How to count unique value in array?

Question

tedy 2013-4-6

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https://ww2.mathworks.cn/matlabcentral/answers/70061-how-to-count-unique-value-in-array

评论： Walter Roberson 2023-10-1

Hello, i have an example array a= [2,2,2,1,1,3,2,1,3,3,2] i do this code

a= [2,2,2,1,1,3,2,1,3,3,2]
unique(a)
histc(a, unique(a))

but, it only result 1,2,3 and 3,5,3 as its frequency.

How to take all the value, but no one duplicate each others. it's like 2,1,3,2,1,3,2 and also with its frequency 3,2,1,1,1,2,1. Thanks in advance

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Answer 1

balaji 2019-6-12

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https://ww2.mathworks.cn/matlabcentral/answers/70061-how-to-count-unique-value-in-array#answer_378841

编辑：balaji 2019-6-12

在 MATLAB Online 中打开

[cnt_unique, unique_a] = hist(a,unique(a))

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Walter Roberson 2023-10-1

@s fry is right: if the array happens to contain only a single value, then unique() would contain only a single value, and hist() would be likely to interpret that value as a bin count instead of as a degenerate list of edges.

Walter Roberson 2023-10-1

If you use the newer histogram then you can specify 'BinEdges' as a name/value option instead of passing positionally and having it guess about your intent.

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Answer 2

Cedric 2013-4-6

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https://ww2.mathworks.cn/matlabcentral/answers/70061-how-to-count-unique-value-in-array#answer_81174

编辑：Cedric 2013-4-6

在 MATLAB Online 中打开

You could build something around the following example:

 >> b = a([diff(a)~=0, false])
 b =
     2     1     3     2     1     3

let me know if you need help to get counts. Hint: try using DIFF as well as FIND.

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tedy 2013-4-6

Thank you cedric. How about the frequency? i code with histc(b, unique(b)), but it result 2,3,2, not 3,2,1,1,1,2,1

Cedric 2013-4-6

编辑：Cedric 2013-4-6

在 MATLAB Online 中打开

Look at the following:

 >> diff(a)
 ans =
     0     0    -1     0     2    -1    -1     2     0    -1

as you can see, places where diff is not null are places where there is a transition in the value. These locations can be found using FIND:

 >> find(diff(a))
 ans =
     3     5     6     7     8    10

Now you can probably imagine a solution based on DIFF again..

 >> c = diff([0, find(diff(a)), numel(a)])
 c =
     3     2     1     1     1     2     1

EDIT: I see that Roger posted a solution while I was writing my comment. His solution for getting counts is based on the same principle, but it is better than mine in the sense that it involves only two additional logicals, whereas mine involves a call the NUMEL.

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