I want to use all the values present in two arrays but if I run for loops (ie) I used two for loops but only the last value of both the arrays are taken. What is the syntax so that I can use all my values of both arrays simultaneously.?

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Kamali M 2018-12-13

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评论： Kamali M 2018-12-14

I want to use all the values present in two or three arrays but if I run for loops (ie) I used three "for" loops but only the last value of both the arrays are taken. What is the syntax so that I can use all my values of both arrays simultaneously.?

for p1= [60 78 132 12 26 10 150 156 172 168 159 174]

for p2=[61 150 16 24 25 8 155 164 173 160 171 169]

for s=[0 1 2 3 4 5 6 7 8 9 10 11]

these are the three loops...at first iteration I need 60,61,11 next iteration 78,150,1 and so on.

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Answer 1

Stephen23 2018-12-13

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编辑：Stephen23 2018-12-13

V1 = [60,78,132,12,26,10,150,156,172,168,159,174];
V2 = [61,150,16,24,25,8,155,164,173,160,171,169];
V3 = [0,1,2,3,4,5,6,7,8,9,10,11];
for k = 1:numel(V1)
    V1(k)
    V2(k)
    V3(k)
end

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Kamali M 2018-12-14

编辑：Stephen23 2018-12-14

在 MATLAB Online 中打开

clc;
close all;
k=2;
a=(2*k)+1;
b=((2*(k*k))+(2*k)+1);
p1=[60,78,132,12,26,10,150,156,172,168,159,174];
p2=[61,150,16,24,25,8,155,164,173,160,171,169];
s=[0,1,2,3,4,5,6,7,8,9,10,11];
 
forj=1:numel(p1)
    
        c=(a*p1(j))+p2(j);
        f=mod(c,b);                   
        h=s(j)-f;
        t=mod(h,b);
                switch(t)
                    case 0
                        x1=0;
                        x2=0;
                        q1=p1+x1;
                        q2=p2+x2;
                        disp (q1)
                        disp (q2)
                        break
                    case 1
                         x1=0;
                         x2=1;
                         q1=p1+x1;
                         q2=p2+x2;
                         disp (q1)
                         disp (q2)
                         break
                    case 2
                         x1=0;
                         x2=2;
                         q1=p1+x1;
                         q2=p2+x2;
                         disp (q1)
                         disp (q2)
                         break
                     case 3
                         x1=-2;
                         x2=0;
                         q1=p1+x1;
                         q2=p2+x2;
                         disp (q1)
                         disp (q2)
                         break
                     case 4
                          x1=1;
                          x2=-1;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                     case 5
                          x1=1;
                          x2=0;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                     case 6
                          x1=1;
                          x2=1;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                      case 7
                          x1=-1;
                          x2=-1;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                      case 8
                          x1=-1;
                          x2=0;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                      case 9
                          x1=-1;
                          x2=1;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                      case 10
                          x1=2;
                          x2=0;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                      case 11
                          x1=0;
                          x2=-2;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break
                      case 12
                          x1=0;
                          x2=-1;
                          q1=p1+x1;
                          q2=p2+x2;
                          disp (q1)
                          disp (q2)
                          break         
                end
                end

This is my code. P1 and p2 enters correctly..but only the first value of s is taken..i want 12 values of s also. so please kindly tell what changes has to be made here.

thanks in advance.

Stephen23 2018-12-14

编辑：Stephen23 2018-12-14

Get rid of all of the break statements. If you want to process all elements of p1, p2, and s, then do not use break to exit the loop after the first element has been processed.

Kamali M 2018-12-14

Thank you so much. It works.

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I want to use all the values present in two arrays but if I run for loops (ie) I used two for loops but only the last value of both the arrays are taken. What is the syntax so that I can use all my values of both arrays simultaneously.?

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I want to use all the values present in two arrays but if I run for loops (ie) I used two for loops but only the last value of both the arrays are taken. What is the syntax so that I can use all my values of both arrays simultaneously.?

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