| Re: Counting the lengths of continuous numbers in an array [message #61824] |
Fri, 08 August 2008 14:41  |
andy.mcneel
Messages: 3
Registered: August 2008
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Junior Member |
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On Aug 8, 4:28 pm, mankoff <mank...@gmail.com> wrote:
> On Aug 8, 5:05 pm, andy.mcn...@gmail.com wrote:
>
>> Howdy folks.
>
>> Say I have the following array: [1, 2, 3, 4, 7, 8, 11, 13, 14, 17,
>> 18, 19]
>
>> How would I tell IDL to go through this array and return the array [4,
>> 2, 1, 2, 3].....the lengths of the continuous groups of numbers in the
>> array?
>
>> Thanks!
>> -Andy McNeel
>
> I'm not sure how robust this is for other arrays, but I can get your
> 2nd array from your first like so, assuming the 1st array is "a". I
> didn't test boundary cases (runlength of 1 on beginning or end) or
> much else...
>
> IDL> aa = [a[0],a,0]
> IDL> b = where( (aa-shift(aa,1)) ne 1)
> IDL> c = b-shift(b,1)
> IDL> print, c[1:*]
> 4 2 1 2 3
>
> -k.
This worked beautifully. Thanks! -Andy
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| Re: Counting the lengths of continuous numbers in an array [message #61828 is a reply to message #61824] |
Fri, 08 August 2008 14:28  |
mankoff
Messages: 131
Registered: March 2004
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Senior Member |
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On Aug 8, 5:05 pm, andy.mcn...@gmail.com wrote:
> Howdy folks.
>
> Say I have the following array: [1, 2, 3, 4, 7, 8, 11, 13, 14, 17,
> 18, 19]
>
> How would I tell IDL to go through this array and return the array [4,
> 2, 1, 2, 3].....the lengths of the continuous groups of numbers in the
> array?
>
> Thanks!
> -Andy McNeel
I'm not sure how robust this is for other arrays, but I can get your
2nd array from your first like so, assuming the 1st array is "a". I
didn't test boundary cases (runlength of 1 on beginning or end) or
much else...
IDL> aa = [a[0],a,0]
IDL> b = where( (aa-shift(aa,1)) ne 1)
IDL> c = b-shift(b,1)
IDL> print, c[1:*]
4 2 1 2 3
-k.
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| Re: Counting the lengths of continuous numbers in an array [message #61968 is a reply to message #61824] |
Mon, 11 August 2008 06:08 |
Brian Larsen
Messages: 270
Registered: June 2006
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Senior Member |
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>> IDL> aa = [a[0],a,0]
>> IDL> b = where( (aa-shift(aa,1)) ne 1)
>> IDL> c = b-shift(b,1)
>> IDL> print, c[1:*]
>> 4 2 1 2 3
While I can't seem to find it this second my tired and jet lagged
brain is telling me that there is an built-in idl function to
calculate array[*]-array[1:*], which should be equivalent to above.
Now to remember the function.
Cheers,
Brian
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