Thank you for your explanations and especially your clear examples of a phenomenon(when list elements are tuples) which takes a few moments of study to truly grasp.
Robert On Tue, 2009-11-03 at 09:53 -0600, Wayne Werner wrote: > 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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