You can filter the elements of an array by
applying one or more conditions to the array. For instance, if you
want to examine only the even elements in a matrix, find the location
of all 0s in a multidimensional array, or replace NaN
values
in a discrete set of data. You can perform these tasks using a combination
of the relational and logical operators. The relational operators
(>, <, >=, <=, ==, ~=
) impose conditions
on the array, and you can apply multiple conditions by connecting
them with the logical operators and
, or
,
and not
, respectively denoted by &
, |
,
and ~
.
To apply a single condition, start by creating a 5-by-5 matrix, A
,
that contains random integers between 1 and 15.
rng(0) A = randi(15,5)
A = 13 2 3 3 10 14 5 15 7 1 2 9 15 14 13 14 15 8 12 15 10 15 13 15 11
Use the relational less than operator, <
,
to determine which elements of A
are less than
9. Store the result in B
.
B = A < 9
B = 0 1 1 1 0 0 1 0 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0
The result is a logical matrix. Each value in B
represents
a logical 1
(true
) or logical 0
(false
)
state to indicate whether the corresponding element of A
fulfills
the condition A < 9
. For example, A(1,1)
is 13
,
so B(1,1)
is logical 0
(false
).
However, A(1,2)
is 2
, so B(1,2)
is
logical 1
(true
).
Although B
contains information about which elements
in A
are less than 9, it doesn't tell you
what their values are. Rather than comparing
the two matrices element by element, use B
to index
into A
.
A(B)
ans = 2 2 5 3 8 3 7 1
The result is a column vector of the elements in A
that
are less than 9. Since B
is a logical matrix, this
operation is called logical indexing. In this
case, the logical array being used as an index is the same size as
the other array, but this is not a requirement. For more information,
see Using Logicals in
Array Indexing.
Some problems require information about the locations of
the array elements that meet a condition rather than their actual
values. In this example, use the find
function
to locate all of the elements in A
less than 9.
I = find(A < 9)
I = 3 6 7 11 14 16 17 22
The result is a column vector of linear indices. Each index
describes the location of an element in A
that
is less than 9, so in practice A(I)
returns the
same result as A(B)
. The difference is that A(B)
uses
logical indexing, whereas A(I)
uses linear indexing.
You can use the logical and
, or
,
and not
operators to apply any number of conditions
to an array; the number of conditions is not limited to one or two.
First, use the logical and
operator, denoted &
,
to specify two conditions: the elements must be less than 9 AND greater
than 2. Specify the conditions as a logical index to view the elements
that satisfy both conditions.
A(A<9 & A>2)
ans = 5 3 8 3 7
The result is a list of the elements in A
that
satisfy both conditions. Be sure to specify each condition with a
separate statement connected by a logical operator. For example, you
cannot specify the conditions above by A(2<A<9)
,
since it evaluates to A(2<A | A<9)
.
Next, find the elements in A
that are less
than 9 AND even numbered.
A(A<9 & ~mod(A,2))
ans = 2 2 8
The result is a list of all even elements in A
that
are less than 9. The use of the logical NOT operator, ~
,
converts the matrix mod(A,2)
into a logical matrix,
with a value of logical 1
(true
)
located where an element is evenly divisible by 2.
Finally, find the elements in A
that are
less than 9 AND even numbered AND not equal to 2.
A(A<9 & ~mod(A,2) & A~=2)
ans = 8
The result, 8, is even, less than 9, and not equal to 2. It
is the only element in A
that satisfies all three
conditions.
Use the find
function to get the index of
the 8
element that satisfies the conditions.
find(A<9 & ~mod(A,2) & A~=2)
ans = 14
The result indicates that A(14) = 8
.
Sometimes it is useful to simultaneously change the values of several existing array elements. Use logical indexing with a simple assignment statement to replace the values in an array that meet a condition.
Replace all values in A
that are greater
than 10 with the number 10.
A(A>10) = 10
A = 10 2 3 3 10 10 5 10 7 1 2 9 10 10 10 10 10 8 10 10 10 10 10 10 10
A
now has a maximum value of 10
.
Replace all values in A
that are not equal
to 10 with a NaN
value.
A(A~=10) = NaN
A = 10 NaN NaN NaN 10 10 NaN 10 NaN NaN NaN NaN 10 10 10 10 10 NaN 10 10 10 10 10 10 10
The resulting matrix has element values of 10 or NaN
.
Replace all of the NaN
values in A
with
zeros and apply the logical NOT operator, ~A
.
A(isnan(A)) = 0; C = ~A
C = 0 1 1 1 0 0 1 0 1 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0
The resulting matrix has values of logical 1
(true
)
in place of the NaN
values, and logical 0
(false
)
in place of the 10s. The logical NOT operation, ~A
,
converts the numeric array into a logical array such that A&C
returns
a matrix of logical 0
(false
)
values and A|C
returns a matrix of logical 1
(true
)
values.
and
| find
| isnan
| Logical Operators:
Short Circuit
| nan
| not
| or
| xor