Finding minimum of sum of columns

hello guys, thank you for all that you do here.

Please i need help on this code as i have been stuck on this for a while.

This is a simplified version of my code:

a = [2 9 2 8 9 0 1];
b = [3 1 6 2 9 2 5];
c = [0 4 8 7 4 8 3];
d = [a;b;c]
if d(1,1)==d(2,1)||d(1,1)==d(3,1)||d(2,1)==d(3,1)    %if at least two of column 1 are the same
return row d that has the minimum sum of columns 4 to 7 (insert code here)
if at least two of the sums of columns 4 to 7 are the same (insert code here);
return row d that has the maximum sum of columns 2 to 3 (insert code here);
if at least two of the sums of columns 2 to 3 are the same (insert code here);
n = d(1,:); %return the first row of d (that is [2 9 2 8 9 0 1])

I would be glad to give more details where needed. Thank you

Answer 1

[John BG]

Hi Chiamaka

this is John BG <mailto:[e-mail] [e-mail]>

input data

      a = [2 9 2 8 9 0 1]; 
      b = [3 1 6 2 9 2 5]; 
      c = [0 4 8 7 4 8 3]; 
      d = [a;b;c];

condition 1.-

if at least two of column 1 are the same

 % if d(1,1)==d(2,1)||d(1,1)==d(3,1)||d(2,1)==d(3,1) 

   if numel(d(:,1))<=numel(unique(d(:,1)))-1

..

    end

request 1.-

return row d that has the minimum sum of columns 4 to 7 (insert code here)

     sum_d47=(sum(d(:,[4:7])))
     s_min=find(sum_d47==min(sum_d47))
     R1=d(s_min',:)

     sum_d47 =
            18
            18
            22
     s_min =
             1
             2
      R1 =
             2     9     2     8     9     0     1
             3     1     6     2     9     2     5

bear in mind that if you apply condition 1 to any of the requests

all results are void because for the supplied data

numel(d(:,1))==numel(unique(d(:,1)))
=
  logical
   1

there are not repeated elements in d(:,1)

request 2.-

if at least two of the sums of columns 4 to 7 are the same return row d that has the maximum sum of columns 2 to 3 (insert code here);

      sum_d47=(sum(d(:,[4:7]),2))
      s_min=find(sum_d47==min(sum_d47))
      if numel(s_min)>1
        R1=d(s_min',:)
        [ai,vi]=max(sum(R1(:,[2 3]),2))
        R2=R1(vi,:)
      end

sum_d47 =
    18
    18
    22
s_min =
     1
     2
R1 =
     2     9     2     8     9     0     1
     3     1     6     2     9     2     5
ai =
    11
vi =
     1
R2 =
     2     9     2     8     9     0     1

request 3.-

if at least two of the sums of columns 2 to 3 are the same (insert code here) n = d(1,:) return the first row of d (that is [2 9 2 8 9 0 1])

      sum_d47=(sum(d(:,[4:7]),2))
      s_min=find(sum_d47==min(sum_d47))
      if numel(s_min)>1
        R1=d(s_min',:)
        sum_r23=sum(R(:,[2 3]),2)
        s_max=find(sum_r23==max(sum_r23))

        if numel(s_max)>1
          R3=d(1,:)
        end
      end

sum_d47 =
    18
    18
    22
s_min =
     1
     2
R1 =
     2     9     2     8     9     0     1
     3     1     6     2     9     2     5
sum_r23 =
    11
     7
s_max =
     1

if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

[Chiamaka Agwuegbo]

Hi John. Thanks alot for your answer. I'll be sure to accept it if it works. I really do appreciate the effort

[John BG]

happy to help, please let me know if you reword the question or would like my answer in a function.

John BG

<mailto:[e-mail] [e-mail]>

Answer 2

[Walter Roberson]

[~, rowidx] =min( sum(d(:,4:7), 2) );
return_value = d(rowidx,:)

[Chiamaka Agwuegbo]

Thank you very much, Walter. Please why did you put "2" as the rowidx?

[Walter Roberson]

2 is not the rowidx: 2 is the dimension number to sum along. Notice it is not inside the d() expression: it is at the level of the sum()

[John BG]

sum(A)

and

sum(A,1)

are the same, dimension 1 being sum along vertical.

When aiming at horizontal summation dimension 2 has to be specified.

[Chiamaka Agwuegbo]

Ok, i see. Thank you