[R] Simplify iterative programming
Stefaan Lhermitte
stefaan.lhermitte at biw.kuleuven.be
Fri Nov 4 10:28:23 CET 2005
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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