You are not getting the structure you want because the indexes are
wrong. They should be something more like this:
i <- 0
for (row in 1:nrow(alajuela_df)){
for (col in 1:ncol(alajuela_df)){
i <- i + 1
df[i,1]=alajuela_df[row,col]
}
}
but I think what you are doing can be written much
If you don't know the name of the attributes in advance, how can you know
the function name to be able to call it? This seems like a very flawed
approach.
Also, I would discourage the use of eval(parse(text = )), it's almost
always not the right way to do what you want to do. In your case,
I would say this is not an error, but I think what you wrote isn't
what you intended to do anyway.
y[1] is a data.frame which contains only the first column of y, which
you assign to x$C, so now x$C is a data.frame.
R allows data.frame to be plain vectors as well as matrices and
data.frames,
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
You could rewrite
1 - cos(x)
as
2 * sin(x/2)^2
and that might give you more precision?
On Wed, Aug 16, 2023, 01:50 Leonard Mada via R-help
wrote:
> Dear R-Users,
>
> I tried to compute the following limit:
> x = 1E-3;
> (-log(1 - cos(x)) - 1/(cos(x)-1)) / 2 - 1/(x^2) + log(x)
> # 0.4299226
You might also be able to rewrite
log(1 - cos(x))
as
log1p(-cos(x))
On Wed, Aug 16, 2023, 02:51 Iris Simmons wrote:
> You could rewrite
>
> 1 - cos(x)
>
> as
>
> 2 * sin(x/2)^2
>
> and that might give you more precision?
>
> On Wed, Aug 16, 2023, 01:50
Hi,
I'm using R in Emacs and I'm interested in programatically knowing the
details of all opened buffers; details such a buffer name, size, mode,
and possibly associated filename. I've been able to write such a
function in Emacs Lisp, but now I'd like to be able to call that
function from R, or
I might try something like this:
FUN1 <- function ()
{
threshold <- 4L
fails <- 0L
internal <- function() {
## do the actual downloading here
tryCatch({
download.file(<...>)
}, error = function() {
fails <<- fails + 1L
if
You probably want `names(test)`.
On Thu, May 25, 2023 at 7:58 PM Evan Cooch wrote:
> Suppose I have the following list:
>
> test <- list(a=3,b=5,c=11)
>
> I'm trying to figure out how to extract the characters to the left of
> the equal sign (i.e., I want to extract a list of the variable
Hi,
I think there are two easy ways to fix this. The first is to use a `switch`
to call the intended function, this should not be a problem since there are
a small number of print functions in **mixR**
```R
print.mixfitEM <- function (x, digits = getOption("digits"), ...)
{
switch(x$family,
You could also do
dim(x) <- c(length(x), 1)
On Sat, Aug 5, 2023, 20:12 Steven Yen wrote:
> I wish to stack columns of a matrix into one column. The following
> matrix command does it. Any other ways? Thanks.
>
> > x<-matrix(1:20,5,4)
> > x
> [,1] [,2] [,3] [,4]
> [1,]16 11
Hello,
I'm trying to demonstrate the behaviour of my R package and how line
directives change that behaviour. So I've got an R chunk like this:
<>=
{
#line 1 "file1.R"
fun <- function() {
pkg::fun()
}
#line 1 "file2.R"
fun()
}
@
but when it is rendered, the line
Hiya!
You can do this by specifying sub="c99" instead of "Unicode":
```R
x <- "fa\xE7ile"
xx <- iconv(x, "latin1", "UTF-8")
iconv(xx, "UTF-8", "ASCII", "c99")
```
produces:
```
> x <- "fa\xE7ile"
> xx <- iconv(x, "latin1", "UTF-8")
> iconv(xx, "UTF-8", "ASCII", "c99")
[1] "fa\\u00e7ile"
>
```
Hi Reed,
I need to stress before giving my answer that no solution can handle
everything. These scenarios will always lead to problems:
* if any of the formal arguments rely on the current state of the call stack
* if any of the formal arguments rely on a variable that is only
defined later in
Murdoch wrote:
> > I'm not an Emacs user, but the ESS-help mailing list (see
> > ess.r-project.org) might be able to help with this.
> >
> > Duncan Murdoch
> >
> > On 10/11/2023 3:43 a.m., Iris Simmons wrote:
> >> Hi,
> >>
> >>
> >&g
Hi Iago,
This is not a bug. It is expected. Patterns may not overlap. However, there
is a way to get the result you want using perl:
```R
gsub("([aeiouAEIOU])(?=[aeiouAEIOU])", "\\1_", "aerioue", perl = TRUE)
```
The specific change I made is called a positive lookahead, you can read
more
Hi Maria,
I had something similar on my Windows work laptop at some point where the
home directory was something containing non ASCII characters. The easy
solution is to copy said directly from the file explorer into
utils::shortPathName, and then set that as the home directory. In my case,
>
Hi Thomas,
If you want to compare the imaginary portions, you could do:
Im(z1) < Im(z2)
If you want to compare the magnitudes, you could do:
Mod(z1) < Mod(z2)
If you want to compare complex numbers, i.e. z1 < z2, well that just
doesn't make sense.
On Mon, Mar 25, 2024, 10:17 Thomas K
Hi Duncan,
I only know about sub() and gsub().
There is no way to have pattern be a regular expression and replacement be
a fixed string.
Backslash is the only special character in replacement. If you need a
reference, see this file:
This happens because "123e" looks like exponential form. This string has no
exponent, so it gets treated as 0 exponent.
If you're interested in converting hex numbers, append 0x:
as.numeric("0x123a")
or use strtoi:
strtoi("123a", 16)
On Wed, May 1, 2024, 15:24 Carl Witthoft wrote:
> Hello.
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