How to count unique value in array?
显示 更早的评论
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
回答（3 个）
[cnt_unique, unique_a] = hist(a,unique(a))
4 个评论
Payman Ferdosali
2021-7-29
This worked perfectly for me! thank you.
s fry
2023-10-1
a = [-1 -1 ]; breaks the above code.
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.
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 个评论
tedy
2013-4-6
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.
类别
在 帮助中心 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息
2023-10-1
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
