Find Number of Elements in an Array

Hello guys. I want to find the number of elements in a string array.

Lets say A = ['KM'; 'KL'; 'MN'; 'KM', 'MM', 'KL'] is my array list.

It should give output as;

[2,2,1,1] since my string array includes 2 KM, 2 KL, 1MN, and 1MM.

How can i do that?

Answer 1

[Lola Davidson]

If you prefer avoiding cell arrays and/or tables, groupcounts (introduced in R2019a) can do this straight away

A = ['KM'; 'KL'; 'MN'; 'KM'; 'MM'; 'KL'];
groupcounts(A)

ans = 4x1

2 2 1 1

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That said, if you like the decoration, you can put it in a table before sending it to groupcounts.

A = ["KM"; "KL"; "MN"; "KM"; "MM"; "KL"];
groupcounts(table(A),1)

ans = 4x3 table
     A      GroupCount    Percent
    ____    __________    _______

    "KL"        2         33.333 
    "KM"        2         33.333 
    "MM"        1         16.667 
    "MN"        1         16.667 

Answer 2

[Adam Danz]

The "A" array provided in the question will result in a dimensions mismatch error. I'm assuming A is an [nx2] char array.

A = ['KM'; 'KL'; 'MN'; 'KM'; 'MM'; 'KL'];

% Convert char array to cell array of strings
Acell = cellstr(A); 
% Find groups of strings
[groups, groupID]= findgroups(Acell(:)); 
% Count members of each group
count = sum(groups(:).' == unique(groups(:)),2); 
% Display results in a table
countTable = table(groupID(:),count(:),'VariableNames',{'Group','Count'}); 

Result

countTable =
  4×2 table
    Group    Count
    _____    _____
    'KL'       2  
    'KM'       2  
    'MM'       1  
    'MN'       1 

[Burak Alakus]

Thank you Mr. Danz. This answer really helped me.

[Adam Danz]

Glad I could help and learn along with ya!

[Adam Danz]

June 5, 2024 - I unaccepted this answer in favor of Lola's better solution using groupcounts that became available in R2019a. Prior to R2019a, I would recommend Steven Lord's answer.

Answer 3

[Andrei Bobrov]

A = {'KM'; 'KL'; 'MN'; 'KM'; 'MM'; 'KL'};

out = varfun(@x,table(A),'GroupingVariables','A')

[Burak Alakus]

Thank you Mr. Bobrov. This answer really helped me.

Answer 4

[Steven Lord]

Yet another way to do this:

A = ['KM'; 'KL'; 'MN'; 'KM'; 'MM'; 'KL'];
[counts, values] = histcounts(categorical(cellstr(A)))

counts = 1x4

2 2 1 1

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values = 1x4 cell array
    {'KL'}    {'KM'}    {'MM'}    {'MN'}