[R] Simplify iterative programming

[Dimitris Rizopoulos]

you could consider something like:

H <- sum(sqrt(outer(A, B, "-")^2 + outer(C, D, "-")^2))


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: "Stefaan Lhermitte" <stefaan.lhermitte at biw.kuleuven.be>
To: <r-help at stat.math.ethz.ch>
Sent: Friday, November 04, 2005 10:28 AM
Subject: [R] Simplify iterative programming


> Dear,
>
> I am looking for the simplification of a formula to improve the
> calculation speed of my program. Therefore I want to simplify the
> following formula:
>
> H = sum{i=0..n-1 ,  [ sum {j=0..m-1 ,  sqrt ( (Ai - Bj)^2 + (Ci -
> Dj)^2) }  ]  }
>
> where:
> A, C = two vectors (with numerical data) of length n
> B, D = two vectors (with numerical data) of length m
> sqrt = square root
> Ai = element of A with index i
> Bj = element of B with index j
> Ci = element of C with index i
> Dj = element of C with index j
>
> I am calculating H in a merging process so A, B will merge in step 2
> into A' and C, D into C':
> A' = {A,B} : vector of length (n+m)
> C' = {C,D} : vector of length (n+m)
>
> Then again I will calculate H with two new vectors X and Y (of 
> length
> p):
> H = sum{i=0..n+m-1 ,  [ sum {j=0..p-1 ,  sqrt ( (A'i - Xj)^2 + 
> (C'i -
> Yj)^2) }  ]  }
>
> These steps are iterated in a loop with always new vectors (e.g. X 
> and
> Y)
>
> Now I'am looking for a simplication of H in order to avoid long
> calculation time.
> I know a computional simplified formula exists for the standard
> deviation (sd) that is much easier in iterative programming. 
> Therefore
> I wondered I anybody knew about analog simplifications to simplify 
> H:
>
> sd = sqrt [ sum{i=0..n-1, (Xi - mean(X) )^2 ) /n } ]  -> simplified
> computation
> -> sqrt [  (n *  sum{i=0..n-1,  X^2 } ) - ( sum{i=0..n-1,  X } ^2 ) 
> /
> n^2 ]
>
> This simplied formula is much easier in iterative programming, since 
> I
> don't have to keep every element of X .
>
> For example if we want to calculate sd of A' with the vectors A and 
> C:
> sd(A')
> = sqrt [  ((n+m) *  sum{i=0..n+m-1,  A'^2 } ) - ( sum{i=0..n+m-1, 
> A' }
> ^2 ) /  (n+m)^2 ]
> = sqrt [  ((n+m)*  (sum{i=0..n,  A^2 } + sum{i=0..m,  C^2 } ) )
>     - ( ( sum{i=0..n-1,  A } + sum{i=0..m-1,  C } )^2 ) /  (n+m)^2 ]
>
> The advantage of this formula is, that I don't have to keep every 
> value
> of A and C to calculate sd(A'). I can do the following replacements:
> sum{i=0..n,  A^2 } = A2
> sum{i=0..m,  C^2 } = C2
> sum{i=0..n-1,  A } = A3
> sum{i=0..m-1,  C } = C3
>
> So sd(A')=
>  sqrt [  ( (n+m)*(A2+ C2) )  - ( (A3 + C3)^2 ) /  (n+m)^2 ]
>
> In this way my computation intensive calculation is replaced by a
> calculation of simple numbers.
>
> Can anybody help me to do something comparable for H? Any other help 
> to
> calculate H easily in an iterative process is also welcome!
>
> Kind regards,
> Stef
>
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