On Wed, Feb 6, 2013 at 9:48 AM, Kang, Yang Jae <kangyang...@gmail.com>wrote:
This line: > >>> a = [[0,0]]*3 > creates a list, a, which contains the list object [0, 0] three times. What's crucial to note is that it contains *the same object* three times, not three different objects with the same value. You can verify this yourself, with the id() function: >>> a = [[0,0]] * 3 >>> a [[0, 0], [0, 0], [0, 0]] >>> id(a[0]), id(a[1]), id(a[2]) (41087144, 41087144, 41087144) >>> All three have the same id, therefore all three point to one and the same object. > >>> a**** > > [[0, 0], [0, 0], [0, 0]]**** > > >>> a[0][0] += 1**** > > >>> a**** > > [[1, 0], [1, 0], [1, 0]] > It should not be so strange, now, that if you modify a[0], this modification will be reflected in a[1] and a[2] as well, since those are all the same object. Now, this can happen with lists, because they are mutable. If we do the same with integers: >>> a = [1] * 3 >>> a [1, 1, 1] >>> id(a[0]), id(a[1]), id(a[2]) (3779904, 3779904, 3779904) >>> a[0] += 1 >>> a [2, 1, 1] >>> You get the expected behaviour now, even though the list again contained the same object three times. This is because integers are *immutable*, meaning you cannot change them. So this line: >>> a[0] += 1 did not modify the object, but created another integer object, with value 2, and assigned that to a[0]. The conclusion is that you should always be mindful of whether your objects are mutable or not, and you should understand what actually happens when you multiply a list by a constant, to avoid these kinds of problems. In this case, what you wanted to do was create three list objects with value [0, 0], not just the one. You may either generate them with a loop or list comprehension, or simply write the list out in its entirety: >>> a = [[0,0] for i in range(3)] >>> id(a[0]), id(a[1]), id(a[2]) (4070896, 41087984, 41061792) >>> a = [[0, 0], [0, 0], [0, 0]] >>> id(a[0]), id(a[1]), id(a[2]) (41087944, 41087824, 41069624) >>> This way, the lists are actually three different lists and not just the same object three times (as you can see, they have different id's) HTH, Hugo
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