how to compare one element in an array with all the other elements in another array? And repeating it for n elements in the first array?

A = [ 1, 2,3,4,5]
B = [ 3,4,2,6,8,0]

Now , take 1st element (i.e = 1) in vector A and compare it with all elements in vector B. And after comparing it with all elements in B and now we need to consider the second element of A ( i.e 2) and do the same. Finish the loop once we iterate with all the elements in the first vector.

[Geoff Hayes]

Aswin - so once you compare 1 with all elements of B, then what? Are you looking to see if 1 is in B or what?

The easiest - and not the most efficient - is just to have two for loops. The outer loop would iterate over the elements of A and the inner loop would iterate over the elements of B

 for j=1:length(A)
    a = A(j);
    for k = 1:length(B)
       b = B(k);
       % do something
    end
 end

[dpb]

And do what with the result of the comparison? How best to do such an operation can depend drastically on what the end result is intended to be.

[Aswin Sandirakumaran]

I would like to compare whether the number in element A is lesser than the element in B, Iterate it until A element < B element. If it satisfies subtract the element A from B,and the obtained Answer should be substituted instead of B. If not we move to the 2nd element of vector B and start do the same.

[dpb]

"Iterate it until A element < B element."

And then? Look for more for the same element of A in B or when satisfy the < condition once move to the next?

"So many possibilities, so little defined..."

Given the input arrays, what is the correct final answer desired?

[Geoff Hayes]

Aswin - I think that you need to show what you expect with a small example.

[Aswin Sandirakumaran]

A=[1,2,3,4,5] B=[6,7,8] then the code should take the 1st element in A (ie. 1) and compare it with B, if ( the first element is smaller (A<B) do B - A and store that value in a new array. Then go for the 2nd Element in A (ie. 2) and compare it with 2nd element in array B ( 2< 7) therefore do B - A and store it in C[2]. and so on

A = [1,2,3,4,5] B = [6,7,8] % C should be the new result after comparing A[i] < B[i] C[i] = B[i] - A[i]; C = [5,5,5] Once done we are left with 4th,5th element in array A. if length of array A exceeds length of array B. Then the elements after length of A = length of array B should comparing from the index 1 of array C. But now it should compare with the new C element A = [4,5] % new A C = [5,5,5] % new C repeat the same process as above while comparing and get new array Final Result: D = [1,0,5]

Answer 1

[dpb]

A = [ 1, 2,3,4,5]
B = [ 3,4,2,6,8,0]

Taking a guess, and what was written literally,

>> for i=1:length(A)
     B(A(i)<B)=B(A(i)<B)-A(i);
   end
>> B
B =
   2     1     1     3     2     0
>>

Above is repetitive and sequential so the test is repeated on B after the prior elements have already done their duty; hence the second iteration and subsequent operate on an already-modified B.

Above could save a comparison if used a temporary logical vector as

for i=1:length(A)
  isLess=(A(i)<B);
  B(isLess)=B(isLess)-A(i);
end

which may also be a little easier to read...

[Aswin Sandirakumaran]

A=[1,2,3,4,5] B=[6,7,8] then the code should take the 1st element in A (ie. 1) and compare it with B, if ( the first element is smaller (A<B) do B - A and store that value in a new array. Then go for the 2nd Element in A (ie. 2) and compare it with 2nd element in array B ( 2< 7) therefore do B - A and store it in C[2]. and so on

A = [1,2,3,4,5] B = [6,7,8] % C should be the new result after comparing A[i] < B[i] C[i] = B[i] - A[i]; C = [5,5,5] Once done we are left with 4th,5th element in array A. if length of array A exceeds length of array B. Then the elements after length of A = length of array B should comparing from the index 1 of array C. But now it should compare with the new C element A = [4,5] % new A C = [5,5,5] % new C repeat the same process as above while comparing and get new array Final Result: D = [1,0,5]

[dpb]

Don't try to create "poof" new variables into the workplace programmatically; use cell arrays to hold the variably-sized results of each step.

Looks to me as though the above example fails in the second step, however, as there are only two elements left in A after the first step so which is <length(B) so length(D) == 2???