[R] Simplify iterative programming

[Kristel Joossens]

Hello Stefaan,

I'm not an expert, but maybe something like this is quite 
straightforward withourtusing for-loops? (It is an idea I wrote down, 
but check of course if this is correct!)

term1 =(matrix(A,ncol=m,nrow=n)-matrix(B,ncol=m,nrow=n,byrow=TRUE))^2
term2 =(matrix(C,ncol=m,nrow=n)-matrix(D,ncol=m,nrow=n,byrow=TRUE))^2
H = sum(sqrt(term1+term2))

Kind regards,
Kristel

Stefaan Lhermitte wrote:
> 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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-- 

_________________________________________________________________
Kristel Joossens
Ph.D. Student
Faculty of Economics and Applied Economics
Research center ORSTAT
K.U. Leuven
Naamsestraat 69
B-3000 Leuven
Belgium
Tel: +32 16 326929
Fax: +32 16 326732
E-mail: Kristel.Joossens at econ.kuleuven.be
Url: http://www.econ.kuleuven.ac.be/Kristel.Joossens/public/

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