Form a Cumulative sum matrix

S= 10x1300
R=10x1300
z1= S1+R1                         x1300
z2=S1+R1+R1+S2+R2                 x1300
z3=S1+R1+R1+S2+R2+R2+S3+R3         x1300
.
.
.
z10=S1+R1+R1+S2+R2+R2+S3+R3+..       +R9+S10+R10+R10   x 1300
Result: Form z, a 10x1300 matrix

S & R is a 10x1300 matrix of random numbers

First row of matrix z= First row of S + first row of R

Second row of matrix z= First row of S + first row of R + first row of R + second row of S + second row of R + second row of R

Each addition adds Previous row of R and the next row of S and R

[M]

And what is your question ?

[Joseph Lee]

Hi, it is stated in title and in text, form a cumulative sum matrix z.

Answer 1

[Stephen23]

>> S = randi(9,10,1300);
>> R = randi(9,10,1300);
>> Z = cumsum(S,1)+2*cumsum(R,1)-R;
>> Z(10,:) = Z(10,:)+R(10,:);

This correctly takes into account that each row Z(n,:) only adds row R(n,:) once, except for the tenth row which is a special case because R(10,:) is added twice.

Here is a complete example:

>> R = repmat((1:10).',1,3)
R =
    1    1    1
    2    2    2
    3    3    3
    4    4    4
    5    5    5
    6    6    6
    7    7    7
    8    8    8
    9    9    9
   10   10   10
>> S = repmat((1:3),10,1)
S =
   1   2   3
   1   2   3
   1   2   3
   1   2   3
   1   2   3
   1   2   3
   1   2   3
   1   2   3
   1   2   3
   1   2   3
>> Z = cumsum(S,1)+2*cumsum(R,1)-R;
>> Z(10,:) = Z(10,:)+R(10,:)
Z =
     2     3     4
     6     8    10
    12    15    18
    20    24    28
    30    35    40
    42    48    54
    56    63    70
    72    80    88
    90    99   108
   120   130   140

And checking against the given examples from rows one and three:

>> S(1,:)+R(1,:)
ans =
   2   3   4
>> S(1,:)+R(1,:)+R(1,:)+S(2,:)+R(2,:)+R(2,:)+S(3,:)+R(3,:)
ans =
   12   15   18

Answer 2

[Andrei Bobrov]

z = cumsum(S + 2*R) - [R(1:end-1,:); zeros(1,size(R,2))];

[Stephen23]

Almost, but does not take into account:

• row N of R is not counted twice in row N of Z.
• row 10 of R is counted twice in row 10 of Z.

[Andrei Bobrov]

Thank you Stephen!

I'm corrected my code.

Answer 3

[ali fadaie]

Yes it is right