Hi,
I am using ggplot and geom_violin to build a violin plot of some with only 2
categories. All is good except that I cannot set up the colors I want or the
violin plots. Either I have same color for both my categories or colors from
probably rainbow(2), which are red and blue. What if I wan
Hi,
Try:
mtext(bquote(Cortisol~(mu*g/ml)))
A.K.
On Monday, March 31, 2014 10:05 PM, Dennis Fisher wrote:
R 3.0.2
OS X
Colleagues
I am trying to label a graphic:
Cortisol (µg/ml)
A simplified version of the code is:
plot(1) ; mtext(bquote("Cortisol ("~mu~"g/ml)"))
This code inserts s
R 3.0.2
OS X
Colleagues
I am trying to label a graphic:
Cortisol (µg/ml)
A simplified version of the code is:
plot(1) ; mtext(bquote("Cortisol ("~mu~"g/ml)"))
This code inserts spaces around the mu:
Cortisol ( µ g/ml)
How can I suppress those spaces?
Dennis
Dennis Fishe
Hi All,
I am trying to model a set of data that's a little bit complicated. Let's
say a person can live 365 more days from time t=0. I have data on 100
people, but I only know who died between day 0 and day 30. Everyone else
would've died between days 31 and day 365.
Is there any way to fit a dis
Hello all,
A simple question. When I use grf from the package 'geoR' , I adopt the
exponential model. For this model is the parameter phi in m or km?
Best
ioanna
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R-help@r-project.org mailing
Hi,
I have some data from 1 year ago that is a measure of the number of days a
person used a certain product. I used fitdist to build a mle for this data
and got a pretty good fit.
My issue is that this data is outdated. I have new data from Januarry of
this year and would like to build a distrib
Hello all,
A simple question. When I use grf from the package 'geoR' , I adopt the
exponential model. For this model is the parameter range in m or km?
Best
ioanna
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R-help@r-project.org mailing
Hi, I'm using a function in the binom library. I'd like to add a title(s) to
the plot generated by "binom.bayes.densityplot".
I get an error message when trying to use the title function
The error message is: "Error in title(main = "my plot") : plot.new has not been
called yet" occurs after run
On 04/01/2014 09:25 AM, Hurr wrote:
How can I label existing axis tick marks with a
simple function of axis value like 1/AxisValue?
It seems like this should be an operation where
I just use the formula.
Hi Hurr,
You can do this with boxed.labels (plotrix) if you know the position of
the axis
On 04/01/2014 02:42 AM, Rich Shepard wrote:
On Mon, 31 Mar 2014, Rich Shepard wrote:
I've read ?plot, ?plot.default, and ?plot.window to learn how to change
the shape of the plot from a square to a rectangle. plot.window suggests
that the aspect (asp) parameter is appropriate but that associate
How can I label existing axis tick marks with a
simple function of axis value like 1/AxisValue?
It seems like this should be an operation where
I just use the formula.
--
View this message in context:
http://r.789695.n4.nabble.com/Label-axis-tick-marks-with-a-simple-function-of-axis-value-tp4
I would like to share the answer I have received from Thomas Lumley
For models of the same size all those criteria reduce to picking the model with
the smallest residual sum of squares, which is what the code does.
-thomas
Thanks Bert and Dennis for the information but although it does not a
On 01/04/14 05:42, Jefferson Ferreira Ferreira wrote:
There are other issues here addressing the same question, but I don't
realize how to solve my problem based on it. So, I have 5 data frames that
I want to merge rows in one unique data frame using rbind, but it returns
the error:
"Error in ro
One more thing, as.yearmon considers dates such as 1-Jan-61 to be of
year 2061 when they may be of year 1961. There's no problem with that if
the only use for them is in aggregate().
Rui Barradas
Em 31-03-2014 21:22, Rui Barradas escreveu:
Hello,
You have several columns of class "character"
Hello,
You have several columns of class "character" that must be converted to
"numeric" before the code I previously posted can work:
dat[,-1] <- lapply(dat[,-1], as.numeric)
Then use aggregate().
Rui Barradas
Em 31-03-2014 20:24, eliza botto escreveu:
Dear Rui,
I noticed that the previo
Converting data from a non-R format to R's internal format is a common
place to run into errors. Use the str() function to examine any data you
read in to make sure R is seeing it the way that you expect. If that looks ok
try using summary() or plot() for more checks that the data import went
as
Dear Rui,
Here is the complete data.
> dput(dat)
structure(list(Date = c("1-Jan-61", "2-Jan-61", "3-Jan-61", "4-Jan-61",
"5-Jan-61", "6-Jan-61", "7-Jan-61", "8-Jan-61", "9-Jan-61", "10-Jan-61",
"11-Jan-61", "12-Jan-61", "13-Jan-61", "14-Jan-61", "15-Jan-61", "16-Jan-61",
"17-Jan-61", "18-Jan-61"
Hello,
Sorry, but I have no idea why the error message, with me it works
correctly. Are you sure that the four columns A, B, C and D are numeric?
What does str(dat) say?
Rui Barradas
Em 31-03-2014 19:40, eliza botto escreveu:
Dear Rui,
Thanks for your reply. But the command seems not to be
Hi,
replace `lst2` with:
#Subset of data
lst1Sub <- lapply(lst1Not1970,function(x) x[c(1:25, 18707:18708)])
lst2 <- lapply(lst1Sub,function(x) {dateSite <- gsub("(.*G.{3}).*","\\1",x);
dat1 <- data.frame(Year=as.numeric(substr(dateSite,1,4)),
Month=as.numeric(substr(dateSite,5,6)),Day=as.numeri
There are other issues here addressing the same question, but I don't
realize how to solve my problem based on it. So, I have 5 data frames that
I want to merge rows in one unique data frame using rbind, but it returns
the error:
"Error in row.names<-.data.frame(*tmp*, value = value) : 'row.names'
> x <- factor(c("whoops","NA",NA,"B"))
> x
[1] whoops NAB
Levels: B NA whoops
> is.na(x)
[1] FALSE FALSE TRUE FALSE
If this doesn't explain it for you, then read about missing values and
factors in "An Introduction to R."
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467
On Mon, Mar 31, 2014 at 11:08 AM, Tierney, Luke wrote:
> The environment in which default arguments are evaluated is the
> environment of the function call itself, not the environment of the
> caller or the lexical enclosure (the same here). So the two 'y' used
> in function(y = y) refer to the sa
On Mar 31, 2014, at 1:29 PM, eliza botto wrote:
> Dear useRs,
> Sorry for such a ridiculous question but i really need to know that what is
> the difference between and NA and how to convert to NA.
> Thankyou very much in advance
> Eliza
is the printed output th
Dear Rui,
Thanks for your reply. But the command seems not to be working. I am getting
the following error.
Error in FUN(X[[1L]], ...) : invalid 'type' (character) of argument
Any idea? :(
Thanks,
Eliza
> Date: Mon, 31 Mar 2014 18:48:08 +0100
> From: ruipbarra...@sapo.pt
> To: eliza_bo...@hotma
Dear useRs,
Sorry for such a ridiculous question but i really need to know that what is the
difference between and NA and how to convert to NA.
Thankyou very much in advance
Eliza
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_
On Mon, 31 Mar 2014, Rich Shepard wrote:
What do I do to retain the specified width and height in the pdf output?
Use the height and width options to pdf(). Thanks to Don McQueen for
pointing me to the solution.
Rich
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R-help@r-project.org mail
Thanks AK.
Great job. I appreciate your effort.
Atem.
On Monday, March 31, 2014 11:28 AM, arun wrote:
Hi,
replace `lst2` with:
#Subset of data
lst1Sub <- lapply(lst1Not1970,function(x) x[c(1:25, 18707:18708)])
lst2 <- lapply(lst1Sub,function(x) {dateSite <- gsub("(.*G.{3}).*","\\1",x);
dat1 <
Hello,
Maybe the following will do.
library(zoo)
ym <- as.yearmon(dat$Date, "%d-%B-%y")
aggregate(dat[,-1], list(ym), FUN = sum, na.rm = TRUE)
Also, please use dput() to post data examples.
Hope this helps,
Rui Barradas
Em 31-03-2014 18:31, eliza botto escreveu:
Dear useRs,
I have the dat
Dear useRs,
I have the data of following format. I have only pasted some part of the data.
The data starts from 1961 and ends up in december 1987.
dat <- read.table(text="Date A B C D1-Jan-61 0.00 1.27 8.128 0.252-Jan-61 6.10
9.144 94.742 15.493-Jan-61 0.00 0.508 1.27 0.004-Jan-61 0.00 0 NA 0.00
On Mon, 31 Mar 2014, Rich Shepard wrote:
I've read ?plot, ?plot.default, and ?plot.window to learn how to change
the shape of the plot from a square to a rectangle. plot.window suggests
that the aspect (asp) parameter is appropriate but that associates the x
axis size to the value of the y axis
Note that the fact that bar() is defined within foo() is irrelevant.
## At the top level/global prompt:
> y <- 2
> bar<- function(y=y)y^2
> bar()
Error in y^2 : 'y' is missing
## but
> bar(y)
[1] 4
This is due to lazy evaluation and promises: The formal argument "y"
is not evaluated until it's
Compiled code gives a better error message (not clear why the
interpreter doesn't do this as well):
> cmpfun(foo)()
Error in bar() :
promise already under evaluation: recursive default argument reference or ear$
The environment in which default arguments are evaluated is the
environment of the
On 31/03/2014 10:40 AM, Philippe Grosjean wrote:
Hello,
I have difficulties to understand this one:
foo <- function (y = 2) {
bar <- function (y = y) y^2
bar()
}
foo()
#! Error in y^2 : 'y' is missing
foo(3)
#! Error in y^2 : 'y' is missing
This is simply a misunderstanding about sc
I've read ?plot, ?plot.default, and ?plot.window to learn how to change
the shape of the plot from a square to a rectangle. plot.window suggests
that the aspect (asp) parameter is appropriate but that associates the x
axis size to the value of the y axis and, plot.window suggests that it is
bett
Hello,
I have difficulties to understand this one:
foo <- function (y = 2) {
bar <- function (y = y) y^2
bar()
}
foo()
#! Error in y^2 : 'y' is missing
foo(3)
#! Error in y^2 : 'y' is missing
Note that this one works:
foo <- function (y = 2) {
bar <- function (y = y) y^2
bar(y)
Hi,
I am very new on R so I will remember to post in plain text next time.
Thank you all for your help. I think I can figure it out now~ Many thanks!:)
Best regards,
Siqi
2014-03-31 11:43 GMT+01:00 Helios de Rosario :
> >>> El día 30/03/2014 a las 15:23, "Si Qi L."
> > escribió:
> > Hi
> >
>
hello,
It's also possible to use the quantile function :
vec0 <- c(0.1,0.2, 0.5, 0.1,0.8, 0.4, 0.9)
which(vec0 >= quantile(vec0, 0.7, type = 1))
SD
2014-03-29 18:02 GMT+01:00 arun :
> Hi,
> Try:
> vec1 <- setNames(c(0.1,0.2, 0.5, 0.1,0.8, 0.4, 0.9), 1:7)
> vec2 <- sort(vec1,decreasing=TRUE)
>
On 31 Mar 2014, at 00:30 , Emilio Torres Manzanera wrote:
> Dear Sir,
> I am not sure about the precision of the cov.wt function. It seems that it
> provides different results when using frequency weights. This discrepancy
> only occurs with the covariance matrix, not with the correlation mat
I don't think cov.wt uses frequency weights. However, I don't think this is
mentioned in its help page.
Here is some information about the difference:
http://stats.stackexchange.com/questions/61225/correct-equation-for-weighted-unbiased-sample-covariance
The frequency version isn't hard to prog
>>> El día 30/03/2014 a las 15:23, "Si Qi L."
> escribió:
> Hi
>
> I have a problem with linear regression. This is my codes:
>
> acc1<‑ lm(data$acc ~ dummy + data$Reqloanamount + data$Code +
> data$Code.1 +
> data$EmpNetMonthlyPay + data$Customer_Age + data$RTI)
> summary(acc1)
>
> These attr
Hi
create index variable in your data e.g.
Model.Update$idx <- rep(1:n, each=5)
or
Model.Update$idx <- c(0,1:nrow(Model.Update)%/%5)
After that you can use aggregate or by with elaborated function, something like
fn <- function(dat) {
fit <- lm(y ~ x + z, data=dat)
result <- predict(fit)
}
Hi
see in line
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Kristi Glover
> Sent: Friday, March 28, 2014 7:04 PM
> To: R-help
> Subject: [R] starta.sampling with many (1000 times) and average them
>
> Hi R Users,
> I was t
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