Hi all,
I'm trying to use the betamix function (betareg package) to create a beta
mixture model but get this error that I don't understand:
"Error in lm.wfit(x, linkfun(y), weights, offset = offset[[1L]]) :
NA/NaN/Inf in 'y'
Error in lm.wfit(x, linkfun(y), weights, offset = offset[[1L]]) :
NA
Thank you David
Now, the problem is to list all the combinations which verify the
condition III (ie every Rapporteur has to have more or less the same
number of demandeur)
Have you any idea ?
Michel
Le 14/10/2014 13:18, david.kaeth...@dlr.de a écrit :
Hello,
here's a draft of a solution.
[I sent this email to Rich and to the list after some private
communication with Rich. However it got held up for being too big,
presumably because of the pdf attachment. So I am re-sending it to the
list *without* the attachment.]
I resorted to actually trying the expedient of installing "c
On Wed, 15 Oct 2014, Rolf Turner wrote:
In all probability this is FAQ 7.22. I.e. use:
print(ellipses(mean=mn, var=vr, r=r, steps=72, thinRatio=NULL,
aspanel=FALSE,col='red', lwd=2))
Rolf,
Thank you. Didn't occur to me to look in the FAQ for this issue.
P. S. See also fortun
In all probability this is FAQ 7.22. I.e. use:
print(ellipses(mean=mn, var=vr, r=r, steps=72, thinRatio=NULL,
aspanel=FALSE,col='red', lwd=2))
cheers,
Rolf Turner
P. S. See also fortune(123).
R. T.
On 15/10/14 11:00, Rich Shepard wrote:
A rather strange situation here and
A rather strange situation here and I've not found the source of the
problem.
The point is to print a ternary plot matrix of compositional data with
ellipses enclosing 95% of the variance in each plot. The ellipses display on
the monitor, dev = x11cairo, but not when sent directly to a file,
On 15/10/14 04:09, Kate Ignatius wrote:
In the sense - it does not work. it works when there are 50 samples
in the file, but it does not work when there is one.
The usual headings are: sample1.at sample1.dp
sample1.fg sample2.at sample2.dp sample2.fg and so on to a max of
sample50.at sampl
Hello,
Have you tried
mapply(f, list_df, list_par, MoreArgs = list(z = fix), SIMPLIFY = FALSE)
?
Hope this helps,
Rui Barradas
Em 14-10-2014 19:42, Carlos Nasher escreveu:
Hi R helpers,
I'm struggling how to apply a function to multiple lists. My function uses
a dataframe, a list of param
Hello,
Maybe something like
data<-c(1,5,10)
func1<-function(x) {
nm <- deparse(substitute(x))
y<-x^2
z<-x^3
out<-data.frame(nm,y,z)
return(out)
} #function
func1(data[1])
Hope this helps,
Rui Barradas
Em 14-10-2014 19:12, Evan Kransdorf escreveu:
Hello Everyone,
I was w
Hi R helpers,
I'm struggling how to apply a function to multiple lists. My function uses
a dataframe, a list of parameters and a fixed value as arguments. Now I
want to apply that function to several dataframes contained in list, with
several lists of parameters (also contained in a list) and the
Hello Everyone,
I was wondering if someone could help me implement a function in R.
I want to pass a vector x to my function, peform some math, then output the
data. However, I want the output for x to be the *name of the vector
I am *using
as input.
For example, data<-c(1,5,10)
> func1<-funct
Prevent graphic menues with:
options(menu.graphics = FALSE)
or and define repositories:
options(repos = c(CRAN = "http://cran.r-project.org";))
On 14 October 2014 17:00, wrote:
> Subscribers,
>
> A version of R is installed in a virtual machine, which has complete
> internet access via the host.
In the sense - it does not work. it works when there are 50 samples
in the file, but it does not work when there is one.
The usual headings are: sample1.at sample1.dp
sample1.fg sample2.at sample2.dp sample2.fg and so on to a max of
sample50.at sample50.dp sample50.fg
using this greps out a
You're right. I don't use regexps in R very much. In most other
languages, a single \ is needed. The R parser is different and I
forgot. Thanks for the heads up.
On Tue, Oct 14, 2014 at 10:01 AM, Ivan Calandra
wrote:
> Shouldn't it be
> grep("\\.at$",colnames(df))
> with double back slash?
>
> Iv
AT and at are not the same. If you want an case insensitive compare
for the characters "at" you need the "ignore.case=TRUE" added. E.g.:
df[,grep(".at",colnames(df),ignore.case=TRUE)
That should match the column name you gave. Which does not match your
initial description which said "ending with
Shouldn't it be
grep("\\.at$",colnames(df))
with double back slash?
Ivan
--
Ivan Calandra
University of Reims Champagne-Ardenne
GEGENA² - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, France
+33(0)3 26 77 36 89
ivan.calan...@univ-reims.fr
https://www.researchgate.net/profile/Ivan_Calandr
Subscribers,
A version of R is installed in a virtual machine, which has complete
internet access via the host.
The following error occurs when a package is selected:
install.packages([packagename], dependencies=TRUE)
--- Please select a CRAN mirror for use in this session ---
Killed
The err
Your question is missing a reproducible example, and you don't say how it does
not work, so we cannot tell what is going on.
Two things do come to mind, though.
A) Data frame subsets with only one column by default return a vector, which is
a different type of object than a single-column data f
For example,
DF will usually have numerous columns with sample1.at sample1.dp
sample1.fg sample2.at sample2.dp sample2.fg and so on
I'm running this code in R as part of a shell script which runs over
several different file sizes so sometimes it will come across a file
with one sample in it:
On Tue, Oct 14, 2014 at 9:23 AM, Kate Ignatius wrote:
> I'm having an issue with grep:
>
> I have numerous columns that end with .at... when I use grep like so:
>
> df[,grep(".at",colnames(df))]
>
> it works fine. When I have one column that ends with .at, it does not
> work. Why is that? As th
I'm having an issue with grep:
I have numerous columns that end with .at... when I use grep like so:
df[,grep(".at",colnames(df))]
it works fine. When I have one column that ends with .at, it does not
work. Why is that? As this is loop with varying number of columns
ending in .at I would like
If you just want to plot the various combinations of a set of
variables/columns, you don't need a list, just another data frame/matrix with
the combinations of the column numbers you want to plot:
> df <- matrix(rnorm(100), 10, 10)
> df <- data.frame(df)
> comb <- expand.grid(7:10, 7:10)
> comb
Hello,
here's a draft of a solution. I hope it's not overly complicated.
# find all possible combinations
combi <- expand.grid(Dem$Nom, Rap$Nom); names(combi) <- c("Dem", "Rap")
# we need the corresponding departments and units
combi$DemDep <- apply(combi, 1, function(x) Dem$Departement[x[1] ==
On Tue, Oct 14, 2014 at 3:36 AM, jpm miao wrote:
> Hi,
>
> I am plotting time series by ggplot2, but I believe that my question
> applies to other plotting tool as well.
>
> I want to make my x-axis the quarterly scale, e.g:
> 2000Q1 2000Q2.
>
>However, scale_x_date and date_format("%
> I want to plot( 11:20 ) in a plot.
> if i just type the code above, the y value will be from 11 to 20, now i
> want the
> value from a given range like 0 to 40, how can i do it?
See the ylim= argument to plot.default; eg
plot(x, y, ylim=c(0,40))
Also look at ?par and note that plot() and
Hello,
Use the argument ylim. See ?par instead of plot.
plot(11:20, ylim = c(0, 40))
abline(h=40)
Hope this helps,
Rui Barradas
Em 14-10-2014 10:33, PO SU escreveu:
Dear helpeRs,
I want to plot( 11:20 ) in a plot.
if i just type the code above, the y value will be from 11 to 20, now
> Thanks Joshua and Sven - I completely forgot about which() .
Also
na.omit(p[p<=0.05])
#and
p[p<=0.05 & !is.na(p)]
S.
***
This email and any attachments are confidential. Any use...{{dropped:8}}
> I'm stuck trying to begin an axis label in ggplot with a superscript.
For a crude work-round, you could try
ggplot(mydata) +
aes(x = x, y = y) +
geom_line() +
ylab(expression(paste(' '^{14}, "C", sep = "")))
S Ellison
***
Thanks Joshua and Sven - I completely forgot about which() .
Pascal - I never new about complete.cases - interesting function.
Thanks,
Rainer
Rainer M Krug writes:
> Hi
>
> I want to evaluate NA and NaN to FALSE (for indexing) so I would like to
> have the result as indicated here:
>
> ,--
Hi,
Perhaps still not as short as you want, but I normally use which():
p <- c(1:10/100, NA, NaN)
p[which(p <= .05)]
[1] 0.01 0.02 0.03 0.04 0.05
Cheers,
Josh
On Tue, Oct 14, 2014 at 8:51 PM, Rainer M Krug wrote:
> Hi
>
> I want to evaluate NA and NaN to FALSE (for indexing) so I would lik
Hi Rainer,
As "complete.cases()" does?
p <- c(1:10/100, NA, NaN)
complete.cases(p)
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE FALSE
Regards,
Pascal
On Tue, Oct 14, 2014 at 6:51 PM, Rainer M Krug wrote:
> Hi
>
> I want to evaluate NA and NaN to FALSE (for indexing)
use:
which(p<=.05)
this will not yield logical, but integer indices without NA
On 14 October 2014 11:51, Rainer M Krug wrote:
> Hi
>
> I want to evaluate NA and NaN to FALSE (for indexing) so I would like to
> have the result as indicated here:
>
> ,
> | > p <- c(1:10/100, NA, NaN)
> | > p
Dear jpm, have attached a simpler example, hope this helps. Regards.
plot(0:100,pch="",xaxt="n",)
axis(1, at=1,lab=c("abcd"),cex.axis=1,font=4)
axis(1, at=20,lab=c("efgh"),cex.axis=1,font=4)
axis(1, at=60,lab=c("ijkl"),cex.axis=1,font=4)
On Tuesday, October 14, 2014 11:48 AM, Franklin Mairu
Hi
I want to evaluate NA and NaN to FALSE (for indexing) so I would like to
have the result as indicated here:
,
| > p <- c(1:10/100, NA, NaN)
| > p
| [1] 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 NA NaN
| > p[p<=0.05]
| [1] 0.01 0.02 0.03 0.04 0.05 NA NA
| > p[sapply(p<=0.05,
Dear help,
I’m stuck trying to begin an axis label in ggplot with a superscript. While I’m
fine using expression to insert them in between normal text in an axis label,
this doesn’t appear to work at the start of an expression. For example:
mydata <- data.frame(x = 1:10, y = 10:1)
# this works
Dear helpeRs,
I want to plot( 11:20 ) in a plot.
if i just type the code above, the y value will be from 11 to 20, now i want
the value from a given range like 0 to 40, how can i do it?
I read ?plot for a while but still can't solve it.
Actually, the qustion is founded when i already plo
On Oct 13, 2014 11:58 PM, "Patricia Seo" wrote:
>
> Hi everyone,
>
> My help request is similar to what was asked by Ken Termiso on April
18th, 2005. Link here:
https://stat.ethz.ch/pipermail/r-help/2005-April/069729.html
>
> Matt Wiener answered with suggesting a vector list where you hand type
e
On Mon, 13 Oct 2014 08:54:51 PM Kate Ignatius wrote:
> Just an update to this:
>
> gtal <- function(d) {
> alleles <- sapply(d, function(.) strsplit(as.character(.), "/"))
> gt <- unlist(lapply(alleles, function(x)
>ifelse(identical(x[[1]], vcf[,3]) & identical(x[[2]], vcf[,3])
Try this code, this may get a solution close to what you need. The advantage
with this code is that you specify the text you want to appear on the xaxis.
The dates have to be supplied as text formats, and located on the xaxis using
the axis and at commands. the at command allows for customised
Hello
I have 2 df Dem and Rap.
I would want to build all the df (dfnew) by associating these two df
(Dem and Rap) in the following way :
For each value of Dem$Nom (dfnew$Demandeur), I associate 2 different
values of Rap$Nom (dfnew$Rapporteur1 and dfnew$Rapporteur2) in such a way
* for each d
Hi,
I am plotting time series by ggplot2, but I believe that my question
applies to other plotting tool as well.
I want to make my x-axis the quarterly scale, e.g:
2000Q1 2000Q2.
However, scale_x_date and date_format("%m/%d") support all time formats
BUT QUARTERs
library(sca
Hi,
Thanks for your help.I got it installed.
Madhvi
On Tuesday 14 October 2014 12:50 PM, Pascal Oettli wrote:
The support for RStudio is located here: https://support.rstudio.com
Regards,
Pascal
On Tue, Oct 14, 2014 at 4:08 PM, madhvi wrote:
Hi,
How to install RStudio after downloading debi
The support for RStudio is located here: https://support.rstudio.com
Regards,
Pascal
On Tue, Oct 14, 2014 at 4:08 PM, madhvi wrote:
> Hi,
> How to install RStudio after downloading debian package
>
> Madhvi
>
> On Tuesday 14 October 2014 12:09 PM, Pascal Oettli wrote:
>>
>> Please reply to the
Hi,
How to install RStudio after downloading debian package
Madhvi
On Tuesday 14 October 2014 12:09 PM, Pascal Oettli wrote:
Please reply to the list, not only to me.
RStudio is for Ubuntu 10.04+ (please note the "+").
About R 3.1.0, you probably will have to compile from the source.
Regards
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