| WHERE problems (longish) [message #35914] |
Tue, 22 July 2003 08:23  |
Benjamin Panter
Messages: 6
Registered: August 2002
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Junior Member |
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Hiya,
This is puzzling me, and I've been through all that I can think of. I
have a look up table called "dust_lookup". It is a 2 x 300ish array and
has wavelengths and the corresponding correction factor. I need to pluck
a few values out, so I'm using where:
print, where(2900. eq reform(dust_lookup[*,0]))
which works absolutly fine for most values: unfortunatly not for all:
If I write a little test:
PRO tester, dust_lookup
print, where(2900. eq reform(dust_lookup[*,0]))
print, where(2920. eq reform(dust_lookup[*,0]))
print, where(2940. eq reform(dust_lookup[*,0]))
print, where(2960. eq reform(dust_lookup[*,0]))
print, where(2980. eq reform(dust_lookup[*,0]))
print, where(3000. eq reform(dust_lookup[*,0]))
print, where(3020. eq reform(dust_lookup[*,0]))
print, where(3040. eq reform(dust_lookup[*,0]))
print, where(3060. eq reform(dust_lookup[*,0]))
print, where(3080. eq reform(dust_lookup[*,0]))
print, where(3100. eq reform(dust_lookup[*,0]))
END
it comes out with
IDL> tester, dust_lookup
10
11
12
13
-1
-1
-1
17
18
19
20
The values which have -1 certainly exist - and were generated in exactly
the same way as the others. I've put the array online if anyone fancies
looking at it - http://www.roe.ac.uk/~bdp/where_problem.idl
Am I being stupid again? What is special about 2980,3000 and 3020??
Cheers,
Ben
--
Ben Panter, Edinburgh
My name (no spaces)@bigfoot which is a com.
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| Re: Where problem [message #69607 is a reply to message #35914] |
Fri, 05 February 2010 04:01  |
d.poreh
Messages: 406
Registered: October 2007
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Senior Member |
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On Feb 4, 8:20 am, JJ <j...@cornell.edu> wrote:
> The value_locate() solution may work for you, but it will really only
> work in pretty specific cases. The X vector must be monotonically
> increasing (though you could first sort the list). Even if the values
> you're looking for are not in X, a valid result will be returned -
> since value_locate() returns the index of the start of the "interval
> into which the given value falls" (this issue addressed by wox).
>
> If you want a simple and robust solution to this kind of problem (and
> many others), I highly recommend Craig Markwardt's cmset_op routine
> (follow the Array/Set link from the Markwardt IDL Library page athttp://www.physics.wisc.edu/~craigm/idl/idl.html).
>
> With cmset_op you would do something like:
>
> index = cmset_op (x, 'and', y, /index)
>
> Of course, if there are multiple entries in your array that have the
> same value, and you want to find the indices of all of those
> locations, you might want to do a where on each element of y.
>
> -JJ
Thanks JJ.
Cheers
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| Re: Where problem [message #69612 is a reply to message #35914] |
Thu, 04 February 2010 08:20 |
JJ
Messages: 36
Registered: January 2007
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Member |
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The value_locate() solution may work for you, but it will really only
work in pretty specific cases. The X vector must be monotonically
increasing (though you could first sort the list). Even if the values
you're looking for are not in X, a valid result will be returned -
since value_locate() returns the index of the start of the "interval
into which the given value falls" (this issue addressed by wox).
If you want a simple and robust solution to this kind of problem (and
many others), I highly recommend Craig Markwardt's cmset_op routine
(follow the Array/Set link from the Markwardt IDL Library page at
http://www.physics.wisc.edu/~craigm/idl/idl.html).
With cmset_op you would do something like:
index = cmset_op (x, 'and', y, /index)
Of course, if there are multiple entries in your array that have the
same value, and you want to find the indices of all of those
locations, you might want to do a where on each element of y.
-JJ
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| Re: Where problem [message #69617 is a reply to message #35914] |
Thu, 04 February 2010 02:30 |
d.poreh
Messages: 406
Registered: October 2007
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Senior Member |
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On Feb 4, 2:22 am, Wox <s...@nomail.com> wrote:
> On Thu, 4 Feb 2010 01:27:26 -0800 (PST), Dave_Poreh
>
> <d.po...@gmail.com> wrote:
>> Folks
>> I can’t solve this problem. Will somebody tells me what is going on?
>> x=findgen(100)
>> y=[30,40,50,80]*1.0
>> index=where(x eq y)
>> but every time gives me:
>>> index=-1
>> Any help highly appreciated
>> Cheers
>> Dave
>
> Two issues:
>
> 1. wrong use of where:
>
> IDL> x=findgen(100)
> IDL> y=[30,40,50,80]
> IDL> b=x eq y
>
> b will have 4 elements (smallest of the x and y dimension) and will
> b[i] will only be 1 when x[i] eq y[i]. So this is not what you want.
>
> What you can do is:
> ind=value_locate(x,y)
> ind2=where(x[ind] ne y,ct)
> if ct ne 0 then ind[ind2]=-1
>
> 2. comparing floating point numbers, see:www.dfanning.com/code_tips/comparearray.html
>
> IDL> x=findgen(100)
> IDL> y=[30,40,50,80]*1.0
>
> I would do something like
>
> small=1e-6
> ind=value_locate(x,y)
> ind2=where(abs(x[ind] - y) gt small,ct)
> if ct ne 0 then begin
> ind[ind2]++
> ind2=where(abs(x[ind] - y) gt small,ct)
> if ct ne 0 then ind[ind2]=-1
> endif
>
> See David's page on what "small" should be.
>
> You could also do something like this
>
> x=rebin(x,n_elements(x),n_elements(y),/sample)
> y=rebin(transpose(y),n_elements(x),n_elements(y),/sample)
> ind=where(abs(x - y) gt small, ct)
> ... and so on ... which is ok for small arrays but not for large
> arrays (memory issues)
Thanks Guys. This is exactly what I want.
Cheers
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| Re: Where problem [message #69669 is a reply to message #35914] |
Thu, 04 February 2010 02:22 |
Wout De Nolf
Messages: 194
Registered: October 2008
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Senior Member |
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On Thu, 4 Feb 2010 01:27:26 -0800 (PST), Dave_Poreh
<d.poreh@gmail.com> wrote:
> Folks
> I can�t solve this problem. Will somebody tells me what is going on?
> x=findgen(100)
> y=[30,40,50,80]*1.0
> index=where(x eq y)
> but every time gives me:
>> index=-1
> Any help highly appreciated
> Cheers
> Dave
Two issues:
1. wrong use of where:
IDL> x=findgen(100)
IDL> y=[30,40,50,80]
IDL> b=x eq y
b will have 4 elements (smallest of the x and y dimension) and will
b[i] will only be 1 when x[i] eq y[i]. So this is not what you want.
What you can do is:
ind=value_locate(x,y)
ind2=where(x[ind] ne y,ct)
if ct ne 0 then ind[ind2]=-1
2. comparing floating point numbers, see:
www.dfanning.com/code_tips/comparearray.html
IDL> x=findgen(100)
IDL> y=[30,40,50,80]*1.0
I would do something like
small=1e-6
ind=value_locate(x,y)
ind2=where(abs(x[ind] - y) gt small,ct)
if ct ne 0 then begin
ind[ind2]++
ind2=where(abs(x[ind] - y) gt small,ct)
if ct ne 0 then ind[ind2]=-1
endif
See David's page on what "small" should be.
You could also do something like this
x=rebin(x,n_elements(x),n_elements(y),/sample)
y=rebin(transpose(y),n_elements(x),n_elements(y),/sample)
ind=where(abs(x - y) gt small, ct)
... and so on ... which is ok for small arrays but not for large
arrays (memory issues)
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| Re: Where problem [message #69670 is a reply to message #35914] |
Thu, 04 February 2010 01:59 |
Foldy Lajos
Messages: 268
Registered: October 2001
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Senior Member |
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On Thu, 4 Feb 2010, Dave_Poreh wrote:
> On Feb 4, 1:33 am, FÖLDY Lajos <fo...@rmki.kfki.hu> wrote:
>> On Thu, 4 Feb 2010, Dave_Poreh wrote:
>>> Folks
>>> I can?t solve this problem. Will somebody tells me what is going on?
>>> x=findgen(100)
>>> y=[30,40,50,80]*1.0
>>> index=where(x eq y)
>>> but every time gives me:
>>>> index=-1
>>> Any help highly appreciated
>>> Cheers
>>> Dave
>>
>> Try 'help, x eq y' and 'print, x eq y'.
>>
>> ('x eq y' is equivalent to '[0.,1.,2.,3.] eq [30.,40.,50.,80.]')
>>
>> regards,
>> lajos
>
> Thanks. What I am trying to do is to extract index of 30,40,50, and
> 80 in the array of x. I mean
> X[index]=30, X[index]=40, and so on.
> What I have to do?
> Cheers
>
IDL> print, value_locate(x,y)
30 40 50 80
regards,
lajos
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| Re: Where problem [message #69671 is a reply to message #35914] |
Thu, 04 February 2010 01:47 |
d.poreh
Messages: 406
Registered: October 2007
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Senior Member |
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On Feb 4, 1:33 am, FÖLDY Lajos <fo...@rmki.kfki.hu> wrote:
> On Thu, 4 Feb 2010, Dave_Poreh wrote:
>> Folks
>> I can?t solve this problem. Will somebody tells me what is going on?
>> x=findgen(100)
>> y=[30,40,50,80]*1.0
>> index=where(x eq y)
>> but every time gives me:
>>> index=-1
>> Any help highly appreciated
>> Cheers
>> Dave
>
> Try 'help, x eq y' and 'print, x eq y'.
>
> ('x eq y' is equivalent to '[0.,1.,2.,3.] eq [30.,40.,50.,80.]')
>
> regards,
> lajos
Thanks. What I am trying to do is to extract index of 30,40,50, and
80 in the array of x. I mean
X[index]=30, X[index]=40, and so on.
What I have to do?
Cheers
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| Re: Where problem [message #69672 is a reply to message #35914] |
Thu, 04 February 2010 01:33 |
Foldy Lajos
Messages: 268
Registered: October 2001
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Senior Member |
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On Thu, 4 Feb 2010, Dave_Poreh wrote:
> Folks
> I can?t solve this problem. Will somebody tells me what is going on?
> x=findgen(100)
> y=[30,40,50,80]*1.0
> index=where(x eq y)
> but every time gives me:
>> index=-1
> Any help highly appreciated
> Cheers
> Dave
>
Try 'help, x eq y' and 'print, x eq y'.
('x eq y' is equivalent to '[0.,1.,2.,3.] eq [30.,40.,50.,80.]')
regards,
lajos
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