Define a vector with a repeated value.

A = [9 2 9 5];

Find the unique values of A.

C = unique(A)

C = 1×3

     2     5     9

Create a table with some repeated data.

Name = {'Fred';'Betty';'Bob';'George';'Jane'};
Age = [38;43;38;40;38];
Height = [71;69;64;67;64];
Weight = [176;163;131;185;131];
A = table(Age,Height,Weight,'RowNames',Name)

A=5×3 table
              Age    Height    Weight
              ___    ______    ______

    Fred      38       71       176  
    Betty     43       69       163  
    Bob       38       64       131  
    George    40       67       185  
    Jane      38       64       131  

Find the unique rows of A. unique returns the rows of A in sorted order by the first variable Age and then by the second variable Height.

C = unique(A)

C=4×3 table
              Age    Height    Weight
              ___    ______    ______

    Bob       38       64       131  
    Fred      38       71       176  
    George    40       67       185  
    Betty     43       69       163  

Find the table rows with unique values in the first variable Age. If you only want one table variable to contain unique values, you can use the indices returned by unique to extract those rows from the table.

[C,ia] = unique(A.Age);
B = A(ia,:)

B=3×3 table
              Age    Height    Weight
              ___    ______    ______

    Fred      38       71       176  
    George    40       67       185  
    Betty     43       69       163  

Define a vector with a repeated value.

A = [9 2 9 5];

Find the unique values of A and the index vectors ia and ic, such that C = A(ia) and A = C(ic).

[C, ia, ic] = unique(A)

C = 1×3

     2     5     9

ia = 3×1

     2
     4
     1

ic = 4×1

     3
     1
     3
     2

Create a 10-by-3 matrix with some repeated rows.

A = randi(3,10,3)

A = 10×3

     3     1     2
     3     3     1
     1     3     3
     3     2     3
     2     3     3
     1     1     3
     1     2     3
     2     3     2
     3     3     2
     3     3     1

Find the unique rows of A based on the data in the first two columns. Specify three outputs to return the index vectors ia and ic.

[C,ia,ic] = unique(A(:,1:2),'rows')

C = 7×2

     1     1
     1     2
     1     3
     2     3
     3     1
     3     2
     3     3

ia = 7×1

     6
     7
     3
     5
     1
     4
     2

ic = 10×1

     5
     7
     3
     6
     4
     1
     2
     4
     7
     7

Use ia to index into A and retrieve the rows that have unique combinations of elements in the first two columns.

uA = A(ia,:)

uA = 7×3

     1     1     3
     1     2     3
     1     3     3
     2     3     3
     3     1     2
     3     2     3
     3     3     1

Find the unique elements in a vector and then use accumarray to count the number of times each unique element appears.

Create a vector of random integers from 1 through 5.

a = randi([1 5],200,1);

Find the unique elements in the vector. Return the index vectors ia and ic.

[C,ia,ic] = unique(a);

Count the number of times each element in C appears in a. Specify ic as the first input to accumarray and 1 as the second input so that the function counts repeated subscripts in ic. Summarize the results.

a_counts = accumarray(ic,1);
value_counts = [C, a_counts]

value_counts = 5×2

     1    46
     2    36
     3    38
     4    39
     5    41

Use the setOrder argument to specify the ordering of the values in C.

Specify 'stable' if you want the values in C to have the same order as in A.

A = [9 2 9 5];
[C, ia, ic] = unique(A,'stable')

C = 1×3

     9     2     5

ia = 3×1

     1
     2
     4

ic = 4×1

     1
     2
     1
     3

Alternatively, you can specify 'sorted' order.

[C, ia, ic] = unique(A,'sorted')

C = 1×3

     2     5     9

ia = 3×1

     2
     4
     1

ic = 4×1

     3
     1
     3
     2

Define a vector containing NaN.

A = [5 5 NaN NaN];

Find the unique values of A.

C = unique(A)

C = 1×3

     5   NaN   NaN

unique treats NaN values as distinct.

Create a vector x. Obtain a second vector y by transforming and untransforming x. This transformation introduces round-off differences in y.

x = (1:6)'*pi;
y = 10.^log10(x);

Verify that x and y are not identical by taking the difference.

x-y

ans = 6×1
10-14 ×

    0.0444
         0
         0
         0
         0
   -0.3553

Use unique to find the unique elements in the concatenated vector [x;y]. The unique function performs exact comparisons and determines that some values in x are not exactly equal to values in y. These are the same elements that have a nonzero difference in x-y. Thus, c contains values that appear to be duplicates.

c = unique([x;y])

c = 8×1

    3.1416
    3.1416
    6.2832
    9.4248
   12.5664
   15.7080
   18.8496
   18.8496

Use uniquetol to perform the comparison using a small tolerance. uniquetol treats elements that are within tolerance as equal.

C = uniquetol([x;y])

C = 6×1

    3.1416
    6.2832
    9.4248
   12.5664
   15.7080
   18.8496

Create a cell array of character vectors.

A = {'one','two','twenty-two','One','two'};

Find the unique character vectors contained in A.

C = unique(A)

C = 1x4 cell
    {'One'}    {'one'}    {'twenty-two'}    {'two'}

Create a cell array of character vectors, A, where some of the vectors have trailing white space.

A = {'dog','cat','fish','horse','dog ','fish '};

Find the unique character vectors contained in A.

C = unique(A)

C = 1x6 cell
    {'cat'}    {'dog'}    {'dog '}    {'fish'}    {'fish '}    {'horse'}

unique treats trailing white space in cell arrays of character vectors as distinct characters.

Use the 'legacy' flag to preserve the behavior of unique from R2012b and prior releases in your code.

Find the unique elements of A with the current behavior.

A = [9 2 9 5];
[C1, ia1, ic1] = unique(A)

C1 = 1×3

     2     5     9

ia1 = 3×1

     2
     4
     1

ic1 = 4×1

     3
     1
     3
     2

Find the unique elements of A, and preserve the legacy behavior.

[C2, ia2, ic2] = unique(A, 'legacy')

C2 = 1×3

     2     5     9

ia2 = 1×3

     2     4     3

ic2 = 1×4

     3     1     3     2