Às 20:55 de 03/07/2023, Rui Barradas escreveu:
Às 20:26 de 03/07/2023, Sorkin, John escreveu:
Jeff,
Again my thanks for your guidance.
I replaced dimnames(myvalues)<-list(NULL,c(zzz))
with
colnames(myvalues)<-zzz
and get the same error,
Error in dimnames(x) <- dn :
length of
t;- tools::file_path_sans_ext(files)
new_names <- paste(new_names, "txt", sep = ".")
setNames(x, new_names)
}
# apply function
# note that my test files are in "~/Temp"
txts <- convertpdf2txt(here::here("~", "Temp"))
names(txts)
Hope this hel
d, reproducible code.
Hello,
I had tested the code with a couple of PDF's and it ran with no errors
or warnings.
That error is telling that a "}" is not balanced but in my code they all
are, RStudio checks it automatically.
Can you try to check in an editor with syntax highligh
ellow")
ggplot(df, aes(x = Time, y = ddC, colour = mM, group = mM)) +
geom_line() +
geom_errorbar(aes(ymin = ddC - SD, ymax = ddC + SD), width = 0.1) +
geom_point(size = 6) +
geom_hline(aes(yintercept = 0)) +
scale_colour_manual(values = COLS) +
ggtitle("
Hello,
Maybe this [1] is relevant.
[1]
https://community.rstudio.com/t/meaning-of-common-message-when-install-a-package-there-are-binary-versions-available-but-the-source-versions-are-later/2431
Hope this helps,
Rui Barradas
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think.
The fundamental machine is intelligence.
Fernando Pessoa, 1926
Revista de Comércio e Contabilidade, nº 4. Lisboa, 25-4-1926.
(Magazine of Commerce and Accounting, nº 4. Lisbon, 25-4-1926)
Rui Barradas
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snet.org.au/~bgreen/Data/Hanson1/
# 84
# http://home.brisnet.org.au/~bgreen/Data/Hanson2/
# 749
# matches the question's number
sum(lengths(success))
# [1] 833
Hope this helps,
Rui Barradas
o)) +
geom_point() +
stat_smooth(
formula = y ~ x,
method = glm,
method.args = list(family = binomial),
se = FALSE
)
Hope this helps,
Rui Barradas
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hough it's no
## longer needed
df |>
dplyr::mutate(Stat = factor(Stat)) |>
ggplot(aes(Year, Flow, group = Stat, colour = Stat))+
coord_cartesian(ylim = c(0, 10)) +
geom_line()+
geom_point()+
scale_color_manual(name = "Stat", values = cb8[4:5])+
theme_bw()
Hope
.frame(
Year = rep((1990:1995),8),
Flow = round(rnorm(48,50,10),3),
Stat = rep(c(rep("Tweed",6), rep("Tay",6)),4),
Metric = rep(c("AMAX","Mean","AMIN","Median"),each=12)
)
ggplot(mock, aes(Year,Flow, group = factor(Stat), co
trix(dat2[match(dat1[,1], dat2[,1]),-1])
Res <- data.frame(ID = dat1[,1], Index = multiplication)
# this is what I find simpler
# the method being called is cbind.data.frame
Res2 <- cbind(dat1[1], Index = multiplication)
identical(Res, Res2)
#> [1] TRUE
Hope this helps,
Rui Barrada
but it's the slowest of the three, OP's, Iris' (the
fastest).
Hope this helps,
Rui Barradas
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and provide commented, minimal, self-contained, reproducible code.
Hello,
That erro
does not mean that the function that *creates* the object needs
to be generic, you only need a new generic to have methods processing
inputs of different classes in ways specific to those classes.
Hope this helps,
Rui Barradas
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geom_line(data = fit, aes(x, ymax), linetype = "dashed", linewidth = 1)
Hope this helps,
Rui Barradas
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PLEASE do read the
bmited and anyone
willing to check them can.
[1] https://cran.r-project.org/web/packages/
Hope this helps,
Rui Barradas
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PLEASE do read t
,
It seems that you have data that in tabular form is one column per
answer, so you would end up with 30 columns, maybe an extra id column.
Can you post sample data? If not, make up the answers and post the
answers of the first 6 individuals or so.
Hope this helps,
Rui Bar
values of the other numeric
columns, just of those two.
# define a vector of columns of interest
cols <- c("Length", "Width", "RAWMAT")
# 1) Simple aggregation with 2 variables, select cols:
aggregate(cbind(Length, Width) ~ RAWMAT, data = my_data[cols], FUN
differently, even though they all
use mean(na.rm = TRUE) to calculate the group stats.
Best wishes,
Ivan
On 04/09/2023 13:46, Rui Barradas wrote:
Às 10:44 de 04/09/2023, Ivan Calandra escreveu:
Dear useRs,
I have just stumbled across a behavior in aggregate() that I cannot
explain. Any help
be replaced by the value in the column which is non zero..
I tried append and merge but fail to get what I want.
Hello,
That's a case for ?pmax:
x1 <- c(116,0,115,137,127,0,0)
x2 <- c(0,159,0,0,0,159,127)
pmax(x1, x2)
#> [1] 116 159 115 137 127 159 127
tudio questions to the RStudio support service, they
answer quickly and the answers are generally good.
It's written at the bottom of the attached image that the workspace was
loaded from file
C:/WSI/.RData
Close RStudio, remove this file and restart. See if it solved it.
Hope th
rs, patients)$estimate
#>prop 1prop 2prop 3prop 4
#> 0.9651163 0.9677419 0.9485294 0.8536585
smokers/patients
#> [1] 0.9651163 0.9677419 0.9485294 0.8536585
plot(smokers/patients, type = "b")
Hope this helps,
Rui Barradas
g
with the p-value of prop.trend.test, makes it easier for the users to determine
the existence of trending and its direction.
If there are any other ways to do this, please let me know.
Thomas Subia
On Thursday, September 7, 2023 at 10:31:27 AM PDT, Rui Barradas
wrote:
graphs, only data.
Can you post the code have you tried?
Hope this helps,
Rui Barradas
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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 = "
<- 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
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
s with
calls to library() when using non-base functionality.
Hope this helps,
Rui Barradas
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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
aes(x = eruptions, y = waiting, label = waiting),
vjust = -1
) +
theme_cowplot()
Hope this helps,
Rui Barradas
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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,
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
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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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
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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w
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 %
"CCl2CO2AlPO4SiO4Cl")
split_chem_elements(mol)
#> [[1]]
#> [1] "C" "Cl" "F"
#>
#> [[2]]
#> [1] "Li" "Al" "H"
#>
#> [[3]]
#> [1] "C" "Cl" "C" "O" "Al&q
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
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]|$)|(?<=.)(?=[A-Z])|(?<=[a-z])(?=[^a-z])"
if(rm.digits) {
stringr::s
n/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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,DoReg)
#> mydata$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$StepType: Second
#> lm model parameter contrast
#>
LL
#> .. ..$ : 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
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
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
[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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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
;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
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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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
_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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, "Neu", "Neu",
"Neu",
"Neu", "Neu", "Neu", "Neu", "Neu", "Neu", "Neu", "Neu",
"Pos",
"Pos", "Pos", "Pos", "Pos", &
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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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
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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Est
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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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
_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
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
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),
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
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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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
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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$data1.csv
Hope this helps,
Rui Barradas
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sting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hello,
Thanks for the data but the code is missing from the attachment.
Can you please post your code? In an attachment or directly in the
e-mail body.
Rui Barradas
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rga\Desktop\R-4.3.3\bin
so that Windows can find R.exe and Rgui.exe without the full path name.
Hope this helps,
Rui Barradas
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#> 3 2024-03-24 6.00
#> 4 2024-03-25 4.476190
#> 5 2024-03-26 6.538462
#> 6 2024-03-27 5.20
Also,
I'm glad to help as always but Ogbos, you have been an R-Help
contributor for quite a while, please post data in dput format. Given
the problem the output of the following is
regards from
Ogbos
On Wed, Mar 27, 2024 at 8:44 AM Rui Barradas wrote:
Às 04:30 de 27/03/2024, Ogbos Okike escreveu:
Warm greetings to you all.
Using the tapply function below:
data<-read.table("FD1month",col.names = c("Dates","count"))
x=data$count
f<-factor(d
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
vol <- 14e6
water_level(l = lev)
#> [1] 7056452
water_level(v = vol)
#> [1] 2480
Hope this helps,
Rui Barradas
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er is not the most serious problem here.
Hoep this helps,
Rui Barradas
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ht
%% 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
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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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
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
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
- 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
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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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(
19/08/21.
Try
as.Date(Atest$ddate, format = "%d/%m/%y")
Hope this helps,
Rui Barradas
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,
#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
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
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
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
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
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
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
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)"
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
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
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,]
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,]
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
#>
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
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
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
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
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, 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,
This is not exactly the
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
;-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
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