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Re: last array index subscript [message #33731 is a reply to message #33730]

Thu, 23 January 2003 12:16

kashyap
Messages: 26
Registered: April 1993

Junior Member

Try

value=(my_array[[2147483647L]])[0]

where the "[[.]]" returns an array and the "(.)[0]" ensures that the
output is a scalar, and 2147483647L = 2L^(31L)-1L is the largest
possible I*4 number you can have and surely no will have an array
bigger than that.

vinay

In article <Pine.LNX.4.33.0301231237280.25514-100000@hapuna.ess.sunysb.edu>,
Chad Bender <cbender@mail.astro.sunysb.edu> wrote:
> Hi --
>
> Is there a way in IDL to directly reference the last element of a vector
> without first determining how many elements the vector contains?
>
> For example, something like:
>
> n=N_Elements(my_array)
> value=my_array[n-1]
>
> except without having to make the call to N_Elements. Granted, avoiding
> the N_Elements call probably doesn't save a lot of time. But I figured
> that with all of the complicated syntax that IDL accepts, there might be a
> direct way to do this. The Bldg Apps guide says you can extract a
> subarray from some element e to the end of the array like this:
>
> subarray=my_array[e:*]
>
> So it seems to me that IDL knows what that last index is. I tried
> something like:
>
> value=my_array[*-1], but (not unexpectedly) this caused a syntax error.
>
> I figure that what I want is probably impossible, but if anyone knows how
> to do it your advice is appreciated.
>
> Thanks
> Chad Bender
>
>

--
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kashyap@head-cfa.harvard.edu 617 495 7173 [CfA/P-143] 617 496 7173 [F]

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[Message index]

		Re: last array index subscript By: notspecified on Thu, 23 January 2003 12:33
		Re: last array index subscript By: kashyap on Thu, 23 January 2003 12:16
		Re: last array index subscript By: James Kuyper on Thu, 23 January 2003 10:40
		Re: last array index subscript By: Doug Rowland on Fri, 31 January 2003 08:30
		Re: last array index subscript By: hieninger on Fri, 31 January 2003 08:24

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