| Re: Filling an array [message #23267] |
Thu, 18 January 2001 16:43 |
Pavel A. Romashkin
Messages: 531
Registered: November 2000
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Senior Member |
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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.
Cheers,
Pavel
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[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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| Re: Filling an array [message #23270 is a reply to message #23267] |
Thu, 18 January 2001 13:40  |
Craig Markwardt
Messages: 1869
Registered: November 1996
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Senior Member |
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JD Smith <jdsmith@astro.cornell.edu> writes:
> 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]]
>
Bravo. I guess the juices just weren't flowing last night :-)
Craig
--
------------------------------------------------------------ --------------
Craig B. Markwardt, Ph.D. EMAIL: craigmnet@cow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
------------------------------------------------------------ --------------
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| Re: Filling an array [message #23274 is a reply to message #23273] |
Thu, 18 January 2001 12:28 |
John-David T. Smith
Messages: 384
Registered: January 2000
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Senior Member |
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"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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| Re: Filling an array [message #23278 is a reply to message #23274] |
Thu, 18 January 2001 09:03 |
Pavel A. Romashkin
Messages: 531
Registered: November 2000
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Senior Member |
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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.
>
> Thank you,
> Pavel
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| Re: Filling an array [message #23284 is a reply to message #23278] |
Wed, 17 January 2001 21:37 |
Craig Markwardt
Messages: 1869
Registered: November 1996
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Senior Member |
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davidf@dfanning.com (David Fanning) writes:
> Pavel A. Romashkin (pavel.romashkin@noaa.gov) writes:
>
>> 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.
>
> Totally impossible. :-(
>
> Cheers,
>
> P.S. Let's just say that usually gets the juices going
> on the usual suspects, and I figured you could use the
> help. :-)
Juices or not, I don't see how this can be done without a loop. Since
the segments specified by B can be of different sizes, I think you are
stuck. However, as I've said in the past: loops aren't bad! If you
can get enough work done in a single iteration then loops are fine.
Craig
--
------------------------------------------------------------ --------------
Craig B. Markwardt, Ph.D. EMAIL: craigmnet@cow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
------------------------------------------------------------ --------------
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| Re: Filling an array [message #23293 is a reply to message #23284] |
Wed, 17 January 2001 17:18 |
davidf
Messages: 2866
Registered: September 1996
|
Senior Member |
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Pavel A. Romashkin (pavel.romashkin@noaa.gov) writes:
> 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.
Totally impossible. :-(
Cheers,
P.S. Let's just say that usually gets the juices going
on the usual suspects, and I figured you could use the
help. :-)
--
David Fanning, Ph.D.
Fanning Software Consulting
Phone: 970-221-0438 E-Mail: davidf@dfanning.com
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Toll-Free IDL Book Orders: 1-888-461-0155
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