How to count unique value in array?
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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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Risposte (3)
balaji
il 12 Giu 2019
Modificato: balaji
il 12 Giu 2019
[cnt_unique, unique_a] = hist(a,unique(a))
4 Commenti
Walter Roberson
il 1 Ott 2023
@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
il 1 Ott 2023
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.
Cedric
il 6 Apr 2013
Modificato: Cedric
il 6 Apr 2013
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.
2 Commenti
Cedric
il 6 Apr 2013
Modificato: Cedric
il 6 Apr 2013
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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