I want to initialize an array that has X two-dimensional elements. For example, if X = 3, I want it to be [[0,0], [0,0], [0,0]]. I know that [0]*3 gives [0, 0, 0], but how do I do this for two-dimensional elements?
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1Most of the time in Python, if you are thinking in terms of "initializing" a container to a bunch of 0s, you are doing it wrong.Karl Knechtel– Karl Knechtel2012-04-11 04:09:30 +00:00Commented Apr 11, 2012 at 4:09
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Well I'm practicing for Google Code Jam and I have to read in a stream of numbers for example 2, 3, 1, 4, 5, 2 and turn it into [[2,3], [1,4], [5,2]]. What else would you suggest?Kobi– Kobi2012-04-11 04:31:30 +00:00Commented Apr 11, 2012 at 4:31
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1I would suggest appending the numbers to empty lists as they are received.Karl Knechtel– Karl Knechtel2012-04-11 04:50:03 +00:00Commented Apr 11, 2012 at 4:50
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4 Answers
Try this:
m = [[0] * 2 for _ in xrange(3)]
In the above code, think of the 3 as the number of rows in a matrix, and the 2 as the number of columns. The 0 is the value you want to use for initializing the elements. Bear in mind this: if you're using Python 3, use range instead of xrange.
For a more general solution, use this function:
def create_matrix(m, n, initial=0):
return [[initial] * n for _ in xrange(m)]
For the particular case of the question:
m = create_matrix(3, 2)
print m
> [[0, 0], [0, 0], [0, 0]]
Alternatively, and if you don't mind using numpy, you can use the numpy.zeros() function as shown in Mellkor's answer.
Comments
I guess numpy.zeros is useful for such. Say,
x=4
numpy.zeros((x,x))
will give you:
array([[ 0., 0., 0., 0.], [ 0., 0., 0., 0.], [ 0., 0., 0., 0.], [ 0., 0., 0., 0.]])
Comments
I believe that it's [[0,0],]*3
answered Apr 11, 2012 at 4:02
AI Generated Response
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AI Generated Response
The way it works is it adds the stuff inside the list as a list to itself multiple times, essentially being [[0,0],]+[[0,0],]+[[0,0],]
DSM
This makes a list containing three references to the same list, so it's not the same as
[[0,0],]+[[0,0],]+[[0,0],]. Try changing one of the elements, and you'll see all three sublists change.
Óscar López
This solution is broken. It creates 3 copies of the same list of two elements, when you try to modify one, all the others will get modified too.
Using the construct
[[0,0]]*3
works just fine and returns the following:
[[0, 0], [0, 0], [0, 0]]
