| Histogram quickie [message #51793] |
Fri, 08 December 2006 10:55  |
Christopher Thom
Messages: 66
Registered: October 2006
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Member |
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Hi all,
I'm a long time where() fan, but trying to learn to wield this histogram
beast. I'm working on an algorithm, and would like a way to divide an
array of values into two bins, such that the sum of each bin is roughly
equal. The values have no fixed distribution, so I expect the bin sizes to
be non-uniform.
This sort of problem seems an ideal place to start earning my histogram
badge, but I have to confess to only being able to think of
brute-force-type solutions. Any suggestions?
cheers
chris
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| Re: Histogram quickie [message #51921 is a reply to message #51793] |
Mon, 11 December 2006 07:49  |
Christopher Thom
Messages: 66
Registered: October 2006
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Member |
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Quoth JD Smith:
> On Fri, 08 Dec 2006 12:55:56 -0600, Christopher Thom wrote:
>
>> Hi all,
>>
>> I'm a long time where() fan, but trying to learn to wield this histogram
>> beast. I'm working on an algorithm, and would like a way to divide an
>> array of values into two bins, such that the sum of each bin is roughly
>> equal. The values have no fixed distribution, so I expect the bin sizes to
>> be non-uniform.
>>
>> This sort of problem seems an ideal place to start earning my histogram
>> badge, but I have to confess to only being able to think of
>> brute-force-type solutions. Any suggestions?
>
> Probably WHERE will serve you well:
>
> IDL> t=total(a,/CUMULATIVE)
> IDL> bin1=where(t lt t[n_elements(a)-1]/2,COMPLEMENT=bin2)
>
> Of course, there are many ways to divide values such that they fall into
> two roughly equal bins (n choose 2), some of which may be better than
> others.
aha! I wasn't aware of the /cumulative flag to total(), despite some
searching. Thanks for the pointer.
cheers
chris
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| Re: Histogram quickie [message #51931 is a reply to message #51793] |
Fri, 08 December 2006 18:09 |
Braedley
Messages: 57
Registered: September 2006
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Member |
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I was thinking the same thing. This never struck me as a histogram
problem. Sure, there are some novel uses for histogram, but this isn't
one of them. If the array needs to be sorted, and you need to retain
the original array, and you want one bin to have the low values and the
other to have the highs, then I can conceivably see a use for histogram
in this problem, but it would be so convoluted that it wouldn't be
worth it (you would still have to go through JD's process anyways).
The only difference I'm going to suggest is to allow for the situation
that I commented about:
t=randomu(seed, x)
t=t[sort(t)]
u=total(t, /cumulative)
bin1=t[where(u le u[x-1]/2.0)]
bin2=t[where(u gt u[x-1[/2.0)]
Braedley
JD Smith wrote:
> On Fri, 08 Dec 2006 12:55:56 -0600, Christopher Thom wrote:
>
>> Hi all,
>>
>> I'm a long time where() fan, but trying to learn to wield this histogram
>> beast. I'm working on an algorithm, and would like a way to divide an
>> array of values into two bins, such that the sum of each bin is roughly
>> equal. The values have no fixed distribution, so I expect the bin sizes to
>> be non-uniform.
>>
>> This sort of problem seems an ideal place to start earning my histogram
>> badge, but I have to confess to only being able to think of
>> brute-force-type solutions. Any suggestions?
>
> Probably WHERE will serve you well:
>
> IDL> t=total(a,/CUMULATIVE)
> IDL> bin1=where(t lt t[n_elements(a)-1]/2,COMPLEMENT=bin2)
>
> Of course, there are many ways to divide values such that they fall into
> two roughly equal bins (n choose 2), some of which may be better than
> others.
>
> JD
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| Re: Histogram quickie [message #51935 is a reply to message #51793] |
Fri, 08 December 2006 14:38 |
JD Smith
Messages: 850
Registered: December 1999
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Senior Member |
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On Fri, 08 Dec 2006 12:55:56 -0600, Christopher Thom wrote:
> Hi all,
>
> I'm a long time where() fan, but trying to learn to wield this histogram
> beast. I'm working on an algorithm, and would like a way to divide an
> array of values into two bins, such that the sum of each bin is roughly
> equal. The values have no fixed distribution, so I expect the bin sizes to
> be non-uniform.
>
> This sort of problem seems an ideal place to start earning my histogram
> badge, but I have to confess to only being able to think of
> brute-force-type solutions. Any suggestions?
Probably WHERE will serve you well:
IDL> t=total(a,/CUMULATIVE)
IDL> bin1=where(t lt t[n_elements(a)-1]/2,COMPLEMENT=bin2)
Of course, there are many ways to divide values such that they fall into
two roughly equal bins (n choose 2), some of which may be better than
others.
JD
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