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Re: Count duplicate elements in an array but keep their order! [message #85751 is a reply to message #85750]

Sat, 31 August 2013 08:03

Josh Sixsmith
Messages: 13
Registered: December 2012

Junior Member

The following might work.
Using the UNIQ function will give the indices, which you can use to get the value counts.
To get the counts for each value you'll then need to use the SHIFT function to offset the indices in order subtract from the original indices.

Assuming that a is already sorted, other wise use SORT(a) first
un = UNIQ(a)

In order to get the proper count for the first value, the unique array will need to be extended by 1 element.

un_extra = intarr(n_elements(un)+1)
un_extra[1:n_elements(un_extra)-1] = un

Add 1 to the array to account for the fact that indices start at 0
un_extra += 1
un_offset = SHIFT(un_extra, -1)

counts = un_offset - un_extra

Remove the last element in counts
counts = count[0:n_elements(counts)-2]
values = a[un]

Hope that helps ,if it works :)

Cheers
Josh

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[Message index]

		Count duplicate elements in an array but keep their order! By: Steve Daal on Fri, 30 August 2013 16:23
		Re: Count duplicate elements in an array but keep their order! By: Josh Sixsmith on Sat, 31 August 2013 08:03
		Re: Count duplicate elements in an array but keep their order! By: David Fanning on Tue, 03 September 2013 05:58
		Re: Count duplicate elements in an array but keep their order! By: Steve Daal on Mon, 16 September 2013 10:51

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