Not a direct answer but you may find lm.fit worth experimenting with.
Also try the high-performance computing task view on CRAN
Cheers,
Andrew
--
Andrew Robinson
Chief Executive Officer, CEBRA and Professor of Biosecurity,
School/s of BioSciences and Mathematics & Statistics
Universit
Hi John,
the negative binomial is a tricky one - there are several different
parameterisations and therefore different interpretations of the parameters.
Joseph Hilbe wrote a whole book on it that might be wroth checking.
Cheers,
Andrew
--
Andrew Robinson
Chief Executive Officer, CEBRA and
Hi Bob,
there may be more efficient ways to go about it but I would use R to scrape the
contents of
http://home.brisnet.org.au/~bgreen/Data/Hanson1/
http://home.brisnet.org.au/~bgreen/Data/Hanson2/
in order to form the URLs of the files, and then loop over the URLs.
Cheers,
Andrew
--
Andrew
@ 1.80GHz 2.00 GHz, 8 cores
16.0 GB
Does anyone know if my script has any issues, or it is normal for sbf() to
run this long?
Thank you!
Andrew
> sessionInfo()
R version 4.3.1 (2023-06-16 ucrt)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 19045)
Matri
packages/methods for feature selection prior to building a classification
model using machine learning models.
Thanks,
Andrew
[[alternative HTML version deleted]]
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https://stat
It says that nser requires the most recent version of magrittr that you do
not have installed. You must update magrittr before attempting to install
nser:
update.packages(oldPkgs = "magrittr")
or at the prompt you were presented before, choose to update magrittr
before installing nser.
On Sun, A
What I meant is that that
mydata[, !grepl("^yr", colnames(mydata)), drop = FALSE]
and
mydata[!grepl("^yr", colnames(mydata))]
should be identical. Some people would prefer the first because the
indexing looks the same as matrix indexing, whereas some people would
prefer the second because it is
one column where the object in the
column is the same character vector.
You could alternatively use
x["V2"]
which should be identical to x[, "V2", drop = FALSE], but some people don't
like that because it doesn't look like matrix indexing anymore.
On Sun, Feb 12
It is not possible, apply() converts its argument to an array. You might be
able to use split() and lapply() to solve your problem.
On Tue, Feb 7, 2023, 07:52 Naresh Gurbuxani
wrote:
>
> > Consider a data.frame whose different columns have numeric, character,
> > and factor data. In apply funct
Try something like
with(df, predict(smooth.spline(x = altitude, y = atm_values), deriv = 1))
Cheers,
Andrew
--
Andrew Robinson
Chief Executive Officer, CEBRA and Professor of Biosecurity,
School/s of BioSciences and Mathematics & Statistics
University of Melbourne, VIC 3010 Australia
Tel:
I would use replicate() to do an operation with random numbers repeatedly:
```
mysim <- replicate(10, {
two.mat <- matrix(rnorm(4), 2, 2)
four.mat <- matrix(rnorm(16), 4, 4)
list(two.mat = two.mat, four.mat = four.mat)
})
```
which should give you a matrix-list. You can slice this mat
t;
>
> I need to have the same number of backticks in the opening and closing
> marker. So I make the pattern more complicated, and it doesn't work:
>
>pattern2 <- "\n([`]{3,})html\n.*?\n\\1\n"
>
> This matches all of x:
>
>> pattern
grep(value = TRUE) just returns the strings which match the pattern. You
have to use regexpr() or gregexpr() if you want to know where the matches
are:
```
x <- "abaca"
# extract only the first match with regexpr()
m <- regexpr("a.*?a", x)
regmatches(x, m)
# or
# extract every match with gregex
R converts floats to strings with ~15 digits of accuracy, specifically
to avoid differentiating between 1 and 1 + .Machine$double.eps, it is
assumed that small differences such as this are due to rounding errors
and are unimportant.
So, if when making your factor, you want all digits, you could wr
I tried this again with R 2.15.3, the oldest version I have installed,
and I still got the same behaviour. It extracts the first exact match,
then the only partial match, then NULL.
On Tue, Jan 24, 2023 at 5:04 PM Rolf Turner wrote:
>
>
> Has something changed, but I missed it?
>
> My recollectio
junk$y extracts the element named "y" because it found an exact match for
the name.
junk$yu extracts nothing because it does not find an exact match and finds
multiple partial matches.
junk$yuc or junk$yur would work because it finds no exact match and then
exactly one partial match.
On Tue, Jan
You'll want to use grep() or grepl(). By default, grep() uses extended
regular expressions to find matches, but you can also use perl regular
expressions and globbing (after converting to a regular expression).
For example:
grepl("^yr", colnames(mydata))
will tell you which 'colnames' start with
Returning the last value of { is the basis of functions not needing a
return statement. Before R invokes a function (specifically a closure), it
creates a new context. When R evaluates a call to return, it looks for a
context to return from and finds the context of function, ending the
context and
/sin(x), -pi/2, pi/2)
works perfectly fine and is a better function definition anyway.
`integrate` in the {stats} package is unlikely to change, so use the
alternate function definition or use {pracma}.
On Sat, Jan 7, 2023 at 10:41 PM Leonard Mada wrote:
>
> Dear Andrew,
>
>
> Th
You're dividing 0 by 0, giving you NaN, perhaps you should try
function(x) ifelse(x == 0, 0, x^3/sin(x))
On Sat, Jan 7, 2023, 22:24 Leonard Mada via R-help
wrote:
> Dear List-Members,
>
> I encounter a problem while trying to integrate the following function:
>
> integrate(function(x) x^3 / sin
I converted `date` to a factor and it seemed to work:
```
library(ggplot2)
library(cowplot)
date <- c("12-29","12-30","01-01")
date <- factor(date, labels = unique(date))
PT <- c(.106,.130,.121)
data <- data.frame(date,PT)
ggplot(data, aes(x=date,y=PT,group=1))+
geom_point(size=4)+
geom_line(
Keep in mind that in thie example you're processing x and placing the
result back in x (so x must already exist). You can write this a bit
more cleanly using the -> variant of the assignment operator as follows:
x |> cos() |> max(pi/4) |> round(3) -> x
Hth,
Andrew.
exists() is for bindings in an environment, not for names in a list. try
"T1A1" %in% names(E$L[[i]])
instead
On Tue, Dec 27, 2022, 12:36 akshay kulkarni wrote:
> Dear members,
> I have the following code:
> > E <- new.env()
> > E$L <- list()
> > i <- 1
> > E$
1. The execution environment for a script is the global environment. Each R
script run from a shell will be given its own global environment. Each R
session has exactly one global environment, but you can have several active
R sessions.
2. Using return in a script instead of a function will throw
these in order
to obtain the correct interval sizes.
Hth,
Andrew.
On 2/12/2022 14:18, Evan Cooch wrote:
Was wondering if there is an 'efficient/elegant' way to do the following
(without tidyverse). Take a string
abaaabbabaaab
Its easy enough to count the number of times the c
try
gregexpr('b+', target_string)
which looks for one or more b characters, then get the attribute
"match.length"
On Fri, Dec 2, 2022, 18:56 Evan Cooch wrote:
> Was wondering if there is an 'efficient/elegant' way to do the following
> (without tidyverse). Take a string
>
> abaaabbabaaab
>
For print(), digits is the minimal number of significant digits. In
your case, rounding the first column to the 3rd decimal place gives at
least 2 sigfigs and rounding the second column to the 2nd decimal
place.
If you want to print all numbers to two significant digits, regardless
of what other n
This seems to be a bug. I tried creating this function in the global
environment:
str.pdMat <- function (object, ...)
{
if (nlme::isInitialized(object)) {
NextMethod()
}
else {
cat(" Uninitialized positive definite matrix structure of class ",
class(object)[
the return list in any way.
On Mon, Nov 7, 2022 at 12:38 PM akshay kulkarni wrote:
>
> Dear Andrew
> It doesn't work:
>
> > lapply(TP,function(x){print(x^2)})
> [1] 1
> [1] 4
> [1] 9
> [1] 16
> [[1]]
> [1] 1
>
> [[2]]
> [
put print() around x^2
On Mon, Nov 7, 2022, 12:18 akshay kulkarni wrote:
> Dear members,
> I have the following code and output:
>
> > TP <- 1:4
> > lapply(TP,function(x){print(x);x^2})
> [1] 1
> [1] 2
> [1] 3
> [1] 4
> [[1]]
> [1] 1
>
> [[2]]
> [1] 4
>
> [[3]]
> [1]
In an R session, run this:
writeLines(normalizePath(R.home("bin")))
Right click your .R file > Open with > Choose another app > Check the box
"Always use this app to open .R files" > Look for another app on this PC
Paste the directory found above, then select "Rgui.exe"
On Fri, Nov 4, 2022, 04:4
partial match attr is for something like
attr(data.frame(), "cla")
which will partially match to "class".
On Sun, Oct 30, 2022, 12:55 Joshua Ulrich wrote:
> For what it's worth, I set these options to warn me about partial matches:
>
> options(warnPartialMatchArgs = TRUE,
> warnPartial
I would suggest using strwrap(), the documentation at ?strwrap has
plenty of details and examples.
For paragraphs, I would usually do something like:
strwrap(x = , width = 80, indent = 4)
On Fri, Oct 28, 2022 at 5:42 PM Leonard Mada via R-help
wrote:
>
> Dear R-Users,
>
> text = "
> What is the
$ does not evaluate its second argument, it does something like
as.character(substitute(name)).
You should be using
lapply(list, function(x) x$a)
or
lapply(list, `[[`, "a")
On Thu, Oct 27, 2022, 12:29 Hilmar Berger wrote:
> Dear all,
>
> I'm a little bit surprised by the behavior of the $ o
n require( 'base' ), what is the "first argument"?
>
> On Mon, Oct 24, 2022 at 12:29 PM Andrew Simmons
> wrote:
> >
> > require(), similarly to library(), does not evaluate its first argument
> UNLESS you add character.only = TRUE
> >
> > req
require(), similarly to library(), does not evaluate its first argument
UNLESS you add character.only = TRUE
require( packages_i_want_to_use[1], character.only = TRUE)
On Mon, Oct 24, 2022, 12:26 Kelly Thompson wrote:
> # Below, when using require(), why do I get the error message "Error
> in
(readLines(FILE))
try(sys.source(FILE, envir = environment()))
})
The part that matters is that the function body is wrapped with
braces. `if` statements inside braces or parenthesis (or possibly
brackets) will continue looking for `else` even after `cons.expr` and
a newline has been fully par
The error comes from the expression not being wrapped with braces. You
could change it to
if (is.matrix(r)) {
r[w != 0, , drop = FALSE]
} else r[w != 0]
or
{
if (is.matrix(r))
r[w != 0, , drop = FALSE]
else r[w != 0]
}
or
if (is.matrix(r)) r[w != 0, , drop = FALSE] else r[w
To install the packages from source, you need to install make, gcc, and g++:
sudo apt install make
sudo apt install gcc
sudo apt install g++
then try installing them again
On Thu, Oct 6, 2022 at 2:54 AM Rhon Calderon, Eric
wrote:
>
> Hi,
>
> I am using R in my HPC terminal. After many tries
he Windows clipboard is capable of holding Unicode text. Any advice
would be gratefully received.
Thanks,
Andrew.
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PLEASE do read
thout the need to set options or
specify the encoding with Rscript. Very nice and kudos to the R
developers for taking the leap!
Thanks everyone for your help.
Cheers,
Andrew.
On 21/09/2022 16:40, Andrew Simmons wrote:
If you're running it from Rscript, you'll have to specify the enco
ay text help on Linux, but I was logged in remotely.
Text-based help is fine when ssh-ing into a machine, but HTML help is
much nicer to read and navigate.
I think I'll just switch over to Rterm for a while, but I can also check
out ess, which I wasn't aware of.
Thanks a lot,
Andre
Hello,
I'm going to quote the tempdir() doc page: "By default, tmpdir will be the
directory given by tempdir(). This will be a subdirectory of the
per-session temporary directory found by the following rule when the R
session is started. The environment variables TMPDIR, TMP and TEMP are
checked
In the first scenario, your object is class AB, then class A with distance
1, then class B with distance 2. This means that method A is preferable
since it is less distance away than method B.
However, in your second function, both methods are a total distance of 3
away, so (as far as I know) it c
y what you wrote and not have any issues. Are you using an older
version of R?
On Wed., Sep. 21, 2022, 14:20 Andrew Hart via R-help,
wrote:
> On 21/09/2022 11:46, Bert Gunter wrote:
> > ?options
> >
> > options(encoding = "utf-8")
> > in a startup file or fu
In general, you should be using inherits(netwotks, "matrix") or
is(networks, "matrix") instead of class() ==
Your function fails because your object has multiple classes so class==
returns multiple logical values so if will fail.
But inherits or is will return one logical value, so if will not ra
g source to read in the R
script. Mind you, if I do
Rscript -e source('myfile.R')
it works properly just like in Rgui.
Once again, Thanks heaps.
Cheers,
Andrew.
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e or the user Rprofile or an
environment variable I can set or some other way to instruct R to always
assume that input stream encodings will be utf-8 unless otherwise
specified? This way, I would only ever have to supply an encoding or
fileEncoding argument to specify "lati
Hello,
I'm working on a function envvars() which I'd like to get, set, and
remove environment variables in a manner similar to options(). At the
R level, I have this function:
envvars <- function (...)
.External2(C_envvars, pairlist(...))
and then at the C level:
#define set_R_Visible(X) (eval
What you're asking for doesn't make sense: 9098 and 09098 are the same
9098L == 09098L
If you mean specifically while printing, you could use sprintf:
cat(sprintf("%05d", 9098))
On Sun., Sep. 11, 2022, 14:58 akshay kulkarni,
wrote:
> Dear Tim,
> So there is no way to coerce
1 and 2 are not valid identifiers in R, so you need to surround them with
backticks to make them valid, the same as quoting a string:
switch(Stst, `1` = print("NO"), `2` = print("YES"))
On Wed., Sep. 7, 2022, 14:35 akshay kulkarni, wrote:
> Dear members,
> The foll
You can specify multiple indexes to replace at once, so you can avoid
a for loop entirely like this:
M <- matrix(nrow = 10, ncol = 10)
M[1:4, 5: 6] <- M[5: 6, 1:4] <- m[1, 2]
M[1:4,7] <- M[ 7, 1:4] <- m[1, 3]
M[1:4, 8:10] <- M[8:10, 1:4] <- m[1, 4]
M[5:6,7] <- M[ 7, 5:6] <- m[2, 3]
M[5:
I like package openxlsx, with the function openxlsx::read.xlsx()
Another common package that people use readxl
On Tue., Aug. 23, 2022, 7:51 p.m. Anas Jamshed,
wrote:
> I have .xlsx files with gene names in first column.How can read and load in
> R?
>
> [[alternative HTML version deleted
?regex is a nice starting point, it's got plenty of details on meta
characters and characters classes, if you need more advanced stuff you'll
probably have to look at the perl regex documentation, I believe it's
linked in ?regex.
I might try something like
grep("^[xz]\\.")
or
grep("^[xz][.]")
You can use getAnywhere
On Sun, Jun 19, 2022, 13:23 Christofer Bogaso
wrote:
> Hi,
>
> I am trying to see the source code of rstandard function. I tried below,
>
> > methods('rstandard')
>
> [1] rstandard.glm* rstandard.lm*
>
> What do I need to do if I want to see the source code of rstandard.l
I think you have to use print(df$price, digits = 14)
On Tue, May 24, 2022, 10:01 maithili_shiva--- via R-help <
r-help@r-project.org> wrote:
> Dear Forum
> In my code, I am trying to compute Value at risk.
> I have an issue as mentioned below:
> I have defined
>
>
> options(digits = 14)
> But my
A sequence where 'from' and 'to' are both integer valued (not necessarily
class integer) will use R_compact_intrange; the return value is an integer
vector and is stored with minimal space.
In your case, you specified a 'from', 'to', and 'by'; if all are integer
class, then the return value is als
It seems like the current version of lubridate is 1.8.0, which does
raise a warning for an invalid timezone, just like as.POSIXct. This is
what I tried:
print(lubridate::parse_date_time("1970-01-01 00:01:00", "ymd
HMS" , tz = "PST"))
print(as.POSIXct("1970-01-01
It should also be noted that format(x, digits = 17) instead of
as.character(x) won't lose any accuracy.
On Thu, Feb 17, 2022, 17:41 Marius Hofert
wrote:
> Dear expeRts,
>
> I'm familiar with IEEE 754. Is there an easy way to explain why even
> just printing of small numbers fails?
>
> 1e-317 # 1
The NOTE saying 'Possibly misspelled words in DESCRIPTION' can probably be
ignored (though I would probably put the name of your package in single
quotes).
The NOTE 'Non-standard files/directories found at top level' means that you
should move the non-standard files to a different location OR add
y, c, and f only exist in the context of mac2
If you want to use them, you'll have to write mac2$y, mac2$c, or mac2$f (or
the [[ versions mac2[["y"]], mac2[["c"]], or mac2[["f"]])
Combining that with index i would then look like mac2$y[[i]] or mac2[[i,
"y"]]
Also, I think you want to use aes_strin
nrow() is just the numbers of rows in your data frame, use seq_len(nrow())
or seq(nrow()) to loop through all row numbers
On Wed, Dec 22, 2021, 12:08 Kai Yang via R-help
wrote:
> Hello R team,I want to use for loop to generate multiple plots with 3
> parameter, (y is for y axis, c is for color a
I've tried a bunch of different R versions, I can't seem to replicate what
you described. Here's the code I used:
R.pattern <- paste0("^R-", .standard_regexps()$valid_R_system_version, "$")
x <- list.files(dirname(R.home()), pattern = R.pattern, full.names = TRUE)
x <- x[file.info(x, extra_cols =
vector of the
> classes which UseMethod() uses, is available as .class2(x) since R
> version 4.0.0. (This also applies to S4 objects when S3 dispatch is
> considered, see below.)"
>
> I think this is the "official" explanation, but I find it rather opaque.
>
This is because + dispatches on the class attribute, which a string like
"test" has set to NULL, so it doesn't dispatch. You can add the class
yourself like structure("test", class = "character") and that should work.
I'm not sure where it's explained, but most primitive functions dispatch on
the
ction converts this to the UTF-8 encoding that
> **should** display as a "degree" symbol, subject to the above
> referenced caveats.
> Hence:
> > intToUtf8(176)
> [1] "°" ##degree character
>
> ## So, for example, you can do:
> > degrF <-paste0(i
ard wrote:
>
> > Thanks, Andrew. I will.
>
> plotmath didn't have the solution; the use of the LaTeX ^ for a superscript
> had a character or number preceeding it. Using 'degree' prints that string
> on the axis.
>
> What does work is using the unicode for t
Excuse my brevity, but take a look at ?plotmath
It has tons of tips for making pretty labels
On Tue, Nov 30, 2021, 14:05 Rich Shepard wrote:
> I want to present the temperature on the Y-axis label as 'Water Temperature
> (oC)' with the degree symbol as a superscript.
>
> My web search found a c
It seems like the headers are misnamed, that should be a comma between
sampdate and param, not a period
On Tue, Nov 30, 2021, 09:35 Rich Shepard wrote:
> A short data file:
> site_nbr,sampdate.param,quant,unit
> 31731,2005-07-12,temp,19.7,oC
> 31731,2007-03-28,temp,9,oC
> 31731,2007-06-27,temp,1
though has the data type as chr.
> >
> > Once changing the type to 'Date', however, the date is reconfigured.
> For instance, 15/01/2010 (15 January 2010), becomes 0015-01-20.
> >
> > I've tried ```data_long$Date <- as.Date(data_long$Date, format =
> "%d/%m.
If you want to use the 'methods' package, you could do something like:
replace <- function (p1, p2, ...)
{
stop(gettextf("cannot 'replace' with arguments of class %s and %s",
sQuote(class(p1)[1L]), sQuote(class(p2)[1L])))
}
methods::setGeneric("replace")
methods::setMethod("replac
First, your signature for names.pm is wrong. It should look something more
like:
names.pm <- function (x)
{
}
As for the body of the function, you might do something like:
names.pm <- function (x)
{
NextMethod()
}
but you don't need to define a names method if you're just going to call
cat in R behaves similarly to cat in unix-alikes, sends text to a stdout.
Usually, that stdout would be a file, but usually in R it is the R Console.
I think it might also help to note the difference between cat and print:
x <- "test\n"
cat(x)
print(x)
produces
> cat(x)
test
> print(x)
[1] "t
You probably want to use cat and print for these lines. These things won't
print when not run at the top level, so if you want them to print, you must
specify that.
On Tue, Nov 2, 2021, 13:18 Rich Shepard wrote:
> I've read ?sink and several web pages about it but it's not working
> properly
> w
If you're thinking about using environments, I would suggest you initialize
them like
x <- new.env(parent = emptyenv())
Since environments have parent environments, it means that requesting a
value from that environment can actually return the value stored in a
parent environment (this isn't an
I would usually use 'tapply'. It splits an object into groups, performs
some function on each group, and then (optionally) converts the input to
something simpler.
For example:
tapply(dat$wt, dat$Year, mean) # mean by Year
tapply(dat$wt, dat$Sex , mean) # mean by Sex
tapply(dat$wt, list(dat$Yea
It might not be random, depending upon a seed being used (usually by
set.seed or RNGkind).
However, it's the best method for generating a random number within a
specified range without weights.
If you want weights, there are many other random number generation
functions, most notably rnorm. You c
You've just got the brackets in the wrong spot:
creditsub <- read.csv(unz(here::here("Data.zip"), "creditcardsub.csv"))
instead of
creditsub <- read.csv(unz(here::here("Data.zip", "creditcardsub.csv")))
Also, you probably want to save that connection and close it manually.
creditsub <- read
You're missing a comma, this should fix it
df[(df$Y>0.2) & (df$X<10),][2, ]
On Thu, Oct 14, 2021, 03:51 Luigi Marongiu wrote:
> Hello,
> I have selected a subset of a dataframe with the vector syntax (if
> this is the name):
> ```
> > df[(df$Y>0.2) & (df$X<10),]
> YX
> 10 0
Hello,
The issue comes that 'apply' tries to coerce its argument to a matrix. This
means that all your columns will become character class, and the result
will not be what you wanted. I would suggest something more like:
sapply(d, function(x) all(x[!is.na(x)] <= 3))
or
vapply(d, function(x) a
Hello,
I think you have to use 'matches' instead of 'starts_with', I believe
starts_with does not accept a regular expression whereas matches does.
On Tue, Oct 5, 2021 at 12:15 PM Anne Zach wrote:
> Dear R users,
>
> I have a dataframe that contains several variables, among which 105
> corresp
t know of any packages that do such a thing.
If such a thing exists in another language, there's probably an R package
with a similar name containing ports of such functions, that might be your
best bet. I hope this helps.
On Tue, Sep 28, 2021, 23:51 Leonard Mada wrote:
> Thank you Andr
I think what you're looking for is 'strwrap', it's in package base.
On Tue, Sep 28, 2021, 22:26 Leonard Mada via R-help
wrote:
> Dear R-Users,
>
>
> Does anyone know any package or library that implements functions for
> word wrapping?
>
>
> I did implement a very rudimentary one (Github link be
aks2, right = FALSE, i = TRUE),
check.names = FALSE
)
I hope this helps.
On Fri, Sep 17, 2021 at 6:26 PM Leonard Mada wrote:
> Hello Andrew,
>
>
> But "cut" generates factors. In most cases with real data one expects to
> have also the ends of the interval: the argument "incl
'warn', but probably not to change the
default of 'include.lowest'. I hope this helps
On Fri, Sep 17, 2021 at 6:01 PM Leonard Mada wrote:
> Thank you Andrew.
>
>
> Is there any reason not to make: include.lowest = TRUE the default?
>
>
> Regarding the NA
Regarding your first point, argument 'include.lowest' already handles this
specific case, see ?.bincode
Your second point, maybe it could be helpful, but since both 'cut.default'
and '.bincode' return NA if a value isn't within a bin, you could make
something like this on your own.
Might be worth
lue), "test"))
}
this won't work for a case like:
wrapper(right(letters))
because now, when you do 'substitute', it will give you the literal symbol
'x', not really what you wanted, because of the non-standard evaluation. Of
course, do whatever you want to make the s
I'd like to point out that base R can handle a list as a data frame column,
it's just that you have to make the list of class "AsIs". So in your example
temp <- list("Hello", 1, 1.1, "bye")
data.frame(alpha = 1:4, beta = I(temp))
means that column "beta" will still be a list.
On Wed, Sep 15,
c("some names"))
x <- `*tmp*`
Another example,
y <- `names<-`(x, value = c("some names"))
now y will be equivalent to x if we did
names(x) <- c("some names")
except that the first will not update x, it will still have its old names.
On Mon, Sep 13, 2021
'is.numeric' is a function that returns whether its input is a numeric
vector. It looks like what you want to do is
VPN_Sheet1 <- VPN_Sheet1[!vapply(VPN_Sheet1$HVA, "is.numeric", NA), ]
instead of
VPN_Sheet1 <- VPN_Sheet1[!is.numeric(VPN_Sheet1$HVA), ]
I hope this helps, and see ?vapply if nece
ting.
> The forced evaluation makes sense for subsetting.
>
>
> On 9/13/2021 9:42 PM, Leonard Mada wrote:
>
> Hello Andrew,
>
>
> I try now to understand the evaluation of the expression:
>
> e = expression(r(x) <- 1)
>
> # parameter named "value" see
# more code
}
border <- function (x, which)
{
which <- match.which(which)
# more code
}
some_other_style <- function (x, which)
{
which <- match.which(which)
# more code
}
I hope this helps.
On Mon, Sep 13, 2021 at 12:17 PM Leonard Mada wrote:
> Hello Andrew,
>
>
>
I think you're trying to do something like:
`padding<-` <- function (x, which, value)
{
which <- match.arg(which, c("bottom", "left", "top", "right"),
several.ok = TRUE)
# code to pad to each side here
}
Then you could use it like
df <- data.frame(x=1:5, y = sample(1:5, 5))
padding(df, "
You could use 'split' to create a list of data frames, and then apply a
function to each to get the means and sds.
cols <- "cfs" # add more as necessary
S <- split(discharge[cols], format(discharge$sampdate, format = "%Y-%m"))
means <- do.call("rbind", lapply(S, colMeans, na.rm = TRUE))
sds <-
10:01 AM Luigi Marongiu
wrote:
> Sorry,
> still I don't get it:
> ```
> > dim(df)
> [1] 302 626
> > # clean
> > df <- lapply(x, function(xx) {
> + xx[is.nan(xx)] <- NA
> + xx
> + })
> > dim(df)
> NULL
> ```
>
> On Thu, Sep 2, 2021
t; > str(df)
> List of 1
> $ sd_ef_rash_loc___palm: logi NA
> ```
> What am I getting wrong?
> Thanks
>
> On Thu, Sep 2, 2021 at 3:30 PM Andrew Simmons wrote:
> >
> > Hello,
> >
> >
> > I would use something like:
> >
> >
>
Hello,
I would use something like:
x <- c(1:5, NaN) |> sample(100, replace = TRUE) |> matrix(10, 10) |>
as.data.frame()
x[] <- lapply(x, function(xx) {
xx[is.nan(xx)] <- NA_real_
xx
})
This prevents attributes from being changed in 'x', but accomplishes the
same thing as you have abov
Hello,
I'm assuming you're reading from an "*.xlsx" file. I'm not sure which
package you're using for this scenario, but my preference is 'openxlsx'. If
this is the package you're using, you could set argument 'detectDates' to
'TRUE', and then it should read them correctly.
FILE <- tempfile(fil
Hello,
I would suggest something like:
date <- seq(as.Date("2020-01-01"), as.Date("2020-12-31"), 1)
time <- sprintf("%02d:%02d", rep(0:23, each = 12), seq.int(0, 55, 5))
x <- data.frame(
date = rep(date, each = length(time)),
time = time
)
x$cfs <- stats::rnorm(nrow(x))
cols2aggregate
Hello,
I think I've found a solution, but it's not producing the same answer as
what you're expecting. I think you might've mixed up row and column a few
times, but you should be able to alter the following to your needs. Also,
see ?.bincode:
EB <- matrix(data = c(
1, 271, 179, 359, 178,
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