How to count unique value in array?
253 ビュー (過去 30 日間)
古いコメントを表示
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
0 件のコメント
回答 (3 件)
balaji
2019 年 6 月 12 日
編集済み: balaji
2019 年 6 月 12 日
[cnt_unique, unique_a] = hist(a,unique(a))
4 件のコメント
Walter Roberson
2023 年 10 月 1 日
@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 年 10 月 1 日
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
2013 年 4 月 6 日
編集済み: Cedric
2013 年 4 月 6 日
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 件のコメント
Cedric
2013 年 4 月 6 日
編集済み: Cedric
2013 年 4 月 6 日
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.
参考
カテゴリ
Help Center および File Exchange で Logical についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!