>> am unable to understand why ? >>> >>> list = [4,0,0,4] >>> >>> def re_sort(list): >>> count=0 >>> tmp_list = list >>> print list >>> for i in list: >>> if not i: >>> tmp_list.remove(i) >>> count+=1 >>> print list >>> for n in range(count): >>> tmp_list.append(0) >>> return tmp_list >>> re_sort(list) >>> >>> ************************ >>> re_sort(list) >>> [4, 0, 0, 4] >>> [4, 0, 4] >>> # Result: [4, 0, 4, 0] # >>> >> -- >> You received this message because you are subscribed to the Google Groups >> "Python Programming for Autodesk Maya" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to python_inside_maya+unsubscr...@googlegroups.com <javascript:>. >> To view this discussion on the web visit >> https://groups.google.com/d/msgid/python_inside_maya/700827e3-cb90-4106-8da4-350e08096dbd%40googlegroups.com >> >> <https://groups.google.com/d/msgid/python_inside_maya/700827e3-cb90-4106-8da4-350e08096dbd%40googlegroups.com?utm_medium=email&utm_source=footer> >> . >> For more options, visit https://groups.google.com/d/optout. >> > -- You received this message because you are subscribed to the Google Groups "Python Programming for Autodesk Maya" group. 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