| How to find second minimum elements in an array in IDL? [message #64677] |
Wed, 14 January 2009 07:26  |
Hu
Messages: 35
Registered: January 2009
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Member |
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Supposing that there is an array X=[9,2,3,5,1,6,8,4,7], how can I find
the first and second minimums (in this array are elements 1 and 2) in
this vector?
I use this to find the first minimum (element 1):
index = where(X eq min(X))
minimum_first=X[index]
But, how can i find the elements 2 ?
thanks
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| Re: How to find second minimum elements in an array in IDL? [message #64724 is a reply to message #64677] |
Thu, 15 January 2009 14:39  |
Michael Galloy
Messages: 1114
Registered: April 2006
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Senior Member |
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On Jan 15, 10:36 am, cmanc...@gmail.com wrote:
> I was curious, so I checked out your routine Mike. It looks good but
> one problem - a for loop! I'm pretty sure you can replace:
>
> nCandidates = 0L
> for bin = 0L, nBins - 1L do begin
> nCandidates += h[bin]
> if (nCandidates ge n) then break
> endfor
>
> with:
>
> max( total( h, /cumulative ) < n, bin )
>
> which should work because max will return the first maximum value. Of
> course, I was too lazy to see if the max(total()) method is actually
> faster (since it involves a couple different compuatations), but oh
> well, sometimes laziness wins :)
It turns out that it probably doesn't matter much.
It's not FOR loops per se that are bad, but the execution of many
statements. For perfectly uniformly distributed data, the FOR loop
above will only loop once -- more times the less uniformly distributed
the data, bounded by the number of bins (i.e. number of data
elements / number of elements required).
Averages were computed for 500 runs of finding the smallest k=100
elements of an n=1000000 element dataset.
For uniform data:
mg_n_smallest(randomu(seed, n), k)
vectorized: 0.035663 seconds
loops: 0.036040 seconds
loops are 1.1% faster
For perverse data:
mg_n_smallest([randomu(seed, k - 1), randomu(seed, n - k + 1) + n /
k], 100)
vectorized: 0.279783 seconds
loops: 0.281627 seconds
vectorized is 0.7% faster
Mike
--
www.michaelgalloy.com
Tech-X Corporation
Associate Research Scientist
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| Re: How to find second minimum elements in an array in IDL? [message #64735 is a reply to message #64677] |
Thu, 15 January 2009 09:36 |
Conor
Messages: 138
Registered: February 2007
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Senior Member |
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On Jan 14, 10:53 am, "mgal...@gmail.com" <mgal...@gmail.com> wrote:
> On Jan 14, 8:26 am, Hu <jha...@gmail.com> wrote:
>
>> Supposing that there is an array X=[9,2,3,5,1,6,8,4,7], how can I find
>> the first and second minimums (in this array are elements 1 and 2) in
>> this vector?
>
>> I use this to find the first minimum (element 1):
>
>> index = where(X eq min(X))
>> minimum_first=X[index]
>
>> But, how can i find the elements 2 ?
>> thanks
>
> For a general approach for finding the n smallest elements of an array
> (using HISTOGRAM and REVERSE_INDICES!), try:
>
> http://michaelgalloy.com/2006/06/02/finding-the-n-smallest-e lements-i...
>
> Mike
> --www.michaelgalloy.com
> Tech-X Corporation
> Associate Research Scientist
I was curious, so I checked out your routine Mike. It looks good but
one problem - a for loop! I'm pretty sure you can replace:
nCandidates = 0L
for bin = 0L, nBins - 1L do begin
nCandidates += h[bin]
if (nCandidates ge n) then break
endfor
with:
max( total( h, /cumulative ) < n, bin )
which should work because max will return the first maximum value. Of
course, I was too lazy to see if the max(total()) method is actually
faster (since it involves a couple different compuatations), but oh
well, sometimes laziness wins :)
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| Re: How to find second minimum elements in an array in IDL? [message #64768 is a reply to message #64677] |
Wed, 14 January 2009 07:53 |
Michael Galloy
Messages: 1114
Registered: April 2006
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Senior Member |
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On Jan 14, 8:26 am, Hu <jha...@gmail.com> wrote:
> Supposing that there is an array X=[9,2,3,5,1,6,8,4,7], how can I find
> the first and second minimums (in this array are elements 1 and 2) in
> this vector?
>
> I use this to find the first minimum (element 1):
>
> index = where(X eq min(X))
> minimum_first=X[index]
>
> But, how can i find the elements 2 ?
> thanks
For a general approach for finding the n smallest elements of an array
(using HISTOGRAM and REVERSE_INDICES!), try:
http://michaelgalloy.com/2006/06/02/finding-the-n-smallest-e lements-in-an-array.html
Mike
--
www.michaelgalloy.com
Tech-X Corporation
Associate Research Scientist
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| Re: How to find second minimum elements in an array in IDL? [message #64813 is a reply to message #64677] |
Fri, 16 January 2009 07:48 |
pgrigis
Messages: 436
Registered: September 2007
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Senior Member |
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cmanc...@gmail.com wrote:
> On Jan 15, 5:39 pm, "mgal...@gmail.com" <mgal...@gmail.com> wrote:
>> On Jan 15, 10:36 am, cmanc...@gmail.com wrote:
>>
>>
>>
>>> I was curious, so I checked out your routine Mike. It looks good but
>>> one problem - a for loop! I'm pretty sure you can replace:
>>
>>> nCandidates = 0L
>>> for bin = 0L, nBins - 1L do begin
>>> nCandidates += h[bin]
>>> if (nCandidates ge n) then break
>>> endfor
>>
>>> with:
>>
>>> max( total( h, /cumulative ) < n, bin )
>>
>>> which should work because max will return the first maximum value. Of
>>> course, I was too lazy to see if the max(total()) method is actually
>>> faster (since it involves a couple different compuatations), but oh
>>> well, sometimes laziness wins :)
>>
>> It turns out that it probably doesn't matter much.
>>
>> It's not FOR loops per se that are bad, but the execution of many
>> statements. For perfectly uniformly distributed data, the FOR loop
>> above will only loop once -- more times the less uniformly distributed
>> the data, bounded by the number of bins (i.e. number of data
>> elements / number of elements required).
>>
>> Averages were computed for 500 runs of finding the smallest k=100
>> elements of an n=1000000 element dataset.
>>
>> For uniform data:
>>
>> mg_n_smallest(randomu(seed, n), k)
>>
>> vectorized: 0.035663 seconds
>> loops: 0.036040 seconds
>> loops are 1.1% faster
>>
>> For perverse data:
>>
>> mg_n_smallest([randomu(seed, k - 1), randomu(seed, n - k + 1) + n /
>> k], 100)
>>
>> vectorized: 0.279783 seconds
>> loops: 0.281627 seconds
>> vectorized is 0.7% faster
>>
>> Mike
>> --www.michaelgalloy.com
>> Tech-X Corporation
>> Associate Research Scientist
>
> I didn't really expect much of a difference. I think this is just a
> personal preference of mine - it looks so much nice when it all fits
> on one line!
But we don't want to encourage people writing
all of their programs in one line, don't we?
Ciao,
Paolo
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| Re: How to find second minimum elements in an array in IDL? [message #64814 is a reply to message #64724] |
Fri, 16 January 2009 07:08 |
Conor
Messages: 138
Registered: February 2007
|
Senior Member |
|
|
On Jan 15, 5:39 pm, "mgal...@gmail.com" <mgal...@gmail.com> wrote:
> On Jan 15, 10:36 am, cmanc...@gmail.com wrote:
>
>
>
>> I was curious, so I checked out your routine Mike. It looks good but
>> one problem - a for loop! I'm pretty sure you can replace:
>
>> nCandidates = 0L
>> for bin = 0L, nBins - 1L do begin
>> nCandidates += h[bin]
>> if (nCandidates ge n) then break
>> endfor
>
>> with:
>
>> max( total( h, /cumulative ) < n, bin )
>
>> which should work because max will return the first maximum value. Of
>> course, I was too lazy to see if the max(total()) method is actually
>> faster (since it involves a couple different compuatations), but oh
>> well, sometimes laziness wins :)
>
> It turns out that it probably doesn't matter much.
>
> It's not FOR loops per se that are bad, but the execution of many
> statements. For perfectly uniformly distributed data, the FOR loop
> above will only loop once -- more times the less uniformly distributed
> the data, bounded by the number of bins (i.e. number of data
> elements / number of elements required).
>
> Averages were computed for 500 runs of finding the smallest k=100
> elements of an n=1000000 element dataset.
>
> For uniform data:
>
> mg_n_smallest(randomu(seed, n), k)
>
> vectorized: 0.035663 seconds
> loops: 0.036040 seconds
> loops are 1.1% faster
>
> For perverse data:
>
> mg_n_smallest([randomu(seed, k - 1), randomu(seed, n - k + 1) + n /
> k], 100)
>
> vectorized: 0.279783 seconds
> loops: 0.281627 seconds
> vectorized is 0.7% faster
>
> Mike
> --www.michaelgalloy.com
> Tech-X Corporation
> Associate Research Scientist
I didn't really expect much of a difference. I think this is just a
personal preference of mine - it looks so much nice when it all fits
on one line!
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