Is it possible to parallel scan a large file into a character vector in 1M
chunks using scan() with the "doMC" package? Furthermore, can I specify the
tasks for each child?
i.e. I'm working on a Linux box with 8 cores and would like to scan in 8M
records at time (all 8 cores scan 1M records at a
Thanks a lot
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-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of zhiji19
Sent: Wednesday, 8 December 2010 2:57 PM
To: r-help@r-project.org
Subject: [R] Summing up Non-numeric column
Dear All
If I have the following dataset
V1 V2
x y
y
Hi,
One way would be
> example <- read.table(textConnection("V1 V2
+ x y
+ y x
+ z b
+ a c
+ b j
+ d l
+ c o"), header = TRUE)
> closeAllConnections()
> example
V1 V2
1 x y
2 y x
3 z b
4 a c
5 b j
6 d l
7 c o
> with(example, length(unique(V1)))
[1] 7
> with(example,
Hi Mike,
The book makes use of .csv files, which are provided, along with all R code and
.RData files.
You have an interesting thought about people pulling data from diverse sources
and making everyday use of R. For this, I would suggest using Excel or Google
Docs Spreadsheets to compile and o
For a substantial calculation like this the algorithms will likely be in C or
Fortran.
You will need to download the source for the stats package from CRAN (as a
tar.gz file), expand it, and look at the source code in the appropriate
sub-directories.
You can get a bit of a road map in R by
>
Dear All
If I have the following dataset
V1 V2
x y
y x
z b
a c
b j
d l
c o
How do I use R command to get the total number of different letter in column
"V1"
column "V1" has 7 different letters.
Thank you
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Brand new to R
Want to understand the algorithm used in yule-walker time series
autoregression model
I assume there is a way to see the source for ar.yw
I also assume that everybody except me knows how
Could someone suggest to me how to find out
I've looked thru some of the documenttion - th
A new fortune is born?
"Sharing LaTeX documents with people using word processors only is no more
difficult than giving driving directions to someone who is blindfolded and has
all 4 limbs tied behind their back. Collaboration with people who insist on
using programs that process their words m
On 12/07/2010 05:29 PM, Paul Miller wrote:
Hello Everyone,
Been learning R over the past several months. Read several books and have
learned a great deal about data manipulation, statistical analysis, and
graphics.
Now I want to learn how to make nice looking documents and about "literate
progr
See inline below
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Paul Miller
> Sent: Tuesday, December 07, 2010 4:29 PM
> To: r-help@r-project.org
> Subject: [R] LaTeX, MiKTeX, LyX: A Guide for the Perplexed
>
> Hello Everyo
If you don't want to use the OS (windows does have a scheduling service) then I
would suggest using the tools in tcltk and possibly tcltk2 (tclTaskSchedule).
They can have a delayed thing in the background so you can still use the R
console, but the download will happen automatically.
--
Greg
On Dec 7, 2010, at 9:07 PM, Carolyn Fitzsimmons wrote:
I've downloaded R and started going through Appendix A in the R-
intro manual.
When I use the plox command I expect to see a graphics window, but
there is
none. I've searched the help archives and found that it might go to
a ps file?
C
For the GUI-addicted people (partly including me), I think LyX +
Sweave is an excellent choice. LyX is based on LaTeX; it can do almost
anything that LaTeX can do, and it takes care of a lot of details when
compiling a LaTeX document. The solution for embedding Sweave in LyX
has been there for a lo
I guess I know the answer but I am not completely clear about the
reason; print.latex() calls show.dvi() to open the DVI file using
system(), and under Windows it is usually better using shell() instead
of system(). The help page says shell() is a "friendly" wrapper of
system() under Windows, and t
I've downloaded R and started going through Appendix A in the R-intro manual.
When I use the plox command I expect to see a graphics window, but there is
none. I've searched the help archives and found that it might go to a ps file?
Can someone help me get the graphics window working?
Kind Rega
Hi:
I've experienced the same behavior as Dr. Friendly when trying to use
latex() in an Sweave code chunk (with results = tex in the chunk header) on
a Win7 system with 64-bit R (everything up to date). Is the answer the same
in that case?
TIA for your assistance,
Dennis
On Tue, Dec 7, 2010 at 9
use split and lapply to make it easier. Substitute in your own
calculations since this is just an example:
> x
id t X1 X2
1 1 1 4 3
2 1 2 9 2
3 1 3 7 3
4 1 4 6 6
5 2 1 6 4
6 2 2 5 3
7 2 3 1 1
8 3 1 9 6
9 3 2 5 5
> # first split the data by 'id'
> x.s <- split(x, x$id)
Jim,
Thanks a lot! works perfectly! now i just use a for loop and rbind to create
one dataset with all daily observations.
Arsenio
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I have a for-loop in my code that calls another .R file:
source("estimation.R")
This file runs through without any problems, so the program completes the loop
one time. However, when the loop starts a second time and it comes time to call
the file "estimation.R" again, program stops and prints
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of madr
> Sent: Tuesday, December 07, 2010 3:02 PM
> To: r-help@r-project.org
> Subject: [R] longer object length is not a multiple of shorter object
> length
>
>
> In datamatrix[,
allianz.com.au> writes:
> Is it possible to write a program such that it downloads a csv from a given
> web address? Would be great if this could be done at a particular time
> during the day as well. Say 9AM monday-friday.
downloading a csv file is easy:
x <- read.csv(url("http://whatever.
Hello Everyone,
Been learning R over the past several months. Read several books and have
learned a great deal about data manipulation, statistical analysis, and
graphics.
Now I want to learn how to make nice looking documents and about "literate
programming." My understanding is that R use
Hi,
On Tue, Dec 7, 2010 at 3:01 PM, madr wrote:
>
> In datamatrix[, "y"] == datamatrix[, "y"][-1] :
suppose datamatrix has 10 rows, you select column "y", and then for
the equality remove the first 'row' (technically element at this point
since you selected just column "y"). So R is being asked
In datamatrix[, "y"] == datamatrix[, "y"][-1] :
longer object length is not a multiple of shorter object length
out = c(FALSE,datamatrix[,'y'] == datamatrix[,'y'][-1])
and I do not know why I get that error, the resulting out matrix is somehow
one row larger than datamatrix...
all I try to do
Dear R-helpers,
I have a basic question on using loops.
I have a panel data set with different variables measured for "n" firms over
"t" time periods. A snapshot of the data is given below
id t X1 X2
1 1 4 3
1 2 9 2
1 3 7 3
1 4 6 6
2 1 6 4
Hi All,
Is it possible to write a program such that it downloads a csv from a given
web address? Would be great if this could be done at a particular time
during the day as well. Say 9AM monday-friday.
Incase you are curious Im just trying to analyse some stocks data.
Thanks,
Sachin
p.s. sorry
Hi Laurent,
On 12/7/10 5:01 PM, Laurent Gatto wrote:
Dear Duke,
I'm not sure if limma can generate colourful Venn diagrams. As an
alternative, I can however recommend Vennerable [1].
Thanks. I also found out that there are so many packages drawing venn
diagram now. I am playing with Venneule
Dear Duke,
I'm not sure if limma can generate colourful Venn diagrams. As an
alternative, I can however recommend Vennerable [1].
Hope this helps,
Laurent
[1] https://r-forge.r-project.org/projects/vennerable/
On 7 December 2010 21:18, Duke wrote:
> Hi all,
>
> I am trying to plot a venn diag
Hi all,
I am trying to plot a venn diagram using vennDiagram in limma package,
but I have no idea how to have it colorful such as the color one created
by the following website: http://www.pangloss.com/seidel/Protocols/venn.cgi.
Any one has any advice?
Thanks,
D.
__
From: jypupp...@hotmail.com
To: r-help-boun...@r-project.org
Subject: R programing help-newton iterations for the square root
Date: Tue, 7 Dec 2010 12:00:01 -0800
NEWTON ITERATIONS FOR THE SQUARE ROOT
Newton iterations to find the root of a real valued function f , i.e. a number
x for
x+1 is not that complicated... Am I missing something here?
Le 12/7/2010 16:55, Gabor Grothendieck a écrit :
On Tue, Dec 7, 2010 at 8:43 AM, madr wrote:
many languages have shorthands for that operation like:
variable += 1
or
++variable
is there something like that in R ?
You can do this:
> From: bruce...@usc.edu
> To: r-help@r-project.org
> Date: Tue, 7 Dec 2010 09:27:21 -0800
> Subject: [R] R problem
>
> I'm using R version 2.12.0
>
> I'm trying to open a stata file (.dta); but I keep getting the message:
>
> Error in read.dta("c:/p4s
Thanks for the confirmation Andy. It would be great if one could
specify a specific column as the unique ID to be returned and/or
display the input row, along with the predictions. Thanks for the
quick reply and looking forward to the patch!
Cheers,
Dennis
On Tue, Dec 7, 2010 at 12:00 PM, Liaw,
The order in the output correspond to the order of the input. I will
patch the code so that it grabs the row names of the input (if exist).
If you specify type="prob", it already labels the rows by the input row
names.
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:
your question is not clear to me .. but your solution is a variation of
> data1$x.1 <- data1$x1 %in% data2$y1
you can play with your conditions to get the result you want
On Tue, Dec 7, 2010 at 10:00 PM, Pete Pete wrote:
>
> Hi,
> consider the following two dataframes:
> x1=c("232","3454","3455
hello list,
i'm a bit puzzled by the error message i get when i copy past this in R:
data<-c(16,13,17,4,15,24,59,18,33,8,42,19,20,4,11,9,3,7,10,3,3,67,4,4,13,16,6,3,3,6,3,4,35,10,16,11,24,7,47,8,8,2,12,3,8,4,3,6,6,10,2,9,3,15,21,13,8,16,2,5,14,9,21,4,9,11,36,4,8,4,6,4,10,13,11,5,4,16,14,3,22,20,5
Hi Bert,
thank you for your suggestion. I'm sure it's a good one. But my intention in first place was to
learn about getting subsets of list nested in lists the fast way (and preferably also the easy way,
but that is only my laziness).
It seems this thread is getting a bit long and also lead
Jonathan -
If I understand correctly,
max(0,floor(log(x,10)))
will return the value you want.
- Phil Spector
Statistical Computing Facility
Department of Statistics
I would prefer:
round(log10(x))
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns
Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
-Original Message
What a brain fart...
Thanks!
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
"Is the room still a room when its empty? Does the room,
the thing itself have purpose? Or do we, what's the wor
Jonathan,
I'd just return the integer part of the common log:
floor(log10(x))
Dave
From:
Jonathan P Daily
To:
r-help@r-project.org
Date:
12/07/2010 01:44 PM
Subject:
[R] More elegant magnitude method
Sent by:
r-help-boun...@r-project.org
I have a need to find the order of number to get a
On Dec 7, 2010, at 2:43 PM, Jonathan P Daily wrote:
I have a need to find the order of number to get a scaling parameter
as a
power of 10. I have a function that works *so far*, but it is ugly and
probably buggy. In the interest of avoiding code-based outliers in my
data, I thought I would as
On 07/12/2010 12:42 PM, Gabor Grothendieck wrote:
On Tue, Dec 7, 2010 at 12:25 PM, Bert Gunter wrote:
Ted:
Inline below...
On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding wrote:
Indeed!
x<- x + 1
(and being generous with unnecessary spaces) uses 10 characters.
`+`(x)<-1
(being mean with
Alexander:
I'm not sure exactly what you want, so the following may be irrelevant...
BUT, noting that data frames ARE lists and IF what you have can then
be abstracted as lists of lists of lists of ... to various depths
AND IF what you want is just to pick out and combined all named
vectors (whic
I have a need to find the order of number to get a scaling parameter as a
power of 10. I have a function that works *so far*, but it is ugly and
probably buggy. In the interest of avoiding code-based outliers in my
data, I thought I would ask if anyone here has a better way.
> scl <- function(x
On Dec 7, 2010, at 2:31 PM, Erik Iverson wrote:
Hello,
To do what you want, see ?toupper :
levels(dat$target) <- toupper(levels(dat$target))
However, for clarity,
dat$target is not a "string variable", it is a factor,
which you can verify with
> str(dat)
And to further clarify, it is a f
Phil Spector wrote:
Jahan -
Try
dat$target = toupper(dat$target)
Note that in this case, the above *will* coerce dat$target
to a character vector, which may or may not be what
is intended.
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Ted:
Inline below...
On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding wrote:
> Indeed!
>
> x <- x + 1
>
> (and being generous with unnecessary spaces) uses 10 characters.
>
> `+`(x)<-1
>
> (being mean with them) uses 9. The "mean" version of the first
> uses only 6: x<-x+1
>
> However, I suppose th
Jahan -
Try
dat$target = toupper(dat$target)
I would not recommend a loop for something
like this.
(You might also want to brush up on your python,
because what you're trying doesn't work in
python either.)
- Phil Spector
Hello,
To do what you want, see ?toupper :
levels(dat$target) <- toupper(levels(dat$target))
However, for clarity,
dat$target is not a "string variable", it is a factor,
which you can verify with
> str(dat)
Factors are enumerated types, and have a discrete
set of 'levels' associated with the
well, I found id the simplest answer is:
d<- approx(density(x),xout=...)
where xout is a vector of the points where the data should be sampled
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Sent from the R help mailing l
On 12/7/2010 1:03 AM, Nick Sabbe wrote:
Hi All.
I often find myself in this situation:
. Based on some vector (or list) of values, I need to calculate a
few new values for each of them, where some of the new values are numbers,
but some are more of descriptive nature (so: character stri
Hello,
Here is the data set I am working with:
dat=read.csv('http://dl.dropbox.com/u/1852742/relexpressions.csv')
names(dat)
#Under the 'target' column, I want to change all of the values into
all capital letters (e.g. fgf2 becomes FGF2). I have taken a loop
approach but I think my Python backgro
i have time series of momentum signal. I want to get the date of each of the
"-1" signal period. for example , the first period of -1 signal begins on
2005-9-21 and ends on 2005-9-28. 2nd period of -1 signal begins on
2005-09-30 and ends on 2005-10-28.
Thx
Cameron
datePx
On Tue, Dec 7, 2010 at 11:30 AM, Pete Pete wrote:
>
> Hi,
> consider the following two dataframes:
> x1=c("232","3454","3455","342","13")
> x2=c("1","1","1","0","0")
> data1=data.frame(x1,x2)
>
> y1=c("232","232","3454","3454","3455","342","13","13","13","13")
> y2=c("E1","F3","F5","E1","E2","H4",
I'm using R version 2.12.0
I'm trying to open a stata file (.dta); but I keep getting the message:
Error in read.dta("c:/p4subset_biology.dta") :
Calloc could not allocate (-603966867 of 4) memory
What does this mean?
The file should contain lots of physiological data.
Thank you for your help
It's impossible to give you really good advice without an example of
your data, the R code you used and the error message.
But here's a simple scatterplot example:
fakedata <- data.frame(x=runif(15), y=runif(15))
plot(fakedata$x, fakedata$y)
or since you mentioned a line:
fakedata <- fakedata[o
On Dec 7, 2010, at 1:49 PM, Greg Snow wrote:
tmpdf <- data.frame( x = c(1,2,3), y=c(2,3,1), a=c(10,20,30) )
mymat <- matrix(0, ncol=3, nrow=3)
mymat[ as.matrix(tmpdf[,c('x','y')]) ] <- tmpdf$a
cbind is also useful for assembly of arguments to the matrix-`[<-`
function:
tmpdf <- data.fram
Hello Gerrit, Gabor,
thank you for your suggestion.
Unfortunately unlist seems to be rather expensive. A short test with one
of my datasets gives 0.01s for an extraction based on my approach and
5.6s for unlist alone. The reason seems to be that unlist relies on
lapply internally and does so rec
Hello,
Matthew's hint is interesting:
Am 07.12.2010 19:16, schrieb Matthew Dowle:
> Hello Alex,
>
> Assuming it was just an inadequate example (since a data.frame would suffice
> in that case), did you know that a data.frames' columns do not have to be
> vectors but can be lists? I don't know
Hi Baptiste!
Grand-grand merci!!! It works!
Thanks!
Alex
2010/12/7 baptiste auguie :
> baptiste
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Hi Everyone,
I'm writing to announce my new R beginner's guide book and answer questions
related to it.
The primary focus of Statistical Analysis with R is helping new users become
accustomed to R and empowering them to apply R to suit their own needs. It
is a beginner's guide written for a broa
On Dec 7, 2010, at 1:27 PM, madr wrote:
I have created a density line
d<- density(X)
now I need to read values from that line
for example what is the value of this line at x = 1, 2, 3 etc...
Learn to search:
http://finzi.psych.upenn.edu/Rhelp10/2010-September/253910.html
--
David Winse
Hi all,
When running a prediction using RF on another data, I get two columns
returned: row number(?) and predicted class. Is there a way of
appending the unique row value from an ID column in the dataframe to
the predictions instead of the row number? I'm assuming that the
returned results follow
tmpdf <- data.frame( x = c(1,2,3), y=c(2,3,1), a=c(10,20,30) )
mymat <- matrix(0, ncol=3, nrow=3)
mymat[ as.matrix(tmpdf[,c('x','y')]) ] <- tmpdf$a
--
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Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> F
There are a few different options depending on what you are trying to do.
If you just need some data pairs (for plotting for example), then the return
from density has a vector of x's and a vector of y's, just use those.
If you have specific x values that you need the height at and they follow a
Thanks! Brian,
As you said, if locale changes, it is fine now. What I add is
Sys.setlocale("LC_TIME", "American")
On Tue, Dec 7, 2010 at 12:14 PM, Prof Brian Ripley wrote:
> On Tue, 7 Dec 2010, Jianhong Wang wrote:
>
> Hi
>>
>> I tried to plot something simple
>>
>> x=c('2010-08-20', '2010-08-
index m as a vector and do the assignment in one step
i <- df$row + (df$col-1)*nrow(m)
m[i] <- df$a
or something along those lines.
-Whit
On Tue, Dec 7, 2010 at 1:31 PM, Cutler, Gene wrote:
> I have a data frame with three columns, x, y, and a. I want to create a
> matrix from these values
On Tue, 7 Dec 2010, Jianhong Wang wrote:
Hi
I tried to plot something simple
x=c('2010-08-20', '2010-08-30')
y = c(1,2)
t = strptime(x, "%Y-%m-%d")
plot(t,y)
At the xlab, the fonts are Chinese. How can I switch it to English? I am
working under Windows 7 64 bits Home Edition and R is Win32
Hello,
I have count data with a large number of zeros. To model this data using
gams I want to try the Tweedie distribution instead. (The zero-inflated
poisson willbe my next step)
How do i decide on the value of the power 'p' parameter in the formula:
family=Tweedie(p, link) ? For a glm, this ca
Like Peter says, this sounds like loess, there are examples on the help page
for scatter.smooth, you could also do this with lattice graphics using
type=c('p','smooth'), or ggplot2 graphics (probably something like geom_smooth
or geom_loess, I don't know ggplot2 that well yet).
If you want to l
I have a data frame with three columns, x, y, and a. I want to create a matrix
from these values such that for matrix m:
m[x,y] == a
Obviously, I can go row by row through the data frame and insert the value a at
the correct x,y location in the matrix. I can make that slightly more
efficient
Hi,
Embarrassingly enough, it was quite straight-forward in the first
versions of grid.table(). You might want to try with version r11 for
example,
source("http://gridextra.googlecode.com/svn-history/r11/trunk/R/tableGrob.r";)
library(grid)
tc = textConnection("
carat VeryLongWordIndee
I have created a density line
d<- density(X)
now I need to read values from that line
for example what is the value of this line at x = 1, 2, 3 etc...
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Sent from the R help m
Indeed!
x <- x + 1
(and being generous with unnecessary spaces) uses 10 characters.
`+`(x)<-1
(being mean with them) uses 9. The "mean" version of the first
uses only 6: x<-x+1
However, I suppose there is merit in the spiritual exercise
of contemplating how `+`(x)<-1 gets worked out!
Ted.
On Tue, Dec 7, 2010 at 1:12 PM, David Winsemius wrote:
>
> On Dec 7, 2010, at 12:42 PM, Gabor Grothendieck wrote:
>
>> On Tue, Dec 7, 2010 at 12:25 PM, Bert Gunter
>> wrote:
>>>
>>> Ted:
>>>
>>> Inline below...
>>>
>>> On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding
>>> wrote:
Indeed!
Hello Alex,
Assuming it was just an inadequate example (since a data.frame would suffice
in that case), did you know that a data.frames' columns do not have to be
vectors but can be lists? I don't know if that helps.
> DF = data.frame(a=1:3)
> DF$b = list(pi, 2:3, letters[1:5])
> DF
a
First, subset 'test' once, e.g.
testT <- test[1:3];
and then use sapply() on that, e.g.
val <- sapply(testT, FUN=function (x) { x$a })
Then you can avoid one level of function calls, by
val <- sapply(testT, FUN="[[", "a")
Second, there is some overhead in "[[", "$" etc. You can use
.subset2(
Hi all,
When running a prediction using RF on another data, I get two columns
returned: row number(?) and predicted class. Is there a way of
appending the unique row value from an ID column in the dataframe to
the predictions instead of the row number? I'm assuming that the
returned results follow
On Dec 7, 2010, at 12:42 PM, Gabor Grothendieck wrote:
On Tue, Dec 7, 2010 at 12:25 PM, Bert Gunter
wrote:
Ted:
Inline below...
On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding
wrote:
Indeed!
x <- x + 1
(and being generous with unnecessary spaces) uses 10 characters.
`+`(x)<-1
(being
Thanks a lot for your responses.
Alabama and Rsolnp packages look promising.
My first tests worked well.
For others searching for alike problems I now use something like:
#equality constraint function
eqFkt<-function(x){
#for sum(x)=sum_x
sm<-sum(x)-sum_x
return(c(sm))
}
ineqFk
On 2010-12-07 09:37, madr wrote:
Well, maybe I didn't write it clear.
I know how to create scatterplot, and how to import data from csv file.
But I do not know how to add this fitting that mentioned in sumbject to a
plot.
I do not know for what function name to look for in R. I played some time
On Tue, Dec 7, 2010 at 12:12 PM, Alexander Senger
wrote:
> Hello Gerrit, Gabor,
>
>
> thank you for your suggestion.
>
> Unfortunately unlist seems to be rather expensive. A short test with one
> of my datasets gives 0.01s for an extraction based on my approach and
> 5.6s for unlist alone. The rea
Hi,
consider the following two dataframes:
x1=c("232","3454","3455","342","13")
x2=c("1","1","1","0","0")
data1=data.frame(x1,x2)
y1=c("232","232","3454","3454","3455","342","13","13","13","13")
y2=c("E1","F3","F5","E1","E2","H4","F8","G3","E1","H2")
data2=data.frame(y1,y2)
I need a new column i
Michael,
The easiest workaround is to assign the result of the latex() command.
myfilename <- latex(x)
print.default(myfilename)
It looks to me like the insides of the dvi.latex function aren't quite right
for Windows.
Rich
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Larry,
You found a data set that kills coxph. I'll have to think about what
to do since on the one hand it's your own fault for trying to fit a very
bad model, and on the other I'd like the routine to give a nice error
message before it dies.
In the data set you sent me the predictor varia
I tried to hide the gory details as the structure of my datasets is
rather complicated. Basically its a long list of lists which in turn
contain character vectors, dates, numerics and dataframes, all named.
While the hierarchy is fixed neither the number of elements nor their
ordering is. But if I
I got simple x,y pairs of data and simple scatterplot and just cannot figure
how to do it , there are many examples but always there is error popping out
please show me an example stripped with additional data just core of what I
need to do to get this damn line
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On Tue, Dec 7, 2010 at 12:25 PM, Bert Gunter wrote:
> Ted:
>
> Inline below...
>
> On Tue, Dec 7, 2010 at 8:42 AM, Ted Harding wrote:
>> Indeed!
>>
>> x <- x + 1
>>
>> (and being generous with unnecessary spaces) uses 10 characters.
>>
>> `+`(x)<-1
>>
>> (being mean with them) uses 9. The "mean
Well, maybe I didn't write it clear.
I know how to create scatterplot, and how to import data from csv file.
But I do not know how to add this fitting that mentioned in sumbject to a
plot.
I do not know for what function name to look for in R. I played some time
with ls, and it didn't create a lin
On Dec 7, 2010, at 11:30 AM, Pete Pete wrote:
Hi,
consider the following two dataframes:
x1=c("232","3454","3455","342","13")
x2=c("1","1","1","0","0")
data1=data.frame(x1,x2)
y1=c("232","232","3454","3454","3455","342","13","13","13","13")
y2=c("E1","F3","F5","E1","E2","H4","F8","G3","E1","H
Hi
I tried to plot something simple
x=c('2010-08-20', '2010-08-30')
y = c(1,2)
t = strptime(x, "%Y-%m-%d")
plot(t,y)
At the xlab, the fonts are Chinese. How can I switch it to English? I am
working under Windows 7 64 bits Home Edition and R is Win32 version 2.11.1
Thanks!
[[alternati
On Tue, Dec 7, 2010 at 8:43 AM, madr wrote:
>
> many languages have shorthands for that operation like:
>
> variable += 1
> or
> ++variable
>
> is there something like that in R ?
You can do this:
> x <- 3
> `+`(x) <- 1
> x
[1] 4
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Statistics & Software Consulting
GKX Group, GKX Associates
Hello there,
could you please help to modify gpar() properties of a cell inside
tableGrob() output.
In the following example I want to have different color for one out of 4 cells
require(gridExtra)
z<-matrix(1:4,2,2)
grid.draw(tableGrob(z))
The only way I found for now how to do it is to d
It took me quite some time to understand the difference between sep and
collapse.
The examples in Phil Spector's book (2008) helped me to get it:
paste(c('X','Y'), 1:5, sep='_')
"X_1" "Y_2" "X_3" "Y_4" "X_5"
paste(c('X','Y'), 1:5, collapse='|') ## sep=" " by default
[1] "X 1|Y 2|X 3|Y 4|X 5"
p
Ivan's advice is good, but understanding clearly what
"character string to separate the results" might mean is
a bit tricky!
Example:
cvec <- c("J","e"," ","m","'","a","p","p","e","l","l","e",
" ","B","e","n","o","i","t")
cstring <- paste(cvec,collapse="")
cstring
# [1] "Je m'
On 07.12.2010 14:43, madr wrote:
many languages have shorthands for that operation like:
variable += 1
or
++variable
is there something like that in R ?
No.
Uwe Ligges
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On Dec 7, 2010, at 10:11 AM, Benoit Wastine wrote:
Hi,
I'm running R 2.11
Does anyone know if it possible to transform one character vector to
one character string ?
?gsub
Also look at the even more powerful gsubfn package. There is also the
stringr package.
--
David Winsemius, MD
Wes
Hi,
If I understand what you mean (no example...), see ?paste and the
collpase argument
Ivan
Le 12/7/2010 16:11, Benoit Wastine a écrit :
Hi,
I'm running R 2.11
Does anyone know if it possible to transform one character vector to
one character string ?
Many thanks
Benoit
--
Ivan CALANDR
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