I can not return real values

조회 수: 4(최근 30일)
huda nawaf
huda nawaf 2012년 9월 14일
Hi I have the result of recursive function is saved in txt file , Below are the results of calling recursive fun. that each call divide the vector into two vectors. the problem is when the function divide the vector , give the new vectors the indices of values in original vectors not the same values.So, I lose the real numbers
for ex. this txtfile
*3 4 5 8 9 10
1 2 3
4 5 6
3
1 2
1 2
3 *
I want code by which can return these vectors
*cluster(1)=[3 4 5 8 9 10]
cluster(2)=[3 4 5]
cluster(3)=[8 9 10]
cluster(4)=[5]
cluster(5)=[3 4]
cluster(6)=[8 9]
cluster(7)=[10]*
thanks in advance
  댓글 수: 2
huda nawaf
huda nawaf 2012년 9월 14일
the first set the returned values(the indices of values) of recursive function, and the second set it is what i need.
the first vector in first set (txtfile)is the input, and the remaining values are the returend indices for the vector after dividing it each call.I saved the input in txtfile too.
thanks

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답변(2개)

Jürgen
Jürgen 2012년 9월 14일
I am not sure if I get it , is not so clearly explained but I would work with the length of the OriginalVector of size(OriginalVector,2) and divide it by two ( you will need to use ceil or fix if you have an odd length of course
so in pseudo code
Length= length(V)
HalfLenght=ceil(Length/2)
V1=V(1:HalfLength) V1=V(HalfLength+1:Length)
repeat this until length result is one
or am I mistaken ?
  댓글 수: 5
Jürgen
Jürgen 2012년 9월 15일
Maybe it is because it is the weekend that I am slow but to be sure I'll try to check if I understand it: in your txt file you have a list of numbers that you read into matlab en put in a vector:x=[3 4 5 8 9 10]; then you divide the vector in parts being: [3 4 5] & [8 9 10]&[5]&[3 4]&[8 9]&10
what do you want to do next? write to the txt?

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Jürgen
Jürgen 2012년 9월 15일
편집: Jürgen 2012년 9월 15일
I think this code does what you want or no what I think that you want :-) Of course much nicer if you make a function of it X=[3 4 5 8 9 10];
L=length(X);
NewL=ceil(L/2);
Xnew1=X(1:NewL);
Xnew2=X(NewL+1:L);
X=Xnew1;
L=length(X);
NewL=ceil(L/2);
Xnew1=X(1:NewL);
Xnew1=X(NewL+1:L);
X=Xnew2;
L=length(X);
NewL=ceil(L/2);
Xnew1=X(1:NewL);
Xnew1=X(NewL+1:L);
  댓글 수: 7
huda nawaf
huda nawaf 2012년 9월 19일
hi , many thanks, I have solved the another problem.

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