Plotting number distribution in Matlab

I create random integers based on the given N input and would like to plot these randomly generated numbers using hist and subplot. When I use disp(A) I see the generated integers just fine, but the plotting doesn't work. I need to plot them in one figure but in different subplots. I'm newbie so it might be something very basic that I'm missing.

My Code:

x=-4:.2:4;

for i = 1:n
   A=round(-100+100*rand());
   disp(A)
    c=hist(A,-4:.2:4);
  subplot(n,n,i)
  bar(x,c(:,i))
end

Thank you for your time and help.

[Alexandra Harkai]

What value are you giving for n?

You may want to write

 subplot(1,n,i)

since for (n,n) it creates n*n subplots where you really only need 1*n or n*1.

What would you like to display in each subplot? What errors/problems do you see when you say 'plotting doesn't work'?

[N/A]

the n is the number of random integers I want to create. I would like to display the distribution of these integers in one figure. I'm unable to see the distribution of the integers. The generated integers are not reflected on the graph. I get the same visual regardless of the input I provide and regardless of the range of the integers

Answer 1

[Image Analyst]

Try this:

edges = -4:.2:4;
n = 9;
rows = ceil(sqrt(n));
for k = 1:n
  A = round(-100+100*rand());
  disp(A)
  subplot(rows, rows, k);
  histObject = histogram(A, edges);
end

Of course I hope you know (but you probably don't) that A is a single number so taking the histogram of it will give only one count in one bin. Even worse, A can go from -100 to 0 and your bins only go from -4 to 4, so your bin's won't capture most of your counts at all!

[Image Analyst]

I know. I told you so, and I told you why. I don't know what you were trying to do. I'm going to "fix" it by making some assumptions that A is an array of 5000 random numbers, and you get this:

n = 9;
rows = ceil(sqrt(n));
for k = 1:n
  A = round(-100+100*rand(1, 5000));
  subplot(rows, rows, k);
  histObject = histogram(A);
  caption = sprintf('Run #%d', k);
  title(caption);
end

[N/A]

Thanks for that code. You stated that A is a single number and taking a histogram of that won't capture most of the bins. is that still valid for the code you provided?

[Image Analyst]

No, because I made A a vector of thousands of numbers so now we WILL have a distribution. I also let the histogram() function decide upon the bin edges to use instead of specifying a range of [-4 to 4] that does not include most of the numbers in the distribution.