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.
## S3 method for class 'hyperframe': with(data, expr, ..., simplify = TRUE, ee = NULL, enclos=NULL)
- A hyperframe (object of class
"hyperframe") containing data.
- An Rlanguage expression to be evaluated.
- Logical. If
TRUE, the return value will be simplified to a vector whenever possible.
- Alternative form of
expr, as an object of class
- An environment in which to search for objects that are
not found in the hyperframe. Defaults to
This function evaluates the expression
expr in each row
of the hyperframe
data. It is a method for the generic
expr should be an Rlanguage expression
in which each variable name is either the name of a column in the
data, or the name of an object in the parent frame
(the environment in which
with was called.)
ee can be used as an alternative
expr and should be an expression object (of
For each row of
data, 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 a hyperframe
h has columns
with(h, A != B) inspects
each row of
data 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 <- with(H, rpoispp(L))