| Re: how to convert a row of data in a column? [message #82904] |
Fri, 25 January 2013 09:01 |
Phillip Bitzer
Messages: 223
Registered: June 2006
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Senior Member |
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On Friday, January 25, 2013 7:57:04 AM UTC-6, pan...@gmail.com wrote:
> Thanks, I solved the problem myself by writing a for loop.
In IDL, for loops are the embodiment of pure evil:
http://www.idlcoyote.com/tips/forloops.html
Well, maybe not so bad sometimes:
http://www.idlcoyote.com/tips/forloops2.html
But, for your problem, you're trying to convert to a row vector into a column vector. A couple ways to that:
IDL> d = TRANSPOSE(set) & help, d
or
IDL> d = REFORM(set, 1, N_ELEMENTS(set))
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| Re: how to convert a row of data in a column? [message #82906 is a reply to message #82905] |
Fri, 25 January 2013 05:44  |
Mats Löfdahl
Messages: 263
Registered: January 2012
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Senior Member |
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Den fredagen den 25:e januari 2013 kl. 14:34:58 UTC+1 skrev pan...@gmail.com:
> I am using an external program which returns me a one dimensional array, .i.e. a row of data like this.
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> 1 2 5 7 8 9 0
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> The problem is that I need to read it as a column of data. I tried to create an array to store it but it doesn't work.
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> (Here set=[1 2 5 7 8 9 0])
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> d=dblarr(1,7)
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> set=d
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> but when I print set on the screen, it is still
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> 1 2 5 7 8 9 0
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> instead of
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> 1
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> 2
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> 5
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> 7
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> 8
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> 9
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> 0
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> what can I do?
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> Many thanks
I'm not sure what lines you actually executed, because after the ones you posted both set and d should be a (1,7) array of zeros, which would print as a column and not as a row.
If you replace the set=d line with d[0]=set, the d vector should print as
IDL> print,d
1.0000000
2.0000000
5.0000000
7.0000000
8.0000000
9.0000000
0.0000000
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