Find which vector elements are less than or equal to a given value.

Create a numeric vector.

a = [1 12 18 7 9 11 2 15];

测试向量的元素小于或等于12.

a <= 12

ans =1x8 logical array1 1 0 1 1 1 1 0

The result is a vector with values of logical1(真的）元素Asatisfy the expression.

使用逻辑值向量作为索引来查看值Athat are less than or equal to12.

A(A <= 12)

ans =1×61 12 7 9 11 2

结果是元素的子集A.

Create a matrix.

a =魔术（4）

A =4×416 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1

替换所有小于或等于的值9带有值10.

A(A <= 9) = 10

A =4×416 10 10 13 10 11 10 10 10 10 10 12 10 14 15 10

结果是一个新矩阵，其最小元素是10.

创建一个顺序的分类阵列。

a =分类（{'大的''medium''小的';'medium'...'小的''大的'}，{'小的''medium''大的'}，，“序数”，1）

A =2x3 categoricallarge medium small medium small large

The array has three categories:'小的','medium'， 和'大的'.

Find all values less than or equal to the category'medium'.

a <='medium'

ans =2x3 logical array0 1 1 1 1 0

A value of logical1(真的) indicates a value less than or equal to the category'medium'.

Compare the rows ofA.

a（1，:) <= a（2，:)

ans =1x3逻辑数组0 0 1

该功能返回逻辑1(真的) where the first row has a category value less than or equal to the second row.

Create a vector of complex numbers.

a = [1+i 2-2i 1+3i 1-2i 5-i];

Find the values that are less than or equal to3.

A（a <= 3）

ans =1×4 complex1.0000 + 1.0000i 2.0000 - 2.0000i 1.0000 + 3.0000i 1.0000 - 2.0000i

lecompares only the real part of the elements inA.

利用absto find which elements are within a radius of3from the origin.

A(abs(A) <= 3)

ans =1×3 complex1.0000 + 1.0000i 2.0000 - 2.0000i 1.0000 - 2.0000i

The result has one less element. The element1.0000 + 3.0000i不在半径内3from the origin.

Create aduration大批。

d =hours(21:25) + minutes(75)

d =1x5持续时间22.25 HR 23.25 HR 24.25 HR 25.25 HR 26.25 HR

测试阵列的小于或等于一个标准日的元素。

D <= 1

ans =1x5逻辑数组1 1 0 0 0