Partially evaluate an expression. — partial

This function partially evaluates an expression, using information from the tbl to determine whether names refer to local expressions or remote variables. This simplifies SQL translation because expressions don't need to carry around their environment - all relevant information is incorporated into the expression.

Usage

partial_eval(call, data, env = caller_env(), vars = deprecated(), error_call)

Arguments

call: an unevaluated expression, as produced by quote()
data: A lazy data frame backed by a database query.
env: environment in which to search for local values
vars: : Pass a lazy frame to data instead.

Symbol substitution

partial_eval() needs to guess if you're referring to a variable on the server (remote), or in the current environment (local). It's not possible to do this 100% perfectly. partial_eval() uses the following heuristic:

If the tbl variables are known, and the symbol matches a tbl variable, then remote.
If the symbol is defined locally, local.
Otherwise, remote.

You can override the guesses using local() and remote() to force computation, or by using the .data and .env pronouns of tidy evaluation.

Examples

lf <- lazy_frame(year = 1980, id = 1)
partial_eval(quote(year > 1980), data = lf)
#> year > 1980

ids <- c("ansonca01", "forceda01", "mathebo01")
partial_eval(quote(id %in% ids), lf)
#> id %in% c("ansonca01", "forceda01", "mathebo01")

# cf.
partial_eval(quote(id == .data$id), lf)
#> id == id

# You can use local() or .env to disambiguate between local and remote
# variables: otherwise remote is always preferred
year <- 1980
partial_eval(quote(year > year), lf)
#> year > year
partial_eval(quote(year > local(year)), lf)
#> year > 1980
partial_eval(quote(year > .env$year), lf)
#> year > 1980

# Functions are always assumed to be remote. Use local to force evaluation
# in R.
f <- function(x) x + 1
partial_eval(quote(year > f(1980)), lf)
#> year > f(1980)
partial_eval(quote(year > local(f(1980))), lf)
#> year > 1981