| Re: Removing if then else loop for efficiency [message #69275] |
Tue, 12 January 2010 08:04  |
penteado
Messages: 866
Registered: February 2018
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Senior Member
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On Jan 12, 11:09 am, Tom Ashbee <tlash...@googlemail.com> wrote:
> On Jan 10, 6:25 pm, pp <pp.pente...@gmail.com> wrote:
>
>
>
>> I think this does the same as your code, but it does not use any
>> loops, and it should be much faster and easier to read:
>
>> function velocities,t,xyvec
>> compile_opt idl2
>> ;Constants
>> N=50
>> R=5d0
>> ;Unpack the input
>> xvec=xyvec[0:N-1]
>> yvec=xyvec[N:(N*2)-1]
>> ;Temporary arrays for x[j], x[i], y[j], y[i]
>> xj=rebin(xvec,N,N)
>> xi=transpose(xj)
>> yj=rebin(yvec,N,N)
>> yi=transpose(yj)
>> ;Repeated terms in the expressions
>> tmp1=(xi-xj)^2+(yi-yj)^2
>> tmp2=R^2/(xj^2+yj^2)
>> tmp3=(xi-xj*tmp2)^2+(yi-yj*tmp2)^2
>> ;Terms of dxdt,dydt present everywhere
>> dxdt=-(yi-yj*tmp2)/tmp3
>> dydt=(xi-xj*tmp2)/tmp3
>> ;Terms present only out of the diagonal
>> tmp4=1d0-identity(N) ;this is 0 in the diagonal, 1 out of it
>> dxdt+=((yj-yi)/tmp1)*tmp4
>> dydt-=((xi-xj)/tmp1)*tmp4
>> ;Put the gamma factor
>> gamm=rebin(gamma(N,2.0d,10.0d)/(2d0*!dpi),N,N) ;this does not seem to
>> be IDL's gamma function
>> dxdt*=gamm
>> dydt*=gamm
>> ;Sum over the rows
>> dxvecdt=total(dxdt,1)
>> dyvecdt=total(dydt,1)
>> ;Pack the results
>> z=[dxvecdt,dyvecdt]
>> return,z
>> end
>
>> You should check that I did not misidentify anything, which would not
>> have been difficult in such convoluted expressions.
>
>> Other points to note:
>
>> 1) Do not use () for array indexes. Use [] instead. That makes it
>> unambiguous that it is an array index, and not a function call.
>
>> 2) When using doubles, as you did, use !dpi instead of !pi.
>
>> 3) Your function has an argument t that is not used anywhere in it. I
>> left it there, so that the argument order does not change.
>
> Hi,
>
> thanks a lot for this; it was very insightful and helpful.
> Unfortunately it's just giving NaNs for z at the moment but I'm
> working on debugging it.
Now that you mention it, I see a reason for the NaNs. The lines
tmp4=1d0-identity(N) ;this is 0 in the diagonal, 1 out of it
dxdt+=((yj-yi)/tmp1)*tmp4
dydt-=((xi-xj)/tmp1)*tmp4
were intended to add to dxdt only in the off-diagonal elements, by
multiplying the diagonal elements of ((yj-yi)/tmp1) by 0 (same for
dydt). But these diagonal elements are some non finite value (some
form of NaN or Infinity), so their product with 0 is not 0, it is some
NaN.
One way to get around this is to replace those 7 lines with:
;Terms of dxdt,dydt present only out of the diagonal
dxdt=((yj-yi)/tmp1)
dydt=-((xi-xj)/tmp1)
;Reset to 0 the diagonal elements
dxdt[0:N*N-1:N+1]=0d0
dydt[0:N*N-1:N+1]=0d0
;Terms of dxdt,dydt present everywhere
dxdt-=(yi-yj*tmp2)/tmp3
dydt+=(xi-xj*tmp2)/tmp3
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