Find Array Elements That Meet a Condition - MATLAB & Simulink (2024)

Apply a Single Condition

To apply a single condition, start by creating a 5-by-5 matrix that contains random integers between 1 and 15. Reset the random number generator to the default state for reproducibility.

rng defaultA = randi(15,5)

A = 5×5 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 = 5x5 logical array 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, you can use B to index into A.

A(B)

ans = 8×1 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 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, you can use the find function to locate all of the elements in A less than 9.

I = find(A < 9)

I = 8×1 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.

Apply Multiple Conditions

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 = 5×1 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 = 3×1 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 element equal to 8 that satisfies the conditions.

find(A<9 & ~mod(A,2) & A~=2)

ans = 14

The result indicates that A(14) = 8.

Replace Values That Meet a Condition

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 = 5×5 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

Next, replace all values in A that are not equal to 10 with a NaN value.

A(A~=10) = NaN

A = 5×5 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

Lastly, replace all of the NaN values in A with zeros and apply the logical NOT operator, ~A.

A(isnan(A)) = 0;C = ~A

C = 5x5 logical array 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.