4 ms·

In case there are Pythonistas here who will indulge a question: one thing I've never figured out how to do elegantly ("idiomatically", I suppose) is create a l

• Goladus 15y ago
  x = [0] * 100
  • zmmmmm 15y ago
    Thanks! At first I wondered about efficiency but it seems like I can allocate lists with tens of millions of elements this way without any spike in CPU so it is obviously implemented efficiently. I knew there would be a better way :-)
  • Ixiaus 15y ago
    This is much better than my own comment - I never even knew about that syntax! Is it efficient though?
    • Goladus 15y ago
      My guess is that it's optimized, but I don't know. Semantics-wise, it's consistent with list addition. >>> [0] + [0] + [0] [0, 0, 0] >>> [0] * 3 [0, 0, 0]
      • Ixiaus 15y ago
        I just read, that syntax is sugar for the list.extend() method so it's still making copies of the list - I wouldn't use it for a list containing complex objects; but that is very idiomatic for simple use cases (like this one?).
    • calpaterson 15y ago
      It isn't syntax - help(list.__mul__)
  • daeken 15y ago
    Just make sure that when you do this, you're not initializing it with mutable objects, e.g. lists. x = [[0]] * 2 print x[0], x[1] # prints [0] [0] x[0].append(1) print x[0], x[1] # prints [0, 1] [0, 1] That's because the objects inside the repeated list are simply references to the same object internally.
  • deleted 15y ago
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• deleted 15y ago
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• Ixiaus 15y ago
  Yeah, that is kludgy (you're creating two lists now!). First thing I would do is ask why I need that; I've never come across a case where I needed a pre-built list of items all initialized to a common or empty value - if you're doing that, I would explore some sort of custom generative process (building the items of your list as you need them). But, just because I can't think of a good use case for this (where I couldn't use generative recursion or iteration) doesn't mean you haven't. Weigh what you are doing. I would do this one of two ways: create a function that will produce an object that will generatively build the list (like xrange, but pass an "initializer" value, like myxrange(0, 100, initialize=0)). See the source for Python's xrange to do this it should be quite easy. Or do this in a while loop (flat, easy to understand, and you aren't being too naughty by producing more than once list): ls = [] cnt = 0 while cnt <= 100: ls.append(0) cnt += 1 Alternatively you can do the above with a for loop but you would have to iterate over a sequence (which means creating a list to do it); the while loop is going to be the most efficacious way of doing what you want I think, excluding building a custom xrange style implementation.
  • natedub 15y ago
    Actually [x for x in xrange(..)] will produce only one list. In Python 2.X, range() returns a list while xrange() an iterable. If you just need an iterable, itertools.repeat(0, 100) will do the trick.
    • Ixiaus 15y ago
      I suppose I didn't say it correctly, you're still maintaining two objects though (xrange the iterable, and the new list being built by the comprehension). The "optimization" I suppose is a moot point because Python's handling of lists is pretty solid and this isn't the 90's anymore, but at least with the while loop you've only got one list object and an integer (which is faster, I would think, to increment than it is to produce the next sequence in the iterable object?). Anywho, splitting hairs. Goladus had the most helpful comment (I even learned something).
      • bendotc 15y ago