On Mon, Nov 7, 2011 at 9:23 PM, Kevin Burton wrote:
> I am trying to upgrade to R 2.14 from R 2.13.1 I have compied all the
> libraries from the 'library' directory in my existing installation (2.13.1)
> to the installed R 2.14. Now I want to uninstall the old installation (R
> 2.13.1) and I get t
No, I have not.
But it is a very nice option. Can't believe that I have overlooked it.
Thanks a lot.
Best regards
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That is exactly what I thought I would do. But the problem persist.
In the document below I have added "\inputencoding utf8", yet LyX
fails to compile.
#LyX 2.0 created this file. For more info see http://www.lyx.org/
\lyxformat 413
\begin_document
\begin_header
\textclass article
\use_default_
If anyone has ever hesitated about posting a question to R-help that
might be less about R and more about scientific computing in general
(algorithms, datasets etc) then you might be interested in a proposed
new site on StackExchange:
http://area51.stackexchange.com/proposals/28815/scientific-comp
*The situation (or an example at least!)*
example<-data.frame(rep(letters[1:10]))
colnames(example)[1]<-("Letters")
example$numb1<-rnorm(10,1,1)
example$numb2<-rnorm(10,1,1)
example$numb3<-rnorm(10,1,1)
example$id<-c("CG234","CG232","CG441","CG128","CG125","CG182","CG232","CG441","CG232","CG125")
I'm testing out quantstrat using a simple one-indicator strategy.
The error I get after applyStrategy(...) is
Error in .xts(e, .index(e1), .indexCLASS = indexClass(e1), .indexFORMAT =
indexFormat(e1), :
index length must match number of observations
In addition: Warning messages:
1: In match.n
On 11-11-15 4:46 AM, Giuseppe wrote:
That is exactly what I thought I would do. But the problem persist.
In the document below I have added "\inputencoding utf8", yet LyX
fails to compile.
Yihui is probably right that this is a question about Lyx, not R, but
one other thing you can try is to
Hi Paul,
You are right. Model selection takes places using the common data sample
across ALL checked models (otherwise you would end up comparing models
estimated on different data!). What the procedure returns are the results
based on the common sample. If you want to have the full-sample results,
Jim,
2011/10/15 jim holtman :
> You might also want to consider the XLConnect package. I have had
> better luck reading/writing Excel files than with xlsReadWrite.
XLConnect looks good but - as the xlsReadWrite author and planing to
release a xlsx/64 bit successor - I'd be interested to learn wh
On 11-11-14 10:25 PM, Tyler Rinker wrote:
Duncan,
Thank you for your reply. I was not clear about the Internet access. I do
have access, just at times I don't, hence the need to produce the manuals from
latex rather than simply using the Internet.
Please pardon my lack of knowledge around
Hello,
I need to analyse an experiment with 3 groups:
Control group
Treated group 1 (drug 1)
Treated group 2 (drug 2)
If I use a Kruskal-Wallis test do analyse the data and if I define the
control group as the reference group does the test do then the following
comparisons?
Control group vs. T
Good morning Rob,
First off, thank you for providing a reproducible example. This is one
of those little tasks that R is pretty great at, but there exist
>\infty ways to do so and it can be a little overwhelming for the
beginner: here's one with the base function ave():
cbind(ave(example[,2:4], e
Oh sorry -- my mistake with ave() -- I only checked the first row
drop = F is an optional argument to the function "[" which tells it to
return one of what it began with, rather than simplifying.
E.g.,
X = matrix(1:9, 3)
is.matrix(X)
TRUE
is.matrix(X[,2:3])
TRUE
is.matrix(X[,3])
FALSE # Ju
On Nov 14, 2011, at 7:20 PM, Carlisle Thacker wrote:
Thanks, Dennis. Yes, I can do that, but that locks the physical
units to locations of the labels. I had hoped that there might be
something a bit more flexible, like a subtitle or more general text.
If you would take the time to descri
On Nov 14, 2011, at 10:49 PM, Miles Yang wrote:
Hi all,
I wrote a r program as below:
x <- 1:10
y <- c(3,3,3,3,3,3,3,3,3,3)
fit <- lm(log(y) ~ x)
summary(fit)
And I expect to get some error message from R, because "y" is
constant.
But, I got the message as below:
You are asking R to tel
On Nov 14, 2011, at 11:49 PM, Sverre Stausland wrote:
Hi R users,
I end up with a list object after running an anova:
lm(speed ~ 1 + dist + speed:dist, data = cars) -> Int
lm(speed ~ 1 + dist, data = cars) -> NoInt
anova(Int, NoInt) -> test
test <- test[c("Df", "F", "Pr(>F)")][2,]
is.list(te
Hi,
I'm new to R and tried a search but couldn't find what I was looking for.
I have some data as a csv file with columns:-
longditude latitude year month rainfall region
What I need to do is produce a monthly time series for each region, where
region is an integer id and where each time point
Hello,
i use ggplot to plot some measures including CIs as horizontal errorbars. I get
an error when the scale limits are narrower than the boundaries of the error
bar and hence the CIs are not plotted.
library(ggplot2)
df <- data.frame(resp=c(1,2), k=c(1,2), se=c(1,2))
ggplot(df, aes(resp,y=k
Thanks Michael,
That second (aggregate) option worked perfectly - the first (cbind)
generated averages for each row between the columns. (rather than between
rows for each column).
I came so close with aggregate yesterday - it is only slightly different to
one my attempts (of admittedly very ma
Dear experts,
I would like to plot a hierarchical clustering of 300 items. I had a
distance matrix with dimension of 300*300. The 300 items were from 7 groups
which I would like to label with 7 different colours in the plot.
>h<-hclust(as.dist(300_distance_matrix))
>plot(h,hang=-1,cex=0.5, col="
Dear experts,
I would like to plot a hierarchical clustering of 300 items. I had a
distance matrix with dimension of 300*300. The 300 items were from 7
groups which I would like to label with 7 different colours in the plot.
h<-hclust(as.dist(300_distance_matrix))
plot(h,hang=-1,cex=0.5, c
It's a big subject and various mechanisms exist, but you should
probably start by looking into the zoo package and the read.zoo()
function.
Hope that helps,
Michael
On Tue, Nov 15, 2011 at 8:03 AM, Chuske wrote:
> Hi,
>
> I'm new to R and tried a search but couldn't find what I was looking for.
Hi Dongli,
Questions about usage of specific contributed packages are best directed toward
the package maintainer/author first, as they are likely the best sources of
information, and they don't necessarily subscribe to or keep up with the daily
deluge of R-help messages.
(In this particular c
Hola,
Alguno ha usado el paquete RODBC para acceder a una BBDD MySQL desde R en
Windows?
Qu'e mas tengo que hacer a parte de:
1) Aniadir el Driver de MySQL a la lista de User DSN en Control panel ->
Administrative tools -> Data Sources(ODBC)
2) Testear que funciona la conexion
2) Ejecutar: ch <-
On Tue, Nov 15, 2011 at 8:20 AM, R. Michael Weylandt
wrote:
> It's a big subject and various mechanisms exist, but you should
> probably start by looking into the zoo package and the read.zoo()
> function.
>
Note that there is an entire vignette on read.zoo, as well. See the
Reading Data in Zoo
Fischer, Felix charite.de> writes:
>
> Hello,
>
> i use ggplot to plot some measures including CIs as horizontal
> errorbars. I get an error when the scale
> limits are narrower than the boundaries of the error bar and
> hence the CIs are not plotted.
>
> library(ggplot2)
> df <- data.frame(
Dear Miles,
Within rounding error, you got the right intercept, log(3); slope, 0;
residuals, all 0; residual standard error, 0; and standard errors of the
intercept and slope, both 0. The R^2 should have been undefined (i.e., 0/0),
but dividing one number that's 0 within rounding error by another
On Nov 15, 2011, at 6:46 AM, R. Michael Weylandt wrote:
Good morning Rob,
First off, thank you for providing a reproducible example. This is one
of those little tasks that R is pretty great at, but there exist
\infty ways to do so and it can be a little overwhelming for the
beginner: here's
On Nov 15, 2011, at 8:23 AM, Carlisle Thacker wrote:
Sorry that I was not clear. I was asking how to add annotation to
levelplot's colorkey, not the levelplot itself. The only entry I
can find from the help pages is via its labels.
Googling did yield this: " draw.colorkey() doesn't supp
I would like to get an hierarchical clustering tree with bootstrap values
indicated on the nodes, as in pvclust. The problem is that I have only
distance matrix instead of the raw data, required for pvclust. Is there a
way to get it?
fit1 <- hclust(dist) # an object of class '"dist"
plot(fit1) # d
Sorry that I was not clear. I was asking how to add annotation to
levelplot's colorkey, not the levelplot itself. The only entry I can
find from the help pages is via its labels.
Googling did yield this: " draw.colorkey() doesn't support a title for
the legend". So I presume there is also n
On 15.11.2011 14:34, Usuario R wrote:
Hola,
Alguno ha usado el paquete RODBC para acceder a una BBDD MySQL desde R en
Windows?
Qu'e mas tengo que hacer a parte de:
1) Aniadir el Driver de MySQL a la lista de User DSN en Control panel ->
Administrative tools -> Data Sources(ODBC)
2) Testear q
Hi Dongli,
Sorry for the delay in following up.
You might want to read the dsDesignManual.pdf document, which is available in
the 'inst/doc' folder in the package source tarball on CRAN, or in the package
'doc' directory in your R installation. Use:
system.file(package = "gsDesign")
to get
Hello,
I created several plot with ggplot2 dev mode.
Now I want to combine the plots in a grid
e.g. 2x2 with a fixed size of the output.
What I am doing at the moment is:
grid.newpage()
pushViewport(viewport(layout = grid.layout(nrow=2, ncol=2,
widths = unit(c(7.5,6.5),
In the solution below, what is the advantage of using "0L".
M0 <- read.csv("M1.csv", nrows = 1)[0L, ]
Thanks!
2011/11/8 Gabor Grothendieck :
> 2011/11/8 Sergio René Araujo Enciso :
>> Dear all:
>>
>> I have two larges files with 2000 columns. For each file I am
>> performing a loop to extract t
Please ask the authors and/or discussion list of dism, it may be a bug in dism
or incompatibility between the maxent you are using and the package, I can't
check sine maxent has a restrictive license. Also please do not cross-post to
multiple mailing lists.
Thanks,
Simon
On Nov 14, 2011, at 1:
On Tue, Nov 15, 2011 at 9:44 AM, Juliet Hannah wrote:
> In the solution below, what is the advantage of using "0L".
>
> M0 <- read.csv("M1.csv", nrows = 1)[0L, ]
>
As mentioned, you will find quite a bit of additional info on the
sqldf home page but to address the specific question regarding the
Part of my problem has to do with getting through the corporate
firewall to access the other program I have to download to use it. I
just tried today and this is what I got:
> xls.getshlib()
Loading required package: tools
--- xls.getshlib running... ---
- download.file from
'http://dl.dropbox.
Yes, that is the point. I would prefer a LyX file instead of text in
email, since the encoding can be different.
You still did not tell us your R version, or better, sessionInfo().
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State Univers
Good Morning.
I wonder if R have a funtion to convert an ARMA process to infinite AR
process, like
ARMAtoMA function wich convert an ARMA process to infinite MA process
--
Diego Fernando Lemus Polanía
Ingeniero Industrial
Universidad Nacional de Colombia
Sede Medellín
[[alternative HTML
Duncan,
Thank you for your patience, time and expertise. You were 100% correct and the
problem has been resolved. I'm adding what I did as a windows user to complete
the list record for future searchers.
To download the inconsolata package (you may approach this several ways this
one see
Dear Ben,
great, works fine! I guess, the error occurs because data outside the scale
limits "is thrown away" as stated in ?coord_cartesian .
Thanks,
Felix
From: Ben Bolker
mailto:bbolker_at_gmail.com?Subject=Re:%20[R]%20break%20error%20bars%20in%20ggplot2>>
Date: Tue, 15 Nov 2011 13:44:44 +
Hello;
I am having a problems with the interpretation of models using ordered or
unordered predictors.
I am running models in lmer but I will try to give a simplified example
data set using lm.
Both in the example and in my real data set I use a predictor variable
referring to 3 consecutive days o
hi R,users
Now I read a data from a txt file
newdata<-read.table("1.txt")
in the 1.txt ,there are several column shown as below
1 3 4 5
2 3 5 6
4 5 6 7
so when I want analysis the second column
anadata<-newdata$V2
but my question I can not use some certain variable to indice the column?
e.g
cmn=2
Has anyone come across the right combinations to print a limited number of
digits? My trial and error approach is taking too much time. Here is what I
have tried:
> op <- options()
> a <- c(1e-10,1,2,3,.5,.25)
> names(a) <- c("A", "B", "C", "D", "E", "F")
> # default
> a
A B
Hi,
Look at ?"["
anadata <- newdata[, cmn]
## i.e., extract all rows (the first argument is empty of the 2 column
anadata <- newdata[, 2]
Hope this helps,
Josh
On Tue, Nov 15, 2011 at 8:02 AM, haohao Tsing wrote:
> hi R,users
> Now I read a data from a txt file
> newdata<-read.table("1.txt")
On Tue, Nov 15, 2011 at 11:23 AM, Joshua Wiley wrote:
> Hi,
>
> Look at ?"["
>
> anadata <- newdata[, cmn]
> ## i.e., extract all rows (the first argument is empty of the 2 column
> anadata <- newdata[, 2]
Or, if this is part of a more general problem and the column names are
not necessarily in s
Hi Kevin,
I am not sure you will find anything other than manual tweaking, that
will vary between no decimals for integers, some for small fractions,
and scientific for very small. You can also look at: ?round
?format. If this is for code/a report, you could make any formatting
you wanted with
Ordered factors use orthogonal polynomial contrasts by default. The .L and
.Q stand for the linear and quadratic terms. Unordered factors use
"treatment" contrasts although (they're actually not contrasts), that are
interpreted as you described.
If you do not know what this means, you need to do s
On Nov 15, 2011, at 11:02 AM, haohao Tsing wrote:
hi R,users
Now I read a data from a txt file
newdata<-read.table("1.txt")
in the 1.txt ,there are several column shown as below
1 3 4 5
2 3 5 6
4 5 6 7
so when I want analysis the second column
anadata<-newdata$V2
but my question I can not use
... In addition, the following may also be informative.
> f <- paste("day", 1:3)
> contrasts(ordered(f))
.L .Q
[1,] -7.071068e-01 0.4082483
[2,] -7.850462e-17 -0.8164966
[3,] 7.071068e-01 0.4082483
> contrasts(factor(f))
day 2 day 3
day 1 0 0
day 2 1
When you print a vector R uses a single
format for the whole vector and tries to
come up with one format that displays all the values
accurately enough. For a matrix (or data.frame)
it uses a different format for each column, so perhaps
you would like the output of:
> matrix(a, nrow=1, dimnames
Thanks. Yes, I meant nrow(dataset)+1 (typo...)
Sammy
On Mon, Nov 14, 2011 at 1:29 AM, Petr PIKAL wrote:
> >
> > dataset[ nrow(dataset), ] <- c ("Male", 5, "bad")
> >
> > The above seems to have worked to append a row in place of a rbind().
> This
>
> No. It overwrites your last row. You maybe m
On Tue, Nov 15, 2011 at 9:00 AM, Catarina Miranda
wrote:
> Hello;
>
> I am having a problems with the interpretation of models using ordered or
> unordered predictors.
> I am running models in lmer but I will try to give a simplified example
> data set using lm.
> Both in the example and in my rea
Thank you. I mainly didn't know about the vector/matrix printing rules.
Kevin
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Tuesday, November 15, 2011 10:43 AM
To: Kevin Burton; r-help@r-project.org
Subject: RE: [R] Controlling the precision of the digits prin
Good afternoon list,
I have the following character strings; one with spaces between the maths
operators and variable names, and one without said spaces.
form<-c('~ Sentence + LEGAL + Intro + Intro / Intro1 + Intro * LEGAL +
benefit + benefit / benefit1 + product + action * mean + CTA + help + me
Hi Michael,
You need to take another look at the examples you were given, and at
the help for ?sub():
The two ‘*sub’ functions differ only in that ‘sub’ replaces only
the first occurrence of a ‘pattern’ whereas ‘gsub’ replaces all
occurrences. If ‘replacement’ contains backreferen
I'm using Bill Browne's MLPowSim to do some sample size estimation for a
multilevel model. It creates an R program to carry out the estimation using
lmer in the lme4 library. When there are predictors with more than two
categories one has to modify the code generated to account for the multino
Hi Michael,
Your strings were long so I made a bit smaller example. Sarah made
one good point, you want to be using gsub() not sub(), but when I use
your code, I do not think it even works precisely for one instance.
Try this on for size, you were 99% there:
## simplified cases
form1 <- c('produ
Hello,
with Sys.time() you get the following string:
"2011-11-15 16:25:55 GMT"
How can I extract the following substrings:
year <- 2011
month <- 11
day_time <- 15_16_25_55
Cheers,
Syrvn
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take a look at the structure of what Sys.time returns.
str(Sys.time)
and now at ?strptime!
> format(Sys.time(),format='%d-%H-%M-%S')
[1] "15-09-55-55"
> format(Sys.time(),format='%Y')
[1] "2011"
> format(Sys.time(),format='%m')
[1] "11"
>
Hope that helps,
Justin
On Tue, Nov 15, 2011 at 9:48
Hi List,
I am new to R, this may be simple.
I want to store directory path as parameter which in turn to be used while
reading and writing data from csv files.
How I can use dir defined in the below mentioned example while reading the
csv file.
Example:
dir <- "C:/Users/Desktop" #location of
Hi all,
I have the mean vector mu<- c(0,0) and variance sigma <- c(10,10), now how
to sample from the bivariate normal density in R?
Can some one suggest me?
I did not fine the function "mvdnorm" in R.
Best
Gyan
[[alternative HTML version deleted]]
Just as an update on this problem:
I have managed to get the variance for the selected columns
Now all I need is the covariance between these 2 selections -
the two target columns are and the aim is that a new column contain a
covariance value between these on each row:
maindata[,c(174:213)] an
Hi Aajit,
try using the ?paste function to combine the variable with your
directly and the filename into one string, and then pass that to
read.csv() or whatever
paste(dir, "/bs_dev_segment_file.csv", sep = '')
HTH,
Josh
On Tue, Nov 15, 2011 at 6:12 AM, aajit75 wrote:
> Hi List,
>
> I am new
Try pasting them together like
paste(dir, "...")
You may need to use the collapse argument. Alternatively, change your working
directory to dir with setwd().
M
On Nov 15, 2011, at 9:12 AM, aajit75 wrote:
> Hi List,
>
> I am new to R, this may be simple.
>
> I want to store directory path
Hi Rob,
Here is one approach:
## define a function that does the calculations
## (the covariance of two vectors divided by the square root of
## the products of their variances is just a correlation)
rF <- function(x, a, b) cor(x[a], x[b], use = "complete.obs")
set.seed(1)
bigdata <- matrix(rno
On Nov 15, 2011, at 11:21 AM, Gyanendra Pokharel wrote:
Hi all,
I have the mean vector mu<- c(0,0) and variance sigma <- c(10,10),
now how
to sample from the bivariate normal density in R?
Can some one suggest me?
I did not fine the function "mvdnorm" in R.
But when you typed ?mvdnorm R sh
file.path() is much better for this than paste(), e.g.
dir <- "C:/Users/Desktop"
pathname <- file.path(dir, "bs_dev_segment_file.csv")
temp_data <- read.csv(pathname)
/Henrik
On Tue, Nov 15, 2011 at 10:08 AM, R. Michael Weylandt
wrote:
> Try pasting them together like
>
> paste(dir, "...")
>
>
Hi R users,
I want to colored points by their value
for example:
x <- c(1,2,3,4)
y <- c(1,2,3,4)
z <- c(2,3,4,9)
y and x are coordinates
z is the value of the coordinates
points(x,y,col= rainbow(z))
something like that
But haven't found any solution at the moment.
Thanks.
Chris
--
View
I was able to solve this problem by going back to nls and obtaining the
initial parameter estimates through optim. When I used nlsList with my
dataset, it took 2 minutes to solve and was not limited by the bounds. Now I
have the bounds working and it takes 45 seconds to solve. Here is the new
code:
Hi all,
I'm trying to estimate model parameters in R for a pretty simple system of
equations, but I'm having trouble. Here is the system of equations (all
derivatives):
eqAlgae <- (u_Amax * C_A) * (1 - (Q_Amin / Q_A))
eqQuota <- (p_max * R_V) / (K_p + R_V) - ((Q_A-Q_Amin)*u_Amax)
eqResource <- -C
Dear Louise
On 15 November 2011 19:03, lstevenson wrote:
> Hi all,
>
> I'm trying to estimate model parameters in R for a pretty simple system of
> equations, but I'm having trouble. Here is the system of equations (all
> derivatives):
> eqAlgae <- (u_Amax * C_A) * (1 - (Q_Amin / Q_A))
> eqQuota
I'm not sure your request completely makes sense: marginal means and variances
are not sufficient to give the joint distribution; even if you can be assured
it is bivariate normal, you still need a correlation. Just a heads up
Michael.
PS - next time would you please use a slightly more nuance
Try either col=z or col=rainbow(max(z))[z] depending on what color scheme you
want.
Michael
On Nov 15, 2011, at 1:47 PM, Chris82 wrote:
> Hi R users,
>
> I want to colored points by their value
>
> for example:
>
> x <- c(1,2,3,4)
> y <- c(1,2,3,4)
> z <- c(2,3,4,9)
>
> y and x are coord
Hi Chris,
On Tue, Nov 15, 2011 at 1:47 PM, Chris82 wrote:
> Hi R users,
>
> I want to colored points by their value
>
> for example:
>
> x <- c(1,2,3,4)
> y <- c(1,2,3,4)
> z <- c(2,3,4,9)
>
> y and x are coordinates
>
> z is the value of the coordinates
>
> points(x,y,col= rainbow(z))
In the ge
Dear all,
I am trying to install a package from bioconductor (biomaRt) for which I
need the RCurl package. I get the following main error message when I try
to install RCurl (and its dependencies).
configure: error: no acceptable C compiler found in $PATH
See `config.log' for more details.
ERROR:
Take advantage of a 20% discount on the most recent R books from Chapman &
Hall/CRC!
We are pleased to offer our latest R books at a 20% discount through our
website. To take advantage of this offer, simply visit www.crcpress.com, choose
your titles and insert code AZL02 in the 'Promotion C
Hello,
Can you please help me with this? I am also stack in the same problem.
Sam
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__
I would like the text plotted with 'mtext' to be alighned like it is for
printing on the console. Here is what I have:
> print(emt)
ME RMSE MAE
MPE MAPE MASE
original -1.034568e+07 1.097695e+08 2.433160e+07 -31.30554 3
Thanks David - this is pretty close to what I am looking for. However,
the output of vec[2] now includes the row number [1] and quotations
marks at the endpoints of the row. Is there an easy way to exclude
those?
Thanks
Sverre
On Tue, Nov 15, 2011 at 8:11 AM, David Winsemius wrote:
>
> On Nov 14
This problem is not for the faint of heart. Doug Bates, author of
nls(...) has said that a general purpose implementaion of R code for
multiresponse nonlinear regression is unlikely in the near future. You
have a large set of issues to deal with here. First, you have a system
of differential eq
I'm getting the following error in a script: "Error: could not find function
"lmer." I'm wondering of my lme4 package is installed incorrectly. Can
someone tell me the installation procedure? I looked at the support docs but
couldn't translate that into anything that would work.
__
Hi Kevin,
On Tue, Nov 15, 2011 at 2:36 PM, Kevin Burton wrote:
> I would like the text plotted with 'mtext' to be alighned like it is for
> printing on the console. Here is what I have:
You don't provide any of the info in the posting guide (OS may be
important here), or a reproducible example,
On Nov 15, 2011, at 2:43 PM, Sverre Stausland wrote:
Thanks David - this is pretty close to what I am looking for. However,
the output of vec[2] now includes the row number [1] and quotations
marks at the endpoints of the row. Is there an easy way to exclude
those?
The usual method is to use
Yet another solution. This time using the LaF package:
library(LaF)
d<-c(1,4,7,8)
P1 <- laf_open_csv("M1.csv", column_types=rep("double", 10), skip=1)
P2 <- laf_open_csv("M2.csv", column_types=rep("double", 10), skip=1)
for (i in d) {
M<-data.frame(P1[, i],P2[, i])
}
(The skip=1 is needed as l
Never mind-I fixed it.
My script is throwing the following error:
"Error in glmer(formula = modelformula, data = data, family = binomial(link =
logit), :
Argument ‘method’ is deprecated.
Use ‘nAGQ’ to choose AGQ. PQL is not available."
I remember hearing somewhere that PQL is no longer av
Error in cor(x[a], x[b], use = "complete.obs") : 'x' must be numeric
This is strange, it works on your example (and you've understood what I'm
trying to do perfectly), but then when I use it on the original data it
comes up with the error above
I've checked str() and the columns are all numeri
Wow, that is a very clever way to do it.
Thank you very much!
Cheers,
Syrvn
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I hadn't considered altering the font. Thank you I will try that.
-Original Message-
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: Tuesday, November 15, 2011 1:53 PM
To: Kevin Burton
Cc: r-help@r-project.org
Subject: Re: [R] Plot alignment with mtext
Hi Kevin,
On Tue, Nov 15,
Is the whole thing a data frame? Then any multi-column subset is also a data
frame. Try adding a as.matrix() wrapper in the definition of rF.
Michael
On Nov 15, 2011, at 3:14 PM, "Rob Griffin" wrote:
> Error in cor(x[a], x[b], use = "complete.obs") : 'x' must be numeric
>
> This is strange,
OK, I think I see the problem. Rather than setting method="nAGQ" I need
nAGQ=1. Doing so throws the following error:
"Warning messages:
1: glm.fit: algorithm did not converge
2: In mer_finalize(ans) : gr cannot be computed at initial par (65)
Error in diag(vcov(fitmodel)) :
error in evaluat
Excellent, as.matrix() didn't work but switched it to as.numeric() around
the definition of both variables in the function and it did work:
rF <- function(x, a, b) cor(as.numeric(x[a]), as.numeric(x[b]), use =
"complete.obs")
maindata$rFcor<-apply(maindata,1,FUN=rF,a=174:213,b=214:253)
Thanks
It's not clear what it means for the differences to be "of increasing
order" but if you simply mean the differences are increasing, perhaps
something like this will work:
library(caTools)
X = cumsum( 2*(runif(5e4) > 0.5) - 1) # Create a random Walk
Y = runmean(X, 30, endrule = "mean", align = "rig
Hi:
Summary:
I am trying to determine the 90th percentile of ambulance response times for
groups of data.
Background:
A fire chief would like to look at emergency response times at the 90th
percentile for 1 kilometer grids in Cape Coral, Florida. I have mapped out
ambulance response times on a
1) tapply will work for quantile, but the syntax was a little off: try this
tapply(Cape $ ResponseTime, Cape $ Grid_ID, quantile, c(0.05, 0.95))
The fourth argument is additional parameters passed to the function
FUN which here is quantile. You could also do this
tapply(Cape $ ResponseTime, Cape
There's a slight variant that might be even more helpful if you need
to line the data up with how you started: ave(). I'll let you work out
the details, but the key difference is that it returns a vector that
has the 90th percentile for each group, each time that group appears,
instead of the summa
David:
You need to re-read ?tapply _carefully_. Note that:
FUN the function to be applied, or NULL. In the case of functions
like +, %*%, etc., the function name must be backquoted or quoted.
Now note that in tapply(whatever, byfactor, mean), "mean" _is_ a function.
However in tapply (what
Given the example:
R> (levs <- quantile(volcano,c(0,0.1,0.5,0.9,0.99,1)))
0% 10% 50% 90% 99% 100%
94 100 124 170 189 195
R> levelplot(volcano,at=levs)
How can I make the key categorical with the size of the divisions equally
spaced in the key? E.g., five equal size rectangles wit
Generally I have several versions of R installed on any PC running
Windows 7 and have never had any problem choosing which version of R
was to be the default one. In the past the updated version did not
always register as the default but there was and is a utility
RSetReg.exe in the bin directory
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