The logical vectors created from logical and relational operations can be used to reference subarrays. Suppose X is an ordinary matrix and L is a matrix of the same size that is the result of some logical operation. Then X(L) specifies the elements of X where the elements of L are nonzero.
This kind of subscripting can be done in one step by specifying the logical operation as the subscripting expression. Suppose you have the following set of data:
x = [2.1 1.7 1.6 1.5 NaN 1.9 1.8 1.5 5.1 1.8 1.4 2.2 1.6 1.8];
The NaN is a marker for a missing observation, such as a failure to respond to an item on a questionnaire. To remove the missing data with logical indexing, use isfinite(x), which is true for all finite numerical values and false for NaN and Inf:
x = x(isfinite(x)) x = 2.1 1.7 1.6 1.5 1.9 1.8 1.5 5.1 1.8 1.4 2.2 1.6 1.8
Now there is one observation, 5.1, which seems to be very different from the others. It is an outlier. The following statement removes outliers, in this case those elements more than three standard deviations from the mean:
x = x(abs(x-mean(x)) <= 3*std(x)) x = 2.1 1.7 1.6 1.5 1.9 1.8 1.5 1.8 1.4 2.2 1.6 1.8
For another example, highlight the location of the prime numbers in Dürer's magic square by using logical indexing and scalar expansion to set the nonprimes to 0. (See The magic Function.)
B=magic(4)
B =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
A = B(:,[1 3 2 4]) % interchange 2nd and 3rd columns
A =
16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1
A(~isprime(A)) = 0
A =
0 3 2 13
5 0 11 0
0 0 7 0
0 0 0 0
The find function determines the indices of array elements that meet a given logical condition. In its simplest form, find returns a column vector of indices. Transpose that vector to obtain a row vector of indices. For example, start again with Dürer's magic square. (See The magic Function.)
k = find(isprime(A))'
picks out the locations, using one-dimensional indexing, of the primes in the magic square:
k =
2 5 9 10 11 13
Display those primes, as a row vector in the order determined by k, with
A(k)
ans =
5 3 2 11 7 13
When you use k as a left-hand-side index in an assignment statement, the matrix structure is preserved:
A(k) = NaN
A =
16 NaN NaN NaN
NaN 10 NaN 8
9 6 NaN 12
4 15 14 1