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
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indhu ravi
2019년 4월 5일
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
2019년 4월 5일
편집: Walter Roberson
2019년 4월 5일
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
채택된 답변
Rick Rosson
2011년 7월 28일
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
댓글 수: 6
Walter Roberson
2021년 5월 16일
I do not see any z in the code?
If A is not a vector, then I suggest
count(k) = sum(A(:) == x(k));
추가 답변 (7개)
Alexander
2014년 6월 4일
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.
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Walter Roberson
2011년 7월 28일
u = unique(A);
fprintf('%d appears %d times\n', [u; histc(A,u)].');
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hello_world
2014년 11월 24일
It gives me the following error:
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
Walter Roberson
2015년 9월 26일
[u, ~, uidx] = unique(A(:));
counts = accumarray(uidx, 1);
fprintf('%d appears %d times\n', [u, counts].');
Joseph Sottosanti
2018년 12월 14일
c = nnz(a==1)
댓글 수: 1
Walter Roberson
2018년 12월 14일
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.
Steven Lord
2019년 4월 5일
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')
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indhu ravi
2019년 4월 7일
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
댓글 수: 3
Zhe Li
2019년 5월 27일
편집: Zhe Li
2019년 5월 27일
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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PetterE
2024년 5월 3일
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.
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