How can I count the number of times a number appear in a vector?

How can I count the number of times a number appear in a vector? A=[ 1 2 3 1 1 6 3 4 2 2 1 1 1 ]

1 appears 7 times 2 appears 3 times 3 appears 3 times and on ... Best Regards

[adi]

you can also do this without for loops:

    bins = 0:100;
    data = rand(1,1000)*100;

    % find the bin index for every data point
    binIndex4data = ceil((data-bins(1))/(bins(2) - bins(1)));

    binIndex = 1:length(bins);

    % equals(i,j) = 1 means that binIndex4data(i) equals binIndex(j).
    equals = bsxfun(@eq,binIndex4data',binIndex);

    % counts how many data point are in each bin (sums every cloumn of
    % equals matrix) 
    hist = sum(equals);

% plot histogram

    increment = bins(2)-bins(1);
    bar(bins+increment/2,hist,increment);

[indhu ravi]

sir actually i have an matrix of about 0 0 0

1 0 1 but i should find the maximum occurence of same numbers i.e 0

1

[Walter Roberson]

Rick showed how to count all the occurances of all of the numbers, and I showed a different approach at https://www.mathworks.com/matlabcentral/answers/12636-how-can-i-count-the-number-of-times-a-number-appear-in-a-vector#comment_312351

Answer 1

[Rick Rosson]

Please try the following:

   A = [ 1 2 3 1 1 6 3 4 2 2 1 1 1 ];
   x = unique(A);
   N = numel(x);

   count = zeros(N,1);

   for k = 1:N
      count(k) = sum(A==x(k));
   end

   disp([ x(:) count ]);

HTH.

Rick

[Rafael Freire]

Thank you Rick!

[simith]

Thank You very much.This really helped me to sort out my problems :)

[Thet Win]

Hi

the error is showing for this line: count(k) = sum(A==x(k));

[Amir Hossein Khosravi]

Hello

Replace line 6 with this line of code to make it work.

count(k) = sum(sum(z == x(k)));

[Walter Roberson]

@Amir Hossein Khosravi

I do not see any z in the code?

If A is not a vector, then I suggest

count(k) = sum(A(:) == x(k));

[Enrique Sánchez]

If A is not a vector, then I suggest

count(k) = sum(A == x(k), 'all');

Answer 2

[Alexander]

I realize this is old, but in case someone else stumbles upon it:

c = numel(find(a==1))

returns c = 7, so 1 occurs 7 times. Can embed into a for loop to find the occurrence of all members.

[Tejas k]

Thankx man :) It really helped me a lot. Other answers were doing something else than what I intend ed to find

[Roshan]

saved my life

[Stephen23]

Simpler and more efficient:

c = nnz(a==1)

Answer 3

[Walter Roberson]

u = unique(A);
fprintf('%d appears %d times\n', [u; histc(A,u)].');

[hello_world]

It gives me the following error:

Error using vertcat
Dimensions of matrices being concatenated are not consistent.

[Walter Roberson]

[u, ~, uidx] = unique(A(:));
counts = accumarray(uidx, 1);
fprintf('%d appears %d times\n', [u, counts].');

Answer 4

[Joseph Sottosanti]

c = nnz(a==1)

[Walter Roberson]

This is good if you are wanting to count a specific value, but other code is needed if each value that occurs has to be tallyed like the original question asked.

Answer 5

[Steven Lord]

You can use histcounts or histogram for this purpose. Let's generate some sample data, 100 integers between 1 and 10.

A = randi(10, 1, 100);

The edges should start at each unique value in A. We need to add one additional edge at the end, otherwise the last bin will count the two largest unique values contained in A, not just the largest. All bins but the last contain their left edge but not their right edge, while the last bin contains both its edges.

U = unique(A);
edges = [U, max(U)+1];

Now let's count. I'll put the results in a table array for easy reading.

counts = histcounts(A, edges);
results = table(U(:), counts(:), 'VariableNames', {'UniqueElements', 'Count'})

If you want to see the results graphically, use histogram instead of histcounts.

h = histogram(A, edges);
counts2 = h.Values;
isequal(counts2, counts) % true

If you want the bins to be centered on integer values (useful if your data is all integer values, as in the case in the original question) you can specify the 'integers' BinMethod instead of building the edges yourself. If you do this with histcounts you'd want to call it with two outputs so you receive both the bin counts and the vector of bin edges. Remember that the vector of bin edges will be one element longer than the vector of bin counts since bin edges are fence posts and bin counts are fence rails.

h = histogram(A, 'BinMethod', 'integers');
[counts3, edges3] = histcounts(A, 'BinMethod', 'integers')

Answer 6

[indhu ravi]

sir for example i have an matrix of 3x3

0 0 0

1 0 1

1 1 1

i should get the maximum occurrence of each row (i.e) 0

1

1

can anyone help me

[Walter Roberson]

mode()

[indhu ravi]

i can't understand

[madhan ravi]

mode(matrix,2)

Answer 7

[Zhe Li]

A simple command that should work in most situations:

>> summary(categorical(A))

gives

1 2 3 4 6

6 3 2 1 1

But this is for display (or quick overview) only -- you cannot use the result without some copying and pasting. So if you want an array of these values you need to use some of the other methods found here.

Answer 8

[PetterE]

The simplest solution I can think of is probably

Au=unique(A);
occurances=sum(A(:)==Au(:)')

this will give you the list of unique numbers in Au and the occurances on the same indicies in the occurances vector.