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