Dear all,
I am using gnome-terminal to run R, and I noticed that standard
PageUp/Down do not work but they do work for other programs within the
same terminal window (e.g. irssi). Scroll bar does not work either.
I run R using GNU-screen, but it doesn't seem to make any difference.
Many thanks
O
How about this ("df" is your input data.frame)
data.frame(ID=df[,1], apply(df[,2:4], 2, function(x) c("00", "AA",
"GG", "CC", "TT")[match(x, c("0/0", "1/1", "2/2", "3/3", "4/4"))]))
Michael
On 5 August 2010 10:55, karena wrote:
>
> Hi, I have a question about the data handling. I have a dataset
Ralf B wrote:
> Besides beauty, is there an actual advantage in terms of run-time
> and/or memory use?
If you look at the actual definition of tapply, I'm sure you realize
that the answer is no:
ans <- lapply(split(X, group), FUN, ...)
inbetween 40-odd lines of "red tape" is something of a g
Tena koe Karena
See ?sub and ?gsub
HTH ...
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of karena
> Sent: Thursday, 5 August 2010 12:56 p.m.
> To: r-help@r-project.org
> Subject: [R] questions about string han
Hi, I've got a quite tricky question.
I have a txt file, named 'temp.txt', as the following:
snp1snp2snp3
AA 00 00
GG GG 00
00 AA 00
I want to read the file into R.
1) when I use 'read.table' without 'header=T' option,
>
I don't know if it has anything to do with these installation warnings:
* installing to library ‘/home/mandova/R/i486-pc-linux-gnu-library/2.11’
* installing *source* package ‘affy’ ...
creating cache ./config.cache
checking how to run the C preprocessor... cc -E
checking for main in -lz... yes
c
Hi, I have a question about the data handling. I have a dataset as following:
ID snp1snp2 snp3
1001 0/0 1/11/1
1002 2/2 3/31/1
1003 4/4 3/32/2
I want to convert the dataset to the following for
Great! Thank you
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Hi all,
I am using RCurl to try and download data from a website, but I'm having
trouble finding out what URL to use. Here is the site:
http://www.invescopowershares.com/products/holdings.aspx?ticker=PGX
See how in the upper right, above the displayed sheet, there's a link to
download the data
read.table("temp.txt",header=T,colClasses = "character")
-
A R learner.
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Hi,
I have two sets of data, an observed data and generated data.
The generated data is obtained from the model where the parameters is estimated
from the observed data.
So I'm not sure which to use either
one-sample test
ks.test(x+2, "pgamma", 3, 2) # two-sided, exact
or
two-sample test
ks.
Fun question.
Here is one solution that has come to mind:
# graphics output test
a <- c(1,3,2,1,4)
b <- c(2,1,1,1,2)
c <- c(4,7,2,4,5)
d <- rnorm(500)
e <- rnorm(600)
op <- par(mfrow=c(2,2))
pie(a)
pie(b)
pie(c)
txt <- capture.output(ks.test(d,e))
txt <- txt[txt != ""]
txt <- paste(txt, collapse
It looks like the test is indicating a far bigger difference than
could be explained by random variation.
Since the sample sizes are equal, have you considered plotting the
ordered values of one against the ordered values of the other
(essentially an empirical QQplot), with a 45 degree line drawn
On Wed, Aug 4, 2010 at 8:33 PM, Steven Kang wrote:
> Hi all,
>
> I am trying to convert all the dates (all days that are not Friday) in data
> frame into dates to next Friday.
>
> The following works but the result is returned as vector rather than the
> original class.
>
> It would be greatly app
Hi all,
I am trying to convert all the dates (all days that are not Friday) in data
frame into dates to next Friday.
The following works but the result is returned as vector rather than the
original class.
It would be greatly apprecited if you could provide any solution to this
problem.
Many th
You have almost achieved your goal.
plot(1:100, axes=F)
axis(1,at=seq(0,100,10)[c(1,3,5,7,11)])
-
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- Original Message
> From: Megh Dal
> To: r-h...@stat.math.ethz.ch
> Sent: Wed, August 4, 2010 4:04:06 AM
> Subject: [R] Spliting a text
>
> Hi, I want to split a text to seperate numerical and non-numerical portions
> of
>that. For example suppose I have a text "abc 3456" and I w
On Aug 4, 2010, at 5:49 PM, Ralf B wrote:
Hi R Users,
I have two vectors, x and y, of equal length representing two types of
data from two studies. I would like to test if they are similar enough
to use them interchangeably. No assumptions about distributions can be
made (initial tests clearly
Jennifer,
Please provide us more information. It would be helpful to see the
actual code you ran along with the actual output. How did you get the
values that your included with your email message? Did you get the
values using the summary function, e.g.
> fit0<-lm(y~x+sex+x*sex)
> summary(fit0)
C
On Wed, Aug 4, 2010 at 2:09 PM, Erik Iverson wrote:
> Hello,
>
>
>> I have a problem which has bitten me occasionally. I often need to
>> prepare graphs for many variables in a data set, but seldom for all.
>> or for any large number of sequential or sequentially named variables.
>> Often I need s
Besides beauty, is there an actual advantage in terms of run-time
and/or memory use?
Ralf
On Wed, Aug 4, 2010 at 3:44 PM, Bert Gunter wrote:
> It's not that it's "bad" -- it's just unnecessarily clumsy. ALmost
> always, tapply/by will do the same thing more simply.
>
> -- Bert
>
> On Wed, Aug 4,
Hello,
I had a question about how to label a axis of a plot.
for example, my plot is
>plot(1:100, axes=F)
>box()
>axis(1)
then, I want my y-axis has six ticks ( at=seq(0,100,10)) , but I don't want
to label all the 11 ticks, I only want to label the 1st, 3rd, 5th, 7th and
11th ticks. But the co
Hi R Users,
I need to produce a simple report consisting of some graphs and a
statistic. Here simplification of it:
# graphics output test
a <- c(1,3,2,1,4)
b <- c(2,1,1,1,2)
c <- c(4,7,2,4,5)
d <- rnorm(500)
e <- rnorm(600)
op <- par(mfrow=c(3,2))
pie(a)
pie(b)
pie(c)
text(ks.test(d,e))
obvious
In addition to the reply from Duncan Murdoch,
help("text")
and
demo("Hershey")
are very informative in this regard.
Simple example with plot for added text:
plot(1,1)
text(0.8, 0.8, "B\\`ad Fr\\'ench", vfont=c("serif", "plain"))
title(main = "B\u{E0}d Fr\u{E9}nch")
>From the help page for
Hi R Users,
I have two vectors, x and y, of equal length representing two types of
data from two studies. I would like to test if they are similar enough
to use them interchangeably. No assumptions about distributions can be
made (initial tests clearly show that they are not normal).
Here some res
What OS are you specifically referring to?
Even the most basic web search surely would have turned up the R
Installation and Administration Manual
http://cran.r-project.org/doc/manuals/R-admin.html#Installing-R-under-Unix_002dalikes
There may be pre-built packages available for your particula
1) When running ks.test, I am getting the following error after the
test presents its result::
'ks.test(x, y) : cannot compute correct p-values with ties'
I wonder what means and what causes it.
2) Also, how do I calculate an effect size from this statistic?
R.
Hello,
how can i install R on a Unix server?
best regards,
Nero
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__
R-help@r-project.org mailin
Thanks to Joshua and Erik for very helpful suggestions. This clarifies
what I had found to be a tricky area of the documentation!
Best wishes,
Anthony Staines
--
Anthony Staines, Professor of Health Systems Research,
School of Nursing, Dublin City University, Dublin 9, Ireland.
Tel:- +353 1 700
Dear Anthony
On Wed, 4 Aug 2010 14:56:58 +0100
Anthony Staines wrote:
> What I would like to do is something like "write a function which
> takes the *name* of a variable, presumably a s a character string,
> from a dataframe, as one argument, and the dataframe, as a second
> argument".
>
I am
Jennifer Hou wrote:
Hello,
I have got a linear model that looks like this:
lm(criterion ~ variable.A*variable.a + variable.B*variable.b + variable.C
*variable.c)
The output computed with stdCoeff() seems to be all right, but it does not show the
coefficients of the interaction of the first
Hello,
I have got a linear model that looks like this:
lm(criterion ~ variable.A*variable.a + variable.B*variable.b + variable.C
*variable.c)
The output computed with stdCoeff() seems to be all right, but it does not show
the coefficients of the interaction of the first pair of variables. Inst
Karen Moore tcd.ie> writes:
>
> I'm dealing with count data that's nested and has spatial dependence.
> I ran a glmm in lmer with a random factor for nestedness. Spatial dependence
> seems to have been accommodated by model. However I can't add a variance
> strcuture to this model (to accommodat
You can't find a way to get the 3 column result by setting the aggregate
function.
data.frame(res[1], res[[2]],check.names = F)
-
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On 04/08/2010 3:40 PM, Jennifer Young wrote:
Hello
Could someone please direct me to the correct commands for adding accents
(grave and aigu) to a letter in a plot title, label, or in added text? I'm
sure there's a handy list somewhere, but I've failed in coming up with the
correct search words
On Wed, Aug 4, 2010 at 3:40 PM, Dimitri Liakhovitski
wrote:
> Thanks a lot, David.
> It works perfectly. Of course, lapply is also a loop!
>
> So, your method is:
> z<-data.frame(nam1=c("bbb..aba","ccc..abb","ddd..abc","eee..abd"),stringsAsFactors=FALSE)
> z$nam2<-unlist(lapply( strsplit(z[[1]],sp
On Aug 4, 2010, at 3:40 PM, Dimitri Liakhovitski wrote:
Thanks a lot, David.
It works perfectly. Of course, lapply is also a loop!
So, your method is:
z<-
data
.frame
(nam1
=
c("bbb..aba","ccc..abb","ddd..abc","eee..abd"),stringsAsFactors=FALSE)
z$nam2<-unlist(lapply( strsplit(z[[1]],spl
On 04/08/2010 2:40 PM, Michael Lachmann wrote:
I just tried it under OSX and linux, and both get stuck. Maybe R on windows
forks the R help server?
It doesn't fork, but it might run in another thread.
But what you suggested works perfectly. Thanks!
Do you by chance know if there was somet
Thanks a lot, David.
It works perfectly. Of course, lapply is also a loop!
So, your method is:
z<-data.frame(nam1=c("bbb..aba","ccc..abb","ddd..abc","eee..abd"),stringsAsFactors=FALSE)
z$nam2<-unlist(lapply( strsplit(z[[1]],split="\\.."), "[", 1))
z$nam3<-unlist(lapply( strsplit(z[[1]],split="\\..
On Wed, Aug 4, 2010 at 12:03 PM, Dimitri Liakhovitski
wrote:
> I am sorry, someone said that strsplit automatically works on a
> column. How exactly does it work?
I was not suggesting it was the ideal choice in this case, merely that
a loop was not required. I tend to use strsplit() more for doi
Hello
Could someone please direct me to the correct commands for adding accents
(grave and aigu) to a letter in a plot title, label, or in added text? I'm
sure there's a handy list somewhere, but I've failed in coming up with the
correct search words to find it.
Thank you muchly!
Jen
___
On Aug 4, 2010, at 3:31 PM, David Winsemius wrote:
On Aug 4, 2010, at 3:03 PM, Dimitri Liakhovitski wrote:
I am sorry, someone said that strsplit automatically works on a
column. How exactly does it work?
For example, if I want to grab just the first (or the second) part of
the string in nam
On Aug 4, 2010, at 3:03 PM, Dimitri Liakhovitski wrote:
I am sorry, someone said that strsplit automatically works on a
column. How exactly does it work?
For example, if I want to grab just the first (or the second) part of
the string in nam1 that should be split based on ".."
x<-data.frame(nam
Hi Anthony,
I don't know if this will help you. A similar method should work for
any function that accepts a formula, which is your main issue I think.
Outside of formulae, you should just be able to pass the arguments
directly (as with the data argument of xyplot).
The only other thought that
In such cases (which I don't think must happen), consider using
?lapply
and also
?unlist
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebre
Hello,
I have a problem which has bitten me occasionally. I often need to
prepare graphs for many variables in a data set, but seldom for all.
or for any large number of sequential or sequentially named variables.
Often I need several graphs for different subsets of the dataset
for a given vari
Hi R-users,
I'm using R 2.11.1, mgcv 1.6-2 to fit a generalized additive mixed model.
I'm new to this package...and just got more and more problems...
1. Can I include correlation and/or random effect into gam( ) also? or only
gamm( ) could be used?
2. I want to estimate the smoothing function s
I am sorry, someone said that strsplit automatically works on a
column. How exactly does it work?
For example, if I want to grab just the first (or the second) part of
the string in nam1 that should be split based on ".."
x<-data.frame(nam1=c("bbb..aba","ccc..abb","ddd..abc","eee..abd"),
stringsAsF
Hi R-users,
Since R.2.11 aggregate can now deal with non-scalar functions, which is very
useful to me.
However, I have a question about how best to process the output.
test <- data.frame(a = rep(c("g1", "g2"), each = 50), b = runif(100))
res <- aggregate(test$b, list(group = test$a), function(
Hello
I do wavelet transform by using this code:
dec=dwt(ld, filter='d8', n.levels=lev, boundary="reflaction");
dec consists of the decomposition coefficients and other
How can I change the coefficients the decomposition manualy?
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Hello
dec=dwt(ld, filter='d8', n.levels=lev, boundary="reflaction");
dwt is an function which returns W, V, filter
dec is an object which consist of W, V, filter and others
How to change values (of W, V, filter...) when i have dec
thank you!
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Hi!I'm doing a little data importing from .cel files,
> setwd("/home/mandova/celfiles")
> mydata<-ReadAffy()
Error in sub("^/?([^/]*/)*", "", filenames, extended = TRUE) :
unused argument(s) (extended = TRUE)
Then I tried
> filenames<-paste("GSM",c(seq(138597,138617,1)),".cel",sep="")
> f
Dear colleagues,
I have a problem which has bitten me occasionally. I often need to
prepare graphs for many variables in a data set, but seldom for all.
or for any large number of sequential or sequentially named variables.
Often I need several graphs for different subsets of the dataset
for a giv
Hi,
I am sorry for that I canât determine which R-sig list to post
questions about package ff.
Now I have made a little progress with this:
read.dbres.ffdf <- function(
res){
data1 <- fetch(res, 0)
if (nrow(data1) == 0){
Hi R-users,
Since R.2.11 aggregate can now deal with non-scalar functions, which is
very useful to me.
However, I have a question about how best to process the output.
test <- data.frame(a = rep(c("g1", "g2"), each = 50), b = runif(100))
res <- aggregate(test$b, list(group = test$a), function(
I just tried it under OSX and linux, and both get stuck. Maybe R on windows
forks the R help server?
But what you suggested works perfectly. Thanks!
Do you by chance know if there was something equivalent in R 2.10.x and/or
2.9.x
Duncan Murdoch-2 wrote:
>
> On 04/08/2010 12:44 PM, Michael Lac
Thank you very much, everyone!
Dimitri
On Wed, Aug 4, 2010 at 2:10 PM, David Winsemius wrote:
>
> On Aug 4, 2010, at 1:42 PM, Dimitri Liakhovitski wrote:
>
>> I am sorry, I'd like to split my column ("names") such that all the
>> beginning of a string ("X..") is gone and only the rest of the text
On Aug 4, 2010, at 1:42 PM, Dimitri Liakhovitski wrote:
I am sorry, I'd like to split my column ("names") such that all the
beginning of a string ("X..") is gone and only the rest of the text is
left.
I could not tell whether it was the string "X.." or the pattern "X.."
that was your goal f
On 04/08/2010 12:44 PM, Michael Lachmann wrote:
I'm trying to get help to print the help pages in html format to the
terminal. This is in order to be able to see the html help files remotely.
If I do
printURL = function(file) {a=readLines(url(file));cat(a,sep="\n")}
options(browser=printURL)
opt
Hi,
You already have great solutions. I just wanted to point out that
A) strsplit() works on the entire column automatically so you would
not need a loop
B) with the argument stringsAsFactors = FALSE, your character data
will not be converted to factor, so you would not need to convert it
back.
On Wed, Aug 4, 2010 at 1:35 PM, Gabor Grothendieck
wrote:
> On Wed, Aug 4, 2010 at 1:28 PM, Greg Snow wrote:
>> Do ?'for' to learn about for loops.
>
> and put that in quotes since its a reserved word:
>
> ?"for"
>
Sorry, I missed the fact that Greg actually did put single quotes
around for. On
Try this:
gsub("X\\.\\.", "", x$names)
On Wed, Aug 4, 2010 at 2:42 PM, Dimitri Liakhovitski <
dimitri.liakhovit...@gmail.com> wrote:
> I am sorry, I'd like to split my column ("names") such that all the
> beginning of a string ("X..") is gone and only the rest of the text is
> left.
>
> x<-data
Have a look at the
stringr
package
It simplifies such things...
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com
Dimitri Liakhovitski wrote:
I am sorry, I'd like to split my column ("names") such that all the
beginning of a string ("X..") is gone and only the rest of the text is
left.
x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd"))
x$names<-as.character(x$names)
(x)
str(x)
Can't figure out h
I am sorry, I'd like to split my column ("names") such that all the
beginning of a string ("X..") is gone and only the rest of the text is
left.
x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd"))
x$names<-as.character(x$names)
(x)
str(x)
Can't figure out how to apply strsplit in this sit
On Wed, Aug 4, 2010 at 10:17 AM, Joshua Wiley wrote:
> Hello,
>
> This will put the results of the Fisher test in a list, with each
> element of the list being the results for mouse type a, b, c, and d.
>
> mice <- rep(letters[1:4],10)
> outcome <- sample(c(0,1),length(mice),replace=T)
> group <-
On Wed, Aug 4, 2010 at 1:28 PM, Greg Snow wrote:
> Do ?'for' to learn about for loops.
and put that in quotes since its a reserved word:
?"for"
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the postin
Do ?'for' to learn about for loops. Look at FAQ 7.21 for the other part of the
question (the most important part of the answer is the end where it tells you
to use lists instead like Patrick suggests).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@i
Hi,
I am running a Cox Mixed Effects Hazard model using the library coxme. I
am trying to model time to onset (age_sym1) of thought problems (e.g.
hearing voices) (sym1). As I have siblings in my dataset, I have
decided to account for this by including a random effect for family
(famid). My
Hello,
This will put the results of the Fisher test in a list, with each
element of the list being the results for mouse type a, b, c, and d.
mice <- rep(letters[1:4],10)
outcome <- sample(c(0,1),length(mice),replace=T)
group <- c(rep("A",length(mice)/2),rep("B",length(mice)/2))
my.data <- data.
> In general, the lapply(split(...)) construction should never be used.
Why? What makes it so bad to use?
__
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PLEASE do read the posting guide http://www.R-project.org/posting
?scale
--
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Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of leepama
> Sent: Tuesday, August 03, 2010 10:55 PM
> To
I'm trying to get help to print the help pages in html format to the
terminal. This is in order to be able to see the html help files remotely.
If I do
printURL = function(file) {a=readLines(url(file));cat(a,sep="\n")}
options(browser=printURL)
options(help_type="html")
then invoking help with
?
Sorry, I found the solution:
library(zoo)
z <- read.zoo(monthly, split = "market")
all.dates <- seq(start(z), as.Date(as.yearmon(end(z)), frac = 1), by = "day") ##
mondays <- all.dates[weekdays(all.dates) == "Monday"] ##
weeks <- na.locf(z, xout = mondays, na.rm = FALSE, maxgap=0)
do.call(rbind, b
Hello!
I have a code for converting monthly values into weekly values:
monthly<-data.frame(month=c(20100301,20100401,20100501,20100601,20100301,20100401,20100501,20100601),monthly.value=c(100,NA,200,300,10,NA,20,30),market=c("Market
A","Market A","Market A","Market A","Market B","Market B","Market
In general, the lapply(split(...)) construction should never be used.
Use tapply() or by() instead, along the lines of
by(dataframe,with(yourdata,list(your columns)), function(...),...)
If you find this complexity annoying, then look into Hadley Wickham's
plyr package for simpler constructions, a
Hello,
I have a data set which is similar to the following data
mice <- rep(letters[1:4],10)
outcome <- sample(c(0,1),length(mice),replace=T)
group <- c(rep("A",length(mice)/2),rep("B",length(mice)/2))
my.data <- data.frame(mice,outcome,group)
my.sort.data <- my.data[order(my.data[,1]),]
I woul
On Aug 4, 2010, at 11:06 AM, Charles C. Berry wrote:
On Wed, 4 Aug 2010, David Winsemius wrote:
Dear list;
I just created a utility function that replicates what I have done
in the past with Excel or OO.org by putting a formula of the form
=sum($A1:A$1) in an upper-corner of a section an
Hi,
Try help("$").
dwt <- function(ld, filter='d8', n.levels=lev, boundary="reflaction")
{
list(W=ld, V=n.levels,filter=filter)
}
dec <- dwt(1:10, filter='d8', n.levels=10, boundary="reflaction")
dec$W
dec$V
dec$filter
dec$W <- sqrt(1:10)
dec
-
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Hi R,
Seems like the maximum seasonal 'q' parameter for the ?arima is 350. Any
way, where we can increase this? Since I am working on 3 year (q=252*3)
and 5 year(q=252*5) returns, I may require this option. Thanks.
> fit=arima(r,c(3,0,0),seasonal = list(order = c(0, 0, 500), period =
NA));
Just a little difference.
ecdf.tbl <- function (.dat) {
.dat <- as.matrix(.dat)
na.m <- is.na(.dat)
.dat[na.m]<-0 # Assign NA a value 0
res <- apply(t(apply(.dat,1,cumsum)),2,cumsum)
res[na.m] <- NA # Assign NA back
return(res)
}
-
Hi,
I'm trying to use RWeka and followed the following example from the
RWeka manual.
## Use some example data.
w <- read.arff(system.file("arff","weather.nominal.arff",
package = "RWeka"))
## Identify a decision tree.
m <- J48(play~., data = w)
m
## Use 10 fold cross-validation.
On Wed, 4 Aug 2010, David Winsemius wrote:
Dear list;
I just created a utility function that replicates what I have done in the
past with Excel or OO.org by putting a formula of the form =sum($A1:A$1) in
an upper-corner of a section and then doing a "fill" procedure by dragging
the lower-rt
Great! I will give it a try ASAP!
Thanks!
Joh
On Wednesday 04 August 2010 16:47:12 baptiste Auguié wrote:
> I added a parse argument to grid.table so that when switched to TRUE
> (default FALSE) all the text strings are interpreted as expressions
> (inspired by ggplot2::geom_text),
>
> d <- dat
Hi Leigh,
Several aspects of your email make it challenging to offer advice. We
are missing part of your data (e.g., 'T' in 1:T), and it is not
terribly clear what 't' in (t-5) to t(19950630) to (t+7) is supposed
to mean (thought I would hazard the guess time, and that it is
represented by your '
I added a parse argument to grid.table so that when switched to TRUE (default
FALSE) all the text strings are interpreted as expressions (inspired by
ggplot2::geom_text),
d <- data.frame("alpha", "beta")
grid.table(d, parse=T)
you'll need revision 258 of gridExtra for this to work (googlecode n
Dear list;
I just created a utility function that replicates what I have done in
the past with Excel or OO.org by putting a formula of the form
=sum($A1:A$1) in an upper-corner of a section and then doing a "fill"
procedure by dragging the lower-rt corner down and to the right. When
divid
Hello Wolfgang.
I'd appreciate if you could help me check whether I am doing the proper
thing to do an arm-level meta-analysis with metafor and what differences
there might be in trying to do the same with lme and lm.
I am following the arm based model described in section 3.2 of the
Salanti's pa
Dear Duncan
On Wed, 04 Aug 2010 08:33:49 -0400
Duncan Murdoch wrote:
> As other have pointed out, in a function you can use substitute(arg)
> to retrieve the expression passed as arg, and
> deparse(substitute(arg)) to turn it into a string that's suitable for
> using as a label. But that's not t
Is this what you mean?
x=c(1,2,2,3,4,5,6,3,2,1)
y=c(2,3,4,2,1,2,3,4,5,6)
matplot(cbind(x,y),type="l")
which(diff(sign(x-y))!=0)+1
[1] 4 8
--
View this message in context:
http://r.789695.n4.nabble.com/Finding-points-where-two-timeseries-cross-over-tp2313257p2313510.html
Sent from the R help ma
On Aug 4, 2010, at 9:46 AM, Leigh E. Lommen wrote:
I have the following array:
head(stocks)
DATE TICKER PERMNO EXCHCD TSYMBOL TRDSTAT SHROUTPRC
RET
1 19950131 EWST 10001 3EWST A 2224 -7.75000
-0.031250
2 19950228 EWST 10001 3EWST A
Thanks, this indeed solved the problem.
Regards,
Dieter
On 4/08/2010 15:21, Shentu wrote:
>
> The reason you see the exra markers is that the first part of the command
> "qplot(DT$N,DT$D,fill=factor(DT$C))" already plots the individual points.
> You didn't see it with "geom_bar(stat = "identity"
I have the following array:
> head(stocks)
DATE TICKER PERMNO EXCHCD TSYMBOL TRDSTAT SHROUTPRC RET
1 19950131 EWST 10001 3EWST A 2224 -7.75000
-0.031250
2 19950228 EWST 10001 3EWST A 2224 7.54688
-0.026210
3 19950331 EWST 10001 3
Thanks David,
I looked briefly through the bigmemory package and it looks very
promising. As I mentioned in my previous post to Steven I am calculating
distance matrices, and if it is my bottleneck (I am nearly sure it is) I
believe that the bigmemory will be particularly useful as it is designed
On Aug 4, 2010, at 9:32 AM, Ally wrote:
Hi,
I'm trying to use grid.polygon() to plot several polygons at once,
with a
view to putting coloured polygons beneath a curve. I'm struggling
just to
get the grid.polygon to plot anything
# PLOT SOME POINTS
x <- 1:100
y <- 1:100
Seems to me it may be worth stating what may be elementary to some on this list:
- If all relevant variables are included in the model and the "true model" is
indeed linear, then all least squares estimated coefficients are unbiased. [
David Ruppert once said about the three kinds of lies: Lie
Hi,
I'm trying to use grid.polygon() to plot several polygons at once, with a
view to putting coloured polygons beneath a curve. I'm struggling just to
get the grid.polygon to plot anything
# PLOT SOME POINTS
x <- 1:100
y <- 1:100*0.5 + 3
plot(x, y, pch = ".")
# PLOT 2
The reason you see the exra markers is that the first part of the command
"qplot(DT$N,DT$D,fill=factor(DT$C))" already plots the individual points.
You didn't see it with "geom_bar(stat = "identity")" simply because the
stacked bars made the previous layer invisible. To see this you can use the
gg
haenl...@gmail.com wrote:
I'm sorry -- I think I chose a bad example. Let me start over again:
I want to estimate a moderated regression model of the following form:
y = a*x1 + b*x2 + c*x1*x2 + e
Based on my understanding, including an interaction term (x1*x2) into the
regression in addition
Another way is
t1<- c("3:00","1:59","3:00","2:00")
t2 <- strptime(t1, format="%H:%M")
t2[-4]>t2[-1]
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Aug 4, 2010, at 8:47 AM, Gabor Grothendieck wrote:
On Wed, Aug 4, 2010 at 8:43 A
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