Thanks Steve.
It's obvious now that you've pointed it out.
Do you happen to know if there is an efficient way to initialize a list
like this without explicitly writing out each element?
Mike
----- Original Message -----
From: "Steve Willoughby" <[EMAIL PROTECTED]>
To: "Mike Meisner" <[EMAIL PROTECTED]>
Cc: <tutor@python.org>
Sent: Friday, July 25, 2008 7:03 PM
Subject: Re: [Tutor] List indexing problem
Mike Meisner wrote:
I need to do some statistical analysis by binning values into an array.
Being new to Python, I tried to use a list of lists. I've extracted
just the minimum code that I'm having trouble with:
What you need to remember is that Python works with *objects*, and
variables are simply *references* to those objects. So if I say
a = [1,2,3]
b = a
b and a both refer to the *SAME* list object (not two lists which happen
to have the same elements).
temp = [[0, 0, 0],[0, 0, 0],[0, 0, 0]]
temp now refers to a list object containing 3 lists, each containing 3
integers.
# initialize to zero
for i in range(20):
IP.append(temp)
Now IP contains 20 copies of references to the *same* list object.
So you were modifying the underlying list object which happens to be
referenced many times in IP.
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