# has

Check if expression contains particular subexpression

## Syntax

## Description

`has(`

returns
logical `expr`

,`subexpr`

)`1`

(true) if `expr`

contains
`subexpr`

. Otherwise, it returns logical `0`

(false).

If

`expr`

is an array,`has(expr,subexpr)`

returns an array of the same size as`expr`

. The returned array contains logical`1s`

(true) where the elements of`expr`

contain`subexpr`

, and logical`0s`

(false) where they do not.If

`subexpr`

is an array,`has(expr,subexpr)`

checks if`expr`

contains any element of`subexpr`

.

## Examples

### Check If Expression Contains Particular Subexpression

Use the `has`

function to check if an expression
contains a particular variable or subexpression.

Check if these expressions contain variable `z`

.

syms x y z has(x + y + z, z)

ans = logical 1

has(x + y, z)

ans = logical 0

Check if `x + y + z`

contains the following subexpressions. Note that
`has`

finds the subexpression `x + z`

even though the
terms `x`

and `z`

do not appear next to each other in the
expression.

has(x + y + z, x + y) has(x + y + z, y + z) has(x + y + z, x + z)

ans = logical 1 ans = logical 1 ans = logical 1

Check if the expression `(x + 1)^2`

contains `x^2`

.
Although `(x + 1)^2`

is mathematically equivalent to the expression
`x^2 + 2*x + 1`

, the result is a logical `0`

because
`has`

typically does not transform expressions to different forms when
testing for subexpressions.

has((x + 1)^2, x^2)

ans = logical 0

Expand the expression and then call `has`

to check if the result
contains `x^2`

. Because `expand((x + 1)^2)`

transforms the
original expression to `x^2 + 2*x + 1`

, the `has`

function finds the subexpression `x^2`

and returns logical
`1`

.

has(expand((x + 1)^2), x^2)

ans = logical 1

### Check If Expression Contains Any of Specified Subexpressions

Check if a symbolic expression contains any of subexpressions specified as elements of a vector.

If an expression contains one or more of the specified subexpressions,
`has`

returns logical `1`

.

syms x has(sin(x) + cos(x) + x^2, [tan(x), cot(x), sin(x), exp(x)])

ans = logical 1

If an expression does not contain any of the specified subexpressions,
`has`

returns logical `0`

.

syms x has(sin(x) + cos(x) + x^2, [tan(x), cot(x), exp(x)])

ans = logical 0

### Find Matrix Elements Containing Particular Subexpression

Using `has`

, find those elements of a symbolic
matrix that contain a particular subexpression.

First, create a matrix.

syms x y M = [sin(x)*sin(y), cos(x*y) + 1; cos(x)*tan(x), 2*sin(x)^2]

M = [ sin(x)*sin(y), cos(x*y) + 1] [ cos(x)*tan(x), 2*sin(x)^2]

Use `has`

to check which elements of `M`

contain
`sin(x)`

. The result is a matrix of the same size as
`M`

, with `1s`

and `0s`

as its elements.
For the elements of `M`

containing the specified expression,
`has`

returns logical `1s`

. For the elements that do
not contain that subexpression, `has`

returns logical
`0s`

.

T = has(M, sin(x))

T = 2×2 logical array 1 0 0 1

Return only the elements that contain `sin(x)`

and replace all other
elements with `0`

by multiplying `M`

by
`T`

elementwise.

M.*T

ans = [ sin(x)*sin(y), 0] [ 0, 2*sin(x)^2]

To check if any of matrix elements contain a particular subexpression, use `any`

.

any(has(M(:), sin(x)))

ans = logical 1

any(has(M(:), cos(y)))

ans = logical 0

### Find Vector Elements Containing Any of Specified Subexpressions

Using `has`

, find those elements of a symbolic
vector that contain any of the specified subexpressions.

syms x y z T = has([x + 1, cos(y) + 1, y + z, 2*x*cos(y)], [x, cos(y)])

T = 1×4 logical array 1 1 0 1

Return only the elements of the original vector that contain `x`

or
`cos(y)`

or both, and replace all other elements with
`0`

by multiplying the original vector by `T`

elementwise.

[x + 1, cos(y) + 1, y + z, 2*x*cos(y)].*T

ans = [ x + 1, cos(y) + 1, 0, 2*x*cos(y)]

### Use `has`

for Symbolic Functions

If `expr`

or `subexpr`

is a
symbolic function, `has`

uses `formula(expr)`

or
`formula(subexpr)`

. This approach lets the `has`

function check if an expression defining the symbolic function `expr`

contains an expression defining the symbolic function `subexpr`

.

Create a symbolic function.

syms x f(x) = sin(x) + cos(x);

Here, `sin(x) + cos(x)`

is an expression defining the symbolic function
`f`

.

formula(f)

ans = cos(x) + sin(x)

Check if `f`

and `f(x)`

contain
`sin(x)`

. In both cases `has`

checks if the expression
`sin(x) + cos(x)`

contains `sin(x)`

.

has(f, sin(x)) has(f(x), sin(x))

ans = logical 1 ans = logical 1

Check if `f(x^2)`

contains `f`

. For these arguments,
`has`

returns logical `0`

(false) because it does not
check if the expression `f(x^2)`

contains the letter `f`

.
This call is equivalent to `has(f(x^2), formula(f))`

, which, in turn,
resolves to `has(cos(x^2) + sin(x^2), cos(x) + sin(x))`

.

has(f(x^2), f)

ans = logical 0

### Check for Calls to Particular Function

Check for calls to a particular function by specifying the function name as the second argument. Check for calls to any one of multiple functions by specifying the multiple functions as a cell array of character vectors.

Integrate `tan(x^7)`

. Determine if the integration is successful by
checking the result for calls to `int`

. Because `has`

finds the `int`

function and returns logical `1`

(`true`

), the integration is not successful.

syms x f = int(tan(x^7), x); has(f, 'int')

ans = logical 1

Check if the solution to a differential equation contains calls to either
`sin`

or `cos`

by specifying the second argument as
`{'sin','cos'}`

. The `has`

function returns logical
`0`

(`false`

), which means the solution does not contain
calls to either `sin`

or `cos`

.

syms y(x) a sol = dsolve(diff(y,x) == a*y); has(sol, {'sin' 'cos'})

ans = logical 0

## Input Arguments

## Tips

`has`

does not transform or simplify expressions. This is why it does not find subexpressions like`x^2`

in expressions like`(x + 1)^2`

. However, in some cases`has`

might find that an expression or subexpression can be represented in a form other than its original form. For example,`has`

finds that the expression`-x - 1`

can be represented as`-(x + 1)`

. Thus, the call`has(-x - 1, x + 1)`

returns`1`

.If

`expr`

is an empty symbolic array,`has`

returns an empty logical array of the same size as`expr`

.

## Version History

**Introduced in R2015b**