On Thu, Sep 06, 2012 at 02:45:20PM -0700, Shayne Hodge wrote:
> I accidentally posted the wrong version of my code, omitting the Q1 loop,
> and, sure enough that loop generates non-integers:
>
> qfracseq <- seq(from = 1, to = 0, by = -.1)
> Q <- 10
> Qseq <- Q*qfracseq
>
> > Qseq [1] 10.0
To bootstrap from a histogram, use
sample(bins, replace = TRUE, prob = counts)
Note that a kernel density estimate is biased, so some bootstrap
confidence intervals have poor coverage properties.
Furthermore, if the kernel bandwidth is data-driven then the estimate
is not functional, so some boo
Hi Tom,
Using the script below I had no trouble. Here's an example
Minke:R ben$ Rscript --no-init-file ./argCatcher.rs ./the=trouble .with=./args
is=the.y argue
./the=trouble
.with=./args
is=the.y
argue
# script starts here
# argumentCatcher.rs
args = commandArgs(trailing = TRUE)
cat(args, se
> toeplitz(1:4)
[,1] [,2] [,3] [,4]
[1,]1234
[2,]2123
[3,]3212
[4,]4321
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.o
Dear R People:
Suppose I have the following:
x <- 1:4
And I want to create the following matrix:
1 2 3 4
2 1 2 3
3 2 1 2
4 3 2 1
Is there a function in place for this, please? I looked at symMatrix
in micEcon and some of the items in Matrix, but they didn't quite do
it.
I can put together s
Wanting a commandline solution (for a problem detailed @
http://mailman.unidata.ucar.edu/mailing_lists/archives/netcdfgroup/2012/msg00279.html
) I turned to Rscript, and whacked out the q'n'd
https://github.com/TomRoche/GEIA_to_netCDF/blob/master/netCDF.stats.to.stdout.r
However it wasn't as q
Hi,
Try this:
list1<-list(a=a,b=b,c=c,d=d,e=e)
dat1<-as.matrix(do.call(rbind,lapply(list1,`[`,1:4)))
dat1[duplicated(dat1)|duplicated(dat1,fromLast=TRUE),]
[,1] [,2] [,3] [,4]
#a 1 2 3 NA
#c 1 2 3 NA
#d 1 2 3 NA
dat1[!(duplicated(dat1)|duplicated(dat1,fromLast=
Hi Weija,
You can also try this instead of the loop solution in my previous reply.
dat1<-read.table(text="
ID PT_ID BASE IDX_DT OBS_DATE OBS_VALUE CATEGORY DAYS_DIFF
LDL_BASE rf
118485 9624295 164.2 2006-11-21 2009-02-17 157.6 2 819 2006-11-20
2.5
118486 9624295 1
On Sep 6, 2012, at 2:45 PM, Shayne Hodge wrote:
> I accidentally posted the wrong version of my code, omitting the Q1 loop,
> and, sure enough that loop generates non-integers:
>
> qfracseq <- seq(from = 1, to = 0, by = -.1)
> Q <- 10
> Qseq <- Q*qfracseq
>
>> Qseq [1] 10.0 9.0
I am new to R and am encountering memory issues while trying to download a
large table from Green Plump, using sqlQuery. Is there any way this FF
package can help me create a large dataframe in R while downloading from the
server?
The FF documentations are very confusing. Thanks for any help
You find code to compute the Cragg-Donald Wald F test statistic to test for
weak instruments (e.g., Stock and Yogo, 2005) below with a working example.
The code reproduces the results from Stata 12 using the same data (see
bottom of this email), but further testing is needed. Use at your own risk.
I accidentally posted the wrong version of my code, omitting the Q1 loop,
and, sure enough that loop generates non-integers:
qfracseq <- seq(from = 1, to = 0, by = -.1)
Q <- 10
Qseq <- Q*qfracseq
> Qseq [1] 10.0 9.0 8.0
> 7.0
[5
Update: seems one way to skin this cat is to add
gp = gpar(fontfamily="mono")
to the viewport() call itself. If anyone has any suggestions for a
robust way to extract this piece information from the key grob itself
(it's nested several levels deep), I'm all ears.
Thanks
Ben
__
For posterity, I found the TeachingDemos::shadowtext option most
agreeable for this problem:
* legend puts a large box around the text which did not seem
possible to shrink, and does not accept vector x, y arguments
* plotrix::boxed.labels did not work with pos=4 (this moved the
text, but
Michi,
Try putting your par() statements BEFORE your barplot() statement. For
example,
oldpar <- par(mar=c(18, 7, 2, 2.1))
oldfont <- par(font=3)
bp <- barplot(data.matrix(geo_barplot_percentage_top10),
...
Jean
Michael Eisenring wrote on 09/06/2012 12:40:27 PM:
>
> Dear R-help members
Try using ave(), as in
dat1WithMean <- within(dat1, Mean <- ave(OBS_VALUE, PT_ID, DAYS_DIFF, FUN
= mean))
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of arun
>
Dontrece,
Try using the all.terms argument to the plot.gam() function.
plot(gam4, shade=TRUE, pages=1, all.terms=TRUE)
Does that work?
Jean
Dontrece Smith wrote on 09/06/2012 06:50:07 AM:
>
> Hello,
> My name is Dontrece Smith. I am creating figures for my GAMs. I change
my
> qualitative v
HI,
For your question to split by two variables, try this:
dat1<-read.table(text="
ID PT_ID BASE IDX_DT OBS_DATE OBS_VALUE CATEGORY DAYS_DIFF
LDL_BASE rf
118485 9624295 164.2 2006-11-21 2009-02-17 157.6 2 819 2006-11-20
2.5
118486 9624295 164.2 2006-11-21 2006-09
As Linus would say:
AGHHH...!
1. Rui's solution clearly violates my conditions 1) and 2) and so does not work.
2. David's violated my UNSTATED condition 4: The order of the columns
cannot be changed. Any matrix "columns" must be expanded "in place"
between their flanking columns of the data fram
On 06/09/2012 2:09 PM, Shayne Hodge wrote:
Hello,
(Apologies for length, wanted to get all the relevant detail in that I know
of).
I've been having a lot of trouble with some code for an inventory analysis
problem I was doing, and finally came to the conclusion that it appears
that choose() is
Hello,
1. With R-2.15.1 on Windows 7 your first plot places all labels when you
maximize the graphics device.
2. To place the labels in your second plot you can change the margin
settings:
op <- par(mar = c(7, 4, 4, 2) + 0.1) # default is c(5, 4, 4, 2) + 0.1
barplot(sales,ylim=c(0,2500),la
Dear Rxperts..
I am trying to install a few packages in R (version 2.14.2) on a *64-bit
Window 7* workstation. Some of the errors are posted below.. Both 32-bit
and 64-bit versions of R were installed. I did not get errors while
installing packages through *32-bit R*, except that it could not upda
Look at the staxlab function in the plotrix package.
On Thu, Sep 6, 2012 at 12:03 PM, David-Arnold wrote:
> All,
>
> I have:
>
> sales <- c(2300,900,155,102,42,10)
> names(sales) <- c("Christmas","Valentine's Day",
>"Mother's Day","Father's Day",
>"Thanksgiving","New Year'
The primary recommendation is to use the "y ~ x" formula rather than
the "~ y | x"
formula. The place to look first is in the help files
?bwplot
?panel.bwplot
Here is a example that matches your data structure
Rich
tmp <- data.frame(y=rnorm(20),
x=rep(letters[1:2], each=10)
On Thu, 6 Sep 2012, Richard M. Heiberger wrote:
The primary recommendation is to use the "y ~ x" formula rather than the
"~ y | x" formula.
Rich,
I missed this.
The place to look first is in the help files
?bwplot
?panel.bwplot
I tend to grab books first. :-)
Here is a example that
Your Q1 may be a tad below 3. When printed it is, by default, rounded
to 7 significant digits. E.g.,
> 3 - 3*10^-16
[1] 3
> options(digits=17)
> 3 - 3*10^-16
[1] 2.9996
choose will truncate, not round, its inputs to make integers out of them
so your 3 - 3*10^-16 is treated a
This is the third time I have seen this message. I don't have an answer for
you. Someone else might, but repeatedly posting won't make an answer magically
appear, and if no one knows what you are talking about you won't see any
response. In that case you may need to translate your references int
Him
>T his is the third time I have seen this message. I don't have an answer for
> you. Someone else might, but repeatedly posting won't make an answer
> magically appear, and if no one knows what you are talking about you won't
> see any response. In that case you may need to translate your
Hello,
(Apologies for length, wanted to get all the relevant detail in that I know
of).
I've been having a lot of trouble with some code for an inventory analysis
problem I was doing, and finally came to the conclusion that it appears
that choose() is returning incorrect values. Specifically:
--
All,
I have:
sales <- c(2300,900,155,102,42,10)
names(sales) <- c("Christmas","Valentine's Day",
"Mother's Day","Father's Day",
"Thanksgiving","New Year's Day")
barplot(sales,ylim=c(0,2500))
But it doesn't place all of the name labels on the plot. So I tried:
sales <- c(23
On 9/6/2012 7:24 PM, Bert Gunter wrote:
... and if it Duncan's suggestion won't do, maybe approaching it via
clustering might be useful.
But do note that, as stated, the problem is not well defined, because
transitivity fails: consider
v <- c(1,2,3,4,5,10)
with a tolerance of <=2. Then 1 is the
Dear R-help members,
with the help of one of you the following R-Code was developed. No I have the
(probably simple) problem that I want to increase font size of "text" and
"ylab" names from 1 to 2. Unfortunately I'm not able to adjust the margin so,
that the plot is appers complete (all names
Help! Weird behavioral difference between R interactive and command-line?
Hi all,
This weird problem has been bugging me for a while.
If I run the R program in an interactive session, it worked;
But if I run the R program on command-line, it stopped with no reason: with
"Execution halted" messa
I've looked again at Deepayan Sakar's 'Lattice' book and not seeing the
information I know must be in there. A clue stick would be much appreciated.
A pair of box-and-whisker plots for two conditions of the same data set
display each box plot horizontally (see attached pH figure). Relations o
On Sep 6, 2012, at 1:40 AM, Sri krishna Devarayalu Balanagu wrote:
> Hello,
>
> Say all the below five vectors should have same elements in any situation.
> How to find the non matching vectors among these five, if so how we can find
> the non matching elements of those vectors?
"non-matching"
This should get you started.
Rich
aa <- c(1,2,3)
bb <- c(1,2,3,4)
cc <- c(1,2,3)
dd <- c(1,2,3)
ee <- c(1,4,5)
outer.all.equal <- function(input.list) {
nn <- names(input.list)
if (is.null(nn)) stop("input must be a named list.")
n <- length(input.list)
result <- vector("list", n*n)
On Sep 6, 2012, at 10:28 AM, Bert Gunter wrote:
> On Thu, Sep 6, 2012 at 10:20 AM, David Winsemius
> wrote:
>
>
>> I guess this means you are not the one performing the d$c <- m step? If you
>> were under control of that step, you can get different (and more to your
>> liking) behavior wit
I would like to fit regression models of the form
y ~ ns(x, df=splineDF)
where splineDF is passed as an argument to a wrapper function.
This works fine if the regression function is lm(). But with lme(),
I get two different errors, depending on how I handle splineDF
inside the wrapper function
Hello,
Try the following.
class(d[3][[1]])
dim(d[3][[1]])
(d2 <- cbind(d[1:2], d[3][[1]]))
dim(d2)
Funny how the original names of 'm' are retrieved.
Rui Barradas
Em 06-09-2012 18:28, Bert Gunter escreveu:
On Thu, Sep 6, 2012 at 10:20 AM, David Winsemius wrote:
I guess this means you ar
> * David Winsemius [2012-09-06 10:30:16 -0700]:
>
>> these are the results of applying a model to the test data.
>> the first column is the ID
>
> In which case you should be using the 'by' argument to `merge`
I already do! see my initial message!
>> 3. sort by the sum/mean of the V3 columns an
On Sep 6, 2012, at 6:42 AM, Sam Steingold wrote:
>> * David Winsemius [2012-09-05 21:02:16 -0700]:
>>
>> On Sep 5, 2012, at 8:51 PM, Sam Steingold wrote:
>>
>>> I have a list of data frames:
>>>
str(data)
>>> List of 4
>>> $ :'data.frame':700773 obs. of 3 variables:
>>> ..$ V1: chr
Thanks to both of you for suggestions.
I settled for the round() approach.
much obliged,
Michael Bach
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-g
On Thu, Sep 6, 2012 at 10:20 AM, David Winsemius wrote:
> I guess this means you are not the one performing the d$c <- m step? If you
> were under control of that step, you can get different (and more to your
> liking) behavior with 'cbind.data.frame':
Correct. d is given to me already, as d
... and if it Duncan's suggestion won't do, maybe approaching it via
clustering might be useful.
But do note that, as stated, the problem is not well defined, because
transitivity fails: consider
v <- c(1,2,3,4,5,10)
with a tolerance of <=2. Then 1 is the same as 2 and 3, 2 and 3 are
the same as
On Sep 6, 2012, at 4:50 AM, Dontrece Smith wrote:
> Hello,
> My name is Dontrece Smith. I am creating figures for my GAMs. I change my
> qualitative variables to 1 or 2 in my dataset, so I would be able to run my
> GAMs. However, R will only display plots for my quantitative variables and
> not m
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of salind
> Sent: Thursday, September 06, 2012 8:56 AM
> To: r-help@r-project.org
> Subject: [R] Interaction plot glm?
>
> Hi,
> I'm a beginner in both statistics and the usage
Dear R-Users,
I was able to use mvpart package for my multi-response data and was able to
generate the decision tree.
I realized that the colors (3) for the legend are being recycled. In my
situation I have 12
dependent variables and would like to have 12 different colors for the
legend and barp
On Sep 6, 2012, at 9:03 AM, Bert Gunter wrote:
> Hi Folks:
>
> Here's the situation:
>
>> m <- cbind(x=letters[1:3], y = letters[4:6])
>> m
> x y
> [1,] "a" "d"
> [2,] "b" "e"
> [3,] "c" "f"
>
> ## m is a 2 column character matrix
>
>> d <- data.frame(a=1:3,b=4:6)
>> d$c <- m
>> d
> a
Hello all,
Evidently my previous message met some filter due to subject line. I am
re-sending my message. I apologize if this was sent out twice.
Based on "Ripley & Thompson, Analyst, 1987", I am trying to do a regression of
my data which assumes a linear
relationship between measurement
Hi Bert,
maybe I'm missing the point, but
dd<-cbind(d,m)
does 1, 2 and 3 as desired:
n <- data.frame(nx=letters[7:9], ny = 7:9)
str(cbind(d,m,n))
t<-letters[7:9]
str(cbind(d,t))
cheers.
Am 06.09.2012 18:03, schrieb Bert Gunter:
> Hi Folks:
>
> Here's the situation:
>
>> m <- cbind(x=letter
print(i).Arms
is not valid R syntax. Did you mean
print(i.Arms)
?
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Sri krishna Devarayalu Balanagu
> Sent: Wednesda
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Michael Bach
> Sent: Thursday, September 06, 2012 12:48 PM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] unique with tolerance
>
> Dear R Users and Developers,
>
> I am tryi
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Sri krishna Devarayalu Balanagu
> Sent: Thursday, September 06, 2012 8:54 AM
> To: r-help@r-project.org
> Subject: [R] Why the error is coming, can anyone help?
>
> Hi everyo
On 06/09/2012 6:48 AM, Michael Bach wrote:
Dear R Users and Developers,
I am trying to do the equivalent of
v <- c(1,2,3,3,2,1,)
vu <- unique(v)
for a vector such as
v2 <- c(1.02, 2.03, 1.00, 3.04, 3.06)
vut <- ...
As indicated in the subject, we need approximately unique values with a defin
thank you very much,
I will be expecting your book soon.
-
Thank you,
with regards,
Punitha
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
On 06-09-2012, at 08:54, Sri krishna Devarayalu Balanagu wrote:
> Hi everyone
> Can anyone help?
>
> x = c("Treatment", "Planned", "Actual", "Demographics", "Outcomes")
> for (i in x) {
> df <- paste(i)[, c("Reference_ID", "Arm_ID")]
> unique(df) -> uniq
> as.data.frame(table(uniq$Reference_ID))
Hi,
May be this is what you wanted:
I named your dput() data as dat1.
source("HuYan.txt")
head(dat1)
dat1$date<- as.Date(dat1$V2,format="%d/%m/%Y %H:%M")
dat2<-aggregate(V3~date+V1,data=dat1,mean)
dat2$date1<-gsub(".*-(.*)-(.*)","\\2/\\1",dat2$date)
dat2$V3<-1
xtabs(V3~V1+date1,data=dat2)
What did you expect 'print(i).Arms' to do? This is causing part of
the problem since the syntax is not correct.
Also what does 'paste(i)[, c("Reference_ID", "Arm_ID")]' do? Does not
seem correct since 'paste' does not return a matrix.
On Thu, Sep 6, 2012 at 2:54 AM, Sri krishna Devarayalu Bala
HI HJ,
No problem.
The gsub() was used to just format your date column to make almost the way you
wanted the results:
dat2$date
# [1] "2012-04-29" "2012-04-30" "2012-05-01" "2012-04-28" "2012-04-29"
#[6] "2012-04-30" "2012-05-01" "2012-05-02" "2012-04-30" "2012-05-01"
gsub(".*-(.*)-(.*)","\\
Hello,
My name is Dontrece Smith. I am creating figures for my GAMs. I change my
qualitative variables to 1 or 2 in my dataset, so I would be able to run my
GAMs. However, R will only display plots for my quantitative variables and
not my qualitative variables. Is there any way to fix this issue? I
Hi Arun and David
Thanks a lot for your reply and sorry for sending the csv file. (p.s. I
can download my csv file from the email I sent, so I'm not sure why this
doesn`t work for other users...)
Anyway, below I used dput() and I am attaching the output from R:
structure(list(V1 = c(532703L
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of oliverjack38
> Sent: Wednesday, September 05, 2012 3:57 PM
> To: r-help@r-project.org
> Subject: [R] Smooth contour of a matrix
>
> Hi R users
>
> I'm looking for a way to d
Dear useR's,
I was comparing results for a logistic regression model between different
library's.
themodel formula is arranged as follows:
response ~ (intercept) + value + group
OR:
glm( response ~ (intercept) + value + group ,
family=binomial(link='logit'))
lrm( response ~ (intercept
This is the named article:
http://ije.oxfordjournals.org/content/37/5/1161.long
maybe it can help you to help me... :-(
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View this message in context:
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Sent from the R help mailing list archive
Dear R Users and Developers,
I am trying to do the equivalent of
v <- c(1,2,3,3,2,1,)
vu <- unique(v)
for a vector such as
v2 <- c(1.02, 2.03, 1.00, 3.04, 3.06)
vut <- ...
As indicated in the subject, we need approximately unique values with a defined
tolerance, i.e. for the v2 vector the resu
Hi R Experts,
Just wanted to know if there's any package in R that can be used for
multi-class multi-label SVM classification(classes are not mutually
exclusive). I am using the e1071 package for multi class classifications
(here there are N mutually exclusive classes) , not sure if the same can b
Hi,
There are many ways to do this. As well as the php way mentioned by
Franz, you could use cgi. This would probably be considered the
traditional way of doing it. As we are java web developers we are
aiming to use Rserve (http://www.rforge.net/Rserve/) and tomcat (this
can also interact
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Sam Steingold
> Sent: Thursday, September 06, 2012 3:43 PM
> To: David Winsemius
> Cc: r-help@r-project.org
> Subject: Re: [R] merge a list of data frames
>
> > * David Winse
Hi,
I'm a beginner in both statistics and the usage of R. I have been stuck for
days with a problem I hope somebody is willing to help me solving.
I'm trying to create some kind of graphical plot etc. of my results of the
following glm of interactions.
I17<-glm(formula = response ~ var1+ var2 +
Hello,
Say all the below five vectors should have same elements in any situation.
How to find the non matching vectors among these five, if so how we can find
the non matching elements of those vectors?
Can anyone help?
a=c(1,2,3)
b=c(1,2,3,4)
c=c(1,2,3)
d=c(1,2,3)
e=c(1,4,5)
identical(a,b,c,d,e
On Sep 4, 2012 5:45 PM, "Bert Gunter" wrote:
>
> I may be out of line here, but I fail to see what on earth this has to do
> with R and why this thread should be continued here.
Perhaps I'll write an R function that hooks into the console prompt and
reminds users every N minutes to take a break?
Hi everyone
Can anyone help?
x = c("Treatment", "Planned", "Actual", "Demographics", "Outcomes")
for (i in x) {
df <- paste(i)[, c("Reference_ID", "Arm_ID")]
unique(df) -> uniq
as.data.frame(table(uniq$Reference_ID)) -> y
i.Arms<- paste(y$Var1, y$Freq, sep="_")
Print_Message("Check for the Total N
Hello
Within my panel function, I am using draw.key to create a custom key for
each panel. I'm trying to use grobWidth to calculate the correct width
for the viewport, but it appears to be disregarding the fontfamily when
computing the width. Attached is an example; any ideas what I'm doing wr
Hi Folks:
Here's the situation:
> m <- cbind(x=letters[1:3], y = letters[4:6])
> m
x y
[1,] "a" "d"
[2,] "b" "e"
[3,] "c" "f"
## m is a 2 column character matrix
> d <- data.frame(a=1:3,b=4:6)
> d$c <- m
> d
a b c.x c.y
1 1 4 a d
2 2 5 b e
3 3 6 c f
## But please note (as
On Sep 6, 2012, at 2:02 AM, HJ YAN wrote:
> Hi Arun and David
>
>
> Thanks a lot for your reply and sorry for sending the csv file. (p.s. I can
> download my csv file from the email I sent, so I'm not sure why this doesn`t
> work for other users...)
I already explained why. You did not read
On Thu, Sep 06, 2012 at 10:50:00AM -0400, Jie wrote:
> Dear All,
>
> Suppose I have two loops and would like to convert it to sapply, how
> to achive the goal?
> Below is an example, the calculation inside the loop is not essential,
> so please do not use tricks for this part.
> a <- 1:5
> b <- 1:
"not practical [...] to rename the column to something unique."
On the contrary, since R is a scripting language it is quite practical.
Depending on the format of your data, it is probably necessary as well.
If all of the files have exactly the same number of rows corresponding to the
same key
Hello,
Use expand.grid.
One line at a time, to make it clearer:
a <- 1:5
b <- 1:10
m <- expand.grid(b, a)[, c(2, 1)]
resu <- matrix(NA, nrow=5, ncol=10)
for (i in 1:5)
{
for (j in 1:10) # --> was 1:5 in your post
{
resu[i,j]=a[i]+b[j] # will be more complicated
}
}
Dear All,
Suppose I have two loops and would like to convert it to sapply, how
to achive the goal?
Below is an example, the calculation inside the loop is not essential,
so please do not use tricks for this part.
a <- 1:5
b <- 1:10
resu <- matrix(NA,nrow=5,ncol=10)
for (i in 1:5)
{
for (j in 1:
On 09/06/2012 05:00 AM, r-help-requ...@r-project.org wrote:
Hi, R experts
I am currently using lmekin() function in coxme package to fit a
mixed effect model for family based genetic data. How can I extract the p
value from a lmekin object? When I print the object in R console, I can
s
> * David Winsemius [2012-09-05 21:02:16 -0700]:
>
> On Sep 5, 2012, at 8:51 PM, Sam Steingold wrote:
>
>> I have a list of data frames:
>>
>>> str(data)
>> List of 4
>> $ :'data.frame': 700773 obs. of 3 variables:
>> ..$ V1: chr [1:700773] "200130446465779" "200070050127778"
>> "2000306337
As the footer says, "provide commented, _minimal_, self-contained, reproducible
code."
This is hardly "minimal" or reproducible.
Anyway, this probably ought to qualify as a plyr FAQ of some sort: don't use
the summarise function as the FUN argument within a function. Define a named or
anonymous
Dear all,
I am encountering problems with the application of ddply within the body of a
self-defined function.
The script is the following:
moncostcarmoto <- function(costtype){
costaux_result <- data.frame()
for (purp in PURPcount){for (per in PERcount){
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