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Re: Where.. Then Loop IDL help [message #93782 is a reply to message #93777]

Fri, 14 October 2016 11:20

smnadoum
Messages: 24
Registered: June 2016

Junior Member

On Friday, October 14, 2016 at 2:25:22 AM UTC-7, Markus Schmassmann wrote:
> 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

Thanks so much.

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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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