subtracting matrices in a special way

a = randi(50,600,9);
mx = randi(50,60,1);
q = bsxfun(@minus,reshape(a(:,7),10,[]),mx);
q = reshape(q,[],1);

After reshaping and cutting the variable 'a' into 10 blocks I want the first mx value to be subtracted from the first block of ten, the second mx value from the second block of 10, the third mx value from the third block of 10 and so on.... How can I implement these changes in my bsxfun function?

thanks

[Azzi Abdelmalek]

What is the problem with your code?

[AA]

• bold * if mx is= 16311440404823233043 and the first ten rows of a in column 7 are 414674632514284849 then I want 16 to be subtracted from 414674632514284849

[AA]

31 should be subtracted from the second block of tens, 14 from the third block of tens, 40 from the fourth block of tens and so on.

[Azzi Abdelmalek]

This is not clear for me

[Stephen23]

After dividing every tenth a value, what happens after mx(10), when there are still 500 a values remaining but no mx values left? Ten blocks of ten makes 100 values, but a(:,7) contains 600... please advise us on what should happen with these remaining values!

[AA]

you are right. I forgot to mention the condition that the rows of variable a are always a multiple of mx/mn. Thus

a = randi(50,600,9);
mx = randi(50,60,1);
mn = randi(50,60,1);
I wish to subtract mx(1) from the first ten consecutive values of a. q(1:10) = a(1:10) - mx(1) q(11:20)=a(11:20)-mx(2) ... q(591:600)=a(591:600)-mx(60).

[Stephen23]

PS: Is there a purpose to creating a huge array a and then using only the seventh column from it?

Answer 1

[Stephen23]

Using the arrays that you specified in your most recent comment:

a = randi(50,600,9);
mx = randi(50,60,1);
mn = randi(50,60,1);

b = reshape(a(:,7),10,[]).';
q = bsxfun(@minus,b,mx);

%q = bsxfun(@rdivide,q,mx-mn);
%q = reshape(q.',[],1);

Answer 2

[Andrei Bobrov]

a = randi(50,600,9);
mx = randi(50,60,1);

a7 = reshape(a(:,7),10,[]);
out0 = bsxfun(@minus,a7,mx(:)');
out = out0(:);

or

out = a(:,7) - mx(ceil((1:size(a,1))/numel(mx)));

[Andrei Bobrov]

corrected