> if I do: > > a = [ [0] * 3 ] * 3 > a[0][1] = 1 > > I get > > a = [[0,1,0],[0,1,0],[0,1,0]]
The language reference calls '*' the "repetition" operator. It's not making copies of what it repeats, it is repeating it. Consider the following code: >>> a = [] >>> b = [] >>> a == b True >>> a is b False >>> >>> a = b = [] >>> a is b True >>> a.append(1) >>> a [1] >>> b [1] Each time you use [], you are creating a new list. So the first code sets a and b to two different new lists. The second one, "a = b = []", only creates a single list, and binds both a and b to that same list. In your example, first you create a list containing [0, 0, 0]; then you repeat the same list three times. >>> a = [[0]*3]*3 >>> a [[0, 0, 0], [0, 0, 0], [0, 0, 0]] >>> a[0] is a[1] True >>> a[0] is a[2] True When you run [0]*3 you are repeating 0 three times. But 0 is not mutable. When you modify a[0] to some new value, you are replacing a reference to the immutable 0 with some new reference. Thus, [0]*3 is a safe way to create a list of three 0 values. When you have a list that contains three references to the same mutable, and you change the mutable, you get the results you discovered. -- Steve R. Hastings "Vita est" [EMAIL PROTECTED] http://www.blarg.net/~steveha -- http://mail.python.org/mailman/listinfo/python-list
