numpy.tile is what I was after.
Thank you!
On Tue, Nov 22, 2011 at 1:13 PM, Olivier Delalleau wrote:
> I can't really figure out if that's the case in your code, but if you need
> to repeat the mask along a new dimension (for instance, the first one), you
> can do:
>
> numpy.tile(mask.mask, [num
I can't really figure out if that's the case in your code, but if you need
to repeat the mask along a new dimension (for instance, the first one), you
can do:
numpy.tile(mask.mask, [number_of_repeats] + [1] * len(mask.mask.shape))
(not sure that's the most elegant way to do it, but it should work
Excellent, thank you.
I just realised this does not work with my data because of the extra
dimension.
I have a mask that matches my 2-dimensional array but my data is for every
hour over a month so the arrays do not match. Is there a way to make them
match or mask each time?
thanks again
This is s
If your new array is x, you can use:
numpy.ma.masked_array(x, mask=mask.mask)
-=- Olivier
2011/11/21 questions anon
> I am trying to mask one array using another array.
>
> I have created a masked array using
> mask=MA.masked_equal(myarray,
> 0),
> that looks something like:
> [1 - - 1,
>
I am trying to mask one array using another array.
I have created a masked array using
mask=MA.masked_equal(myarray,
0),
that looks something like:
[1 - - 1,
1 1 - 1,
1 1 1 1,
- 1 - 1]
I have an array of values that I want to mask whereever my mask has a a '-'.
how do I do this?
I
