. Start with R-intro.pdf that comes with R. There are also many texts
online that cover this and lots of other graphics stuff.
Hope this helps,
Rui Barradas
Às 14:49 de 29/10/20, Joy Kissoon escreveu:
runif (100,
1000, 1)
__
R-help@r-projec
s::Mode(x),
so = apply(x, 1, Mode),
times = 10
)
print(mb, unit = "relative", order = "median")
[1] https://stackoverflow.com/a/8189441/8245406
Hope this helps,
Rui Barradas
Às 17:12 de 31/10/20, Luigi Marongiu escreveu:
Thank you. The problem was not finding the mode
t_PT.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=pt_PT.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] compiler_4.0.3
Hope this helps,
Ru
class(), str() or dim()
return when applied to both.
See also this StackOverflow post:
https://stackoverflow.com/questions/1169456/the-difference-between-bracket-and-double-bracket-for-accessing-the-el
Hope this helps,
Rui Barradas
Às 13:26 de 04/11/20, Engin Yılmaz escreveu:
Dear
I use
Hello,
Or maybe
logical_idx <- max_usage_hours_per_region$Region %in% input$Region
Another option is ?match
Hope this helps,
Rui Barradas
Às 15:41 de 07/11/20, Jeff Newmiller escreveu:
This looks odd...
max_usage_hours_per_region[input$Region,]
This would only work if you
Hello,
Try removing I() from I(log(y)). But it's hard to say without a
reproducible example, please post the output of
dput(dat)
or, if dat is big, the output of
dput(head(dat, 20))
Hope this helps,
Rui Barradas
Às 22:11 de 11/11/20, Marcelo Laia escreveu:
Hi,
I am running
/treeManipulation.html
Hope this helps,
Rui Barradas
Às 21:59 de 11/11/20, April Ettington escreveu:
I've been using groupOTU to color paraphyletic clades in my tree based on
lists of tips, but I have multiple clades I want to highlight. Is there
some way to use ggplot to indicate mul
sition[w, ]
Hope this helps,
Rui Barradas
Às 09:59 de 13/11/20, ani jaya escreveu:
Dear r list,
I try to locate any local max value and location of data that follow 7
moving window condition, meaning that this data is largest and
centered in 7 values to the left and 7 values to the right.
Hello,
You can compute the exact result with package Rmpfr.
See ?mpfr and [1].
library(Rmpfr)
two <- mpfr(2, precBits = 64)
two^64 - 1
#1 'mpfr' number of precision 64 bits
#[1] 18446744073709551615
[1] https://www.mpfr.org/
Hope this helps,
Rui Barradas
Às 01:44 de 14/
Hello,
I forgot to suggest package gmp. See the second example in
?gmp::bigz
Hope this helps,
Rui Barradas
Às 05:50 de 14/11/20, Rui Barradas escreveu:
Hello,
You can compute the exact result with package Rmpfr.
See ?mpfr and [1].
library(Rmpfr)
two <- mpfr(2, precBits = 64)
two^64
len(q), na.rm = na.rm)
}
# Tests
set.seed(2020)
x1 <- rpois(10, lambda = 4)
x2 <- rpois(100, lambda = 10)
n_l(x1, q = 4)
n_l(x2, q = 21)
Hope this helps,
Rui Barradas
Às 12:45 de 16/11/20, Ablaye Ngalaba escreveu:
Hello,
please, I need help in programming R.
Here i
devtools::install_github('abnormally-distributed/cvreg')
Hope this helps,
Rui Barradas
Às 09:00 de 08/12/20, Ishaqbaba via R-help escreveu:
Hello Sir,
Hope this email finds you hale, healthy and safe.
I have been having problem in installing this:
install.packages("remotes&
Does this answer the question?
Hope this helps,
Rui Barradas
Às 12:22 de 20/12/20, Martin Møller Skarbiniks Pedersen escreveu:
Hi,
I posted this on the Google Group for ggplot2 but got no response.
https://groups.google.com/g/ggplot2/c/441srnt6RZU
So I hope someone can help me here in
Hello,
Is it this?
plot.background = element_rect(colour = "yellow2", fill = "yellow2")
The small white space goes away.
Hope this helps,
Rui Barradas
Às 17:00 de 20/12/20, Martin Møller Skarbiniks Pedersen escreveu:
On Sun, 20 Dec 2020 at 17:43, Rui Barradas wrote:
erflow.com/questions/1886/dynamically-select-data-frame-columns-using-and-a-vector-of-column-names
[2]:
https://stackoverflow.com/questions/1169456/the-difference-between-bracket-and-double-bracket-for-accessing-the-el
Hope this helps,
Rui Barradas
Às 21:48 de 27/12/20, Jim Lemon escreveu:
and"
MyT2 <- table(MyData[[V1]], MyData[[V2]])
MyT2
MyChi <- chisq.test(MyT2)
MyChi
name2 <- c(V1, V2, "Frequency")
dMyT <- as.data.frame(MyT2)
names(dMyT) <- name2
library(ggplot2)
ggplot(data = na.omit(dMyT), aes(get(V1), Frequency, fill = get(V2))) +
geom_bar(position
)
Note: your df.wide is not a data.frame, the transpose coerces it to
matrix. In this case it doesn't matter because it was just an example of
expected output but in other, real use cases you must be careful.
df.wide <- as.data.frame(df.wide)
would solve it.
Hope this helps,
Rui
= subset(daT1, variable == "y1"),
aes(y = value)) +
geom_line(data = subset(daT1, variable == "y2"),
aes(y = value*diff(rng) + rng[1])) +
facet_wrap(~group) +
scale_y_continuous(sec.axis = sec_axis(~ .*0.0001))
Hope this helps,
Rui Barradas
Às 01:0
l be separating the instruction you wrote from the NULL
instruction. A simpler example could be
mean(1:5);
This is *two* instructions:
mean(1:5); NULL
Hope this helps,
Rui Barradas
Às 22:33 de 08/01/21, Yuan Chun Ding escreveu:
Hi R users,
I am analyzing miRNA sequence data for a special network
element on its own:
xlim(1, 6)
Or
lims(x = c(1, 6))
But with these two, the labels are not like you want them, they are 2 by 2.
Hope this helps,
Rui Barradas
Às 10:34 de 10/01/21, Chris Evans escreveu:
[I must try to remember that swapping between Rstudio and my Emailer is a
recipe for hi
web and have only their URLs posted. This way a
reader has the option to download them or not.
Can you post sample data? Please post the output of `dput(df)`. Or, if
it is too big the output of `dput(head(df, 20))`. (`df` is the name of
your dataset.)
Hope this helps,
Rui Barradas
Às 11:30
1998-01-03
#4 1998 4 0 332 1998-01-04
#5 1998 5 0 302 1998-01-05
#6 1998 6 0 329 1998-01-06
Hope this helps,
Rui Barradas
Às 06:48 de 16/01/21, Jibrin Alhassan escreveu:
Hi Barradas
Sorry for the delay. Below is a section of my data. I have up to 1826
covering 1998 to 2002
year
, assign to df1, without creating df2.
Hope this helps,
Rui Barradas
Às 15:12 de 17/01/21, Jibrin Alhassan escreveu:
Hi Barradas,
Thanks for your assistance. It has brought me closer to what I am
looking for. I tried your code as shown below:
> df1 <- read.table("SWSdata_1998_2002&q
7] LC_PAPER=pt_PT.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=pt_PT.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] compiler_4.0.3
Hope this help
ormal")
f$estimate
p <- pretty(range(data))
x <- seq(from = min(p), to = max(p), by = 0.1)
hist(data, freq = FALSE)
lines(x, y = dlnorm(x,
meanlog = f$estimate["meanlog"],
sdlog = f$estimate["sdlog"])
)
Hope this helps,
Rui Barra
original
sd(b$t) # bootstrapped estimate of the SE of the sample prop.
hist(b$t, freq = FALSE)
Hope this helps,
Rui Barradas
Às 21:57 de 22/01/21, Marna Wagley escreveu:
Hi All,
I was trying to estimate standard error (SE) for the proportion value using
some kind of randomization process (bo
ction(x, xbar, s){
dnorm(x, mean = xbar, sd = s)
}
hist(b$t, freq = FALSE)
curve(f(x, xbar = b$t0, s = sd(b$t)), from = 0, to = 1, col = "blue",
add = TRUE)
curve(f(x, xbar = b$t0, s = sd(b$t)/sqrt(R)), from = 0, to = 1, col =
"red", add = TRUE)
Hope this helps,
Hello,
Base R:
colrs <- rainbow(length(unique(ROI$Z)))
roi_wide <- reshape(ROI, v.names = "Y", timevar = "Z", idvar = "X",
direction = "wide")
matplot(roi_wide, type = "l", lwd = 3, lty = "solid", col = colrs)
Hope this h
DF: normal or logistic
# *
if(logistic) pbivlogis(x,y,rho) else pbivnorm(x,y,rho,recycle=TRUE)
}
Hope this helps,
Rui Barradas
Às 06:14 de 25/01/21, Steven Yen escreveu:
Dear All
Below are calls to functions to calculate bivariate and univariate
logistic probabilities.It works for the fo
1L, 1L, 1L, 2L, 2L, 2L, 2L), species =
c("red deer",
"red deer", "red deer", "red deer", "red deer", "red deer",
"red deer", "red deer")), class = "data.frame", row.names = c(NA,
-8L))
Hope this
it a try would have their job
made easier.
To the OP: the code before "Data in dput format:" is a proposed
solution, not the structure(list(etc)), like Michael says.
If you run that statement, you will see the data as you've posted it,
even with the original column names.
Hope t
tion in [1]
[1] https://CRAN.R-project.org/package=investr
Hope this helps,
Rui Barradas
Às 09:11 de 26/01/21, Luigi Marongiu escreveu:
Hello,
I have a series of x/y and a model. I can interpolate a new value of x
using this model, but I get funny results if I give the y and look for
the correspo
Hello,
From help('forecast::fitted.Arima'):
h The number of steps to forecast ahead.
So you have the default h = 1 step ahead forecast for your model.
Hope this helps,
Rui Barradas
Às 12:13 de 28/01/21, Md. Moyazzem Hossain escreveu:
Dear R-experts,
I hope that all
ay. It will now return a matrix of indices with R
= 1000 rows and 19 columns.
Hope this helps,
Rui Barradas
Às 19:29 de 28/01/21, Marna Wagley escreveu:
Hi Rui,
I am sorry for asking you several questions.
In the given example, randomizations (reshuffle) were done 1000 times,
and its 1000
Hello,
Yes, sorry for my previous post, I had forgotten about boot.array.
That's a much better solution for your problem.
Rui Barradas
Às 20:29 de 28/01/21, Marna Wagley escreveu:
Thank you Rui,
This is great. How about the following?
SimilatedData<-boot.array(b, indices=T)
see
Hello,
Maybe this?
n_l <- function(Y, l, na.rm = FALSE) sum(Y == l, na.rm = na.rm)
set.seed(2020)
q <- 6
y <- sample(q, 10, TRUE)
l <- 4
n_l(y, l)
#[1] 3
Hope this helps,
Rui Barradas
Às 14:27 de 29/01/21, Ablaye Ngalaba escreveu:
Hello,
please, I need to calculate t
Hello,
Please cc the list, R-Help is threaded and your doubt and answers might
be of interest to others.
With a vector Y, you want 0 in all Y != l and 1 in all Y == l?
n_l <- function(Y, l) as.integer(Y == l)
Hope this helps,
Rui Barradas
Às 11:26 de 30/01/21, Ablaye Ngalaba escre
forecast(model1)
plot(fc)
Hope this helps,
Rui Barradas
Às 23:31 de 01/02/21, Md. Moyazzem Hossain escreveu:
Dear Rui Barradas
Thank you very much for your reply.
However, still now, I have a confusion whether I get the fitted value
for the year 2000, 2001, ..., 2020 or 2001, 2002, ..., 202
Hello,
Thanks for the links, they are very helpful.
Rui Barradas
Às 11:36 de 02/02/21, Partho Sarkar escreveu:
In case further clarification is needed, this from Rob Hyndman, author
of the Forecast package, may be helpful:
"fitted produces one-step in-sample (i.e., training data) "
20 secs", date_labels = "%M:%S") +
theme(axis.text.x = element_text(angle = 90, vjust = 0.5, hjust=1,
size = 5))
Hope this helps,
Rui Barradas
Às 15:05 de 12/02/21, Informatique escreveu:
Hello,
i'm using Rstudio from a few day, and i have some information in a CVS file,
take
s", date_labels = "H:%M:%S")
Can you post as data sample the output of dput?
dput(head(listeMesuresPropres, 20)) # or 30
Hope this helps,
Rui Barradas
Às 07:35 de 14/02/21, Informatique escreveu:
Hello,
the time are in the first table like that : 10:24:00
and i use this f
the entire ggplot code, not just the scale_x_datetime
instruction but the rest of the plotting code?
Hope this helps,
Rui Barradas
Às 13:25 de 14/02/21, Informatique escreveu:
Hello,
Thank's for your help.
i when i try this one
geom_line() +
scale_x_datetime(date_breaks =
alue)
liste_data_long$value <- as.numeric(liste_data_long$value)
library(ggplot2)
ggplot(liste_data_long, aes(x = DateHeure, y = value, color = variable)) +
geom_line() +
geom_point() +
scale_x_datetime(date_breaks = "60 secs", date_labels = "%H:%M:%S")
Hope this
Hello,
Sorry, my typo now, see inline.
Às 16:56 de 14/02/21, Rui Barradas escreveu:
Hello,
Please always cc the r-help mailing list, this might be helpful to
others in the future.
1. You have a data file using the continental Europe convention of
marking the decimals with a comma
p + geom_histogram(
mapping = aes(y = ..density..),
position = position_dodge(),
bins = 10) +
scale_y_continuous(labels = scales::label_percent())
Hope this helps,
Rui Barradas
Às 07:43 de 25/02/21, Bogdan Tanasa escreveu:
Thanks a lot Petr !
shall i uses "dodge" also for
t is
predict(m, newdata = newd)
#[1] Yes
#Levels: No Yes
Hope this helps,
Rui Barradas
Às 14:42 de 27/02/21, Jeff Reichman escreveu:
R User Forum
Is there a better way than grabbing individual cell values from a model
output to make predictions. For example the output from the following Naï
Hello,
Maybe define an infix operator?
`%!%` <- function(x, y) {
stopifnot(ncol(x) == length(y))
t(t(x)/y)
}
x <- matrix(1:20, ncol = 2)
s <- 1:2
x %!% s
x %!% 1:4
Hope this helps,
Rui Barradas
Às 11:00 de 03/03/21, Steven Yen escreveu:
I have a 10 x 2 matrix x. Like to d
Hello,
I forgot about sweep:
sweep(x, 2, s, '/')
sweep(x, 2, 1:4, '/')
Hope this helps,
Rui Barradas
Às 11:12 de 03/03/21, Rui Barradas escreveu:
Hello,
Maybe define an infix operator?
`%!%` <- function(x, y) {
stopifnot(ncol(x) == length(y))
t(t(x)/y)
}
= col_character(),
state = col_character(),
fips = col_character(),
cases = col_double(),
deaths = col_double()
)
us_counties <- read_csv(url(urlfile), col_types = cols_spec)
Hope this helps,
Rui Barradas
Às 15:24 de 03/03/21, Gayathri Nagarajan escreveu:
Hi Team
I have a tibble like the b
Hello,
This is a known issue with renderTable. Show the results with
renderDataTable instead.
Hope this helps,
Rui Barradas
Às 15:24 de 03/03/21, Gayathri Nagarajan escreveu:
Hi Team
I have a tibble like the below :
class(us_counties)
[1] "tbl_df" "tbl"
Hello,
Please read the posting guide at the end of this and every R-Help mail.
You should post the output of
dput(data)
or, if the data set 'data' is too big, the output of
dput(head(data, 20))
for us to be able to help you.
Rui Barradas
Às 18:29 de 10/03/21, Areti Panopoulo
uot;) +
geom_point() +
scale_x_datetime(breaks = myDat$datetime) +
scale_y_continuous(labels = NULL) +
ylab("event") +
theme(axis.text.x = element_text(angle = 60, vjust = 1, hjust=1),
axis.ticks.y = element_blank())
Hope this helps,
Rui Barradas
Às 13:36 de 16/03/21, S
Hello,
Yes, you can ask R-Help questions on R code, that's what it is intended for.
But please read the posting guide, it's link is at the bottom of this
and of any R-Help mail.
Good luck, enjoy R!
Rui Barradas
Às 18:19 de 16/03/21, Jonathan Lim escreveu:
Hi
I found your
1)
plot(1:45, high, type = "l", col = "red")
points(1:45, R, col = "blue")
points(1:45, highpred$Prediction, col = "cyan", pch = 3)
[1] https://CRAN.R-project.org/package=qpcR
Hope this helps,
Rui Barradas
Às 06:51 de 18/03/21, Luigi Marongiu escreveu:
Hello,
R-Help is for questions on R code.
RStudio has its own help forum that you can access from the Help menu,
Help > RStudio Community Forum
Hope this helps,
Rui Barradas
Às 18:07 de 22/03/21, Parkhurst, David escreveu:
I�m just starting to learn and use R Studio in my Mac. Now I f
Hello,
R is case sensitive, the column name is YEAR_END_Date, neither of
Year_END_Date
YEAR_END_DATE
matches that name. Try
select(PLC, format(YEAR_END_Date,format = "%B %d, %Y"), EPS)
Hope this helps,
Rui Barradas
Às 19:11 de 24/03/21, e-mail ma015k3113 via R-help escreveu:
Hello,
In the following code, the fixed parts of the plot are drawn first,
assigning the plot to p. Then geom_hline is conditionally added to p and
the result returned to caller.
This may be a problem if the conditional geom needs to be in a specified
order in the plot. Function plotLineFunc2
-matches("^gender_.*_std$"),
-matches("^race_.*_std$")
) %>%
rename_with(
.cols = matches("^gender|^race"),
~sub("mean$", "prop", .x)
) %>%
all.equal(need)
#[1] TRUE
Hope this helps,
Rui Barradas
Às 13:47 de 26/03/21, Paul Mil
s, will be used.
p + geom_dotplot(binaxis='y', stackdir='center', dotsize=1) +
scale_y_continuous(
trans = scales::pseudo_log_trans(base = exp(1))
)
Hope this helps,
Rui Barradas
Às 23:49 de 31/03/21, Mahmood Naderan-Tahan escreveu:
Hi,
With the following command
great
software with no real Windows Server 2019 compatibility problems.
Hope this helps,
Rui Barradas
Às 19:49 de 07/04/21, Tammy Regalado escreveu:
What versions of Windows can R Language be installed on? Specifically, is
Windows Server 2019 a supported pla
}
To keep extending a vector or list object in a loop is inefficient, this
creates the list with the right length beforehand.
Hope this helps,
Rui Barradas
Às 13:21 de 09/04/21, Wolfgang Grond escreveu:
Greg,
here I get the error message:
Error my_function(val) :
cannot find function
Hello,
Maybe something like
ok <- sapply(mydata, is.numeric)
mydata <- mydata[ok]
to keep the numeric columns only.
Hope this helps,
Rui Barradas
Às 04:25 de 10/04/21, Steven Yen escreveu:
I have data of mixed types in a data frame - date and numeric, as shown
in summary below. Ho
vector("list", length = nrow(datatable))
for(i in seq_along(datatable$column)){
net_name <- datatable$column[i]
NET <- get(net_name, envir = .GlobalEnv)
result3[[i]] <- my_function(NET)
}
names(result3) <- paste("subnet", datatable$column, sep = "_")
4.
were right.
result <- logical(nrow(df1))
inx <- findInterval(df1$Time, df2$start)
not_zero <- inx != 0
result[not_zero] <- df1$Time[not_zero] <= df2$end[ inx[not_zero] ]
Hope this helps,
Rui Barradas
Às 12:06 de 10/04/21, Kulupp escreveu:
Dear all,
I have two data frames (df1 a
] = 100/is.na(CLOSE_SHARE_PRICE[i])
}
}
Hope this helps,
Rui Barradas
Às 13:51 de 13/04/21, e-mail ma015k3113 via R-help escreveu:
Dear All,I have a dataframe with 4 variables and I am trying to calculate how
many shares can be purchased with £100 in the first year when the company was
listed
Hello,
Typo, inline.
Às 17:06 de 13/04/21, Rui Barradas escreveu:
Hello,
A close parenthesis is missing in the nd if.
Should be "the 2nd if".
Rui Barradas
for (i in 1:(nrow(PLC_Return)-1)){
if (i == 1){
NUMBER_OF_SHARES[i] = 100/is.na(CLOSE_SHARE_PRICE[i])
} els
/error-package-or-namespace-load-failed-for-xlconnect
Hope this helps,
Rui Barradas
Às 18:42 de 14/04/21, Bernard McGarvey escreveu:
I installed the "XLConnect" package which appears to be successful and then tried to load the
"XLConnect" library and got an error as shown
quot;) +
geom_edge_link() +
geom_node_point(aes(colour = Acronym), size = 8) +
scale_color_manual(name = "Project / Projekt",
values = c("blue", "red"))
Hope this helps,
Rui Barradas
Às 15:57 de 15/04/21, Wolfgang Grond escreveu:
Dear all,
I'm
Hello,
That error message means you need to run
install.packages("rlang")
before running
library(farff)
Hope this helps,
Rui Barradas
Às 20:02 de 17/04/21, Neha gupta escreveu:
Hi, suddenly the packages I installed not working. It gives me the error:
Error: package or name
very, very long time to process.
Revise the problem?
Hope this helps,
Rui Barradas
Às 13:35 de 19/04/21, Shah Alam escreveu:
Dear All,
I would like to know that is there any problem in *expand.grid* function or
it is a limitation of this function.
I am trying to create a combination of
Hello,
The following will probably not solve the problem but I would
1. use pattern = "\\.wma$" in list.files;
2. use sub, not gsub, in
sub("\\.wma$", ".mp3", i)
Hope this helps,
Rui Barradas
Às 05:29 de 20/04/21, Jim Lemon escreveu:
Hi John,
If the progra
tact the
maintainer("vars")
?
Hope this helps,
Rui Barradas
Às 18:31 de 25/04/21, Sun Yong Kim escreveu:
vars package
From: John Kane
Sent: Sunday, April 25, 2021 12:30 PM
To: Sun Yong Kim
Cc: r-help@r-project.org
Subject: Re: [R] Query
What package
Hello,
For column J, ave/seq_along seems to be the simplest. For column I, ave
is also a good option, it avoids split/lapply.
xx$I <- ave(xx$NUMBER_OF_YEARS, xx$COMPANY_NUMBER, FUN = function(x){
c(rep(1, length(x) - 1), max(length(x)))
})
xx$J <- ave(xx$NUMBER_OF_YEARS, xx$COMPANY_NUMBER,
Hello,
Right, thanks. I should be
xx$I <- ave(xx$NUMBER_OF_YEARS, xx$COMPANY_NUMBER, FUN = function(x){
c(rep(1, length(x) - 1), length(x)) ### ???
})
Hope this helps,
Rui Barradas
Às 19:46 de 30/04/21, Bert Gunter escreveu:
There is something wrong here I believe --
Hello,
Can you post sample data? For instance, the output of
dput(head(dovrez, 20))
dput(head(rqa_df_USD, 20))
Or maybe you could rbind the data.frames with one column telling which
of Res or LAM the values come from.
Hope this helps,
Rui Barradas
Às 19:38 de 02/05/21, Baki UNAL via R
"x1.conserv" "x2.black" "x2.conserv"
Hope this helps,
Rui Barradas
Às 18:20 de 08/05/21, Jeff Newmiller escreveu:
Regular expression patterns are not vectorized... only the data to be searched are. Use
one of the many websites dedicated to tutoring regular expres
e = FALSE)
}
jj <- jindex[!i1]
x.label <- x.label[jj]
}
Or even simpler
if(any(!i1 & !i2)){
words <- jindex[!i1 & !i2]
pattern <- paste(words, collapse = "|")
jindex <- grep(pattern = pattern, x.label, value = FALSE)
jj <- jindex[!i1]
x.label
)
[...]
par(op)
dev.off()
#
The comments only line is your last code line.
The result is attached.
Hope this helps,
Rui Barradas
Às 19:39 de 09/05/21, varin sacha via R-help escreveu:
Dear R-experts,
I am trying to get the 8 graphs like the ones in this
alue[flag]) ) %>%
select(-flag)
identical(df2$MinValue, df3$MinValue)
#[1] TRUE
Or, simpler:
df4 <- df %>%
group_by(Department,Class) %>%
mutate(MinValue = min(Value[Value != 0]) )
identical(df2$MinValue, df4$MinValue)
#[1] TRUE
Hope this helps,
Rui Barradas
Às 12:11 de 11
(old_par) # at the end, put the graphics device pars back.
Hope this helps,
Rui Barradas
Às 09:33 de 12/05/21, varin sacha via R-help escreveu:
Dear Experts,
My R code was perfectly working since I decide to add a 5th correlation
coefficient : hoeffdings' D.
fter a google search, I guess I n
;
found2 <- sapply(df1, function(x) grepl(pattern, x))
which(found2, arr.ind = TRUE)
Hope this helps,
Rui Barradas
Às 18:07 de 15/05/21, Tuhin Chakraborty escreveu:
Hi,
How can I find the location of string data in my 2D dataset? spec(Dataset)
will reveal the columns that contain the strings. Bu
ngs(is.na(as.numeric(x)))
}
found <- sapply(df1, f)
which(found, arr.ind = TRUE)
Hope this helps,
Rui Barradas
Às 06:31 de 16/05/21, Tuhin Chakraborty escreveu:
Thank you everyone, for the very helpful suggestions. I understand that my
question is not altogether clear. So let me share an example.
RUE)
res <- as.data.frame(res)
res$col_name <- names(df1)[ res$col ]
With a big data set the first res is a numeric matrix and it's access
and extraction is faster, matrix operations are generally faster than
data.frame operations.
Hope this helps,
Rui Barradas
Às 08:30 de 16/05/21, Rui Barrad
Hello,
I too don't see why grouping is needed but here it goes.
df1 %>%
group_by(Department, Class) %>%
mutate(Value = Value + 5*(Date == "4.01.2020"))
Hope this helps,
Rui Barradas
Às 17:01 de 26/05/21, Christopher W Ryan via R-help escreveu:
Is the grouping
Hello,
ifelse needs a logical condition, not the value. Try grepl.
CRC$MMR.gene <- ifelse(grepl("MLH1"|"MSH2",CRC$gene.all), "Yes", "No")
Hope this helps,
Rui Barradas
Às 05:29 de 27/05/21, Kai Yang via R-help escreveu:
Hi List,
I wrote the
;- ''
c[i] <- ''
d[i] <- ''
})
Hope this helps,
Rui Barradas
Às 17:28 de 30/05/21, Kai Yang via R-help escreveu:
Hello List,I have a data frame which having the character columns:
| a1 | b1 | c1 | d1 |
| a2 | b2 | c2 | d2 |
| a3 | b3 | c3 | d3 |
| a4 | b4
uot;exchangeable"
)
plot_model(m5, type = "pred", terms = c("Pa", "Pt"))
Hope this helps,
Rui Barradas
Às 23:10 de 31/05/21, Luis Fernando García escreveu:
Dear all,
I want to plot a Generalized Estimating Equation (GEE) model, with
interactions and the confidence
to post in HTML format.
Hope this helps,
Rui Barradas
Às 22:07 de 02/06/21, Kai Yang via R-help escreveu:
Hi List,
I use paste function to concatenate 3 character columns together.
when I run table to see that, I found 3 categories. How can I write script to
trim NA in 2nd and 3rd group and
Hello,
Why not write a function?
odd <- function(x, numeric = TRUE){
i <- x %% 2 == 1
if(numeric) x[i] else i
}
odd(1:100)
Hope this helps,
Rui Barradas
Às 19:17 de 02/06/21, nelpar escreveu:
I don't understand. --
7%%2=1
9%%2=1
11%%2=1
What aren't these numbers
Hello,
This is cross-posted from StackOverflow [1]. Cross posting is not well
seen on R-help and the SO post is better explained (at least the data
seem to be more complete). You should have waited for an answer there.
Hope this helps,
Rui Barradas
Às 15:03 de 04/06/21, Madison Bell
- read_csv(
all_epi_files,
col_types = cols(
TestAssay = col_character(),
Var1 = col_character(),
Var2 = col_character(),
Freq = col_double()
)) %>%
mutate(AssayNum = str_extract(TestAssay, "\\d+")) %>%
group_by(AssayNum) %>%
group_map(~ epi_analysis(.x
Hello,
This is not reproducible, we don't have access to ttclasses.csv.
Can you post sample data? Please post the output of
dput(ttclasses)
Or, if it is too big, the output of
dput(head(ttclasses, 20))
Hope this helps,
Rui Barradas
Às 18:03 de 07/06/21, Biplab Nayak escreveu:
Hi A
)
)
# Define server logic to plot
server <- function(input, output) {
output$gradePlot <- renderPlot({
grade_ad = input$assessment
boxplot(ttclasses$score[ttclasses$assessment==grade_ad],
frame.plot=FALSE, horizontal=TRUE, col="magenta",
main=grade_ad)
ttcla
Hello,
Please cc the list.
R-Help is not a code writing service, it's a mailing list for doubts on
R code. I'm sorry but without answers to the below questions I am not
going to answer.
What have you tried? What went wrong?
Hope this helps,
Rui Barradas
Às 04:01 de 08/06/
Hope this helps,
Rui Barradas
Às 18:57 de 09/06/21, Enrico Gabrielli escreveu:
Hello
I just registered on the list.
I am an agricultural technician and I am collaborating on a research
project on agroforestry and Brown Marmorated Stink Bug (Halyomorpha
halys, abbreviated BMSB).
Through kob
e = +str_detect(Utterance, coreWordsPat)) %>%
select(ID, Utterance, Core),
df %>%
mutate(Fringe = str_remove_all(Utterance, coreWordsPat),
Fringe = +(nchar(trimws(Fringe)) > 0)) %>%
select(ID, Fringe),
by = "ID"
)
Hope this helps,
Rui Barradas
Às 18:02
with another with the new values.
mydf2 <- data.frame()
mydf2
#data frame with 0 columns and 0 rows
rbind(mydf2, data.frame(data_POSIX = c(day1, day2), value = NA))
# data_POSIX value
#1 2018-02-01NA
#2 2018-02-02NA
But once again assignment will fail, and for the same reason.
Ho
ts_ctry %>%
slice_max(order_by = Count, n = top_count)
Hope this helps,
Rui Barradas
Às 17:10 de 27/06/21, Jeff Reichman escreveu:
R-help Forum
I am attempting to create a stacked bar chart but I have to many categories.
The following code works and I end up plotting all 134 countries but
Hello,
Either I'm not understanding or isn't this just any of
aggregate(count ~ ., data = test, FUN = length)
test %>% count(group1, group2, name = "Count")
?
Hope this helps,
Rui Barradas
Às 23:27 de 02/07/21, Yuan Chun Ding escreveu:
Hi R users,
In this test fi
"ABC" "ABC" "DDG" "DDG" "DDG" "DDG" "DDG" "ABB" "ABB"
Hope this helps,
Rui Barradas
Às 14:27 de 06/07/21, Evan Cooch escreveu:
Suppose I have a file with the the following structure - call the two
space
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