| interpolate [message #2125] |
Sat, 28 May 1994 02:54  |
vshvetsk
Messages: 13
Registered: March 1994
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Junior Member |
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I have a following question that kind of left me in a dark ( too late to think )
I have a matrix,let's say 3x3 and let's say I want to interpolate the middle
cell value bu using other cells: that is M22 is result of interpolating all
data around it: would I says:
result=Interpolate(([1,2,3],[1,2,3]),[2],[2])?
And what if I wanted just to do Linear Inerpolation by row only?
Could someone please help? Thank you
Victor Shvetsky
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| Re: INTERPOLATE [message #80399 is a reply to message #2125] |
Wed, 06 June 2012 03:21  |
Mats Löfdahl
Messages: 263
Registered: January 2012
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Senior Member |
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Den onsdagen den 6:e juni 2012 kl. 11:34:57 UTC+2 skrev Mats Löfdahl:
> Den onsdagen den 6:e juni 2012 kl. 10:42:40 UTC+2 skrev Laura:
>> Hi,
>>
>> I´ve got a temporal serie, temperatures, and I´ve got some 0's I would like to interpolate so that the serie is as complete as posible, with no blank spaces, e.g. no ceros. Is the Interpolate function the correct one to employ in this case?? I´m having problems understanding the location array, I thought I had only to look for the 0's in my serie, for example by
>>
>> index= where (data eq 0, count)
>>
>> and then apply the interpolation function, I know I´m missing something but I simply don´t get it.
>>
>> The example that is on help contents:
>>
>> p = FINDGEN(4,4)
>> PRINT, INTERPOLATE(p, [.5, 1.5, 2.5], [.5, 1.5, 2.5], /GRID)
>>
>> and prints the 3 by 3 array:
>>
>> 2.50000 3.50000 4.50000
>> 6.50000 7.50000 8.50000
>> 10.5000 11.5000 12.5000
>>
>> corresponding to the locations:
>>
>> (.5,.5), (1.5, .5), (2.5, .5),
>> (.5,1.5), (1.5, 1.5), (2.5, 1.5),
>> (.5,2.5), (1.5, 2.5), (2.5, 2.5)
>>
>> what I know is that
>>
>> p[0,0]= 0.0000 or p[0,3]=12
>>
>> but what I don´t get is why
>>
>> p[0.5,0.5]=0.00000
>>
>> or
>>
>> print, p[0.9,0.9]=0.0000
>> Does that mean that p[0,0],p[0.5,0.5], and p[0.9,0.9] are equivalent??? How do I construct the location array?
>>
>> Thank you
>>
>> Laura
>
> Yes, p[0,0], p[0.5,0.5], and p[0.9,0.9] are equivalent because floating point indexes are turned into integers. So you are accessing the same element in the p array.
>
> As for your own example, interpol might be easier to use than interpolate. Try this:
>
> data=[1.,2.,3.,4.,0.,6.,7.,0.,9.]
> data2=data
> indx0=where(data eq 0.)
> indx1=where(data ne 0.)
> data2[indx0]=interpol(data[indx1],indx1,indx0)
> print,data
> print,data2
>
>
> It will print out:
>
> 1.00000 2.00000 3.00000 4.00000 0.00000 6.00000
> 7.00000 0.00000 9.00000
>
> and
>
> 1.00000 2.00000 3.00000 4.00000 5.00000 6.00000
> 7.00000 8.00000 9.00000
>
> which I believe is what you wanted?
Sorry, you said you have a temporal sequence, so you have a time coordinate as well. Try this instead:
data=[1.,2.,3.,4.,0.,6.,7.,0.,9.]
t=[.4,.5,.6,.7,.8,.9,1.0,1.1,1.2]
data2=data
indx0=where(data eq 0.)
indx1=where(data ne 0.)
data2[indx0]=interpol(data[indx1],t[indx1],t[indx0])
plot,t,data,psym=4,yrange=[-.1,10]
oplot,t,data2,psym=1
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| Re: INTERPOLATE [message #80400 is a reply to message #2125] |
Wed, 06 June 2012 02:34 |
Mats Löfdahl
Messages: 263
Registered: January 2012
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Senior Member |
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Den onsdagen den 6:e juni 2012 kl. 10:42:40 UTC+2 skrev Laura:
> Hi,
>
> I´ve got a temporal serie, temperatures, and I´ve got some 0's I would like to interpolate so that the serie is as complete as posible, with no blank spaces, e.g. no ceros. Is the Interpolate function the correct one to employ in this case?? I´m having problems understanding the location array, I thought I had only to look for the 0's in my serie, for example by
>
> index= where (data eq 0, count)
>
> and then apply the interpolation function, I know I´m missing something but I simply don´t get it.
>
> The example that is on help contents:
>
> p = FINDGEN(4,4)
> PRINT, INTERPOLATE(p, [.5, 1.5, 2.5], [.5, 1.5, 2.5], /GRID)
>
> and prints the 3 by 3 array:
>
> 2.50000 3.50000 4.50000
> 6.50000 7.50000 8.50000
> 10.5000 11.5000 12.5000
>
> corresponding to the locations:
>
> (.5,.5), (1.5, .5), (2.5, .5),
> (.5,1.5), (1.5, 1.5), (2.5, 1.5),
> (.5,2.5), (1.5, 2.5), (2.5, 2.5)
>
> what I know is that
>
> p[0,0]= 0.0000 or p[0,3]=12
>
> but what I don´t get is why
>
> p[0.5,0.5]=0.00000
>
> or
>
> print, p[0.9,0.9]=0.0000
> Does that mean that p[0,0],p[0.5,0.5], and p[0.9,0.9] are equivalent??? How do I construct the location array?
>
> Thank you
>
> Laura
Yes, p[0,0], p[0.5,0.5], and p[0.9,0.9] are equivalent because floating point indexes are turned into integers. So you are accessing the same element in the p array.
As for your own example, interpol might be easier to use than interpolate. Try this:
data=[1.,2.,3.,4.,0.,6.,7.,0.,9.]
data2=data
indx0=where(data eq 0.)
indx1=where(data ne 0.)
data2[indx0]=interpol(data[indx1],indx1,indx0)
print,data
print,data2
It will print out:
1.00000 2.00000 3.00000 4.00000 0.00000 6.00000
7.00000 0.00000 9.00000
and
1.00000 2.00000 3.00000 4.00000 5.00000 6.00000
7.00000 8.00000 9.00000
which I believe is what you wanted?
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