On 11/3/2009 7:20 AM Robert Berman said...
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
Start out slower...
a = (1,2)
a[1]=3
You can't change a tuple. You can create a new tuple, or you can change
the content of an item help in a tuple, but you can't change the content
of the tuple itself. Try the following then see if it helps answer your
questions...
b = [1]
a = (1,b)
b[0]=2
print a
a[1][0]=3
print a
Emile
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.
Robert
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