| Re: Padding arrays - vector subscripts not working [message #71575] |
Sat, 03 July 2010 14:01  |
penteado
Messages: 866
Registered: February 2018
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On Jul 3, 5:52 pm, James <donje...@gmail.com> wrote:
> Paulo, your explanation makes perfect sense. I guess I was hoping
> there was a more elegant way than calculating every single 1d index
> that will be used, but your method looks good. Thank you!
You do not need to calculate every 1D index. Only the first index is
needed. The single assignment
X[start_index]=Y
will assign all the 12 elements of Y to 12 elements of X (from
start_index to start_index+11).
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| Re: Padding arrays - vector subscripts not working [message #71576 is a reply to message #71575] |
Sat, 03 July 2010 13:52  |
James[2]
Messages: 44
Registered: November 2009
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Member |
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On Jul 2, 1:01 pm, Paulo Penteado <pp.pente...@gmail.com> wrote:
> On Jul 2, 3:45 pm, James <donje...@gmail.com> wrote:
>
>> By the way, I would like my final program to work on arrays with any
>> number of dimensions, so I'd rather avoid a kludge like X[diff[0],
>> diff[1], diff[2]] = Y.
>
> As I wrote above, X[diff[0],diff[1],diff[2]] is not the same as
> X[diff]. What you seem to want is the equivalent of X[diff[0],
> diff[1], diff[2]] = Y, but working for any number of dimensions. That
> is, assign the elements of Y to a contiguous piece of X that starts at
> some location you calculate with the N-dimensional indexes. So it is
> just a matter of converting from those N indexes the a 1D index of
> that element:
>
> IDL> strides=[1L,product(size(x,/dimensions),/integer,/cumulative )]
> IDL> start_index=total(strides*diff,/integer)
>
> Then you can do X[start_index]=Y
Paulo, your explanation makes perfect sense. I guess I was hoping
there was a more elegant way than calculating every single 1d index
that will be used, but your method looks good. Thank you!
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| Re: Padding arrays - vector subscripts not working [message #71583 is a reply to message #71576] |
Fri, 02 July 2010 13:01 |
penteado
Messages: 866
Registered: February 2018
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Senior Member
Administrator |
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On Jul 2, 3:45 pm, James <donje...@gmail.com> wrote:
> By the way, I would like my final program to work on arrays with any
> number of dimensions, so I'd rather avoid a kludge like X[diff[0],
> diff[1], diff[2]] = Y.
As I wrote above, X[diff[0],diff[1],diff[2]] is not the same as
X[diff]. What you seem to want is the equivalent of X[diff[0],
diff[1], diff[2]] = Y, but working for any number of dimensions. That
is, assign the elements of Y to a contiguous piece of X that starts at
some location you calculate with the N-dimensional indexes. So it is
just a matter of converting from those N indexes the a 1D index of
that element:
IDL> strides=[1L,product(size(x,/dimensions),/integer,/cumulative )]
IDL> start_index=total(strides*diff,/integer)
Then you can do X[start_index]=Y
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| Re: Padding arrays - vector subscripts not working [message #71584 is a reply to message #71583] |
Fri, 02 July 2010 12:29 |
penteado
Messages: 866
Registered: February 2018
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Senior Member
Administrator |
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On Jul 2, 3:51 pm, James <donje...@gmail.com> wrote:
> Well, I see that this was already asked by hradliv in 2007 and
> answered by JD Smith, among others. Still, it seems strange that the
> natural method I tried to use doesn't work.
It is not strange.
X[1,1,1] = Y
and
X[diff] = Y with diff=[1,1,1]
are different things. X[diff] is the same as X[[1,1,1]].
To see the difference, consider this case (where I defined X so that
its elements are not all equal)
IDL> X = bindgen(5,5,5)
IDL> print,X[1,1,1]
31
IDL> print,X[31]
31
Both of these print lines are accessing the same (one) element of X,
the one at the (3D) indexes 1,1,1, which, as it happens, is the 32nd
element of the array:
IDL> print,array_indices(X,31)
1 1 1
Now,
IDL> diff=[1,1,1]
IDL> print,X[diff]
1 1 1
X[diff], which is the same as X[[1,1,1]], is a 3-element array, where
each element has the element of X with the (1D) index 1, that is, this
is 3 copies of the second element of X:
IDL> print,array_indices(X,1)
1 0 0
IDL> print,X[1,0,0]
1
X[diff] is the same as X[1,0,0], repeated 3 times.
Now, looking at your assignments:
X[1,1,1]=Y
is assigning 12 elements of X (Y has 12 elements), sequentially,
starting at X[1,1,1]. Which is working only because X has 125
elements, and this assignment is using 12 elements, starting at the
32nd, thus it fits inside X (it is assigning from to the 32nd to the
43rd element of X the 12 values in Y)
X[diff], since it is X[[1,1,1]] is attempting to assign the 12 values
of Y into 3 values (since X[[1,1,1]] is a 3-element array), thus it
fails. Only if you tried to assign to X[diff] a 3-element array (or a
scalar, which would be repeated), would this have worked.
If this is seeming too abstract, redo these examples with small 2D
arrays (which print nicely), and look at the whole X array, and how it
changes after assignments.
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| Re: Padding arrays - vector subscripts not working [message #71651 is a reply to message #71575] |
Tue, 06 July 2010 11:38 |
James[2]
Messages: 44
Registered: November 2009
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Member |
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OK, I tried using your method, but it seemed to only take the first
dimension of Y. Here's the function I used:
function pad, array, dims
;error checking, etc. removed for more compact newsgroup post
type = size(array, /type)
out = make_array(dims, type=type)
;calculate where to put the original array in the new
sz = size(array, /dimensions)
diff = (dims - sz)/2
strides = [1L, product(dims, /integer, /cumulative)]
start = total(strides*diff, /integer)
out[start] = array
return, out
end
And here is an example of its output:
IDL> test = replicate(4,3,3)
IDL> padtest = pad(test, [5,6])
IDL> print, padtest
0 0 0 0 0
0 4 4 4 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
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