How to count similar elements and store this value separately?

I have a data set that is one column and looks like:

[0000000011111022222033333330444]

I am trying to count the number of elements for a specific number. The list goes up to 170, increasing by 1 integer. And a 0 breaks up the previous number and the next number. I am attempting to count the occurrence of the numbers. For example, for 1 it would be 5 since shows up 5 times before moving onto 2.

Ideally, I am hoping to store this information separately to look like [5,2], something to show that 2 shows up 5 times.

loop_counter = 0;
m = 0;
y = zeros(length(x))
for i = 1:length(x)
    if x(i,1) == 1
        loop_counter = loop_counter + 1;
    else
        y(i, 1) = loop_counter;
        m = m + 1;
    end
end

Answer 1

[Image Analyst]

But 2 shows up 4 times, not 5. Why don't you simply use histcounts:

data = [0,0,0,0,0,0,0,0,1,1,1,1,1,0,2,2,2,2,0,3,3,3,3,3,3,3,0,4,4,4]
[counts, edges] = histcounts(data)
result = [(1:length(counts)-1)', counts(2:end)']

You'll see

result =
     1     5
     2     4
     3     7
     4     3

[Graduate Student]

Histcounts didn't work, maybe I'm applying it incorrectly? Attached is what I am working with.

[Image Analyst]

Attaching your data in advance is always a good idea. Prevents wasting time. But at least better late than never. So, this works:

s = load('sweep_ind.mat')
sweep_ind = s.sweep_ind
edges = .5 : 1 : max(sweep_ind)+ 0.5;
[counts, edges] = histcounts(sweep_ind, edges)
result = [(1:length(counts))', counts']

[Graduate Student]

Thanks! This works!

Answer 2

[KSSV]

   A = '000000001111102222033333330444' ;
   num = 2 ;
   iwant = [length(strfind(A,num2str(num))),num]

[Graduate Student]

Is there something similar to strfind for columns? I ran the code and the error I received was: Input string must have one row.