# Re: Filling an array

```"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