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.
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,
Actually a better solution would be to make PID into a factor. They can
always be coerced to a number, but will display with your meaningful labels.
Duncan Murdoch
On 25/10/2023 3:38 p.m., Duncan Murdoch wrote:
I don't see it documented, but it appears that the gee() function
assumes the id
I don't see it documented, but it appears that the gee() function
assumes the id variable can be coerced to a number. Your ids are in
PID, and are strings like "HIPS004", etc. Change that to "004" or a
numeric 4 and the error goes away.
Duncan Murdoch
On 25/10/2023 3:23 p.m., Sorkin, John
Colleagues,
I am receiving several error messages from the gee function. I don't understand
the ides the error messages are trying to impart, and I don't know how to debug
or correct the error. The error messages follow:
> fitgee <- gee(HipFlex ~
>
Às 00:22 de 25/10/2023, Sorkin, John escreveu:
Colleagues,
I have written an R function (see fully annotated code below), with which I
want to process a dataframe within levels of the variable StepType. My program
works, it processes the data within levels of StepType, but the usual headers
Dear John,
Printing inside the function is problematic. Your function itself does
NOT print the labels.
Just as a clarification:
F = factor(rep(1:2, 2))
by(data.frame(V = 1:4, F = F), F, function(x) { print(x); return(NULL); } )
# V F
# 1 1 1
# 3 3 1
# V F
# 2 2 2
# 4 4 2
# F: 1 <- this
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
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
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
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,
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