Find indexes in cell array
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Hi all,
i have 2 arrays one containing all my values (even duplicates) and one containing the unique elements.
My goal is to find the index of my values based on the index of the unique elements
R = [2 ,1 , 1 , 2 , 4 , [1,3] , [1,3] , 4 , [1,3] , [2,5] , [2,5] , 4]
C= [2 , 1 , 4 , [1,3] , [2,5]]
Vectors [1,3] should be treated as 1 unique element, not as 1 and 3
If its easier we can work with letters
R1=['2B' , '1A' , '1A' , '2B' , '1D' , '1A+1C' ,'1A+1C' , '1D' , '1A+1C' , '1B+1E' , '1B+1E', '1D']
C1=['1A' , '2B' , '1A+1C' , '1D' , '1B+1E']
In both cases i expect to get an array
idx = [2, 1, 1, 2, 4, 3, 3, 4, 3, 5, 5, 4]
The order is important, idx should be in the same order of the original R/R1 vector.
I tried find and a bit of arrayfun but couldnt get it to work
Thank you so much in advance
采纳的回答
R1={'2B' , '1A' , '1A' , '2B' , '1D' , '1A+1C' ,'1A+1C' , '1D' , '1A+1C' , '1B+1E' , '1B+1E', '1D'}
C1={'1A' , '2B' , '1A+1C' , '1D' , '1B+1E'}
[~,Locb] = ismember(R1,C1)
3 个评论
Jon
2022-5-10
Note curly braces to make a cell array, if you use square brackets all of the elements will just be concatenated into one long character array
Davide Bordin
2022-5-13
Jon
2022-5-13
Glad that it helped, good luck with your project
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