Re: IDL QUESTIONS (INTERPOLATION) [message #78364 is a reply to message #78362] |
Mon, 21 November 2011 08:13   |
Jeremy Bailin
Messages: 618 Registered: April 2008
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Senior Member |
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On 11/21/11 8:59 AM, zolile mtumela wrote:
> Hi all,
> I have interpolated some data, which I think I did well, the data
> looks like this
> time (s) velocity
> 24785555 -8.90000
> 24785675 -6.90000
> 24785795 -5.00000
> 24785915 -15.4000
> 24786035 -1.80000
> 24786155 21.5000
> 24786275 -2.70000
> 24786395 1326.90
> 24786515 0.100000
> 24786635 69.6000
> 24786755 -624.900
> 24786875 263.700
> 24786995 30.3000
> 24787115 157.000
> 24787235 488.300
> 24787355 25.6000
> 24787475 40.6000
> 24787595 32.5000
> 24787715 21.2000
> 24787835 13.6000
> 24787955 92.4000
> 24788075 37.0000
> 24788195 7.50000
> 24788315 4.80000
> 24788435 8.30000
> 24788555 43.4000
> 24788675 26.2000
> 24788795 36.7000
> 24788915 17.6000
> 24789035 -39.2000
> 24789155 -76.5000
> 24789275 -31.7000
> 24789395 -31.3000
> 24789515 -20.0000
> 24789635 1.10000
> I want to create time axis in hours, I need help on time convetion, I
> wrote a program like this
> File = Dialog_pickfile(Filter = '*.txt')
>
> ;read the data
> rows = File_lines(file)
> data = Fltarr(2,rows)
> openr,Lun,file,/Get_lun
> ReadF,lun, data
> b =floor(min(data[0,*])); start time
> e = ceil(max(data[0,*])); end time
> x = Scale_vector(Findgen(e-b+1),b,e)
> y = interpol(data[1,*], data[0,*],x); interpolating data
> ;draw the plots
> plot,data[0,*],data[1,*],linestyle=2 ; for original data
> oplot,x,y,color=120 ; newly interpolated velocity data
>
> end
> I want to covert that time in two hour in the plot.
>
> Thank you in advance for ur time
> Zolile
>
>
>
>
Divide by 3600.0?
-Jeremy.
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