[Tutor] Being beaten up by a tuple that's an integer thats a tuple that may be an unknown 'thing'.

[Andre Engels]

On Tue, Nov 3, 2009 at 4:20 PM, Robert Berman <bermanrl at 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
>
> 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?
>
> 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.
>

Tuples are immutable types. Thus it is not possible to change one of the
values of a tuple (or even of changing both of them). The only thing you can
do, is create a new tuple, and put that in the same place in the list. In
your example, when you do l1[2][0] = 3, you try to change the tuple l1[2],
which is impossible.

To do what you want to do, you have to create a new array with the same
second but different first value, and put that array in l1[2], that is:

l1[2] = (3, l1[2,1])




-- 
André Engels, andreengels at gmail.com
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