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)data |
A hyperframe (object of class |
expr |
An R language expression to be evaluated. |
... |
Ignored. |
simplify |
Logical. If |
ee |
Alternative form of |
enclos |
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
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.
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.
Adrian Baddeley Adrian.Baddeley@curtin.edu.au
and Rolf Turner r.turner@auckland.ac.nz
# generate Poisson point patterns with intensities 10 to 100
H <- hyperframe(L=seq(10,100, by=10))
if(require(spatstat.core)) {
X <- with(H, rpoispp(L))
} else {
X <- with(H, runifrect(rpois(1, L)))
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