ismembertol
Find set members of data within tolerance
Syntax
Description
LIA = ismembertol(A,B,tol)1 (true)
where the elements of A are within tolerance of
the elements in B. Otherwise, the array contains
logical 0 (false). Two values, u and v,
are within tolerance if
abs(u-v) <= tol*max(abs([A(:);B(:)]))
That is, ismembertol scales the tol input
based on the magnitude of the data.
ismembertol is similar to ismember.
Whereas ismember performs exact comparisons, ismembertol performs
comparisons using a tolerance.
[___] = ismembertol(___, uses
additional options specified by one or more Name-Value pair arguments
using any of the input or output argument combinations in previous
syntaxes. For example, Name,Value)ismembertol(A,B,'ByRows',true) compares
the rows of A and B and returns
a logical column vector.
Examples
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 = 6×1 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 = 6×1 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 = 5×1 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 = 5×1 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 = 1×15 logical array
1 0 1 0 1 1 1 1 1 1 0 1 1 1 0
LocAllB=1×15 cell array
{2×1 double} {[0]} {2×1 double} {[0]} {3×1 double} {2×1 double} {[4]} {3×1 double} {3×1 double} {2×1 double} {[0]} {2×1 double} {4×1 double} {2×1 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 = 1×2 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:
Specify
ByRowsastrue, since the point coordinates are in the rows ofAandB.Specify
OutputAllIndicesastrueto return the indices for all points inAthat have an x-coordinate within tolerance of the query points inB.Specify DataScale as
[1 Inf]to use an absolute tolerance for the x-coordinate, while ignoring the y-coordinate.
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 = 6×1 logical array
1
1
1
1
1
1
LocAllB=6×1 cell array
{ 94×1 double}
{223×1 double}
{195×1 double}
{212×1 double}
{187×1 double}
{ 89×1 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
Input Arguments
Name-Value Arguments
Output Arguments
Extended Capabilities
Version History
Introduced in R2015a
