; y 1.056885 0.8704518
Hope this helps,
Rui Barradas
Às 13:28 de 04/10/2024, Steven Yen escreveu:
OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way
to do the covariance matrix with “apply”. The matrix I need really
contains the deviation products divided by the degr
zero, it's
only one value therefore it does not vary. A similar reasonong can be
applied to cov(x[1], x[2]), etc.
Hope this helps,
Rui Barradas
Às 12:14 de 04/10/2024, Steven Yen escreveu:
Hello
I have a vector:
set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23
Hello,
If you have a numeric matrix or data.frame, try something like
cov(mtcars)
Hope this helps,
Rui Barradas
Às 10:15 de 04/10/2024, Steven Yen escreveu:
On 10/4/2024 5:13 PM, Steven Yen wrote:
Pardon me!!!
What makes you think this is a homework question? You are not
obligated to
i <- index(dt_ts) >= from & index(dt_ts) <= to
dt_ts[i]
Also, instead of copying&pasting the data, you can attach a file with
extension .txt.
Hope this helps,
Rui Barradas
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n question and if
you have doubts translating the Python code to R code, ask us more
specific questions on those doubts.
Hope this helps,
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w
(y ~ x, data = df),
lm(y ~ x + I(x^2), data = df)))
mydt[[2L]][1L] |> class()
#> [1] "list"
mydt[[2L]][[1L]] |> class()
#> [1] "lm"
Hope this helps,
Rui Barradas
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Hello,
There is a CRAN Task View: Epidemiology that should be or have what you
are looking for.
[1] https://CRAN.R-project.org/view=Epidemiology
Hope this helps,
Rui Barradas
Às 06:29 de 19/09/2024, Aleena Shaji escreveu:
Dear R Support Team,
I hope this email finds you well.
I am
creates a mean at each position for three subjects,
replacing instead of the value of the single, the group mean.
But when NA appears, all the group gets NA.
Perhaps there is a different way to obtain the same result.
On Mon, 16 Sept 2024 at 11:35, Rui Barradas wrote:
Às 08:28 de 16/09/20
mean(.x, na.rm = TRUE)))
# same result, summarise's new argument .by avoids the need to group_by
db10 %>%
summarise(across(starts_with("cp"), ~ mean(.x, na.rm = TRUE)), .by =
groupid)
Can you post the expected output too?
Hope this helps,
Rui Barradas
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4 4 3 2 1 3 2112
5 1 NA NA NA NA NA NA NA NA NA NA NA NA
6 2 5 5 10 10 9 10 10 10 NA 109 10", header = TRUE)
df1
library(dplyr)
df1 %>%
mutate(across(starts_with("cp"), ~ +(is.na(.) & id != 1), .names =
"
red$se)
# with more points ahead
predict(model, n.ahead = 2, newxreg = c(10, 12))
Hope this helps,
Rui Barradas
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ata_POSIX, "%Y-%m-%d") status hs
#> 1 2024-01-02 D 51.2
# the formats in the OP but extracted from the date/time and used in the
formula that follows.
year <- format(mydf$data_POSIX, "%Y")
month <- format(mydf$data_POSIX, "%m")
day <
-Original Message-
From: R-help On Behalf Of Rui Barradas
Sent: Wednesday, August 28, 2024 4:19 AM
To: Francesca PANCOTTO ; r-help@r-project.org
Subject: Re: [R] Fill NA values in columns with values of another column
[External Email]
Às 11:23 de 27/08/2024, Francesca PANCOTTO via R-help escreveu
21
#> 2172
#> 2272
#> 31 103
#> 32 103
#> 4144
#> 4244
#> 5195
#> 5295
#> 6156
#> 6256
#> 7127
#> 7227
#> 8168
#> 8268
Hope this helps,
R
#> version.string R version 4.4.1 (2024-06-14 ucrt)
#> nickname Race for Your Life
Hope this helps,
Rui Barradas
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own above. Guess I need to read more
about invisible.
On 8/11/2024 10:09 PM, Rui Barradas wrote:
Às 09:51 de 11/08/2024, Steven Yen escreveu:
Hi
In the following codes, I had to choose between printing (= TRUE) or
deliver something for grab (ei, vi). Is there a way to get both--that
is, to
ing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
Hello,
Maybe change the end of the code to return a bigger list.
Hello,
.lm.fit is an order of magnitude faster than lm.fit but the Description
section warns on its use, see the examples in help("lm.fit").
Hope this helps,
Rui Barradas
Às 21:08 de 10/08/2024, Yuan Chun Ding via R-help escreveu:
You are right. I also just thought abou
hat you are saying is hardly (not) possible.
If you ever call that code with joint12 set to TRUE, do you reset to
FALSE afterwards?
Can you give a small working example with code and data showing this
behavior?
Hope this helps,
Rui Barradas
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ommunist", "minority", "religious", "social",
"no",
"primary", "middle", "high", "veryh", "somewhath", "notveryh",
"verym", "somewhatm", "notverym&
plot.margin = unit(c(0.2, 0, 0.1, 0), "cm"))
p2 <- df %>%
filter(nm != "A") %>%
ggplot(aes(x = date)) +
geom_col(aes(y = val0), na.rm = TRUE, fill = "white") +
geom_line(aes(y = val1)) +
ylab("") +
facet_wrap(~ nm, scales = "fre
re, 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.
Hello,
Try to remove
scales="free_y"
from facet_wrap(). With scales="f
ot your posted origin date,
# see the examples on Windows and Excel
# dates in help("as.Date")
as.Date(19024, origin = "1970-01-01")
#> [1] "2022-02-01"
Hope this helps,
Rui Barradas
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;-rowSums(tmp)
tem2 <-tmp[row0!=0,]
tmp3 <- cor.test(tem2[,1],tem2[,2])
r[i, j] <- tmp3$estimate
P[i, j] <- tmp3$p.value
}
}
}
r<-as.data.frame(r)
P<-as.data.frame(P)
From: R-help On Behalf Of Yuan Chun Ding via
R-help
Sent: Thursday, July 2
38 0.68452834
#> g2 0.7979717 NA 0.4070838 0.06758329
#> g3 0.4070838 0.40708382NA 1.0000
#> g4 0.6845283 0.06758329 1.000 NA
You can put these two results in a list, like Hmisc::rcorr does.
lst_rcorr <- list(r = r, P = P)
Hope this helps,
Rui Bar
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hello,
This is not exactly the
Às 17:43 de 18/07/2024, Rui Barradas escreveu:
Às 16:27 de 18/07/2024, SIBYLLE STÖCKLI via R-help escreveu:
Hi
I am using ggplot to visualise y for a two-factorial group (Bio: 0 and
1) x
= 6 years. I was able to adapt the colour of the lines (green and red)
and
the linetype (solid and
d provide commented, minimal, self-contained, reproducible code.
Hello,
To have one legend only, the labels must be the same. Try using
labels=c("ÖLN", "BIO")
in
scale_linetype_manual(values=c("dashed", "solid"), labels=c("ÖLN", "BIO"))
H
algebra
x <- cbind(1, (Dat$Gender == "Male")) %*% coef(Model)
pred1 <- exp(x)/(1 + exp(x))
# use the fitted line equation
y <- coef(Model)[1L] + coef(Model)[2L] * (Dat$Gender == "Male")
pred2 <- exp(y)/(1 + exp(y))
head(predict(Model, type="response"))
head(p
bject it returns and the following
should work.
# this is 'x', a named character vector
coef(fit)
#
fit |> coef() |> names() |> grep("somewhat|very", x = _)
Hope this helps,
Rui Barradas
Às 10:26 de 12/07/2024, Steven Yen escreveu:
Thanks. In this case below, what
p="")
SS[i] <- sum(get(e))
}
SS
#> [1] 55 54 52 49 45
Or all in one instruction with the assistance of ?ls.
# ls(pattern = "^adds") |> mget() |> lapply(sum)
ls(pattern = "^adds") |> mget() |> sapply(sum)
#> adds1 adds2 adds3 adds4 adds5
#>
es)
matrix(nrow = nsims, ncol = 1L + ngrps, dimnames = list(NULL, nms))
}
NSims <- 4
Grps <- 5
create_matrix(NSims, Grps)
#> NSims Value1 Value2 Value3 Value4 Value5
#> [1,]NA NA NA NA NA NA
#> [2,]NA NA NA NA NA NA
#> [3,]
x) - 1L)))
colnames(x) <- nms
x
}
NSims <- 4
Grps <- 5
DiffMeans <- matrix(nrow=NSims,ncol=1+Grps)
names_cols(DiffMeans)
#> NSims Value1 Value2 Value3 Value4 Value5
#> [1,]NA NA NA NA NA NA
#> [2,]NA NA NA NA NA NA
#> [3,]
turn value of lapply
OUTPUT <- lapply(INPUT, \(f) {
mydata <- read.csv(f)
boprobit(eqs, mydata, wt=weight, method="BHHH",
tol=0, reltol=0, gradtol=1e-5, Fisher=TRUE)
})
# assign the output list's names
names(OUTPUT) <- paste0("bop", seq.int(m))
Hope th
line of the output)?
You are right, in the case I posted there were unwanted characters.
Most of the tests I ran there were no additional, unwanted charcters,
though.
This is definitely unstable, that's all I can say.
Hope this helps,
Rui Barradas
Thank you again
Tanguy
1
---
4) GUI: Rgui
Output:
[1] "plot(AirPassengers)" "က \005ⷀǏǭ"
[1] "plot(AirPassengers)"
[1] "plot(AirPassengers)"
[1] "plot(AirPassengers)"
[1] "plot(AirPassengers)"
[1] "plot(AirPassengers)"
[1] "plot(AirPassengers)"
152.8
4705 2012-07-15 0 11.4 665 1179 145.1
4706 2012-07-15 1 9.7 657 1170 145.1
*Jibrin Adejoh Alhassan (Ph.D)*
Department of Physics and Astronomy,
University of Nigeria, Nsukka
On Mon, Jun 17, 2024 at 9:23 AM Rui Barradas wrote:
Às 09:12 de 17/06/2024, Jibrin Alhassan escreveu
71 -9 999.9
6 2012-01-01 5 4.2 368 71 -7 999.9
Many thanks.
*Jibrin Adejoh Alhassan (Ph.D)*
Department of Physics and Astronomy,
University of Nigeria, Nsukka
On Mon, Jun 17, 2024 at 8:14 AM Rui Barradas wrote:
Às 07:53 de 17/06/2024, Jibrin Alhassan escreveu:
Part of it is pasted below
e whole data points. I have tried following
your tested solution but was unsuccessful. My regards.
*Jibrin Adejoh Alhassan (Ph.D)*
Department of Physics and Astronomy,
University of Nigeria, Nsukka
On Sun, Jun 16, 2024 at 8:33 AM Rui Barradas wrote:
Às 21:42 de 15/06/2024, Jibrin Alhassan escr
a has 38735
print(df2)
Error: object 'df2' not found
My data is an hourly data but desire to have the date as
yearmonthday hour
2012 08 01 01
2012 08 01 02
2012 0801 03 etc
Thanks.
*Jibrin Adejoh Alhassan (Ph.D)*
Department of Physi
er needed
df1 <- df1[-(1:2)]
# relocate the new date column
df1 <- df1[c(ncol(df1), 1:(ncol(df1) - 1L))]
head(df1)
#> Date HR IMF SW SSN Dst f10.7
#> 1 2012-08-01 0 3.4 403 132 -9 154.6
#> 2 2012-08-01 1 3.7 388 132 -10 154.6
#> 3 2012-08-01 2 3.7 383 132 -10 1
o see where the error occurred.
Ding
From: Rui Barradas
Sent: Wednesday, June 12, 2024 11:29 AM
To: Yuan Chun Ding ; CALUM POLWART
Cc: r-help@r-project.org
Subject: Re: [R] my R code worked well when running the first 1000 lines of R
code
Hello, Inline. Às 19: 03 de 12/06/2024, Yuan Chun Ding
,
#standard_deviation = sd(value),
.by = c(dat, measure),
.groups = "drop"
)
This is only a guess, the question cannot really be answered.
Hope this helps,
Rui Barradas
but still not resolved the problem.
I will restart from the
19/08/21.
Try
as.Date(Atest$ddate, format = "%d/%m/%y")
Hope this helps,
Rui Barradas
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Às 18:40 de 02/06/2024, Rui Barradas escreveu:
Às 18:34 de 02/06/2024, Leo Mada via R-help escreveu:
Dear Shadee,
If you have a data.frame with the following columns:
n = 100; # population size
x = data.frame(
Sex = sample(c("M","F"), n, T),
Country = sample(c(
mented, minimal, self-contained, reproducible code.
Hello,
The following is simpler.
r2 <- xtabs(~ ., x) |> as.data.frame()
r2[-4L] # or r2[names(r2) != "Freq"]
Hope this helps,
Rui Barradas
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- Italy
Meteo Section
Snow Section
Via del Colle Ameno 5
60126 Torrette di Ancona, Ancona (AN)
Uff: +39 071 806 7743
E-mail: stefano.so...@regione.marche.it
---Oo-oO
____
Da: Rui Barradas
Inviato: martedì 28 maggio
es a sequence with all the years
years <- Reduce(`:`, years)
# coerce to "Date"
from <- ISOdate(years, 5L, 2L, tz = "Etc/GMT-1")
to <- ISOdate(years, 10L, 30L, tz = "Etc/GMT-1")
# this logical index keeps only the dates between May, 2nd and Nov 1st.
kee
of the format = "%m-%d-%Y" argument.
Let scale_x_date take care of formating the date as you want it
displayed. Any of the two below is a valid date format.
ggplot(data = yyy[1:30,], aes(jdate, Sum)) +
geom_point() +
# scale_x_date(date_labels = "%b %d, %Y")
scale_x_d
Às 09:08 de 21/04/2024, Rui Barradas escreveu:
Às 08:55 de 21/04/2024, Hans W escreveu:
As we all know, in R indices for vectors start with 1, i.e, x[0] is not a
correct expression. Some algorithms, e.g. in graph theory or
combinatorics,
are much easier to formulate and code if 0 is an allowed
i <- i + 1L
NextMethod()
}
as_zerobased <- function(x) {
class(x) <- c("zerobased", class(x))
x
}
x <- 1:10
y <- as_zerobased(x)
y[0]
#> [1] 1
y[1]
#> [1] 2
y[9]
#> [1] 10
y[10]
#> [1] NA
Hope this helps,
Rui Barradas
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%% 2L) == 1L}() |>
which()
data.frame(Col = i, Unbalanced = Unbalanced)
}) |>
do.call(rbind, args = _)
}
# read the data disregardin g quoted strings
df1 <- read.csv(fl, quote = "")
# determine which strings have unbalanced quotes and
# where
unbalance
er is not the most serious problem here.
Hoep this helps,
Rui Barradas
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ht
vol <- 14e6
water_level(l = lev)
#> [1] 7056452
water_level(v = vol)
#> [1] 2480
Hope this helps,
Rui Barradas
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es' not found
# note that the row names are still tapply's names vector
# and that the columns order is not Date/count. Both are fixed
# after the calculations.
res
You can see that the error message is on the pipe. Please, let me know
where I am missing it.
Thanks.
On Wed, Mar 27, 2024
regards from
Ogbos
On Wed, Mar 27, 2024 at 8:44 AM Rui Barradas wrote:
Às 04:30 de 27/03/2024, Ogbos Okike escreveu:
Warm greetings to you all.
Using the tapply function below:
data<-read.table("FD1month",col.names = c("Dates","count"))
x=data$count
f<-factor(d
#> 3 2024-03-24 6.00
#> 4 2024-03-25 4.476190
#> 5 2024-03-26 6.538462
#> 6 2024-03-27 5.20
Also,
I'm glad to help as always but Ogbos, you have been an R-Help
contributor for quite a while, please post data in dput format. Given
the problem the output of the following is
rga\Desktop\R-4.3.3\bin
so that Windows can find R.exe and Rgui.exe without the full path name.
Hope this helps,
Rui Barradas
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sting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hello,
Thanks for the data but the code is missing from the attachment.
Can you please post your code? In an attachment or directly in the
e-mail body.
Rui Barradas
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ec <- sprintf("data%d.csv", 1:24)
data_list <- sapply(file_names_vec, read.csv, simplify = FALSE)
# access the 1st data.frame
data_list[[1L]]
# same as above
data_list[["data1.csv"]]
# same as above
data_list$data1.csv
Hope this helps,
Rui Barradas
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and provide commented, minimal, self-contained, reproducible
ht:
Whenever I have problems updating or installing packages from whithin
RStudio I close RStudio, write a script with the install.packages() call
and run it from a command window.
R -q -f "instscript.R"
This many times works better and it also works with Bioconductor's
BiocMa
aes(Sepal.Length, Sepal.Width, color = Species)) +
geom_point()
g + ylab(expression(paste(frac(
additive~HCO[3]^"-",
true~HCO[3]^"-"
Hope this helps,
Rui Barradas
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ror, an error without a condition message
and no call expression. I find this stranger, a call like the following
is expected.
tryCatch(stop("error"), error = function(e) e) |> str()
List of 2
$ message: chr "error"
$ call : language doTryCatch(return(expr), name, pa
10) %+replace%
theme(
panel.spacing = unit(0, "lines"),
strip.background = element_blank(),
strip.placement = "outside",
# this line was added by me, remove if not wanted
strip.text.x.bottom = element_text(face = "bold", size = 10),
d linear regression for France", x = "Year", y
= "PISA score in mathematics") +
scale_y_continuous(limits=c(470,500),oob=scales::squish)
#
Le lundi 11 décembre 2023 à 23:38:06 UTC+1, Ben Bolker
a écrit :
On 2023-12-11 5:27 p.m., Daniel Nordlund w
_smooth(method = "lm", formula = y ~ x) +
labs(
title = "Standard linear regression for France",
x = "Year",
y = "PISA score in mathematics"
) +
ylim(470, 500)
#> Warning message:
#> In max(ids, na.rm = TRUE) : no non-missing arguments to
Às 16:30 de 07/12/2023, Rui Barradas escreveu:
Às 16:21 de 07/12/2023, Sorkin, John escreveu:
Colleagues,
I have a matrix of character data that represents date and time. The
format of each element of the matrix is
"2020-09-17_00:00:00"
How can I convert the elements into a valid R
as.POSIXct
Don't forget the underscore in the format.
as.POSIXct("2020-09-17_00:00:00", format = "%Y-%m-%d_%H:%M:%S")
Hope this helps,
Rui Barradas
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e http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hello,
Your link points to a GitHub repository, the package can be installed with
devtools::install_github(repo = "Sibada/sibadaR")
Hope this helps
Rui Barradas
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Às 17:57 de 30/11/2023, Rui Barradas escreveu:
Às 17:38 de 30/11/2023, Robert Baer escreveu:
I am having trouble using back ticks with the R extractor function
'predict' and an lm() model. I'm trying too construct some nice
vectors that can be used for plotting the two type
value (y-tilde uses 'p')
# usual trick is to assign x to actual x-var name in middle dataframe
arguement
CI.p = predict(mod2, newdata = newd, interval = 'prediction')# fail
Hope this helps,
Rui Barradas
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, "Neu", "Neu",
"Neu",
"Neu", "Neu", "Neu", "Neu", "Neu", "Neu", "Neu", "Neu",
"Pos",
"Pos", "Pos", "Pos", "Pos", &
_D - se_D, ymax = mean_D + se_D), fill =
"grey", alpha = 0.5) +
geom_line(aes(y = mean_D, color = C)) +
geom_point(aes(y = D, color = C)) +
scale_color_manual(name = "Concentration", values = clrs)
Hope this helps,
Rui Barradas
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code.
Hello,
Yes, CRAN is down.
I know last week there was an anouncement about a maintenance scheduled
but I cannot place that e-mail right now and don't remember the date
exactly so I cannot say for sure this is what is happening.
But it is probably a scheduled maintenance.
Rui Bar
ile, rename it .txt?
See [1], section General Instructions for more on this
[1] https://www.r-project.org/mail.html#instructions
Hope this helps,
Rui Barradas
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;character" to class "Date".
Then the grouped sums are computed.
suppressPackageStartupMessages(
library(dplyr)
)
dt1 %>%
mutate(date = as.Date(date, "%m/%d/%Y")) %>%
summarise(EnergykWh = sum(EnergykWh), .by = date)
#> date EnergykWh
#> 1 2016-0
geom_hline(yintercept = 130) +
scale_y_continuous(
name = "Blood glucose (mg/dL)",
breaks = seq(100, 230, by = 20)
) +
scale_shape_manual(
#name = "Conditions",
labels = c("Missed meds", "Missed exercise"),
values = c(20, 4),
na.tran
[1] 15091
# keep the rows with values not NA
df_long <- df_long[complete.cases(df_long), , drop = FALSE]
# check the dimensions again
dim(df_long)
# [1] 15091
Hope this helps,
Rui Barradas
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date, dfuture
# , lty = "dashed"
, lwd=2
, col = "black")
# if lines() is used for both the interpolated and extrapolated
# values you will have a gap between both fitted and predicted lines
# but it is closer to what you want
# get the fitted values first (in
row. But that is a very special case, the general case would be to
extract the column.
Hope this helps,
Rui Barradas
If you have another row in your x, say:
x <- data.frame(A=c(1,4), B=c(2,5), C=c(3,6))
then your code
x$C <- y[1]
returns an error.
If y has the same number of rows as x
LL
#> .. ..$ : chr [1:2] "Mean" "S"
# nc is just a convenience, avoids repeated calls to ncol
nc <- ncol(agg)
cbind(agg[-nc], agg[[nc]])
#> A MeanS
#> 1 a 14.5 9.082951
#> 2 b 15.5 9.082951
#> 3 c 16.5 9.082951
# all is well
cbind(agg[-nc], a
,DoReg)
#> mydata$StepType: First
#> lm model parameter contrast
#>
#> Contrast S.E.LowerUppert df Pr(>|t|)
#> 1 2.99114 1.956013 -1.05518 7.037461 1.53 23 0.1399
#> ----
#> mydata$StepType: Second
#> lm model parameter contrast
#>
n/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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Às 19:35 de 18/10/2023, Leonard Mada escreveu:
Dear Rui,
On 10/18/2023 8:45 PM, Rui Barradas wrote:
split_chem_elements <- function(x, rm.digits = TRUE) {
regex <- "(?<=[A-Z])(?![a-z]|$)|(?<=.)(?=[A-Z])|(?<=[a-z])(?=[^a-z])"
if(rm.digits) {
stringr::s
about the workaround).
Q: My question focused if there is anything like is.numeric, but to
parse each element of a vector.
Sincerely,
Leonard
On 10/18/2023 6:53 PM, Rui Barradas wrote:
Às 15:59 de 18/10/2023, Leonard Mada via R-help escreveu:
Dear List members,
What is the best way to test
"CCl2CO2AlPO4SiO4Cl")
split_chem_elements(mol)
#> [[1]]
#> [1] "C" "Cl" "F"
#>
#> [[2]]
#> [1] "Li" "Al" "H"
#>
#> [[3]]
#> [1] "C" "Cl" "C" "O" "Al&q
Hello,
Given your date format, try
format = "%d.%m.%Y %H:%M"
Test with your date time:
x <- "2.11.2017 13:30"
as.POSIXct(x, format = "%d.%m.%Y %H:%M")
#> [1] "2017-11-02 13:30:00 WET"
as.POSIXct(su_seviyeleri_data$kayit_zaman, format = "%d.%m.%Y %
x27;t like it but
ifelse(rep(T, length(c(1,2,3))), c(1,2,3), c(5,6))
maybe you should use
max(length(c(1, 2, 3)), length(5, 6)))
instead, but it's still ugly.
Hope this helps,
Rui Barradas
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in Users/Admin
There were a couple of R files in there which I have since deleted but I am
still getting the same issue
Thankyou,
George
________
From: Rui Barradas
Sent: 10 October 2023 12:06
To: George Loftus ; r-help@r-project.org
Subject: Re: [R] Text showing when
ained, reproducible code.
Hello,
Try deleting file
/Users/admin/.RData
It is restoring the previous session and this is many times a source for
problems.
Hope this helps,
Rui Barradas
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Às 10:09 de 06/10/2023, Chris Evans via R-help escreveu:
The reason I am asking is that I would like to mark areas on a plot
using geom_polygon() and aes(fill = variable) to fill various polygons
forming the background of a plot with different colours. Then I would
like to overlay that with poi
.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hello,
RStudio is an IDE for R, not R itself.
That is a RStudio error and RStudio technical support [1] is better
suited to solve your problem.
[1] https://community.rstudio.com/
Hope this helps,
aes(x = eruptions, y = waiting, label = waiting),
vjust = -1
) +
theme_cowplot()
Hope this helps,
Rui Barradas
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R-help@r-project.org mai
My sympathies for your loss.
Jim Lemon was a dedicated contributor to the R community and his answers
were always welcome.
Jim will be missed.
Rui Barradas
Às 23:36 de 04/10/2023, Jim Lemon escreveu:
Hello,
I am very sad to let you know that my husband Jim died on 18th September. I
s with
calls to library() when using non-base functionality.
Hope this helps,
Rui Barradas
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t)
Hope this helps,
Rui Barradas
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, min
<- matrix(1, 4, 4)
z <- lower.tri(diag(4), TRUE)
z[] <- apply(z, 2, as.integer)
H(x)
H(y)
H(z)
Hope this helps,
Rui Barradas
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PLEASE d
cale in order to
make the x axis more readable.
Without the formula and method arguments, geom_smooth will print a
message, they are now made explicit.
suppressPackageStartupMessages({
library(dplyr)
library(ggplot2)
})
d_sum %>%
mutate(md = paste("2023", md, sep = "
graphs, only data.
Can you post the code have you tried?
Hope this helps,
Rui Barradas
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PLEASE do read the posting guide http://www.R
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