| Re: extracting subarray (*,*,index) gives (1,1,index). WHY ??? [message #6868] |
Fri, 23 August 1996 00:00 |
lmudge
Messages: 9
Registered: October 1994
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Junior Member |
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In article <3219CC9A.53D7@neurologie.uni-duesseldorf.de>,
knorr@neurologie.uni-duesseldorf.de says...
> i want to extract a line from a 3D image by doing
> line = image(*,*,z)
> i expect line to be a 1d vector with dimension dim(z)
> or to be more IDLish, i expect the following to hold
> size(line) == (z)
> but instead I get
> size(line) == (1,1,z)
Try line = reform(image(*,*,z))
this should work.
Leith.
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| Re: extracting subarray (*,*,index) gives (1,1,index). WHY ??? [message #6870 is a reply to message #6868] |
Thu, 22 August 1996 00:00  |
Graeme K Harkness
Messages: 3
Registered: August 1996
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Junior Member |
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Uwe Knorr wrote:
>
> Dear reader,
>
> i want to extract a line from a 3D image by doing
> line = image(*,*,z)
> i expect line to be a 1d vector with dimension dim(z)
> or to be more IDLish, i expect the following to hold
> size(line) == (z)
> but instead I get
> size(line) == (1,1,z)
> I hope you know what I mean.
>
> Thanx for giving me some of your time,
>
> Uwe Knorr.
>
I think I know what you mean. You should use the
REFORM
function to convert the 1x1xz sized array into a 1-d one of size z
Graeme
-----------------------------------------------
| Graeme K Harkness |
| Department of Physics & Applied Physics |
| University of Strathclyde |
| GLASGOW G4 0NG UK |
| Tel: +44-141-552 4400 x3354 |
| Fax: +44-141-552 2891 |
| graeme@phys.strath.ac.uk |
-----------------------------------------------
|
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| Re: extracting subarray (*,*,index) gives (1,1,index). WHY ??? [message #6881 is a reply to message #6870] |
Wed, 21 August 1996 00:00 |
peter
Messages: 80
Registered: February 1994
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Member |
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Uwe Knorr (knorr@neurologie.uni-duesseldorf.de) wrote:
: Dear reader,
: i want to extract a line from a 3D image by doing
: line = image(*,*,z)
: i expect line to be a 1d vector with dimension dim(z)
: or to be more IDLish, i expect the following to hold
: size(line) == (z)
: but instead I get
: size(line) == (1,1,z)
: I hope you know what I mean.
Bit confusing here, line=image(*,*,z) will give you a (nx,ny,1) result.
You probably meant you do line=image(x,y,*), which gives a (1,1,nz)
result. If you want a (nz) result, do line=reform(image(x,y,*)).
Reform, used like this, strips collapses leading dimensions of length 1.
Peter
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| Re: extracting subarray (*,*,index) gives (1,1,index). WHY ??? [message #6887 is a reply to message #6881] |
Tue, 20 August 1996 00:00 |
a2652099
Messages: 10
Registered: August 1996
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Junior Member |
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In article <3219CC9A.53D7@neurologie.uni-duesseldorf.de>
knorr@neurologie.uni-duesseldorf.de (Uwe Knorr) wrote:
> i want to extract a line from a 3D image by doing
> line = image(*,*,z)
A line? Wouldn't that rather be image(a,b,*) ?
> i expect line to be a 1d vector with dimension dim(z)
> or to be more IDLish, i expect the following to hold
> size(line) == (z)
> but instead I get
> size(line) == (1,1,z)
> I hope you know what I mean.
If I do, I think using the REFORM command would help you, eg.
line = REFORM( image(a,b,*), z ),
where z is the 3rd dimension of your array. This converts your (1,1,z)
array into the (z) array.
Alex
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| Re: extracting subarray (*,*,index) gives (1,1,index). WHY ??? [message #6888 is a reply to message #6887] |
Tue, 20 August 1996 00:00 |
Thomas A. McGlynn
Messages: 23
Registered: March 1996
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Junior Member |
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Uwe Knorr wrote:
>
> Dear reader,
>
> i want to extract a line from a 3D image by doing
> line = image(*,*,z)
> i expect line to be a 1d vector with dimension dim(z)
> or to be more IDLish, i expect the following to hold
> size(line) == (z)
> but instead I get
> size(line) == (1,1,z)
> I hope you know what I mean.
>
> Thanx for giving me some of your time,
>
> Uwe Knorr.
>
I'm a little confused by your message. I'm guessing
what you want is something like:
x = fltarr(30,30,30)
y = x(1,1,*)
where you'd like y to be a one dimensional vector of length 30.
If so then you need to rewrite the second statement as
y = reform(x(1,1,*))
while will get rid of the extra leading dimensions.
What you describe though is:
y = x(*,*,1)
which will return y as a two-dimensional array and suppress
the third dimension automatically. I.e., we'd have
size(y) == (2,30,30,4,900)
Hope this helps,
Tom McGlynn
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