Alan
Thanks but I've been a bit daft and described the wrong problem which is easy
to solve the long way. Starting again ...
Given a (numpy) array how do you create a dictionary of lists where the list
contains the column indexes of non-zero elements and the dictionary key is the
row index. The easy way is 2 for loops ie.
import numpy
from collections import defaultdict
A =
[[1 6 1 2 3]
[4 5 4 7 0]
[2 0 8 0 2]
[0 0 0 3 7]]
dict = defaultdict(list)
I = A.shape[0]
J = A.shape[1]
for i in xrange(0, I, 1):
for j in xrange(0, J, 1):
if a[i,j] > 0:
dict[i].append(j)
I want to find a faster/efficient way to do this without using the 2 for loops.
Thanks!
Btw, I posted this on the numpy list too to make sure that there aren't any
numpy functions that would help.
Dinesh
--------------------------------------------------------------------------------
Message: 5
Date: Sun, 21 Sep 2008 09:15:00 +0100
From: "Alan Gauld" <[EMAIL PROTECTED]>
Subject: Re: [Tutor] array and dictionary
To: tutor@python.org
Message-ID: <[EMAIL PROTECTED]>
Content-Type: text/plain; format=flowed; charset="iso-8859-1";
reply-type=original
"Dinesh B Vadhia" <[EMAIL PROTECTED]> wrote
> Hi! Say, I've got a numpy array/matrix of the form:
>
> [[1 6 1 2 3]
> [4 5 4 7 0]...
> [2 1 0 5 6]]
>
> I want to create a dictionary of rows (as the keys) mapped
> to lists of non-zero numbers in that row
Caveat, I dont know about numpy arrays.But assuming they
act like Python lists
You can get the non zeros with a comprehension
nz = [n for n in row if n != 0]
you can get the row and index using enumerate
for n,r in enumerate(arr):
So to create a dictionary, combine the elements somethng like:
d ={}
for n,r in enumerate(arr):
d[n] = [v for v in r if v !=0]
I'm sure you could do it all in one line if you really wanted to!
Also the new any() function might be usable too.
All untested....
HTH,
--
Alan Gauld
Author of the Learn to Program web site
http://www.freenetpages.co.uk/hp/alan.gauld
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