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

[Wayne Werner]

On Tue, Nov 3, 2009 at 9:20 AM, 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
>

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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