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 ismember to find the elements of x that are in y. The ismember function performs exact comparisons and determines that some of the matrix elements in x are not members of y.

lia = ismember(x,y)

lia = 6x1 logical array

   0
   1
   1
   1
   1
   0

Use ismembertol to perform the comparison using a small tolerance. ismembertol treats elements that are within tolerance as equal and determines that all of the elements in x are members of y.

LIA = ismembertol(x,y)

LIA = 6x1 logical array

   1
   1
   1
   1
   1
   1

By default, ismembertol looks for elements that are within tolerance, but it also can find 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; 0.82 0.91 1.00];
B = log10(10.^A);

Use ismember to find the rows of A that are in B. ismember performs exact comparisons and thus determines that most of the rows in A are not members of B, even though some of the rows differ by only a small amount.

lia = ismember(A,B,'rows')

lia = 5x1 logical array

   0
   0
   0
   0
   1

Use ismembertol to perform the row comparison using a small tolerance. ismembertol treats rows that are within tolerance as equal and thus determines that all of the rows in A are members of B.

LIA = ismembertol(A,B,'ByRows',true)

LIA = 5x1 logical array

   1
   1
   1
   1
   1

Create two vectors of random numbers and determine which values in A are also members of B, using a tolerance. Specify OutputAllIndices as true to return all of the indices for the elements in B that are within tolerance of the corresponding elements in A.

rng(5)
A = rand(1,15);
B = rand(1,5);
[LIA,LocAllB] = ismembertol(A,B,0.2,'OutputAllIndices',true)

LIA = 1x15 logical array

   1   0   1   0   1   1   1   1   1   1   0   1   1   1   0

LocAllB=1×15 cell array
  Columns 1 through 5

    {2x1 double}    {[0]}    {2x1 double}    {[0]}    {3x1 double}

  Columns 6 through 10

    {2x1 double}    {[4]}    {3x1 double}    {3x1 double}    {2x1 double}

  Columns 11 through 15

    {[0]}    {2x1 double}    {4x1 double}    {2x1 double}    {[0]}

Find the average value of the elements in B that are within tolerance of the value A(13). The cell LocAllB{13} contains all the indices for elements in B that are within tolerance of A(13).

A(13)

ans = 0.4413

allB = B(LocAllB{13})

allB = 1×4

    0.2741    0.4142    0.2961    0.5798

aveB = mean(allB)

aveB = 0.3911

By default, ismembertol 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;
ismembertol(x, exp(log(x)), 10^-6, 'DataScale', 1)

ans = logical
   1

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;
ismembertol(x, exp(log(x)), 10^-6, 'DataScale', 1)

ans = logical
   0

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

Y = [0.1 10^10];
ismembertol(Y, exp(log(Y)))

ans = 1x2 logical array

   1   1

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

A = rand(1000,2);
B = [(0:.2:1)',0.5*ones(6,1)];
[LIA,LocAllB] = ismembertol(B, A, 0.1, 'ByRows', true, ...
    'OutputAllIndices', true, 'DataScale', [1,Inf])

LIA = 6x1 logical array

   1
   1
   1
   1
   1
   1

LocAllB=6×1 cell array
    { 94x1 double}
    {223x1 double}
    {195x1 double}
    {212x1 double}
    {187x1 double}
    { 89x1 double}

Plot the points in A that are within tolerance of each query point in B.

hold on 
plot(B(:,1),B(:,2),'x')
for k = 1:length(LocAllB)
    plot(A(LocAllB{k},1), A(LocAllB{k},2),'.')
end