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Split and count unique string in cell array

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I have a cell array in the form of:
A =
B25
A35
L35 J23
K32 I25
B25 ...
where cetain elements repeat. I need to count how many unique elements there are and then list number of occurences of each element. For the above example it would be something like:
B25 ... 2
L35 ... 1
K32 ... 1 etc.
I tried using different combinations of strplit, regexp and unique, but some returned errors, others returned an array with the whole row counted as unique, so for the example above it would say there are 4 unique elements instead of 6 because L35 J23 is counted as 1, not 2. There is a hint that converting to categorical might help, but I am not sure how to utilize its functions in order to get the desired result.

Risposta accettata

Stephen23
Stephen23 il 25 Mar 2022
A = {'B25';'A35';'L35 J23';'K32 I25';'B25'};
B = regexp(A,'\S+','match');
T = cell2table([B{:}].');
S = groupsummary(T,'Var1')
S = 6×2 table
Var1 GroupCount _______ __________ {'A35'} 1 {'B25'} 2 {'I25'} 1 {'J23'} 1 {'K32'} 1 {'L35'} 1

Più risposte (2)

Mohammed Hamaidi
Mohammed Hamaidi il 25 Mar 2022
A loop solution:
C=unique(A);nc=length(C);
B=char(A);nb=length(B);
D=zeros(nc,1);
for i=1:nc
for j=1:nb
if strcmp(B(j,:),char(C{i}))
D(i)=D(i)+1;
end
end
end
for i=1:nc
disp([char(C{i}) ' ' num2str(D(i))])
end
  1 Commento
Josipe Jurcic
Josipe Jurcic il 25 Mar 2022
Thanks for your reply.
It throws this error:
Index in position 1 exceeds array bounds. Index must not exceed 6.

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Simon Chan
Simon Chan il 25 Mar 2022
Use function groupsummary
A = {'B25';'A35';'L35 J23';'K32 I25';'B25'};
T = table(A);
groupsummary(T,'A')
ans = 4×2 table
A GroupCount ___________ __________ {'A35' } 1 {'B25' } 2 {'K32 I25'} 1 {'L35 J23'} 1
  3 Commenti
Simon Chan
Simon Chan il 25 Mar 2022
Just add a few things as follows:
A = {'B25';'A35';'L35 J23';'K32 I25';'B25'; 'L35 J10'};
B = cellfun(@(x) strsplit(x),A,'uni',0); % Split them
C = cat(2,B{:})'; % Combine as a column
T = table(C);
groupsummary(T,'C')
ans = 7×2 table
C GroupCount _______ __________ {'A35'} 1 {'B25'} 2 {'I25'} 1 {'J10'} 1 {'J23'} 1 {'K32'} 1 {'L35'} 2
Josipe Jurcic
Josipe Jurcic il 25 Mar 2022
Thanks for your reply.
This works as well. Thanks again.

Accedi per commentare.

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