How to fill just such successive elements in row of matrix?

Hello!

i have matrix A (733*3600) randomly generated containing such values,i have a condition , if the value of element (i,j)=k, i want to fill another matrix B starting from element(i,j) until another specific element with a condition to not exceed array bounds.

for example i have k=50

in matrix B , i will fill 500 succesive elements starting from element with value k=50 in matrix A.

for the succesive elements , i want to stop until the limit of number of columns.

i already have a code to not exceed array bounds in this link https://www.mathworks.com/matlabcentral/answers/1799345-how-to-put-condition-in-rows-and-columns-for-not-exceed-array-bounds?s_tid=srchtitle , i tried to fill elements one by one ,but it is hard to apply it now with number of columns so large.

i hope that you could help me!

thanks

[dpb]

"to fill another matrix B starting from element(i,j) until another specific element "

What about multiple locations match in a given row? Take first, last, all matches and start from there or what???

[Safia]

@dpb the based matrix contains in each row just one value, the rest all zeros. i want to fill another matrix starting by the position of this element until specific column. i posted my code in this link may could more understand.

https://www.mathworks.com/matlabcentral/answers/1815380-can-someone-check-my-code

Answer 1

[Torsten]

A = [  0    50    0     0    0
0     0    50    0     0
0     0    50    0     0
0     0    0    50     0];
S = linspace(1,4,4);
T = 2;
B = zeros(size(A));
for i = 1:size(A,1)
    jstart = find(A(i,:)==50,1);
    jend = min(jstart+numel(S)-1,size(A,2));
    B(i,jstart:jend) = (50*S(1:jend-jstart+1))/T;
end
B

B = 4×5

0 25 50 75 100 0 0 25 50 75 0 0 25 50 75 0 0 0 25 50

[Safia]

@Torsten it works well! thank you very much! you always help me with efficient solution.

Answer 2

[Walter Roberson]

Consider for example,

A(i : min(i+k-1, end), j)

This would refer to at most k consecutive rows starting at row i but would stop at the array boundary.

[Safia]

@Walter Roberson i will explain you the code may could more understand

i created a vector from 1 to 720 element.

S=linspace(1,720,720)

i already have a matrix A(1826*3600) randomly generated contains such values.

i wrote this code to fill another matrix P.

for i=1:1826
    for j=1:3600
        for k=1:720
            if A(i,j)==50
                P(i,j)=(50*S(k))/3600;
            else P(i,j)=0;
            end
        end
    end
end

for matrix P , i want to fill just 720 columns, starting by the one having the specific value in matrix A, and we will stop until the bounds of array.

Answer 3

[Matt J]

Is this what you want?

A=[9, 50, 18 3;
   11 7 50 10;
   1 2 3 50;
   50 3 4 5]

A = 4×4

9 50 18 3 11 7 50 10 1 2 3 50 50 3 4 5

B=cumsum(A==50,2)*50

B = 4×4

0 50 50 50 0 0 50 50 0 0 0 50 50 50 50 50

C=B+~B.*A

C = 4×4

9 50 50 50 11 7 50 50 1 2 3 50 50 50 50 50

[Safia]

@Matt J Hello!

Matrix A is already generated, in each row there is just one value and other elements are "0". for example

A [ 0 50 0 0 0

0 0 50 0 0

0 0 50 0 0

0 0 0 50 0]

Now i want to fill specific columns in matrix B. So i generated a vector contains the number of elements i want to fill. for example 4 elements

S=linspace(1,4,4)
for i=1:4
    for j=1:5
        for k=1:4
            if A(i,j)==50
                B(i,j)=(50*S(k))/T;  %T is a value
            else B(i,j)=0;
            end

the result will be like this for T=2

S=[1 2 3 4]

B [ 0 25 50 75 100

0 0 25 50 75

0 0 25 50 75

0 0 0 25 75]

i want that elements can't exceed the bounds of matrix.