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Re: Filling an array [message #23274 is a reply to message #23273] Thu, 18 January 2001 12:28 Go to previous messageGo to previous message
John-David T. Smith is currently offline  John-David T. Smith
Messages: 384
Registered: January 2000
Senior Member
"Pavel A. Romashkin" wrote:
>
> Thanks David and Craig. If Craig says "no", this means something.
> Although I'd wait with the verdict until I hear from JD :-) After all,
> the loop I have with Total, although it goes through all elements of B,
> is taking only 0.03 s on my machine for B with ~2500 points converted to
> C with ~50k points, which is acceptable since it is not executed many
> times repeatedly. In contrast with looping using Replicate that was
> taking 5.5 s :-(
>
> Cheers,
> Pavel
>
> "Pavel A. Romashkin" wrote:
>>
>> If I have
>>
>> a = findgen(10)
>> b = fix(100* randomu(10, 10))
>> ; N_elements(a) is equal to n_elements(b)
>> c = findgen(total(b))
>>
>> how can I fill C with values from A using B as a running index, so that
>>
>> c[0 : b[0]-1] = a[0]
>> c[b[0] : b[0]+b[1]-1] = a[1]
>>
>> etc, without looping through "n_elements(b)-1" iterations?
>> I have a fast solution with a loop and indexing using total(/cumulative)
>> and a very slow one with loop and replicate, but I can't come up with a
>> loop-free one.

Despite the fact that getting a job and writing a thesis should be my
foremost priorities....

tmp=histogram(total(b,/CUMULATIVE)-1,/BINSIZE,MIN=0,REVERSE_ INDICES=ri)
c=a[ri[0:n_elements(ri)-n_elements(b)-2]-ri[0]]

JD

P.S. Let's just say (TM) we all knew it had to use histogram.

P.P.S. A loop free solution is not guaranteed to be fastest in all
cases. It *is* guaranteed to elicit various
histogram/median/rebin/reform/## curses from newsgroup readers.
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