Compare segments of a vector

Hello good day.

I have a problem, which I hope you can give me support or advice: I have the following column vector of 20 elements (the real one is 1000 elements): A=[6.9563; 6.9563; 6.9563; 6.9563; 6.9563; 7.4917; 5.8343; 5.3075; 5.8713; 7.1312; 11.2175; 11.2175; 11.2175; 11.2175; 11.2175; 10.4278; 10.9766; 11.2685; 9.4748; 11.3789] The question is how to compare (greater than>) the first 5 elements with the second 5 elements of vector A. 6.9563 6.9563 6.9563 6.9563 6.9563 Compare with 7.4917 5.8343 5.3075 5.8713 7.1312 and after third 5 elements with the fourth 5 elements of the vector A 11.2175 11.2175 11.2175 11.2175 11.2175 Compare with 10.4278 10.9766 11.2685 9.4748 11.3789 And so on automatically, if the vector A is 1000 elements

I hope you understand me and help me. Greetings.

[David Fletcher]

I'd probably extract every other 5 five elements from A into B compact A down and then just compare with A>B.

[Jos (10584)]

What should the output look like? Can you give a numeric example with integers (whole numbers) to make life easier.

[Ricardo Gutierrez]

A=[2 2 2 2 2 3 6 3 4 5 1 1 1 1 1 2 3 4 5 6] compare 2 2 2 2 2 with 3 6 3 4 5 and then compare 1 1 1 1 1 with 2 3 4 5 6 The expected result is B=[1 2 3 4 5 2 3 4 5 6] The second 5 values are greater than the first. the fourth 5 values are greater than the third. The real column vector have 600 values.

[Jos (10584)]

I fail see the relationship between the values of B and the values of A ...

[Ricardo Gutierrez]

You're right I put them back A=[2 2 2 2 2 3 6 3 4 5 1 1 1 1 1 2 3 4 5 6]........ 2 2 2 2 2 3 6 3 4 5 in the second 5 values; at least one value is greater than 2 therefore the values pass:3 6 3 4 5.....in the following values 1 1 1 1 1 2 3 4 5 6 the last 5 values at least one value is greater than 1 therefore the values pass:2 3 4 5 6....so it would be in vector B....B=[3 6 3 4 5 2 3 4 5 6 ]

Answer 1

[Fangjun Jiang]

What if none of the element in second 5 elements is greater than the first 5 elements? The code below might be easier to understand.

A=[2 2 2 2 2 3 6 3 4 5 1 1 1 1 1 0 0 0 0 0];
A=transpose(reshape(A,10,[]));
B=A(:,6:10);
A=A(:,1:5);
C=B>A;
ind=any(C,2);
B(~ind,:)=A(~ind,:)
B =

     3     6     3     4     5
     1     1     1     1     1

[Ricardo Gutierrez]

Excelent!!!!!

Answer 2

[David Fletcher]

ind=repmat(logical([0 0 0 0 0 1 1 1 1 1]),1,100)

b=a(ind)
a(ind)=[]
c=a>b

Answer 3

[Fangjun Jiang]

This example should give you an idea

A=transpose(reshape(rand(20,1),5,[]))
B=diff(A)
C=B>0

C should tell you the relation

Answer 4

[Jos (10584)]

A  = [2 2 2 2 2 3 6 3 4 5 1 1 1 1 1 2 3 4 5 6]
AA = reshape(A,5,[]) 
dA = diff(AA,[],2) 
ix = all(dA>0)
ix = 2*find(ix(1:2:end))
B = reshape(AA(:,ix),1,[])

[Ricardo Gutierrez]

Your code shows as a result:..B = 2 2 2 2 2 1 1 1 1 1...and must be:...B = 3 6 3 4 5 2 3 4 5 6....thank you very much for your work and time

[Jos (10584)]

Oh, I subtracted a 1 from ix by mistake! Corrected