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Re: Filling an array
I knew it. What the heck will I write to the newsgroup the next time
for? I will just write directly to JD :-) I will give it a shot timing
wise tomorrow, but I am sure it will be faster.
This one was the last drop in the bucket. I am reading and practicing
with the darn Histogram thing tomorrow until I can write my entire code
with a long single call to Histogram.
P.S. Lets just say (TM) that I tried to rig up Histogram for this, but
was distracted before I got too far. Not that I am saying I'd come up
with a solution, if I wasn't :-(
JD Smith wrote:
> "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-1] = a
> > > c[b : b+b-1] = a
> > >
> > > 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....
> 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.