Introduction and custom function
In a general case with floating point precision issues, one would be advised to use a tolerance value for comparisons against suspected zero values and that tolerance must be a very small value. A little robust method would use a tolerance that uses eps
in it. Now, since MATLAB basically performs subtractions with setdiff
, you can use eps
directly here by comparing for lesser than or equal to it to find zeros
.
This forms the basis of a modified setdiff for floating point numbers
shown here -
function [C,IA] = setdiff_fp(A,B)
%//SETDIFF_FP Set difference for floating point numbers.
%// C = SETDIFF_FP(A,B) for vectors A and B, returns the values in A that
%// are not in B with no repetitions. C will be sorted.
%//
%// [C,IA] = SETDIFF_FP(A,B) also returns an index vector IA such that
%// C = A(IA). If there are repeated values in A that are not in B, then
%// the index of the first occurrence of each repeated value is returned.
%// Get 2D matrix of absolute difference between each element of A against
%// each element of B
abs_diff_mat = abs(bsxfun(@minus,A,B.')); %//'
%// Compare each element against eps to "negate" the floating point
%// precision issues. Thus, we have a binary array of true comparisons.
abs_diff_mat_epscmp = abs_diff_mat<=eps;
%// Find indices of A that are exclusive to it
A_ind = ~any(abs_diff_mat_epscmp,1);
%// Get unique(to account for no repetitions and being sorted) exclusive
%// A elements for the final output alongwith the indices
[C,IA] = intersect(A,unique(A(A_ind)));
return;
Example runs
Case1 (With integers)
This will verify that setdiff_fp
works with integer arrays just the way setdiff
does.
A = [2 5];
B = [9 8 8 1 2 1 1 5];
[C_setdiff,IA_setdiff] = setdiff(B,A)
[C_setdiff_fp,IA_setdiff_fp] = setdiff_fp(B,A)
Output
A =
2 5
B =
9 8 8 1 2 1 1 5
C_setdiff =
1 8 9
IA_setdiff =
4
2
1
C_setdiff_fp =
1 8 9
IA_setdiff_fp =
4
2
1
Case2 (With floating point numbers)
This is to show that setdiff_fp
produces the correct results, while setdiff
doesn't. Additionally, this will also test out the output indices.
A=1.:0.1:1.5
B=[A+0.1 5.5 5.5 2.6]
[C_setdiff,IA_setdiff] = setdiff(B,A)
[C_setdiff_fp,IA_setdiff_fp] = setdiff_fp(B,A)
Output
A =
1.0000 1.1000 1.2000 1.3000 1.4000 1.5000
B =
1.1000 1.2000 1.3000 1.4000 1.5000 1.6000 5.5000 5.5000 2.6000
C_setdiff =
1.2000 1.4000 1.6000 2.6000 5.5000
IA_setdiff =
2
4
6
9
7
C_setdiff_fp =
1.6000 2.6000 5.5000
IA_setdiff_fp =
6
9
7