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.
______________________________________________
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