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Re: setintersection assumes sets have no repetitions? [message #83457 is a reply to message #83310]

Tue, 05 March 2013 10:44

penteado
Messages: 866
Registered: February 2018

Senior Member
Administrator

I just noticed that the documentation changed, to say that indices_a
and indices_b require the intersection elements to be unique, while
positions does not.

I understand the reasons to assume the sets have unique elements (both
from the mathematical concept and from practical applications like
identifiers from databases). But since the positions keyword was there
to handle repetitions, I got confused when the code crashed on arrays
with repetitions.

The use I had with repeated elements happened when I was working with
matches of my list of sources in the sky (each with its unique
identifier) with other such lists (with their own unique identifiers),
by finding which objects in one list were within a certain distance of
each object in the other list. Occasionally, one object in my list
would have more than one match, and, as such, its identifier would
appear more than once in the list of results.

I eventually resorted to using histogram, through my wrapper that puts
the reverse indices into hashes or lists:

a=[1,2,3,4,5]
b=[1,2,2,2]
h=histogram_pp(a,binsize=1,reverse_hash=rh)
print,rh
;5: 4
;1: 0
;3: 2
;2: 1
;4: 3
ib=lonarr(n_elements(b))
foreach el,b,i do ib[i]=rh[el]
print,ib
; 0 1 1 1
print,b[ib]
; 1 2 2 2

histogram_pp is at
http://www.ppenteado.net/idl/pp_lib/doc/histogram_pp.html

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[Message index]

		Re: setintersection assumes sets have no repetitions? By: David Fanning on Sat, 23 February 2013 22:23
		Re: setintersection assumes sets have no repetitions? By: Jeremy Bailin on Sat, 23 February 2013 13:15
		Re: setintersection assumes sets have no repetitions? By: penteado on Tue, 05 March 2013 10:44

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