How can the nested if statement in the last for loop be rewritten to be shorter?

 clear; close all

 n = 4000;

 % (A) Use rand to generate randomly distributed points:
 Rand_Number = rand(n, 2);
 x = Rand_Number(:,1)*2-1;
 y = Rand_Number(:,2)*2-1;

 hold on
 figure(1)
 subplot(3,1,1)
 plot(x,y,'r.')
 title('Random Points Generated Using "rand"')
 xlabel('x')
 ylabel('y') 

 subplot(3,1,2)
 count = zeros(1,20);
 for i=1:length(x)
     count = histcounts(x, -1:0.1:1);
 end
 X=-.95:.1:1;
 bar(X,count)
 title('Distribution of Points')
 xlabel('x')
 ylabel('y')

 subplot(3,1,3)
 hist(x)
 title('Histogram of Points')
 xlabel('x')
 ylabel('y')

 hold off
 delta_x = 0.1;
 for j=1:200
     tmp = rand(n,1);
     for i = 1:length(x)
         if tmp < 0.5
             x(i)=x(i)+delta_x;
         else
             x(i)=x(i)-delta_x;
         end
     end
 end
 xmove=x
 countmove = zeros(1,20);

 for i=1:length(x)
     if xmove(i)>= -5.0 && xmove(i)<=-4.5
         count(1) = count(1)+1;
     elseif xmove(i)>-4.5 && xmove(i)<-4.0
         count(2) = count(2)+1;
     elseif xmove(i)>-4.0 && xmove(i)<-3.5
         count(3) = count(3)+1;
     elseif xmove(i)>-3.5 && xmove(i)<-3.0
         count(4) = count(4)+1;
     elseif xmove(i)>-3.0 && xmove(i)<-2.5
         count(5) = count(5)+1;
     elseif xmove(i)>-2.5 && xmove(i)<-2.0
         count(6) = count(6)+1;
     elseif xmove(i)>-2.0 && xmove(i)<-1.5
         count(7) = count(7)+1;
     elseif xmove(i)>-1.5 && xmove(i)<-1.0
         count(8) = count(8)+1;
     elseif xmove(i)>-1.0 && xmove(i)<-0.5
         count(9) = count(9)+1;
     elseif xmove(i)>-0.5 && xmove(i)<0.0
         count(10) = count(10)+1;
     elseif xmove(i)>0.0 && xmove(i)<0.5
         count(11) = count(11)+1;
     elseif xmove(i)>0.5 && xmove(i)<1.0
         count(12) = count(12)+1;
     elseif xmove(i)>1.0 && xmove(i)<1.5
         count(13) = count(13)+1;
     elseif xmove(i)>1.5 && xmove(i)<2.0
         count(14) = count(14)+1;
     elseif xmove(i)>2.0 && xmove(i)<2.5
         count(15) = count(15)+1;
     elseif xmove(i)>2.5 && xmove(i)<3.0
         count(16) = count(16)+1;
     elseif xmove(i)>3.0 && xmove(i)<3.5
         count(17) = count(17)+1;
     elseif xmove(i)>3.5 && xmove(i)<4.0
         count(18) = count(18)+1;
     elseif xmove(i)>4.0 && xmove(i)<4.5
         count(19) = count(19)+1;
     elseif xmove(i)>4.5 && xmove(i)<=5.0
         count(20) = count(20)+1;
     end
 end
 figure(2)
 subplot(2,1,1)
 plot(xmove,y,'r.')
 title('Moved Random Points Generated')
 xlabel('x')
 ylabel('Number of Points')

 subplot(2,1,2)
 Y = -4.75:0.5:4.75;
 bar(Y,countmove)
 title('Distribution of Points')
 xlabel('x')
 ylabel('Number of Points')

[Adam]

This is the 3rd time you have asked what amounts to the same question! Surely you can adapt one of the two solutions provided to your other questions to solve this.

https://uk.mathworks.com/matlabcentral/answers/323068-how-do-you-rewrite-the-nested-if-statement-as-a-loop

https://uk.mathworks.com/matlabcentral/answers/323076-how-do-you-rewrite-the-nested-if-statement-as-a-loop

People on this forum are always willing to help, but we do so out of good will, we don't get paid or expect to get paid, but in general people don't appreciate having the time they give wasted by having someone ask the same question in multiple places, get two answers, both of which you accepted and then see you ask the same thing a 3rd time.

[Stephen23]

@Jillian Sweatt: How is this question different from the last two times that you asked the same question with the same code?:

https://www.mathworks.com/matlabcentral/answers/323076-how-do-you-rewrite-the-nested-if-statement-as-a-loop

https://www.mathworks.com/matlabcentral/answers/323068-how-do-you-rewrite-the-nested-if-statement-as-a-loop

You already accepted two good answers: if they do not work for you then make a comment in the relevant question, rather than creating a new question thread.

[John BG]

May be Jillian wants to find an alternative way of compressing this sorting, but doesn't want to edit an already accepted answer that works.

Editing already accepted answers may turn confusing, and adding comments the kind of 'yes but what if ..' that may be a different question, then, simplify: issue a new question.

So the question originator leaves previous Q-A1 and Q-A2 as they are, and now wants a different way to solve same question without disappointing the suppliers of the previous answers.

John BG

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

[John BG]

To Jillian Sweatt:

if you don't like or change mind or find that a too abridged question requires some explanation, you can perfectly remove the Accepted Answer marking, even edit your question, and ask for a different way to have it solved.

The people who have supplied the previously Accepted Answer do not lose credit points, yet if you decide to remove the status of Accepted Answer from a previously accepted answer, you can perfectly do so,

John BG

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

[Adam]

Maybe they should put their question on Cody if they want 3 or more different ways to solve the same problem!

An answer being accepted suggests it fits the problem, otherwise it shouldn't be accepted. Most of us are more concerned with having wasted time if we answer a question that is also being asked in numerous different places to. Whether the answer was accepted or not doesn't really make any difference, but someone asking the same question 3 times having already accepted 2 answers is clearly just wasting people's time even if in this case the answers were largely trivial.

[Walter Roberson]

Please confirm that you do not want to count any value which is exactly -4, -3.5, -3, -2.5, -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, or 4.5, but that you do want to count values which are exactly -4.5 or 5.0 ?

Answer 1

[Jan]

Seriously? The answer given in http://www.mathworks.com/matlabcentral/answers/323076-how-do-you-rewrite-the-nested-if-statement-as-a-loop can be copied and adjusted easily. Again you do not need a loop, but one single command:

count = histcounts(x, -5:0.5:5)

Are you looking for something else?