As you seem to have a need for this sort of capability (e.g. bquote),
see Section 6: "Computing on the Language" in the R Language
Definition manual. Actually, if you are interested in a concise
(albeit dense) overview of the R Language, you might consider going
through the whole manual.
Cheers,
Bert
On Wed, Oct 25, 2023 at 3:57 AM Shu Fai Cheung <shufai.che...@gmail.com> wrote:
>
> Dear Iris,
>
> Many many thanks! This is exactly what I need! I have never heard
> about bquote(). This function will also be useful to me on other
> occasions.
>
> I still have a lot to learn about the R language ...
>
> Regards,
> Shu Fai
>
>
> On Wed, Oct 25, 2023 at 5:24 PM Iris Simmons <ikwsi...@gmail.com> wrote:
> >
> > You can try either of these:
> >
> > expr <- bquote(lm(.(as.formula(mod)), dat))
> > lm_out5 <- eval(expr)
> >
> > expr <- call("lm", as.formula(mod), as.symbol("dat"))
> > lm_out6 <- eval(expr)
> >
> > but bquote is usually easier and good enough.
> >
> > On Wed, Oct 25, 2023, 05:10 Shu Fai Cheung <shufai.che...@gmail.com> wrote:
> >>
> >> Hi All,
> >>
> >> I have a problem that may have a simple solution, but I am not
> >> familiar with creating calls manually.
> >>
> >> This is example calling lm()
> >>
> >> ``` r
> >> set.seed(1234)
> >> n <- 10
> >> dat <- data.frame(x1 = rnorm(n),
> >> x2 = rnorm(n),
> >> y = rnorm(n))
> >>
> >> lm_out <- lm(y ~ x1 + x2, dat)
> >> lm_out
> >> #>
> >> #> Call:
> >> #> lm(formula = y ~ x1 + x2, data = dat)
> >> #>
> >> #> Coefficients:
> >> #> (Intercept) x1 x2
> >> #> -0.5755 -0.4151 -0.2411
> >> lm_out$call
> >> #> lm(formula = y ~ x1 + x2, data = dat)
> >> ```
> >>
> >> The call is stored, "lm(formula = y ~ x1 + x2, data = dat)", and names
> >> are not evaluated.
> >>
> >> I want to create a similar call, but only one of the elements is from a
> >> string.
> >>
> >> ```r
> >> mod <- "y ~ x1 + x2"
> >> ```
> >>
> >> This is what I tried but failed:
> >>
> >> ```r
> >> lm_out2 <- do.call("lm",
> >> list(formula = as.formula(mod),
> >> data = dat))
> >> lm_out2
> >> #>
> >> #> Call:
> >> #> lm(formula = y ~ x1 + x2, data = structure(list(x1 =
> >> c(-1.20706574938542,
> >> #> 0.27742924211066, 1.08444117668306, -2.34569770262935, 0.42912468881105,
> >> #> 0.506055892157574, -0.574739960134649, -0.546631855784187,
> >> -0.564451999093283,
> >> #> -0.890037829044104), x2 = c(-0.477192699753547, -0.998386444859704,
> >> #> -0.77625389463799, 0.0644588172762693, 0.959494058970771,
> >> -0.110285494390774,
> >> #> -0.511009505806642, -0.911195416629811, -0.83717168026894,
> >> 2.41583517848934
> >> #> ), y = c(0.134088220152031, -0.490685896690943, -0.440547872353227,
> >> #> 0.459589441005854, -0.693720246937475, -1.44820491038647,
> >> 0.574755720900728,
> >> #> -1.02365572296388, -0.0151383003641817, -0.935948601168394)), class
> >> = "data.frame", row.names = c(NA,
> >> #> -10L)))
> >> #>
> >> #> Coefficients:
> >> #> (Intercept) x1 x2
> >> #> -0.5755 -0.4151 -0.2411
> >> ```
> >>
> >> It does not have the formula, "as a formula": y ~ x1 + x2.
> >> However, the name "dat" is evaluated. Therefore, the call stored does
> >> not have the name 'dat', but has the evaluated content.
> >>
> >> The following fits the same model. However, the call stores the name,
> >> 'mod', not the evaluated result, y ~ x1 + x2.
> >>
> >> ```r
> >> lm_out3 <- lm(mod, data = dat)
> >> lm_out3
> >> #>
> >> #> Call:
> >> #> lm(formula = mod, data = dat)
> >> #>
> >> #> Coefficients:
> >> #> (Intercept) x1 x2
> >> #> -0.5755 -0.4151 -0.2411
> >> ```
> >>
> >> The following method works. However, I have to do a dummy call,
> >> extract the stored call, and set formula to the result of
> >> as.formula(mod):
> >>
> >> ```r
> >> lm_out3 <- lm(mod, data = dat)
> >> lm_out3
> >> #>
> >> #> Call:
> >> #> lm(formula = mod, data = dat)
> >> #>
> >> #> Coefficients:
> >> #> (Intercept) x1 x2
> >> #> -0.5755 -0.4151 -0.2411
> >>
> >> call1 <- lm_out3$call
> >> call1$formula <- as.formula(mod)
> >> lm_out4 <- eval(call1)
> >> lm_out4
> >> #>
> >> #> Call:
> >> #> lm(formula = y ~ x1 + x2, data = dat)
> >> #>
> >> #> Coefficients:
> >> #> (Intercept) x1 x2
> >> #> -0.5755 -0.4151 -0.2411
> >> ```
> >>
> >> Is it possible to create the call directly, with only 'mod' evaluated,
> >> and other arguments, e.g., 'dat', not evaluated?
> >>
> >> Regards,
> >> Shu Fai
> >>
> >> ______________________________________________
> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
