Your code works!
strangelines.txt was created, and it's a text file with just spacebars ...
Seems like a few thousand lines of complete blanks (not 1 non-blank entry).
One thing, when I ran your code there was an error message;
> setwd("C:/Users/admin/Desktop/hons/Thesis")
> con <- file("dataset
Ok, I think I've got it this time.
The problem was that you have two different types of data in the same data
structure,
the first row are the result's column names, then the actual numeric data.
First, in what follows I've called your data.frame 'df1',
df1 <- structure(list(A2 = structure(c(9L,
Hi,
I tried to use naivebayes in package 'e1071'.
when I use following parameter, only one predictor, there is an error.
> m <- naiveBayes(iris[,1], iris[,5])
> table(predict(m, iris[,1]), iris[,5])
Error in log(sapply(attribs, function(v) { :
Non-numeric argument to mathematical function
On 5/4/2012 9:27 PM, Duncan Murdoch wrote:
On 12-05-04 10:33 PM, Joshua Wiley wrote:
On Fri, May 4, 2012 at 7:17 PM, Duncan
Murdoch wrote:
On 12-05-04 7:40 PM, Spencer Graves wrote:
[snip]
This is almost enough to drive a person to join the "I hate
MicroSoft" fan club.
I think that
On 12-05-04 10:33 PM, Joshua Wiley wrote:
On Fri, May 4, 2012 at 7:17 PM, Duncan Murdoch wrote:
On 12-05-04 7:40 PM, Spencer Graves wrote:
[snip]
This is almost enough to drive a person to join the "I hate MicroSoft"
fan club.
I think that would just confirm my membership in the "I
See the post by Frank Harrell at:
http://groups.google.com/group/medstats/browse_thread/thread/cbff7871179e9508?pli=1
or google
regrouping to satisfy proportional odds
On Tue, May 1, 2012 at 2:14 AM, 80past2 wrote:
> Hi everyone, I'm a bit new here (and new to R), and I was trying to do an
> O
On Fri, May 4, 2012 at 7:17 PM, Duncan Murdoch wrote:
> On 12-05-04 7:40 PM, Spencer Graves wrote:
[snip]
>> This is almost enough to drive a person to join the "I hate
>> MicroSoft" fan club.
>
> I think that would just confirm my membership in the "I hate Emacs" club.
I don't see how th
On 12-05-04 7:40 PM, Spencer Graves wrote:
On 5/4/2012 5:41 AM, Duncan Murdoch wrote:
On 12-05-04 12:41 AM, Spencer Graves wrote:
On 5/3/2012 9:28 PM, Joshua Wiley wrote:
How are you using R? Any special front ends that might be causing
this? Can you try it in unsuffered consequences?
On Thu, 3 May 2012 10:50:46 -0800
John Kane wrote:
> Thanks Jeff and Sarah.
>
> I was thinking mainly of using the base path and paste routine which
> is something I do in Windows
>
> It will take me a while to figrue out relative paths.
>
Relative paths are not a mystery nor are they so
On Fri, May 4, 2012 at 3:37 PM, knavero wrote:
> "However, I do wonder why it still complains of the vector length even though
> I nulled out the other columns. It's an interesting error to run into.
> Probably looks at FUN before nulling out the other columns was my theory. "
>
> Referring to jus
On 5/4/2012 5:41 AM, Duncan Murdoch wrote:
On 12-05-04 12:41 AM, Spencer Graves wrote:
On 5/3/2012 9:28 PM, Joshua Wiley wrote:
How are you using R? Any special front ends that might be causing
this? Can you try it in unsuffered consequences?
I'm running R 1.15.0; sessionInfo() ap
structure(list(A2 = structure(c(9L, 4L, 4L, 3L, 5L, 7L, 5L, 7L,
6L, 1L, 1L, 1L, 3L, 4L, 5L, 5L, 2L, 5L, 3L, 4L, 4L, 8L, 4L, 3L,
4L, 5L, 4L, 3L), .Label = c("4.957", "4.958", "4.959", "4.96",
"4.961", "4.962", "4.963", "4.964", "x"), class = "factor"),
A2.1 = structure(c(6L, 2L, 2L, 2L, 2L,
Hello, again.
marc212 wrote
>
> I have a pretty good computer with a lot of RAM and it just wont run this
> code. My R crashes...
>
> Running this on a sample does not work either.
>
> Any advice? Thank you for all the help!
>
> Marc
>
What I'm saying is to use dput on a data sample.
For ins
Hello,
marc212 wrote
>
> I am not opposed to for loops just do not know how to implement them.
>
>
> I am sorry for all the questions I am trying to learn. The posted code
> does not work for me in my set I am getting a :
> Error in `dimnames<-.data.frame`(`*tmp*`, value = list(c("A1", "A1.1"
"However, I do wonder why it still complains of the vector length even though
I nulled out the other columns. It's an interesting error to run into.
Probably looks at FUN before nulling out the other columns was my theory. "
Referring to just a straight up read.zoo in this case ^
--
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I am not opposed to for loops just do not know how to implement them.
I am sorry for all the questions I am trying to learn. The posted code does
not work for me in my set I am getting a :
Error in `dimnames<-.data.frame`(`*tmp*`, value = list(c("A1", "A1.1", :
invalid 'dimnames' given for d
Thanks Richard, that works great on the test data,
I'll try it out on the full dataset now and let you know how it goes.
Thanks a lot!
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Hello,
marc212 wrote
>
> I have the following:
> Time A1 A1 B1 B1 C1 C2
> x y x y x y
> 0 5 6 6 7 7 9
> 1 3 4 4 3 9 9
> 2 5 2 6 4 7 4
>
> I wan
Gee Bert, thanks for the really helpful tip. But if you read my post properly
you'll note that I do know how ANOVA's work.
> The anova of *G* /AB01 /would be some thing like: y=V, fixed=S, Random= L
> &
> L*S...
I didn't want to show a full model formula in case it led people do the
wrong path to
Hey Gabor, just trying to understand this here..sorry for the noob question:
DF1 <- read.table(URL, skip = 1, header = TRUE, sep = ",", fill = TRUE,
as.is = TRUE)
I'm not to familiar with as.is, however I quickly read the R documentation
on that. From my understanding it converts character to
Thank you for the suggestion Gabor. It's definitely more elegant than what I
had above. Instead of going from character representation to POSIXct to
chron, it looks at the character representation and goes straight to chron.
It's good. However, I do wonder why it still complains of the vector lengt
Thanks, I know about it but i wat to find several local maxima, so in other
words I need a way to identify the places in the surface where both slopes
are equal to 0 and the second derivative is negative.
On Fri, May 4, 2012 at 9:28 AM, David Winsemius wrote:
>
> On May 3, 2012, at 6:09 PM, Dieg
On Fri, May 04, 2012 at 11:11:47AM -0700, marc212 wrote:
> I have something like 28 rows and 6000 columns. How would I configure this
> with for loops.
The transformation may be splitted into simpler pieces using a for
loop over the 28 rows. Try the following.
orig <- rbind(
"0"=c(5, 6, 6, 7
On Fri, May 04, 2012 at 07:43:32PM +0200, Kehl Dániel wrote:
> Dear Petr,
>
> thank you for your input.
> I tried to experiment with (probably somewhat biased) truncated means
> like in the following code.
> How I got the 225 as a truncation limit is a good question. :)
>
> REPS1 <- REPS2 <- 100
Here is a function I use to look at the size of objects:
##start
my.ls <- function (pos = 1, sorted = FALSE, envir = as.environment(pos))
{
.result <- sapply(ls(envir = envir, all.names = TRUE),
function(..x) object.size(eval(as.symbol(..x),
envir = envir)))
if
OK, not all, but most lines have the same length. Perhaps you could
write the lines with a different line size to a separate file to have
a closer look at those lines. Modifying the previous code (again not
tested):
con <- file("dataset.txt", "rt")
out <- file("strangelines.txt", "wt")
#
The following constructs the data.frame that I think the original
poster asked for.
I don't understand the graph, so I didn't attempt it.
I agree with Bert that this might not make sense. Specifically, the distinction
between AB01 and AB02 is not modeled, and that is probably the critical factor.
I have something like 28 rows and 6000 columns. How would I configure this
with for loops.
Thank you.
Marc
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_
This is quite some time ago, but I just had the same problem and the folder
in the most recent version of TextWrangler.
Right-click (well, Ctrl-click) on the application and choose 'Show Package
Contents'. Then you can find the folder Language Moudles and include the
plist script for R syntax highl
Hi,
I need fit the France model :
y = A{1 - exp[-b(t-T) - c(sqrt(t) - sqrt(T))]}
parameters: A, b, T, c
variable: t (time)
resp: y
I tried:
time = 1:48
resp = rnorm(48, 200, 10)
dados = data.frame(resp, time)
attach(dados)
f = function(x, A, b, T, c)
A*(1-exp(-b*(x-T) - c*(sqrt(x) - sqrt(T
Without going deeply into your analysis, 2 comments:
1) Use the anova command to test two nested models using:
anova(model1, model2, test="Chisq")
2) glm's are non-trivial models (at least to me), be sure to google for
some tutorials in order to understand what you are looking at...
Cheers,
Tal
Hello Sarah,
thanks for your quick answer. This is exactly what I was looking for,
embarrassingly simple if I might add. Sometimes R is like a huge
workshop with unlabeled tool magazines: You know the tool exists, but
not where it is. And if you find it, the instructions are often quite
terse.
Th
Dear Petr,
thank you for your input.
I tried to experiment with (probably somewhat biased) truncated means
like in the following code.
How I got the 225 as a truncation limit is a good question. :)
REPS1 <- REPS2 <- 1000
N1 <- 10
N2 <- 3
N <- N1+N2
x1 <- rep(0,N1)
x2 <- rnorm(N2,300,10
Dear R users
Recently I received advice from this fine group on gee() and sample weights
One suggestion was to use geeglm()
I hope someone can help me to solve a problem that arises when converting a
code from gee to geeglm.
*Here is a code that I wrote with the original data, not weighted:
Rob:
On Fri, May 4, 2012 at 9:18 AM, robgriffin247 wrote:
> Hi,
> I need to create a data frame containing the results of a number of ANOVA's
> but I'm having some trouble setting it up (some being enough for me to spend
> 3 days trying with no progress and be left staring in to the abyss which
>
What about ?axis as a place to start?
Are you sure that the email address that your message appears to be
coming from is identical to the one you used when you signed up?
That's a frequent cause of moderation.
Sarah
On Fri, May 4, 2012 at 1:09 PM, Robert Latest wrote:
> Hello,
>
> is it possibl
Hello,
is it possible to replace the text of tick marks in a plot?
Specifically, I'd like to have a ppnorm plot in which the theoretical
quantiles are not expressed in terms of standard deviations, but in
actual percentages. Anybody who's seen a probability plot in MINITAB
knows what I'm talking a
Hi,
You did a really good job providing a reproducible example, except
that you didn't mention which package sem() comes from. (sem, I'm
assuming).
I don't know how you came up with your covariance matrix, but it
*isn't* symmetric:
> isSymmetric(S.Seed.BB)
[1] FALSE
> S.Seed.BB[6, 2]
[1] 37.758
WARNING: COMPLETELY OFF TOPIC -- Nothing to do with R.
I thought readers of this list might enjoy the following. The link to
the full article is at the bottom. I hope this is not "too"
inappropriate.
---
Overconfidence in crime statistics doesn’t pay. In a new study, a team
of criminologists
As it says, you need to supply a file name to be outputted to.
write.xls(, file = "abc.xls")
Michael
On Fri, May 4, 2012 at 11:31 AM, PaulJr wrote:
> Hello R users,
>
> I want to export to an xls or .csv some predictions I produced with the
> auto.arima and forecast functions.
>
> A detail
Hi Folks,
I'm running 32-bit R 2.14 in RStudio on my Win 7 x64 system with 8GB
RAM. I'm getting memory problems as R wants to swallow more than the
4GB limit.
I think I'm stuck at 4GB as I have to use 32-bit R for a number of
packages (ODBC, etc). However, I doubt I really need to be using that
Em 04-05-2012 11:00, jeff6868 escreveu:
Date: Thu, 3 May 2012 06:45:59 -0700 (PDT)
From: jeff6868
To:r-help@r-project.org
Subject: [R] add an automatized linear regression in a function
Message-ID:<1336052759474-4606047.p...@n4.nabble.com>
Content-Type: text/plain; charset=us-ascii
Dear R users
Hi,
I need to create a data frame containing the results of a number of ANOVA's
but I'm having some trouble setting it up (some being enough for me to spend
3 days trying with no progress and be left staring in to the abyss which
some people call a weekend, and what I will call 2 quiet days in the
Hello,
diegogiri wrote
>
> Hi, i'm a new r user and I will take you a simple (not for me)question.
> I have a r script file like this:
>
> text <- c(" ")
>
> I have a csv file like:
> NAME
> alfa
> beta
> gamma
>
> how can I replace with a loop statment metadata "" o
Hi,
I have a data set with 999 observations, for each of them I have data on
four variables:
site, colony, gender (quite a few NA values), and cohort.
This is how the data set looks like:
> str(dispersal)
'data.frame': 999 obs. of 4 variables:
$ site : Factor w/ 2 levels "1","2": 1 1 1 1 1 1
Hello, I tried to do a 'sem' analysis for data of how blueberry consumption
by birds is influenced by a pollution gradient, using distance and
vegetation structural and composition variables, but I got the following
error message:
Error in sem.default(ram = ram, S = S, N = N, param.names = pars, v
Hi David,
I've tried using sep="\t" but it doesn't work, unfortunately.
Thanks for your help.
-
Isaac
Research Assistant
Quantitative Finance Faculty, UTS
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Hi, i'm a new r user and I will take you a simple (not for me)question.
I have a r script file like this:
text <- c(" ")
I have a csv file like:
NAME
alfa
beta
gamma
how can I replace with a loop statment metadata "" of script file with
the values names in the csv file?
Hello R users,
I want to export to an xls or .csv some predictions I produced with the
auto.arima and forecast functions.
A detail of all my work is presented below. I loaded a package called
dataframes2xls and tried to use the function write.xls without any success.
Can anybody help me figure
You should to download this package and install it:
http://cran.r-project.org/bin/windows/base/R-2.15.0patched-win.exe
You can find it in R website > Download R for Windows > base > Other builds
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Hello,
lunarossa wrote
>
> I have two datasets, the first has this shape (each word is a column)
> Name address phone .. ..
>
> The second one has the following shape
> Name request
>
> I need a contingency table with for example phone and request.
>
>
> The people registered in these datase
Here is a working snippet.
library(epitools)
mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12
,8,13,20,11,58) ,ncol=2)
colnames(mat) <- c("treatmentA","treatmentB")
row.names(mat) <- paste("Cond",rep(1:10,1))
dimnames(mat) <- list("Condition" = row.names(mat), "instrument" =
colna
I have the same problem installing R on a new machine, my codes are nor
working anymore (same error message as you had).
Unfortunately, I do not understand what you mean by R patched and especially
"updating R-patched" and google cannot help me this time... May you explain
what I have to do?
thank
Here is a working snippet.
library(epitools)
mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12
,8,13,20,11,58) ,ncol=2)
colnames(mat) <- c("treatmentA","treatmentB")
row.names(mat) <- paste("Cond",rep(1:10,1))
dimnames(mat) <- list("Condition" = row.names(mat), "instrument" =
colna
Hi,
I tried to use naivebayes in package 'e1071'.
when I use following parameter, only one predictor, there is an error.
m<- naiveBayes(iris[,1], iris[,5])
table(predict(m, iris[,1]), iris[,5])
Error in log(sapply(attribs, function(v) { :
Non-numeric argument to mathematical function
How
I have two datasets, the first has this shape (each word is a column)
Name address phone .. ..
The second one has the following shape
Name request
I need a contingency table with for example phone and request.
The people registered in these datasets are present in both datasets, BUT in
the fi
Dear community,
I'd like to fix a mixed model. I have unbalance data, what should i use:
lme in nlme package , or lmer in lme4.
Thanks, u...@host.com as u...@host.com
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Dear R-help,
I'm trying to apply machine learning methods, such as Random Forest,
Boosted Trees or Multivariate Adaptive Regression Splines for supervised
classification issues.
In a epidemiological study, i'm dealing with high dimensional
cluster-correlated data, each cluster corresponding
When I try to adjust a mixed model with random effects I can make this order
without problem
> lm.FA<-lme(absFA~trait*condition,random=~1|individual)
But if I try to fit a model in which the response (absFA) is not the same in
all individuals at different levels of "trait" factor , but varies ran
one solution is to set NAs to 0, e.g.,
m <- matrix(1:3, 3, 3)
x <- list(m, m+3, m+6)
x[[1]][1] <- NA
x. <- lapply(x, function (x) {x[is.na(x)] <- 0; x} )
Reduce("+", x.)
I hope it helps.
Best,
Dimitris
On 5/4/2012 11:19 AM, Evgenia wrote:
I have a list ( in my real problem a double list y
Jan, thank you.
> table(line_sizes)
line_sizes
01 97 256
1430 2860 46869069 1430
-
Isaac
Research Assistant
Quantitative Finance Faculty, UTS
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Here is a working snippet.
library(epitools)
mat <- matrix(c(10,15,60,25,98, 12,10,70,28,14, 9,11,68,10,12
,8,13,20,11,58) ,ncol=2)
colnames(mat) <- c("treatmentA","treatmentB")
row.names(mat) <- paste("Cond",rep(1:10,1))
dimnames(mat) <- list("Condition" = row.names(mat), "instrument" =
colna
Em 04-05-2012 11:00, jeff6868 escreveu:
Date: Thu, 3 May 2012 06:45:59 -0700 (PDT)
From: jeff6868
To:r-help@r-project.org
Subject: [R] add an automatized linear regression in a function
Message-ID:<1336052759474-4606047.p...@n4.nabble.com>
Content-Type: text/plain; charset=us-ascii
Dear R users
...
as in:
outer(paste("A",A,sep=""),paste("B".B,sep=""), FUN =paste,sep="_")
-- Bert
On Fri, May 4, 2012 at 7:59 AM, Bert Gunter wrote:
> ?outer
>
> Bert
>
> Sent from my iPhone -- please excuse typos.
>
> On May 4, 2012, at 6:25 AM, "R. Michael Weylandt"
> wrote:
>
>> do.call(function(x,y)
You cannot avoid the hyphen I think, but if you say (for example):
"prefix.string=foo/x"
then your files start with 'x-' (so 'graph' becomes 'foo/x-graph') which may be
better for you than the hyphen at the beginning of the file name.
Rgds,
Rainer
On Friday 04 May 2012 17:23:58 julia.jacob...@
I guess that is hard-coded in Sweave, so you probably cannot control
it unless you (partially) rewrite the driver. And just FYI, you can
try the knitr package, which does not add the hyphen for you, but the
name of the option 'prefix.string' has been changed to 'fig.path'
(http://yihui.name/knitr/o
dear Szymon,
it is a bug (in the new version), thanks. It depends on the flat
underlying relationship you are trying to estimate with a small sample..
I will correct it as soon as possible. Meanwhile you can use
o1<-glm(gpp ~ temp)
os1<-segmented(o1, seg.Z=~temp, psi=15, control=seg.control(n.
On May 4, 2012, at 10:52 AM, Saurav Pathak wrote:
On 05/04/2012 10:39 AM, David Winsemius wrote:
On May 3, 2012, at 7:10 PM, Saurav Pathak wrote:
Hi,
For some reason I have been unable to use the predict function
when I desire the standard error to be calculated too. For
example, when
I agree with you. I used this "trick" to take the desired results but I
posted wondering If there was any other solution.
Thanks
Evgenia
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On May 4, 2012, at 9:43 AM, wwreith wrote:
Am I correct in assuming that the output below essentially
translates to
"Males have a mean time that is significantly lower than Females"?
Is this
the correct way to interpret the fact that the coefficient is
negative?
I wouldn't be using exac
Dear Sweave users,
Could you help me to find a way to place Sweave output files in a subdirectory
of the currentfolder without giving them a subname?
If the option "prefix.string=foo/" is used, all files are placed in this
folder, but begin with an hyphen-minus, which makes it difficult to work
You didn't mention it, but did you use something like
options(digits=20)
before displaying that data? In any case,
> 1.4000244 == 1.4
[1] TRUE
because R uses the IEEE-754 double precision floating point
arithmetic that all modern computers support. That gives
you 5
?outer
Bert
Sent from my iPhone -- please excuse typos.
On May 4, 2012, at 6:25 AM, "R. Michael Weylandt"
wrote:
> do.call(function(x,y) paste0("A",x,"_","B",y),expand.grid(x = A,y = B))
>
> seems to be a place to start. Robust generalization seems a hair
> tricky -- I'll mull on it.
>
> Mi
On May 4, 2012, at 2:38 AM, Alok Jadhav wrote:
> Hi,
>
> I am trying to query a Sybase database on my new windows 7 machine. I am
> using native sybase driver "Adaptive server Enterprise" following is example
> code
>
> conn <- sprintf("driver=Adaptive server
> Enterprise;server=PHKSESMD01;d
On 05/04/2012 10:39 AM, David Winsemius wrote:
On May 3, 2012, at 7:10 PM, Saurav Pathak wrote:
Hi,
For some reason I have been unable to use the predict function when I
desire the standard error to be calculated too. For example, when I
try the following:
l<- loess(d~x+y, span=span, se=
On May 4, 2012, at 1:34 AM, iliketurtles wrote:
Dear Experienced R Practitioners,
I have 4GB .txt data called "dataset.txt" and have attempted to use
*ff,
bigmemory, filehash and sqldf *packages to import it, but have had no
success. The readLines output of this data is:
Ther alignment o
On May 3, 2012, at 7:10 PM, Saurav Pathak wrote:
Hi,
For some reason I have been unable to use the predict function when
I desire the standard error to be calculated too. For example, when
I try the following:
l<- loess(d~x+y, span=span, se=TRUE)
p<- predict(l, se=TRUE)
I don't know
On May 3, 2012, at 6:09 PM, Diego Rojas wrote:
If a run a LOESS model and then produce a smoothed surface: Is there
any
way to determine the coordinates of the local maxima on the surface?
?predict# it has a loess method.
[[alternative HTML version deleted]]
--
David Winsem
Hi Istvan,
Your OS and version of R (eg sessionInfo() ) would also be useful, as
would sending your reply to the R-help list and not just to me.
Sarah
-- Forwarded message --
From: Istvan Nemeth
Date: Fri, May 4, 2012 at 10:20 AM
Subject: Re: [R] read-in, error???
To: Sarah Gos
Em 4/5/2012 06:39, Bjørn-Helge Mevik escreveu:
rakeshnb writes:
[snipped]
When in doubt, read the documentation. :)
An obvious candidate for a fortune entry!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLE
Hi Istvan,
That's most unusual, and quite unlikely (and much larger than the
usual floating-point rounding errors).
Please provide a reproducible example. I assume you got the data from here:
http://www.itl.nist.gov/div898/strd/anova/SmLs07.dat
What did you do with it then? How did you delete th
Dear Users!
I encountered with some problem in data reading while I challenged R (and
me too) in a validation point of view.
In this issue, I tried to utilize some reference datasets (
http://www.itl.nist.gov/div898/strd/index.html).
And the result departed a bit from my expectations. This dataset
Solution: have package mgcv loaded when you predict...not just for the fit.
:) Silly mistake...
Thanks Simon!
Ben
On Thu, May 3, 2012 at 3:56 PM, Ben quant wrote:
> Hello,
>
> I don't understand what went wrong or how to fix this. How do I set
> qr=TRUE for gam?
>
> When I produce a fit using
Hi -
So when I run the following, I get a strange formatting output with MikTeX, and
I am unsure if the behavior is due to R, Hmisc, or MikTeX or both:
dfr <- data.frame(x=rnorm(400),y=sample(c('male','female'),400,TRUE))
latex(describe(dfr))
What happens is that the x column is summarized in
Am I correct in assuming that the output below essentially translates to
"Males have a mean time that is significantly lower than Females"? Is this
the correct way to interpret the fact that the coefficient is negative?
Assume the variale sex is treated as a factor with Female =0 and Male=1.
survm
do.call(function(x,y) paste0("A",x,"_","B",y),expand.grid(x = A,y = B))
seems to be a place to start. Robust generalization seems a hair
tricky -- I'll mull on it.
Michael
On Fri, May 4, 2012 at 9:10 AM, Johannes Radinger wrote:
> Hi,
>
> it is easiest to explain what I want to do by an example
Hi,
it is easiest to explain what I want to do by an example:
lets assume there are two factors/variables:
A <- c(1,2,3)
B <- c(1,3,3)
Now I would like to generate a list of strings that should look like
("A1_B1","A1_B2","A2_B1","A2_B2"). So actually the string
contains all possible combinations
simply change the specification in the subset command to >0
John Kane
Kingston ON Canada
-Original Message-
From: shankarla...@gmail.com
Sent: Thu, 3 May 2012 16:51:50 -0400
To: jrkrid...@inbox.com
Subject: Re: [R] Identifying the particular X or Y in a sorted l
Which version of gam are you using (i.e. which package and version number?)
prediction with fitted gam objects should call predict.gam, and I'm not
quite sure why this is not happening here (you do have the mgcv or gam
loaded while trying to predict, I suppose?).
On 03/05/12 22:56, Ben quant
On 12-05-04 12:41 AM, Spencer Graves wrote:
On 5/3/2012 9:28 PM, Joshua Wiley wrote:
How are you using R? Any special front ends that might be causing
this? Can you try it in unsuffered consequences?
I'm running R 1.15.0; sessionInfo() appears below. I get this
from Rgui i386 and
Hi,
I have two variables ranging both from 0 to 1 (n=500 each).
Now I am interested in plotting them both in one plot (using ggplot2).
So far I used ecdf() (from an example I found with google) to get
values for the cumulatice distribution function which gives a relative
curve. I also want to do
Thanks Petr, I'll take a look at that as well.
Cheers,
Gavin.
-Original Message-
From: Petr PIKAL [mailto:petr.pi...@precheza.cz]
Sent: 04 May 2012 13:09
To: Gavin Blackburn
Cc: r-help@r-project.org
Subject: Re: [R] colours in a pdf
Hi
One option for substantial distinguishable range
Ok great, thanks for the help. I don't mind sectioning data off into smaller
groups in this early stage, but the more I can get in one group the better.
Cheers,
Gavin.
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: 04 May 2012 13:04
To: Gavin Bla
Hi
One option for substantial distinguishable range of colours is jet.colors
from matlab package.
Regards
Petr
> Hi,
>
> Thanks for the help. Extending the palette to 16 or 20 would be a big
> help. The largest number of files I've had to handle in a single group
is
> 42 and I wouldn't ex
I'm not sure you'll be able to come up with 42 categorial colors:
perhaps facetting / small-multiples here? Colorspace (on CRAN) will
let you make palettes manually, so perhaps that's worth looking at.
Michael
On Fri, May 4, 2012 at 7:49 AM, Gavin Blackburn
wrote:
> Hi,
>
> Thanks for the help.
Hi,
Thanks for the help. Extending the palette to 16 or 20 would be a big help. The
largest number of files I've had to handle in a single group is 42 and I
wouldn't expect it to get much bigger than that.
I'll take a look at RColorBrewer.
Cheers,
Gavin.
-Original Message-
From: R. M
How many colors are you looking for? There are limits to how many the
eye can make out, but perhaps the RColorBrewer package would be a
place to start. Also check out: http://colorbrewer2.org/
To see all the builtin colors, you can simply use the colors()
function, but your viewer won't be able to
Hi,
I'm plotting PDFs and have a problem. If I have more than 8 sources of data the
colours are repeated. These plots are used to remove poor data from the sets so
it would be helpful if I could expand the colour range. Is there any way to do
this?
The plots are coloured by defining a vector c
Inelegant, but this is one way:
Reduce(function(e1, e2){e1[is.na(e1)] <- 0; e2[is.na(e2)] <- 0; (e1 + e2)}, x)
I.e., set the NAs to 0 before adding in the reduce function.
Michael
On Fri, May 4, 2012 at 5:19 AM, Evgenia wrote:
> I have a list ( in my real problem a double list y[[1:24]][[1:15
quantile(myvec, 0.9)
See ?quantile for the 9 different empirical quantile methods provided.
Michael
On Thu, May 3, 2012 at 2:20 PM, Drew Duckett wrote:
> If i am not mistaking the quantile function and percentile function and the
> same other than the way they are expressed in R. I think where
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