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#1
August 15th, 2003, 11:04 PM
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Bidimensional arrays in Python
Hi, my problem is simple : I'm trying to make some 2x2 matrix operations, similar to the C code like
Code:
int a[4][15] (not necessarily int). And honestly, I have no clue about how to do that in Python. So how can I do that ? I COULD make a "flat", one-dimensional array, and take each x'th element, but I'd be surprised if nobody ever worked with Python and matrices.
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#2
August 16th, 2003, 01:40 AM
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Hello,
I'm not trained in Python but isn't : Code:
>>> a = [[1,2,3],[4,5,6],[7,8,9]] >>> a[0] [1, 2, 3] >>> a[0][0] 1 >>> a[2][2] 9 what you're looking for ? Just an array of arrays  Julien.
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#3
August 16th, 2003, 07:37 AM
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Also, note that what you will be creating is not an array, but a list. A mutable list. You can't directly create an empty array, but you can of course create an array filled with default values:
Code:
[[0 for i in range(2)] for i in range(2)] will return: Code:
[[0, 0], [0, 0]] For more info, see Python Tutorial: List Comprehensions Last edited by percivall : August 16th, 2003 at 07:40 AM.
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#5
August 18th, 2003, 07:42 AM
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Another way of doing the same thing:
Code:
[ [0] * 2 ] * 2 This is essentially the same as the list comprehension in the earlier post, but a little less typing.
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#6
August 19th, 2003, 08:48 AM
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Both do the same thing but making a list of 0 is pretty pointless
. The first list compression does allow you to do allot more than the recusion though..Code:
[[i for i in range(10)] for i in range(2)] This will at least produce two list of numbers 0-9, but then why this would be useful I just don't know, ah maybe one day I will find a use for it Thanks for the info though, Take care guys, Mark.
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#7
August 19th, 2003, 09:18 AM
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List comprehension is most powerful when combined with a function, such as:
Code:
[(lambda i: [i, chr(i)])(i) for i in range(65, 123)] This is of course also possible with a regular list comprehension... Code:
[[i, chr(i)] for i in range(65, 123)] Okay, not very useful. Anyway, it lets you do stuff in a nice way. Last edited by percivall : August 19th, 2003 at 09:29 AM.
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#8
August 19th, 2003, 04:03 PM
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I'd be enclined to use the second
. Smaller and and nicer to look at, plus since it does less (doesnt create a lambda) it should be slightly more efficent (in theory). Does give some idea of whats possible. Good for setting up lists .good to know thanks, Mark.
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#9
September 4th, 2003, 08:08 PM
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Quote:
Actually, you don't want to do this, because of the following: Code:
>>> a = [[0]*2]*2 >>> a [[0, 0], [0, 0]] >>> a[0][0] = 1 >>> a [[1, 0], [1, 0]] >>> see note 2 on this page for more details: http://www.python.org/doc/current/lib/typesseq.html
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