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Re: IDL Way to Remove Rows of an Array [message #76121 is a reply to message #75986]

Tue, 17 May 2011 10:12

Gray
Messages: 253
Registered: February 2010

Senior Member

On May 17, 12:59 pm, JDS <jdtsmith.nos...@yahoo.com> wrote:
>> JD Smith would probably say that it's faster to construct your own
>> index arrays, right? Maybe like this?
>
> Probably not in this case. Typically it's faster to roll your own index lists when the list you are constructing is smaller than the list IDL has to construct, for example if you are further sub-indexing it. The rule of thumb is to try not to make IDL do lots of work you will then promptly throw away.
>
> In this case, your list over columns, and the list IDL would create using "*" are identical. So I'd probably simply go with '*'. If you have a 'reject' list instead of 'keep' (as in David's example) a closely related but somewhat simpler/faster method is:
>
> d=size(array,/DIMENSIONS)
> array=array[*,where(~histogram(reject,MIN=0,MAX=d[1]-1),/NUL L)]

Since David's example converts the reject list to a keep list, then he
could use your method and skip the steps of constructing the keep
list.

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		IDL Way to Remove Rows of an Array By: David Fanning on Tue, 10 May 2011 05:27
		Re: IDL Way to Remove Rows of an Array By: David Fanning on Tue, 17 May 2011 10:35
		Re: IDL Way to Remove Rows of an Array By: Gray on Tue, 17 May 2011 10:12

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