# eval.hyper

##### Evaluate an Expression in Each Row of a Hyperframe

An expression, involving the names of columns in a hyperframe, is evaluated separately for each row of the hyperframe.

- Keywords
- manip, spatial, programming

##### Usage

`eval.hyper(e, h, simplify = TRUE, ee = NULL)`

##### Arguments

- e
- An Rlanguage expression to be evaluated.
- h
- A hyperframe (object of class
`"hyperframe"`

) containing data. - simplify
- Logical. If
`TRUE`

, the return value will be simplified to a vector whenever possible. - ee
- Alternative form of
`e`

. See Details.

##### Details

This function evaluates the expression `e`

in each row
of the hyperframe `h`

.
The argument `e`

should be an Rlanguage expression
in which each variable name is either the name of a column in the
hyperframe `h`

, or the name of an object in the global environment.
For each row of `h`

, the expression will be evaluated
so that variables which are column names of `h`

are
interpreted as the entries for those columns in the current row.

For example, if `h`

has columns called `A`

and `B`

,
then `eval.hyper(A != B, h)`

inspects each row of `h`

in turn,
tests whether the entries in columns `A`

and `B`

are
equal, and returns the $n$ logical values.

##### Value

- Normally a list of length
$n$ (where $n$ is the number of rows) containing the results
of evaluating the expression for each row.
If
`simplify=TRUE`

and each result is a single atomic value, then the result is a vector or factor containing the same values.

##### See Also

##### Examples

```
# generate Poisson point patterns with intensities 10 to 100
H <- hyperframe(L=seq(10,100, by=10))
X <- eval.hyper(rpoispp(L),H)
```

*Documentation reproduced from package spatstat, version 1.12-5, License: GPL (>= 2)*