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Re: Where.. Then Loop IDL help [message #93777 is a reply to message #93772]

Fri, 14 October 2016 02:25

Markus Schmassmann
Messages: 129
Registered: April 2016

Senior Member

On 10/14/2016 07:08 AM, Cheryl wrote:
> On Thursday, October 13, 2016 at 9:39:33 PM UTC-7, Cheryl wrote:
>> I am new to IDL world, and was wondering if you can help me create a loop.
>>
>> I need to create an array by replacing all values in 'a' that are greater than 5 by 20.
>>
>> c= where(a gt 5) then [c] = 20
>>
>> Is this the correct way to do it?
>>
>> Thanks
>
> I think i found a way to do it
>
> a[where(a gt 5,/null)]=20 is it correct?? How can I create an array for this?
> I tried to do :c= a[where(a gt 5,/null)]=20, but I keep getting a syntax error
if you want to retain a unchanged, then

b=a
b[where(b gt 5,/null)]=20

otherwise, just

a[where(a gt 5,/null)]=20

if a is already an array, so will b, and a remains an array

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[Message index]

		Where.. Then Loop IDL help By: smnadoum on Thu, 13 October 2016 21:39
		Re: Where.. Then Loop IDL help By: smnadoum on Thu, 13 October 2016 22:08
		Re: Where.. Then Loop IDL help By: Markus Schmassmann on Fri, 14 October 2016 02:25
		Re: Where.. Then Loop IDL help By: smnadoum on Fri, 14 October 2016 11:20

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