On Tue, Nov 3, 2009 at 9:20 AM, Robert Berman <berma...@cfl.rr.com> wrote:
> > In [69]: l1=[(0,0)] * 4 > > In [70]: l1 > Out[70]: [(0, 0), (0, 0), (0, 0), (0, 0)] > > In [71]: l1[2][0] > Out[71]: 0 > This calls the element at index 2 which is: (0,0) - a tuple, then calls element [0] from that tuple, which is 0 when you try to assign an item into a tuple, you get the same problem: In [1]: x = (1,2,3) In [2]: type(x) Out[2]: <type 'tuple'> In [3]: x[0] Out[3]: 1 In [4]: x[0] = 0 --------------------------------------------------------------------------- TypeError Traceback (most recent call last) /home/wayne/Desktop/<ipython console> in <module>() TypeError: 'tuple' object does not support item assignment And from your example: In [6]: l1 = [(0,0)] *4 In [7]: type(l1[2]) Out[7]: <type 'tuple'> > In [72]: l1[2][0] = 3 > --------------------------------------------------------------------------- > TypeError Traceback (most recent call last) > > /home/bermanrl/<ipython console> in <module>() > > TypeError: 'tuple' object does not support item assignment > > First question, is the error referring to the assignment (3) or the index > [2][0]. I think it is the index but if that is the case why does l1[2][0] > produce the value assigned to that location and not the same error message. > > Second question, I do know that l1[2] = 3,1 will work. Does this mean I > must know the value of both items in l1[2] before I change either value. I > guess the correct question is how do I change or set the value of l1[0][1] > when I specifically mean the second item of an element of a 2D array? > When you use l1[2] = 3,1 it converts the right hand side to a tuple by implication - putting a comma between values: In [8]: x = 3,1 In [9]: type(x) Out[9]: <type 'tuple'> so when you say l1[2] = 3,1 you are saying "assign the tuple (3,1) to the list element at l1[2]" which is perfectly fine, because lists are mutable and tuples are not. > > I have read numerous explanations of this problem thanks to Google; but no > real explanation of setting of one element of the pair without setting the > second element of the pair as well. > > For whatever glimmers of clarity anyone can offer. I thank you. > Hopefully this helps, Wayne p.s. If you want to be able to change individual elements, you can try this: In [21]: l1 = [[0,0] for x in xrange(4)] In [22]: l1 Out[22]: [[0, 0], [0, 0], [0, 0], [0, 0]] In [23]: l1[2][0] = 3 In [24]: l1 Out[24]: [[0, 0], [0, 0], [3, 0], [0, 0]] I don't know if there's a better way to express line 21, but you can't do it the other way or you'll just have the same list in your list 4 times: In [10]: l1 = [[0,0]]*4 In [11]: l1 Out[11]: [[0, 0], [0, 0], [0, 0], [0, 0]] In [12]: l1[2][0] = 3 In [13]: l1 Out[13]: [[3, 0], [3, 0], [3, 0], [3, 0]]
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