sort 2 matrices for minimum numbers sum and divide them

I have r = [ 1 3 4 5 ......n]

and x = [ 5 8 9 4 ......n]

I will like to do this computation

when m = 1

sort the lowest number of r and divide it with the lowest number of x till the end n

for example =[ 1/4, 3/5, 4/8, 5/9.......n]

when m = 2

= [(1 + 3) / (5 + 4) , (4 +5)/(8+9) , ...........n]

when m =3

=[ (1 + 3 + 4)/( 4 + 5 + 8),................n]

Thanks for your help in advance

Tino

[Stephen23]

And what should be the last term if n is not exactly divisble by m ?

[Tino]

Hi Stephen

if the last term is not divisible it is ignored. That is to say the computation stops when it is not divisible by mHope to hear from you soonest

Tino

Answer 1

[Stephen23]

This is MATLAB, so don't waste your time writing inefficient loops.

Learn to write simpler vectorized code, just like experienced MATLAB users do:

>> r = sort([1,3,4,5,10]); % sorted!
>> x = sort([5,8,9,4,10]); % sorted!

>> m = 1;
>> n = m*fix(numel(r)/m);
>> sum(reshape(r(1:n),m,[]),1) ./ sum(reshape(x(1:n),m,[]),1)
ans =
   0.25000   0.60000   0.50000   0.55556   1.00000

>> m = 2;
>> n = m*fix(numel(r)/m);
>> sum(reshape(r(1:n),m,[]),1) ./ sum(reshape(x(1:n),m,[]),1)
ans =
   0.44444   0.52941

>> m = 3;
>> n = m*fix(numel(r)/m);
>> sum(reshape(r(1:n),m,[]),1) ./ sum(reshape(x(1:n),m,[]),1)
ans =
   0.47059

[Tino]

Brilliant Stephen

Just one more help

if I have a data with various n columns and wishes to partition those column into 2 to perform this computation how do I go about it

for instance

1

3

4

5

6

7

8

9

2

6

and wish to divide this one column into halfs (r and x) to carry out the computation

How can I do this intitial code

Thanks again

Tino

[Andrei Bobrov]

@Tino: See code after word 'add' in my answer.

[Stephen23]

"wish to divide this one column into halfs (r and x) to carry out the computation"

Most likely you would use indexing. But your example is not very clear (e.g. you mention "I have a data with various n columns" and then show data that only has one column) and you did not provide any example output data, so it is not clear exactly how you want that data "divided" up.

Please show the expected output data for your example column of values.

[Tino]

for instance when I get a data input which can contain several columns ( nth column) for instance let us 3 column data

1 4 1

2 1 3

3 1 2

4 1 1

5 1 2

6 2 3

each data column is divided by 2 to get

r and x

for the computation above

for column one

r = [ 1 2 3]

x = [4 5 6]

computation

for column two

r = [4 1 1]

x = [1 1 2]

computation

for column three

r = [1 3 2]

x = [1 2 3]

computation

I hope this is clear

Thanks

[Stephen23]

@Tino: it seems that you want to loop over the columns, e.g. where M is your matrix:

A = [1,4,1;2,1,3;3,1,2;4,1,1;5,1,2;6,2,3];
S = size(A);
m = 1;
n = m*fix(S(1)/2/m);
for k = 1:S(2)
    r = A(1:n,k);
    x = A(1+n:2*n,k);
    Z = sum(reshape(r,m,[]),1) ./ sum(reshape(x,m,[]),1)
end

Giving:

Z =
         0.25          0.4          0.5
Z =
            4            1          0.5
Z =
            1          1.5      0.66667

[Tino]

Thanks for your help Stephen

How do I make it

0.25 4 1

0.4 1 1.5

0.5 0.5 0.66667

You ve been a big help

Hoping to hear from you soon

[Stephen23]

>> S = size(A);
>> m = 1;
>> n = m*fix(S(1)/2/m);
>> r = A(1:n,:)
>> x = A(1+n:2*n,:)
>> Z = sum(reshape(r,m,[]),1) ./ sum(reshape(x,m,[]),1);
>> Z = reshape(Z,[],S(2)).'
Z =
   0.25000   0.40000   0.50000
   4.00000   1.00000   0.50000
   1.00000   1.50000   0.66667

Or in one fell swoop:

>> Z = sum(permute(reshape(r,m,[],S(2)),[3,2,1]),3) ./ ...
       sum(permute(reshape(x,m,[],S(2)),[3,2,1]),3)
Z =
   0.25000   0.40000   0.50000
   4.00000   1.00000   0.50000
   1.00000   1.50000   0.66667

Answer 2

[Raj]

I am assuming n is a multiple of m. In that case this works:

r=sort(r);
x=sort(x);
m=input('enter value of m:');
if m==1
   for ii=1:numel(r)
       Answer(1,ii)=r(1,ii)/x(1,ii);
   end
   Answer
elseif m==2
       Answer(1,1)=(r(1,1)+r(1,2))/(x(1,1)+x(1,2));
    for ii=3:2:numel(r)  
            Answer(1,(ii+1)/2)=(r(1,ii)+r(1,ii+1))/(x(1,ii)+x(1,ii+1));        
    end
    Answer
elseif m==3
       Answer(1,1)=(r(1,1)+r(1,2)+r(1,3))/(x(1,1)+x(1,2)+x(1,3));
       n=0;
    for ii=4:3:numel(r)  
            Answer(1,(ii-n)/2)=(r(1,ii)+r(1,ii+1)+r(1,ii+2))/(x(1,ii)+x(1,ii+1)+x(1,ii+2)); 
            n=n+1;
    end
    Answer
else
    disp('Invalid Value of m')
end

There may be better and optimized way of doing this also.

P.S: How about adding a 'Homework' tag next time and showing what you attempted in addition to 'Hope to hear from you soonest'? Everybody wil not be as free as i am today! Cheers!!

Answer 3

[Pullak Barik]

Hi!

I assume that r and x are just vectors, and not matrices with more than one dimension.

I guess the following function will work for you-

function result = sort_sum_and_divide(r, x, m)
    r = sort(r);
    x = sort(x);
    n = length(r);
    if(length(r) == length(x)) %To check if the input is wrong
        i = 1:m:n;
        if(i(end) + m - 1 > n)
            i(end) = []; %To drop the elements at the end if they can not be grouped
        end
        result = zeros(1, length(i)); %preallocation of result array
        for idx = 1:length(i)
            result(idx) = sum(r(i(idx):i(idx)+m-1))./sum(x(i(idx):i(idx)+m-1));
        end
    else
        disp('Length of r and x are not the same');
    end
end

[Stephen23]

@Pullak Barik: note that concatenation onto the result array like that is not considered good practice, and detrimentally affects efficiency:

https://www.mathworks.com/help/matlab/matlab_prog/techniques-for-improving-performance.html

The MATLAB documentation recommends preallocating arrays before the loop:

https://www.mathworks.com/help/matlab/matlab_prog/preallocating-arrays.html

[Pullak Barik]

Ya, I could preallocate my result array. Thanks.

Answer 4

[Andrei Bobrov]

m = 3;
r = [ 1 3 4 5 30];
x = [ 5 8 9 4 78];
out = funt(r,x,3);

function out = funt(r,x,m)
    ad = nan(mod(-numel(r),m),1);
    a = sort(cat(3,[r(:);ad],[x(:);ad]));
    b = sum(reshape(a,m,[],2),'omitnan');
    out = b(:,:,1)./b(:,:,2);
end

add

rx = [1
3
4
5
6
7
8
9
2
6];

m = 2;
rrxx = sum(sort(reshape([reshape(rx,[],2);nan(mod(-numel(rx)/2,m),2)],...
                                                      m,[],2)),'omitnan');
out = rrxx(:,:,1)./rrxx(:,:,2);

[Tino]

Thanks Andrei Just one more help

if I have a data with various n columns and wishes to partition those column into 2 to perform this computation how do I go about it

for instance

1

3

4

5

6

7

8

9

2

6

and wish to divide this one column into halfs (r and x) to carry out the computation accross several columns

How can I do this intitial code

Thanks again

Tino

[Tino]

Hi Andrei

Your code is givng me the right answer

I have r = [ 1 3 4 5 ......n]

and x = [ 5 8 9 4 ......n]

I will like to do this computation

when m = 1

sort the lowest number of r and divide it with the lowest number of x till the end n

for example =[ 1/4, 3/5, 4/8, 5/9.......n]

when m = 2

= [(1 + 3) / (5 + 4) , (4 +5)/(8+9) , ...........n]

when m =3

=[ (1 + 3 + 4)/( 4 + 5 + 8),................n]

Thanks for your help in advance

Tino