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

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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
  3 Comments
indhu ravi
indhu ravi on 5 Apr 2019
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

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Accepted Answer

Rick Rosson
Rick Rosson on 28 Jul 2011
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
  6 Comments
Walter Roberson
Walter Roberson on 16 May 2021
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
Enrique Sánchez on 26 Sep 2021
If A is not a vector, then I suggest
count(k) = sum(A == x(k), 'all');

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More Answers (6)

Alexander
Alexander on 4 Jun 2014
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.

Walter Roberson
Walter Roberson on 28 Jul 2011
u = unique(A);
fprintf('%d appears %d times\n', [u; histc(A,u)].');
  2 Comments
hello_world
hello_world on 24 Nov 2014
It gives me the following error:
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
Walter Roberson
Walter Roberson on 26 Sep 2015
[u, ~, uidx] = unique(A(:));
counts = accumarray(uidx, 1);
fprintf('%d appears %d times\n', [u, counts].');

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Joseph Sottosanti
Joseph Sottosanti on 14 Dec 2018
c = nnz(a==1)
  1 Comment
Walter Roberson
Walter Roberson on 14 Dec 2018
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.

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Steven Lord
Steven Lord on 5 Apr 2019
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')

indhu ravi
indhu ravi on 7 Apr 2019
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

Zhe Li
Zhe Li on 27 May 2019
Edited: Zhe Li on 27 May 2019
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.

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