Re: [ferret_users] How to do @rsum and @iin in reverse?

[Ingo Bethke <Ingo.Bethke@xxxxxxxxxxxxxx>]

Hi William,

At the moment I can't see how your algorithm can work (e.g. why is it 
possible to reproduce the second part of an array from the first part 
and it's sum?)

Alternatively, you can first flip the array using SAMPLEI and then 
compute @rsum or @iin, e.g.

yes? let test={1,2,3,4,5,6}
yes? let ind_rev=1+`test,return=iend`-x[x=1:`test,return=iend`]
yes? let test_rev=samplei(test,ind_rev)
yes? list test_rev[i=@rsum]

Cheers, Ingo

William S. Kessler wrote:
> This may be a really stupid question. How does one perform @rsum or 
> @iin in reverse (i.e. sum from the high-index end of the list)?
>
> The answer I would have given before I actually looked at it in detail 
> would have been: subtract the running sum or integral from the 
> definite sum over the whole range. And indeed I have done this many 
> times. But the result is not quite as expected, as the following 
> example shows:
>
> yes? let test = {1,2,3,4,5,6}
> yes? list test, test[i=@rsum], test[i=@sum] - test[i=@rsum]
> X: 0.5 to 6.5
> Column 1: TEST is {1,2,3,4,5,6}
> Column 2: TEST[X=@RSU] is {1,2,3,4,5,6} (running sum on X)
> Column 3: EX#3 is TEST[I=@SUM]-TEST[I=@RSUM]
> TEST TEST EX#3
> 1 / 1: 1.000 1.00 20.00
> 2 / 2: 2.000 3.00 18.00
> 3 / 3: 3.000 6.00 15.00
> 4 / 4: 4.000 10.00 11.00
> 5 / 5: 5.000 15.00 6.00
> 6 / 6: 6.000 21.00 0.00
>
> The second column does what we want in a running sum: the first value 
> is the value of the function over interval 1. The second value is the 
> sum of the function at intervals 1+2. The last value is the sum over 
> the whole range, as it should be. But I had hoped and expected that 
> the third column would be the reverse: The last value (i=6) should be 
> the last value of the function (6), the second-to-last should be the 
> sum of the two last values (6+5=11), and the first value should be the 
> same sum as if the operation were done the other way (21). But it 
> isn't. The whole thing is shifted by 1 interval, the value in position 
> 6 is 0, and the value in position 1 is not the whole sum.
>
> Am I missing something obvious???? What is the proper way to take a 
> running sum or integral in reverse? It should be possible to get the 
> same qualities of the answer doing the operation in either direction.
>
> Billy K