Is it possible, in the expr
expression of the with()
function, to access the data
argument di开发者_JS百科rectly? Here's what I mean conceptually:
> print(df)
result qid f1 f2 f3
1 -1 1 0.0000 0.1253 0.0000
2 -1 1 0.0098 0.0000 0.0000
3 1 1 0.0000 0.0000 0.1941
4 -1 2 0.0000 0.2863 0.0948
5 1 2 0.0000 0.0000 0.0000
6 1 2 0.0000 0.7282 0.9087
> with(df, subset(.data, select=f1:f3)) # Doesn't work
Of course the above example is kind of silly, but it would be handy for things like this:
with(subset(df, f2>0), foo(qid, vars=subset(.data, select=f1:f3)))
I tried to poke around with environment()
and parent.frame()
etc., but didn't come up with anything that worked.
Maybe this is really a question about eval()
, since that's how with.default()
is implemented.
Using parent.frame()
:
# sample data:
set.seed(2436502)
dfrm <- data.frame(x1 = rnorm(100), x2 = rnorm(100), g1 = sample(letters, 100, TRUE))
# how to use it:
with(subset(dfrm, x1<0), {
str(parent.frame(2)$data)
"Hello!"
})
# 'data.frame': 47 obs. of 3 variables:
# $ x1: num -0.836 -0.343 -0.341 -1.044 -0.665 ...
# $ x2: num 0.362 0.727 0.62 -0.178 -1.538 ...
# $ g1: Factor w/ 26 levels "a","b","c","d",..: 11 4 15 19 8 13 22 15 15 23 ...
How the magic works
Using ls()
you can inspect parent.frames
:
with(subset(dfrm, x1<0), {
print(ls())
print(ls(parent.frame(1)))
print(ls(parent.frame(2)))
print(ls(parent.frame(3)))
})
# [1] "g1" "x1" "x2"
# [1] "enclos" "envir" "expr"
# [1] "data" "expr"
# [1] "dfrm"
As you can see:
parent.frame(3)
is base environment (in this case),parent.frame(2)
is environment ofsubset
functionparent.frame(1)
is environment of{
function (see?Paren
)
I tend to invert it, i.e. put the with()
outside and have subset()
work on its data:
R> data(mtcars)
R> with(subset(mtcars, gear==4), lm(mpg ~ wt)) # no data arg
Call:
lm(formula = mpg ~ wt)
Coefficients:
(Intercept) wt
42.49 -6.86
This is also a silly example because lm(mpg ~ wt, data=mtcars, subset=gear==4)
obviously does the same but you get the drift.
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