An expression, involving the names of columns in a hyperframe, is evaluated separately for each row of the hyperframe.
# S3 method for hyperframe
with(data, expr, ...,
simplify = TRUE,
ee = NULL, enclos=NULL)A hyperframe (object of class "hyperframe")
containing data.
An R language expression to be evaluated.
Ignored.
Logical. If TRUE, the return value
will be simplified to a vector whenever possible.
Alternative form of expr, as an object of class
"expression".
An environment in which to search for objects that are
not found in the hyperframe. Defaults to parent.frame().
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.
This function evaluates the expression expr in each row
of the hyperframe data. It is a method for the generic
function with.
The argument expr should be an R language expression
in which each variable name is either the name of a column in the
hyperframe data, or the name of an object in the parent frame
(the environment in which with was called.)
The argument ee can be used as an alternative
to expr and should be an expression object (of
class "expression").
For each row of data, the expression will be evaluated
so that variables which are column names of data are
interpreted as the entries for those columns in the current row.
For example, if a hyperframe h has columns
called A and B, then with(h, A != B) inspects
each row of data in turn,
tests whether the entries in columns A and B are
equal, and returns the \(n\) logical values.
# NOT RUN {
# generate Poisson point patterns with intensities 10 to 100
H <- hyperframe(L=seq(10,100, by=10))
X <- with(H, rpoispp(L))
# }
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