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

By default, uniquetol looks for unique elements that are within tolerance, but it also can find unique rows of a matrix that are within tolerance.

Create a numeric matrix, A. Obtain a second matrix, B, by transforming and untransforming A. This transformation introduces round-off differences to B.

A = [0.05 0.11 0.18; 0.18 0.21 0.29; 0.34 0.36 0.41; 0.46 0.52 0.76];
B = log10(10.^A);

Use unique to find the unique rows in A and B. The unique function performs exact comparisons and determines that all of the rows in the concatenated matrix [A;B] are unique, even though some of the rows differ by only a small amount.

unique([A;B],'rows')

ans = 8×3

    0.0500    0.1100    0.1800
    0.0500    0.1100    0.1800
    0.1800    0.2100    0.2900
    0.1800    0.2100    0.2900
    0.3400    0.3600    0.4100
    0.3400    0.3600    0.4100
    0.4600    0.5200    0.7600
    0.4600    0.5200    0.7600

Use uniquetol to find the unique rows. uniquetol treats rows that are within tolerance as equal.

uniquetol([A;B],'ByRows',true)

ans = 4×3

    0.0500    0.1100    0.1800
    0.1800    0.2100    0.2900
    0.3400    0.3600    0.4100
    0.4600    0.5200    0.7600

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

x = (1:5)'*pi;
y = 10.^log10(x);

Combine x and y into a single vector, A. Use uniquetol to reconstruct A, treating the values that are within tolerance as equal.

A = [x;y]

A = 10×1

    3.1416
    6.2832
    9.4248
   12.5664
   15.7080
    3.1416
    6.2832
    9.4248
   12.5664
   15.7080

[C,IA,IC] = uniquetol(A);
newA = C(IC)

newA = 10×1

    3.1416
    6.2832
    9.4248
   12.5664
   15.7080
    3.1416
    6.2832
    9.4248
   12.5664
   15.7080

You can use newA with == or functions that use exact equality like isequal or unique in subsequent code.

D1 = unique(A)

D1 = 6×1

    3.1416
    3.1416
    6.2832
    9.4248
   12.5664
   15.7080

D2 = unique(newA)

D2 = 5×1

    3.1416
    6.2832
    9.4248
   12.5664
   15.7080

Create a cloud of 2-D sample points constrained to be inside a circle of radius 0.5 centered at the point $$(\frac{1}{2},\frac{1}{2})$$.

x = rand(10000,2); 
insideCircle = sqrt((x(:,1)-.5).^2+(x(:,2)-.5).^2)<0.5;
y = x(insideCircle,:);

Find a reduced set of points, such that each point of the original dataset is within tolerance of a point.

tol = 0.05;
C = uniquetol(y,tol,'ByRows',true);

Plot the reduced set of points as red dots on top of the original data set. The red dots are all members of the original data set. All the red dots are at least a distance tol apart.

plot(y(:,1),y(:,2),'.')
hold on
axis equal
plot(C(:,1), C(:,2), '.r', 'MarkerSize', 10)

Create a vector of random numbers and determine the unique elements using a tolerance. Specify OutputAllIndices as true to return all of the indices for the elements that are within tolerance of the unique values.

A = rand(100,1);
[C,IA] = uniquetol(A,1e-2,'OutputAllIndices',true);

Find the average value of the elements that are within tolerance of the value C(2).

C(2)

ans = 0.0318

allA = A(IA{2})

allA = 3×1

    0.0357
    0.0318
    0.0344

aveA = mean(allA)

aveA = 0.0340

By default, uniquetol uses a tolerance test of the form abs(u-v) <= tol*DS, where DS automatically scales based on the magnitude of the input data. You can specify a different DS value to use with the DataScale option. However, absolute tolerances (where DS is a scalar) do not scale based on the magnitude of the input data.

First, compare two small values that are a distance eps apart. Specify tol and DS to make the within tolerance equation: abs(u-v) <= 10^-6.

x = 0.1;
uniquetol([x, exp(log(x))], 10^-6, 'DataScale', 1)

ans = 0.1000

Next, increase the magnitude of the values. The round-off error in the calculation exp(log(x)) is proportional to the magnitude of the values, specifically to eps(x). Even though the two large values are a distance eps from one another, eps(x) is now much larger. Therefore, 10^-6 is no longer a suitable tolerance.

x = 10^10;
uniquetol([x, exp(log(x))], 10^-6, 'DataScale', 1)

ans = 1×2
1010 ×

    1.0000    1.0000

Correct this issue by using the default (scaled) value of DS.

format long
Y = [0.1 10^10];
uniquetol([Y, exp(log(Y))])

ans = 1×2
1010 ×

   0.000000000010000   1.000000000000000

Create a set of random 2-D points, then use uniquetol to group the points into vertical bands that have a similar (within tolerance) x-coordinate. Use these options with uniquetol:

A = rand(1000,2);
DS = [1 Inf];
[C,IA] = uniquetol(A, 0.1, 'ByRows', true, ...
    'OutputAllIndices', true, 'DataScale', DS);

Plot the points and average value for each band.

hold on
for k = 1:length(IA)
    plot(A(IA{k},1), A(IA{k},2), '.')
    meanAi = mean(A(IA{k},:));
    plot(meanAi(1), meanAi(2), 'xr')
end