Dear Bert,
Many thanks for your suggestion! I am reading the section to
understand more about this topic. It is highly relevant to what I plan
to work on.
Regards,
Shu Fai
On Thu, Oct 26, 2023 at 5:38 AM Bert Gunter wrote:
>
> As you seem to have a need for this sort of capability (e.g. bquote)
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,
B
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 wrote:
>
> You can try ei
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 wrote:
> Hi All,
>
> I have a pro
Sorry for a typo, regarding the first attempt, lm_out2, using
do.call(), I meant:
'It does have the formula, "as a formula": y ~ x1 + x2.
However, the name "dat" is evaluated. ...'
Regards,
Shu Fai
On Wed, Oct 25, 2023 at 5:09 PM Shu Fai Cheung wrote:
>
> Hi All,
>
> I have a problem that may h
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
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