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.
eval.hyper(e, h, simplify = TRUE, ee = NULL)
- An Rlanguage expression to be evaluated.
- A hyperframe (object of class
"hyperframe") containing data.
- Logical. If
TRUE, the return value will be simplified to a vector whenever possible.
- Alternative form of
e. See Details.
This function evaluates the expression
e in each row
of the hyperframe
e should be an Rlanguage expression
in which each variable name is either the name of a column in the
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
interpreted as the entries for those columns in the current row.
For example, if
h has columns called
eval.hyper(A != B, h) inspects each row of
h in turn,
tests whether the entries in columns
equal, and returns the $n$ logical values.
- Normally a list of length
$n$ (where $n$ is the number of rows) containing the results
of evaluating the expression for each row.
simplify=TRUEand each result is a single atomic value, then the result is a vector or factor containing the same values.
# generate Poisson point patterns with intensities 10 to 100 H <- hyperframe(L=seq(10,100, by=10)) X <- eval.hyper(rpoispp(L),H)