incrementing along a diagonal
David Novakovic
david at distip.com.au
Wed Oct 11 18:34:07 CDT 2006
Johannes Loehnert wrote:
>> I'm just wondering if there is a way that i can increment all the values
>> along a diagonal?
>>
>
> Assume you want to change mat.
>
> # min() only necessary for non-square matrices
> index = arange(min(mat.shape[0], mat.shape[1]))
> # add 1 to each diagonal element
> matrix[index, index] += 1
> # add some other stuff
> matrix[index, index] += some_array_shaped_like_index
>
>
> HTH, Johannes
>
>
Thank you very much for the prompt reply, I'm just having a problem with
this method:
This method appears to only work if the matrix is mxm for example:
>>> zeros((5,5))
array([[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0]])
>>> x = zeros((5,5))
>>> index = arange(min(x.shape[0],x.shape[1]))
>>> x[index,index] += 1
>>> x
array([[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 0, 1, 0],
[0, 0, 0, 0, 1]])
>>>
This is very nice, exactly what i want, but it doesnt work for mxn
matricies:
>>> x = zeros((5,3))
>>> x
array([[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0]])
>>> index = arange(min(x.shape[0],x.shape[1]))
>>> x[index,index] += 1
>>> x
array([[1, 0, 0],
[0, 1, 0],
[0, 0, 1],
[0, 0, 0],
[0, 0, 0]])
>>>
So the min part is right for mxn matrices - but perhaps there is a way
to use the index differently. I'm very new to numpy, so excuse my
noobness :)
Just for reference, this is the line of perl i'm trying to port:
(my $dummy = $ones->diagonal(0,1))++; # ones is a matrix created with
zeroes()
Yes, i know it is horribly ugly, but in this case, diagonal returns a
list of references to the values in the original matrix, so values can
be changed in place. I much prefer python - hence the port, but it seems
like a hard thing to replicate. Perhaps there could be a function that
returns an iterator over the values in a matrix and returns the index's.
like:
for index in diag_iter(matrix,*axes):
matrix[index] +=1
Once again, cheers - i hope we can figure something out :)
Dave Novakovic
PS: If anyone would care to link me to the subscription page for the
mailing list so you dont have to CC me all the time :)
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