What is the recommended way of requiring a certain version when loading
a package (or, indeed, from R itself)?
Perl has the
require module version
use module version
require version
use version
constructs which is kind of what I am looking for (especially 'use'
which is evaluated at compile t
Hi
r-help-boun...@r-project.org napsal dne 16.07.2010 05:02:52:
> On Thu, Jul 15, 2010 at 10:45 PM, Murat Tasan wrote:
> > hi all - i'm just wondering what sort of code people write to
> > essentially performa an aggregate call, but with different functions
> > being applied to the various column
Try
RSiteSearch("coefficient of determination")
which seems to list some promising candidates.
Allan
On 15/07/10 16:47, Alireza Ehsani wrote:
Hello there
how can i calculate coefficient of determination with R?
Kind regards
Ali Reza Ehsani
Ph.d. student
[[alternative HTML version
Hi Allan,
When you create an R package you can specify in the DESCRIPTION file
that your package depends on a certain R version and versions of
packages. For example:
Package: automap
Version: 1.0-7
Date: 2010/05/04
Title: Automatic interpolation package
Author: Paul Hiemstra
Maintainer: Pau
On Fri, 2010-07-16 at 14:25 +1200, Peter Alspach wrote:
> Tena koe Stephen
>
> You'll need to use loess(w[,"bkf_depths"]~w[,"measure"])) and predict
> with a sufficiently dense sequence of w[,'measure'] to interpolate to
> an 'accuracy' that meets your requirements. Actually, if I understand
> yo
On Thu, Jul 15, 2010 at 5:51 PM, Farley, Robert wrote:
> I'm trying to plot a series of densities using/comparing differing weights.
> I see the reference to weights and subscripts, but I don't understand how
> to implement that. My data are of the form:
>
> I, J, Actual, Distance, Subset, Weight
Your specific example can just use 'prod'.
The 'prod' function uses the '...' construct
which I think is what you are looking for.
However, in concurrence with a previous answer,
it is almost surely going to be a good idea
to change your way of thinking about the problem.
One word in the R wor
Hi!
I want to create a discrepancy matrix.
I have got a data.frame Q:
number colour date
2 343b1503
3 678 g 1701
4 347b1904
5 345b2001
6 123 g 1809
Now i want to cre
> So distributing code to other people is preferably done using R packages,
> which gives you this option.
However (as far as I am aware), note that this option is checked at
package build time, not at load time.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statist
On 16/07/10 09:32, Paul Hiemstra wrote:
Hi Allan,
When you create an R package you can specify in the DESCRIPTION file
that your package depends on a certain R version and versions of
packages. For example:
[...]
Depends: R (>= 2.7.0), methods, sp (>= 0.9-4), gstat (>= 0.9-58)
Thanks Paul,
On Wed, Jul 14, 2010 at 1:32 AM, Ian Bentley wrote:
> I've got a couple of more changes that I want to make to my plot, and I
> can't figure things out. Thanks for all the help.
>
> I'm using this R script
>
> library(ggplot2)
> library(lattice)
> # Generate 50 data sets of size 100 and assign th
Hi -
I am trying to map a two dimensional area A to another two dimensional area
B using a function. For instance A = {(x,y), 0https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
Unless I am missing something this should do it
a<- cbind(runif(10),runif(10))
b <- cbind(a[,1]+a[,2], a[,1]/a[,2])
On Jul 16, 2010, at 7:00 AM, Ravi Ramaswamy wrote:
Hi -
I am trying to map a two dimensional area A to another two
dimensional area
B using a function. For instance A =
I have the following data structure:
n=5
mydata <- data.frame(id=1:n, x=rnorm(n), y=rnorm(n), id=1:n,
x=rnorm(n), y=rnorm(n))
print(mydata)
producing the following represention
id x y id.1 x.1y.1
1 1 0.5326855 -2.076337031 0.7930274 -1.0530558
2 2 0.78889
On 16-Jul-10 11:27:18, Ralf B wrote:
> I have the following data structure:
>
> n=5
> mydata <- data.frame(id=1:n, x=rnorm(n), y=rnorm(n), id=1:n,
> x=rnorm(n), y=rnorm(n))
> print(mydata)
>
> producing the following represention
>
> id x y id.1 x.1y.1
> 1 1 0.
its maybe not as simple as Teds solution but points to a more general
approach
reshape(mydata,direction="long",varying=names(mydata),v.names=c("id","x","y"))
Am 16.07.2010 13:27, schrieb Ralf B:
> I have the following data structure:
>
> n=5
> mydata <- data.frame(id=1:n, x=rnorm(n), y=rnorm(n),
I think this is simpler but still not all that clean.
===
xx <- structure(list(ID = 1:9, Age = c(10L, 10L, 10L, 11L, 11L, 11L,
10L, 10L, 11L), School = c(1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 2L
), Grade = c(98L, 97L, 92L, 90L, 80L, 70L, 80L, 7
On Thu, Jul 15, 2010 at 6:38 PM, chakri_amateur wrote:
[...]
> I want to extract the largest connected component (alias sub-graph) of the
> network. My input network is a large network of >1000 vertices and >15000
> arcs. From this, I want to take out only the largest cluster.
>
>> If you decompos
On Thu, Jul 15, 2010 at 11:08 PM, Dennis Murphy wrote:
> Hi:
>
> I sincerely hope there's an easier way, but one method to get this is as
> follows,
> with d as the data frame name of your test data:
>
> d <- d[order(with(d, Age, School, rev(Grade))), ]
> d$Count <- do.call(c, mapply(seq, 1, as.ve
> Could anyone advise me how can I implement classification and
> regression tree analyses for a circular response (angles)?
Rpart allows for user-defined splitting rules. You need to define a
function which is given y and an ordered x, and returns the 'goodness of
split' for each split point. Y
Hi
I want to draw a plot from observed and predicted data and also shows threshold
and data before threshold are identified with different color from data after
threshold.
Suppose:
abserved data are 0 or 1
predicted data= 0 to 1
threshold=0.5
Thanks alot
[[alternative
> Can anyone tell me what is the difference between these two standard
> errors and how should I interpret the confidence intervals and std.err
> given these differences?
help(survfit.object) will give you the answer. The std in the object is
for the cumulative hazard, the printout uses a Taylor
Dear list,
I am running R2.11.1 on 32 bit windows. I am receiving messages as
follows ...
> require(car)
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
The package kohonen was another example.
This failure appears to be fatal and not only affec
Hi all,
when running KL on a small data set, everything is fine:
require("flexmix")
n <- 20
a <- rnorm(n)
b <- rnorm(n)
mydata <- cbind(a,b)
KLdiv(mydata)
however, when this dataset increases
require("flexmix")
n <- 1000
a <- rnorm(n)
b <- rnorm(n)
mydata <- cbind(a,b)
KLdiv(mydata)
KL se
Hi List,
Does R have a package that will do filtered back-projection? I'm a medical
physicist trying to produce an image of a dose cloud within a phantom using
software other than Matlab and wondered if R has this capability. I looked
through the imaging pages on the project site but didn't see
On 16/07/2010 9:50 AM, Philip Whittall wrote:
Dear list,
I am running R2.11.1 on 32 bit windows. I am receiving messages as
follows ...
> require(car)
Loading required package: car
Error: package 'car' was built before R 2.10.0: please re-install it
The package kohonen was another example.
Thi
Hi Duncan,
Many thanks for the response. I did indeed re-install car and this is
what occurred ...
> utils:::menuInstallLocal()
package 'car' successfully unpacked and MD5 sums checked
> require(candisc)
Loading required package: candisc
Loading required package: car
Error: package 'car' was buil
On 2010-07-16 7:56, Ralf B wrote:
Hi all,
when running KL on a small data set, everything is fine:
require("flexmix")
n<- 20
a<- rnorm(n)
b<- rnorm(n)
mydata<- cbind(a,b)
KLdiv(mydata)
however, when this dataset increases
require("flexmix")
n<- 1000
a<- rnorm(n)
b<- rnorm(n)
mydata<- cbin
On 16/07/2010 10:41 AM, Philip Whittall wrote:
Hi Duncan,
Many thanks for the response. I did indeed re-install car and this is
what occurred ...
> utils:::menuInstallLocal()
package 'car' successfully unpacked and MD5 sums checked
> require(candisc)
Loading required package: candisc
Loading re
I should have added that this problem is not serious under 2.10 as you
only get a Warning and everything works
> require(candisc)
Loading required package: candisc
Loading required package: car
Loading required package: heplots
Warning message:
package 'car' was built under R version 2.9.2 and
Hello to all!
I am new with R and I need your help.
I'm trying to read a file which contests are similar to this:
ABCTLengths
14.00.001525878918c(1,2,3)
11.00.001716613824c(1,1,4)
So all the columns are numeric values, except Lengths, which is supp
Hi.
I am a beginner. What is the difference between:
x<-2
and
x=2
Thanks.
Linda
__
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-guide.html
and provide co
On 16/07/2010 10:51 AM, Philip Whittall wrote:
I should have added that this problem is not serious under 2.10 as you
only get a Warning and everything works
> require(candisc)
Loading required package: candisc
Loading required package: car
Loading required package: heplots
Warning message:
pa
Hi Duncan,
That fixed it. Many thanks indeed, I now know what to do if it happens
again,
Philip
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: 16 July 2010 16:00
To: Philip Whittall
Cc: r-help@r-project.org
Subject: Re: [R] Packages built before R 2.10
On 16/07/2010 10:57 AM, linda.s wrote:
Hi.
I am a beginner. What is the difference between:
x<-2
and
x=2
It depends on the context. "x <- 2" always assigns the value 2 to the
variable named x. Typed as an isolated command, that's what "x = 2"
does too. However, when used as an argument
hey, guys, all these methods work perfectly. thank you!!
--
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.
Is there a package in R that can run rank-ordered logit?
Thanks,
Suresh
__
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-guide.html
and provide commented,
I made a plot, but after I made a second plot, the previous plot was
gone. How can I save all the plots in a file (I do not manually copy
and paste them one by one)?
Thanks.
Linda
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo
We need to know how you're doing this, with a minimal example that we
can run. Most graphics devices accept a file or filename argument, so
that's one way. If you're using the pdf device, multiple plots will
create multiple pages in the final output..
linda.s wrote:
I made a plot, but after
names(miceTrainSample)
[1] "b_double" "KierA2""KierFlex" "Q_VSA_POS" "pID50"
In the above code, how do I delete "pID50" column to store the resulting
object without indicating column "5". The code below does the trick, but I
wish to delete the column by specifying "-pID50" instead of "5".
Dear R People:
What is the best way to change a named numeric (in which the names are
dates) to a time series, please?
> str(y1)
Named int [1:730] 102 145 147 120 132 125 137 103 128 130 ...
- attr(*, "names")= chr [1:730] "2006-01-01" "2006-01-02"
"2006-01-03" "2006-01-04" ...
>
Thanks in adv
Here is the original question:
What is the best way to change a named numeric to a time series
> str(y1)
Named int [1:730] 102 145 147 120 132 125 137 103 128 130 ...
- attr(*, "names")= chr [1:730] "2006-01-01" "2006-01-02"
"2006-01-03" "2006-01-04" ...
>
y2 <- zoo(y1,order=names(y1))
Perfec
Did you try:
library(MASS)
?polr
?
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 (English)
-
Hi Addi,
On Fri, Jul 16, 2010 at 3:22 PM, Addi Wei wrote:
>
> names(miceTrainSample)
> [1] "b_double" "KierA2" "KierFlex" "Q_VSA_POS" "pID50"
>
> In the above code, how do I delete "pID50" column to store the resulting
> object without indicating column "5". The code below does the trick, b
On Fri, Jul 16, 2010 at 11:26 AM, Erin Hodgess wrote:
> Dear R People:
>
> What is the best way to change a named numeric (in which the names are
> dates) to a time series, please?
>
>> str(y1)
> Named int [1:730] 102 145 147 120 132 125 137 103 128 130 ...
> - attr(*, "names")= chr [1:730] "200
I must have missed something simple, but still, i don't know what.
I obtained my basic data as follows:
x <- sprintf("SELECT m_id,sale_date,YEAR(sale_date) AS
sale_year,WEEK(sale_date) AS sale_week,return_type,0.0001 +
DATEDIFF(return_date,sale_date) AS elapsed_time FROM
`merchants2`.`risk_input`
I guess your data frame is a little different from the reference, so your
as.logical doesn't work.
attach(Q)
FUN <- function(X, Y) {abs(X - Y)}
round(outer(rank(date)[colour=="b"],rank(date)[colour=="g"],FUN) +
outer(rank(number)[colour=="b"],rank(number)[colour=="g"],FUN))
detach(Q)
-
A
Hi,
First: it's kind of hard to play along w/o some reproducible data. To
that end, you can paste into an email the output of:
dput(moreinfo)
If there are lots of rows in `moreinfo`, just give us the first ~10-20
dput(head(moreinfo, 20))
Anyway:
> At this point, each row in m_id_default_res
Anyway, I have often wished that something like
new.mt.sample <- miceTrainSample[, -"pID50"]
would return miceTrainSample without the pID50 column. Here are three
alternative ways to do it.
# Method 1: Assign NULL to the column
new.mt.sample <- miceTrainsSample
new.mt.sample$pID50 <- NULL
# Me
Hello Nikhil and Wu,
Thank you very much for your reply!
What I want is to calculate the student's score column-wise by ignoring the
specific values such as zeros for example only using c(2,1) in column 8 in the
tmp1 and generate tmp2. I changed the zeros to NAs and modified my stud fun to
stude
Here is a way of creating a separate list of variable length vectors
that you can use in your processing:
> # read into a dataframe
> x <- read.table(textConnection("ABCTLengths
+ 14.00.001525878918c(1,2,3)
+ 14.00.001525878918c(1,2,6,7,8,3)
+ 14
subset(miceTrainSample, select = -plD50)
On Fri, Jul 16, 2010 at 11:22 AM, Addi Wei wrote:
>
> names(miceTrainSample)
> [1] "b_double" "KierA2" "KierFlex" "Q_VSA_POS" "pID50"
>
> In the above code, how do I delete "pID50" column to store the resulting
> object without indicating column "5".
Hi Steve,
Thanks
Here is a tiny subset of the data:
> dput(head(moreinfo, 40))
structure(list(m_id = c(171, 206, 206, 206, 206, 206, 206, 218,
224, 224, 227, 229, 229, 229, 229, 229, 229, 229, 229, 233, 233,
238, 238, 251, 251, 251, 251, 251, 251, 251, 251, 251, 251, 251,
251, 251, 251, 251, 251,
I have several data sets, which begin early in 2002 and run until
yesterday. They do not have observations every day. For example:
> xd1[1:10]
2002-02-25 2002-02-26 2002-02-28 2002-03-01 2002-03-04 2002-03-05 2002-03-07
7 8 1 9 12 3 5
Hello,
I am a new R user having transitioned over from S-plus recently. I have a
question that is probably very trivial but I am having trouble finding a
solution. In S-plus, graphic pages are created as tabs when multi-page
figures are created. I have shown the R code for xpose.VPC (a function
wi
Hello,
This does what you want. The simple solution is shorter but requires
that there is only one value you wish to exclude (e.g., 0). The
second works for any number of values you wish to exclude, but is
subsequently longer. Also there is no need to create your own
function to 'studentize', ?
My understanding is that polr will do ordered logit but I am not sure if it
is also suited for rank ordered logit (or is there no such distinction)
I am thinking of following two situations
1. there is an ordered response (say small,medium,large coffee) and each
individual selects one of these op
Hi szhan,
I think Joshua gives all you wants -- scale is a really good function.
You can also make your own function work by setting an argument na.rm.
tmp1[tmp1==0]<-NA
student<- function(x){
x<-(x-mean(x,na.rm=T))/sd(x,na.rm=T)
return (x)
}
tmp4<-apply(tmp1, 2, student)
-
On Fri, Jul 16, 2010 at 12:38 PM, Erin Hodgess wrote:
> I have several data sets, which begin early in 2002 and run until
> yesterday. They do not have observations every day. For example:
>> xd1[1:10]
> 2002-02-25 2002-02-26 2002-02-28 2002-03-01 2002-03-04 2002-03-05 2002-03-07
> 7
On 7/16/2010 6:00 AM, r-help-requ...@r-project.org wrote:
Message: 5
Date: Thu, 15 Jul 2010 03:36:21 -0700 (PDT)
From: Bart Joosen
To:r-help@r-project.org
Subject: [R] Access web content from within R
Message-ID:<1279190181074-2289953.p...@n4.nabble.com>
Content-Type: text/plain; charset=us-ascii
I am sure this is a very basic question:
I have 600,000 categorical variables in a data.frame - each of which is
classified as "0", "1", or "2"
What I would like to do is collapse "1" and "2" and leave "0" by itself,
such that after re-categorizing "0" = "0"; "1" = "1" and "2" = "1" --- in
the en
iterations <- 100
nvars <- 4
combined <- rbind(scaleMiceTrain, scaleMiceTest)
reducedSample <- combined
reducedSample <- subset(reducedSample, select = -pID50)
reducedSample <- subset(reducedSample, select = -id)
for (i in 1:iterations)
{
miceSample <- sample(combined[,-c(1,2)],nvars, repl
This is not reproducible, and does not look minimal. You'll get better
answers, and probably solve many issues on your own, if you construct
small examples that illustrate the same problem you're having with your
real data.
Addi Wei wrote:
iterations <- 100
nvars <- 4
combined <- rbind(scale
Sorry about that. Still new to this... The code below should be
reproducible.All R2 should just be 1, and I should write 1 to
R2outputKKNN.txt 10 timesnothing is happening. Appreciate the efforts
to help!
for (i in 1:10)
{
adata = 1:5
bdata = 6:10
lm <- lm(adata
Do you want to replace specific values of a data set?
df <- sample(c(0,1,2),600,replace=T)
table(df)
df[df==2]<-1
table(df)
-
A R learner.
--
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Sent from t
Hi Albyn,
Thank you very much for the suggestion.
I managed to install Sage on Windows (via a Linux VM), but I cannot find
any documentation on how to use R from Sage. Maybe I should use the web
interface of Sage to avoid having to install R on the VM.
Best regards,
David
On 10-07-14 10:2
I have a text processing problem I'm hoping someone can help me solve. This
issue it this.
I have a character string in which I need to delete a variable number of
characters from the string. The string itself contains the number of
characters to be deleted. The number of characters to be d
Hi:
Hadley's solution is certainly preferred here due to its relative
simplicity. I just wanted to correct an error from my earlier post.
On Thu, Jul 15, 2010 at 2:08 PM, Dennis Murphy wrote:
> Hi:
>
> I sincerely hope there's an easier way, but one method to get this is as
> follows,
> with d
Hi:
Here's a simple example (divide through by 10 for your case):
x <- 1:10
y <- rnorm(10, x, s = 0.3)
clr <- ifelse(x <= 5, 'red', 'blue')
plot(x, y, col = clr)
HTH,
Dennis
On Fri, Jul 16, 2010 at 6:08 AM, azam jaafari wrote:
> Hi
>
> I want to draw a plot from observed and predicted data an
Hi Davis,
Please try ??regex
gsub("(\\-|\\+)([0-9]+)(\\w*)(\\w)", replacement="\\4", x)
-
A R learner.
--
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Sent from the R help mailing list archive at
Your code between calls to sink() does not generate any output. Hence,
nothing will be diverted to the file. To illustrate this point,
consider
for(i in 1:10) i
This produces no output. However,
for(i in 1:10) print(i)
produces output as expected.
-Matt
On Fri, 2010-07-16 at 13:34 -0400, Ad
Marsh,
you are absolutely right, I forget the link to the db.
Actually its a webpage for each monograph number:
http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/198
http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/731
...
As I have a list with the numbers of interest (l<- c(198, 731,
Hi:
See ? levels. Here's a toy example:
> x <- factor(sample(0:2, 10, replace = TRUE))
> x
[1] 1 2 1 0 2 2 2 2 2 1
Levels: 0 1 2
> levels(x) <- c(0, 1, 1)# Change level 2 to 1
> x
[1] 1 1 1 0 1 1 1 1 1 1
Levels: 0 1
HTH,
Dennis
On Fri, Jul 16, 2010 at 10:18 AM, CC wrote:
> I am sure t
I would have thought this to be relatively elementary, but I can't find it
mentioned in any of my stats texts.
Please consider the following:
library(fitdistrplus)
fp = fitdist(y,"exp");
rate = fp$estimate;
sd = fp$sd
fOneWeek = exp(-rate*7); #fraction that happens within a week - y is
m
Hi,
On Fri, Jul 16, 2010 at 5:18 PM, CC wrote:
> I am sure this is a very basic question:
>
> I have 600,000 categorical variables in a data.frame - each of which is
> classified as "0", "1", or "2"
>
> What I would like to do is collapse "1" and "2" and leave "0" by itself,
> such that after re-c
On Wed, Jul 14, 2010 at 10:12 AM, Neil123 wrote:
>
> Hi,
>
> I have the following dataset and I would like to create a dotplot with
> confidence limits:
>
> CAT1 CAT2 MEAN Lower
> Upper
> 1 1 1 0.619
I've seen a few threads about this, but none that seem to answer my problem
I have a list of .txt files in a directory that I am reading into R and row
binding together. I am using the following code to do so:
# Directory where files are found
my.txt.file.directory <- "C:/Jared/Data/Kenya/Wildeb
I think I have identified the problem such that when I identify the
structure of some of the files that I am reading in, columns are labeled as
"Factors". In other files, the same columns are labeled as numeric
values. Is there a way to assign the data structure to these columns in the
dataf
Dear R-list members,
I´m using the package "survey" and I need to find a function for
multinomial logistic regression in a complex design. The functions that
I see are only for dicotomic and ordinal variables.
Thank you!
Rosario Austral
De: "r-help-requ...@r-
Hi Jared:
Is it possible you're using read.table with variable names in line 1 but not
using header = TRUE as an argument?
HTH,
Dennis
On Fri, Jul 16, 2010 at 12:41 PM, Erik Iverson wrote:
>
>
>
>
>> I think I have identified the problem such that when I identify the
>> structure of some of t
Dear David,
> I managed to install Sage on Windows (via a Linux VM), but I cannot find any
> documentation on how to use R from Sage. Maybe I
> should use the web interface of Sage to avoid having to install R on the VM.
In the Sage command line, you can type
sage: r?
to get some very basic in
On Mon, Jul 12, 2010 at 8:16 AM, Jun Shen wrote:
> Dear Deepayan,
>
> Thank you for taking the time to look into this issue.
>
> I have a data object called "Data", please find it at the end of the
> message. Then I can run the code below separately in the console.
[...]
> ##
On Fri, Jul 16, 2010 at 1:59 PM, Davis, Brian wrote:
> I have a text processing problem I'm hoping someone can help me solve. This
> issue it this.
>
> I have a character string in which I need to delete a variable number of
> characters from the string. The string itself contains the number
Hello Joshua and Wu,
Thank you for your excellent solutions.
Joshua
- Original Message -
From: "Joshua Wiley"
To: "Shuhua Zhan"
Cc: r-help@r-project.org
Sent: Friday, July 16, 2010 12:39:22 PM GMT -05:00 US/Canada Eastern
Subject: Re: [R] how to skip a specific value when using apply() f
Try this:
library(XML)
readHTMLTable("http://extranet.pheur.org/4DLink1/4DCGI/Web_View/mono/198";,
which = 5)
On Fri, Jul 16, 2010 at 4:08 PM, Bart Joosen wrote:
>
> Marsh,
>
> you are absolutely right, I forget the link to the db.
> Actually its a webpage for each monograph number:
>
> http://
Hi - I am something of a newbie and am a little perplexed. When (trying to)
modify a table I issue the following commands with subsequent errors
sqldf("alter table Korea drop column code", dbname = "mydb")
error in statement: near "drop": syntax error
or
sqldf("alter table Korea rename column
Hello Everyone,
Â
This is my first attempt to do something in R. As a precursor to a Willingness
to Pay analysis, I want to conduct a Mixed Conditional Logit analysis but am
unsure how to proceed because of some nesting within my data.
Â
Below is some data and code that illustrate what Iâm t
Following is a function that I wrote (It is working well). It's a simple one,
nothing complicated. The only question that I have is a qutation mark issue,
I guess.
#
funcname <- function(trait.file){#line1
setwd('/root/sub
Hello,
I am trying to find a direct way to write a nested if of sorts to find data
for a specific time range for a specific day (across a range of days) and
have exhausted my abilities with the manuals I have at hand. I have a good
deal of data of this approximate form:
day time price
1
I am feeling rather dumb right now.
I created what I thought was a data.frame as follows:
aaa <- lapply(split(moreinfo,list(moreinfo$m_id),drop = TRUE), fun_m_id)
m_id_default_res <- do.call(rbind, aaa)
print("==")
m_id_default_res
print("==
Hello,
I'm trying to do some differential equation modeling in R using the
package 'deSolve.' Briefly, I'm trying to use the law of mass action
(the details of which aren't really important) to structure a vector
of rate equation which will be passed into an ODE function and solved
with associate
Did you ever find a solution for this?
--
View this message in context:
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On Fri, Jul 16, 2010 at 2:46 PM, PeterTucker wrote:
>
> Hi - I am something of a newbie and am a little perplexed. When (trying to)
> modify a table I issue the following commands with subsequent errors
>
> sqldf("alter table Korea drop column code", dbname = "mydb")
> error in statement: near "d
Hi:
I'm guessing that your time variable is not defined with a class that
respects (time) ordering, so your first if statement is meaningless. In the
absence of further information, either create a meaningful time/date
variable or at the very least, an ordered factor.
Given what you've provided,
Hi Karena,
On Fri, Jul 16, 2010 at 11:23 AM, karena wrote:
> So, as we all know, when using "read.csv", we need to use qutation mark out
> side the filename which we wanna read in. At first, for line 8, I wrote:
> read.csv("trait.file"), at last line, I wrote:
> funcname(folder/hyper.csv), but
You need to understand the difference between a variable and a value.
In your case that doesn't work, you are supplying a value, "trait.value".
but you have no file named "trait.value".
In the case that does work you are supplying a variable, trait.value, and
the value contained in that variable
On 2010-07-16 13:38, Jared Stabach wrote:
I've seen a few threads about this, but none that seem to answer my problem
I have a list of .txt files in a directory that I am reading into R and row
binding together. I am using the following code to do so:
# Directory where files are found
my.txt.f
Yes, this is problem. Thanks for the suggestion. I had deleted the N/A
values later in the script but should have taken care of them in this first
step.
Thanks very much
Jared
On Fri, Jul 16, 2010 at 4:11 PM, Peter Ehlers wrote:
> On 2010-07-16 13:38, Jared Stabach wrote:
>
>> I've seen a f
Hello
I have a table of this kind:
functionx1 x2 x3
2.232 1 1 1.00
2.242 1 1 1.01
2.732 1 1 1.02
2.770 1 2 1.00
1.932 1 2 1.01
2.132 1 2 1.02
3.2
Hello,
Is there an way to easy comment of sections of code? I was thinking
something along the lines of
\dontrun{
codeline 1
codeline k
}
but that could be used in regular script files. When I am still
working on a script, I often want to being using what is done, but I
would like the par
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