On Thu, Jul 21, 2011 at 3:20 PM, Madana_Babu wrote:
> Hi all,
>
> Currently i am trying to this on R which is running on multicore processor.
> I am not sure how to use mclapply() function on this task. Can anyone help
> me.
>
>
> # Setting up directory
> setwd("/XXX////2011/07/20"
Dear R-users,
 I have about 32 variables and am attempting to analysis any interactions Â
using the package deal. I am getting the error
"[Heuristic [Auto Search Error in phi[[y]] : attempt to select less than one
element"
Could you please assist in solving this, my code is as follows:
k
Hi:
This worked for me:
t <- 0:100
DBH <- 10:100
TxDBH<-expand.grid(t, DBH)
rm(t, dbh)
time<-TxDBH[,1]
dbh<-TxDBH[,2]
M <- 4000*(1-exp(-time*(1.104-(0.67*0.7)-0.163*log(dbh))^2))
df <- data.frame(time, dbh, M)
library('lattice')
wireframe(M ~ time + dbh, data = df)
# Check the versions of R and
Hi:
Here's one way using uniform(0, 1) pseudo-random numbers, but there
are many ways you could go about this.
# each row comprises a set of three pseudo-random numbers
u <- matrix(runif(30), nrow = 10)
# divide each element in a row by its row sum
v <- t(apply(u, 1, function(x) x/sum(x)))
rowSum
To Jason: Still thank you, your reply gives me some clue about how to do
this.
To Duncan: Thank you for rely. It helped.
Karena
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Could this possibly be a homework question?
cheers,
Rolf Turner
On 22/07/11 11:18, karena wrote:
Hi,
I want to generate multiple sets of random numbers.
The requirement is that:
1) each set have 3 random numbers;
2) the sum of the three number is always 1.
how to do this?
thanks Rainer,
it worked!
> From: rainer.schuerm...@gmx.net
> To: r-help@r-project.org; fin_e...@hotmail.com
> Subject: Re: [R] for/if loop in R
> Date: Fri, 22 Jul 2011 03:23:19 +0200
>
> For me, this works:
>
> > Now, I want to add a 4th column, trend
> pricedata$trend <- 0
>
> > which can
Hi,
It is slightly clunky but:
mod2 <- glm(Y ~ offset(I(1.3 * X1)) + X2, family = "binomial")
should do the trick. See ?offset for details.
Cheers,
Josh
On Thu, Jul 21, 2011 at 7:42 PM, Johan Jackson
wrote:
> Hello all,
>
> I am using a glm() and would like to fix one of the regression coef
Hello all,
I am using a glm() and would like to fix one of the regression coefficients
to be a particular value and see what happens to the fit of the model. E.g.:
mod1 <- glm(Y ~ X1 + X2,family='binomial')
mod2 <- glm(Y~[fixed to 1.3]X1 + X2,family='binomial')
The beta for X1 is freely estimate
Hi,
I'm trying to plot an equation in two variables to get a feel
for sensitivity to its parameters. I've run expand.grid to get made-up
vectors of the combinations of the two independent variables, and am trying
to plot the output of the dependent, M, against both the dependent in a 3d
space.
t
Clue: make a column representing the desired classification, and then use any
of the methods you considered to do the summarizing.
One possible approach:
myframe$wend <- as.factor(ifelse(weekday( myframe$dates) %in%
c("Saturday","Sunday"), "Weekend", "Weekday"))
(Sorry for reposting. Please delete previous msgs. Thanks!)
Hi, all
Here I created a data frame like
mydates<- seq(as.Date("2010-05-29"), length = 43, by = "day")
myvalues<-runif(43,0,1)
myframe<-data.frame(dates=mydates, day=weekdays(dates), value=myvalues)
dates day valu
Sorry for reposting. The previous message not showing up.
Hi, all
Here I created a data frame like
mydates<- seq(as.Date("2010-05-29"), length = 43, by = "day")
myvalues<-runif(43,0,1)
myframe<-data.frame(dates=mydates, day=weekdays(dates), value=myvalues)
dates day value
1
Hi, all
Here I created a data frame like
mydates<- seq(as.Date("2010-05-29"), length = 43, by = "day")
myvalues<-runif(43,0,1)
myframe<-data.frame(dates=mydates, day=weekdays(dates), value=myvalues)
dates day value
1 2010-05-29 Saturday 0.14576143
2 2010-05-30Sunday 0
For me, this works:
> Now, I want to add a 4th column, trend
pricedata$trend <- 0
> which can have 2 values 0 or 1. if return>1%, trend=1 else trend=0.
pricedata$trend <- ifelse( pricedata$return > .01, 1, 0 )
Rgds,
Rainer
On Thursday 21 July 2011 19:39:15 financial engineer wrote:
>
> hi,
>
Hi, all
Here I created a data frame like
mydates<- seq(as.Date("2010-05-29"), length = 43, by = "day")
myvalues<-runif(43,0,1)
myframe<-data.frame(dates=mydates, day=weekdays(dates), value=myvalues)
dates day value
1 2010-05-29 Saturday 0.14576143
2 2010-05-30Sunday 0
On 21/07/2011 7:18 PM, karena wrote:
Hi,
I want to generate multiple sets of random numbers.
The requirement is that:
1) each set have 3 random numbers;
2) the sum of the three number is always 1.
how to do this?
There are lots of ways. What is the context? It would determine the
first two
Hi,
I've done some searching where others have had trouble with this error (or
"warning" actually), but I'm unable to solve my problem. I have a data
sheet with 13 columns and 36 rows. Each column has exactly the same number
of rows. I've created glms and now want to do cross-validation on 2 of
Works perfectly, thank you.
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https://s
Hi David and Tom
I was having the same problem and was reading through your threads. I
finally tried:
summary(fit, rmean=TRUE)$table[5]
This produced the mean.
Hope this works for you as well.
Casey
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hi,
Can someone please help me figure out where I am making a mistake in my for/if
loop:
I have a data frame (112 rows) called pricedata with 3 columns: date, prices,
return.
Now, I want to add a 4th column, trend, which can have 2 values 0 or 1. if
return>1%, trend=1 else trend=0.
so, this
Hi, I can't for the life of me find how to do this in base R, but I'd be
surprised if it's not possible.
I'm just trying to replace multiple columns at once in a data frame.
#load example data
data(api)
#this displays the three columns and eight rows i'd like to replace
apiclus1[ apiclus1$meals
On 07/21/2011 08:04 PM, Jason Edgecombe wrote:
On 07/21/2011 07:18 PM, karena wrote:
Hi,
I want to generate multiple sets of random numbers.
The requirement is that:
1) each set have 3 random numbers;
2) the sum of the three number is always 1.
how to do this?
Try this:
df<-data.frame(a
On 07/21/2011 07:18 PM, karena wrote:
Hi,
I want to generate multiple sets of random numbers.
The requirement is that:
1) each set have 3 random numbers;
2) the sum of the three number is always 1.
how to do this?
Try this:
df<-data.frame(a=runif(10),b=runif(10))
df$c=df$a+df$b
d
Hi, all
Here is what I tried to do.
> strptime('20110101 0900',"%Y%m%d %H%M")
[1] "2011-01-01 09:00:00"
> strptime('20110101 900',"%Y%m%d %H%M")
[1] NA
If I have a 3-digit hour like '900', please show me how to convert it to
09:00:00.
Thank you in advance.
Wonjae
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Hello all,
I'm trying to find all combinations of 4 numbers that satisfy 4 criteria,
inside of a matrix (62 x 25). I've found a way to do this using for loops,
but it is extremely slow because it involves checking every possible
combination of numbers (567300) to see if the criteria are satisfied
Steve: I would guess that the problem relates to the large number of tied
values of -1 in your dependent y variable. You could randomly jitter
these y = -1 by adding a random uniform number between, say, [ -0.01,
0.01] and see if the rq() converges to a solution. Then you would know
that w
Hi,
I want to generate multiple sets of random numbers.
The requirement is that:
1) each set have 3 random numbers;
2) the sum of the three number is always 1.
how to do this?
thank you,
karena
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Dear list,
I'd like to use the function smooth.map{maps} to smooth out aggregated
spatial data. The first argument to this function must be a map object.
I have a shape file which I read with readShapeSpatial{maptools} and I got
an object of class "SpatialPolygonsDataFrame".
Is it possible to cr
Brian
Thanks for the quick response! You have found a solution. I tried this
and it worked well. For my current purpose it is perfectly fine to use
jitter and the result is just as useful. For the one problem subset (of
160 subsets) the actual regression won't be super useful, but at least
it w
On 11-07-21 5:17 PM, pilchat wrote:
thank you guys for your reply.
i was sure that it was related to changes in the generation of ps
files with the latest R release.
now the question is: how can i restore the old behavior in R2.13?
Peter told you.
Duncan Murdoch
thanks
gaetano
On 7/21/1
Dear all,
I need your help as I was not able to find out the solution.
The thing is-
I am having a code which is reading file with this code-
df=read.table("Case2.pileup",fill=T,sep="\t",colClasses="character")
but as am making a tool so that user can use it and can do analysis on his
file.But
Hi
I am using the quantreg package for median regression for a large series
of subsets of data. It works fabulously for all but one subset. When it
reaches this subset, R takes the command and never responds. I end up
having to kill R and restart it.
It appears to be something with the parti
Hi all,
Currently i am trying this on R which is running on multicore processor. I
am not sure how to use mclapply() function on this task. Can anyone help me.
# Setting up directory
setwd("/XXX////2011/07/20")
library(sqldf)
# Data is available in the form of multiple stru
Hi:
I *think* this is what you're after, but I get dizzy trying to read
convoluted loops. Try this instead;
# Function to find the total SBA2 for all dbh2 larger than the given one:
foo <- function(d) {
d <- d[order(d$dbh2), ]
d <- transform(d, BAL2 = sum(SBA2) - cumsum(SBA2))
d
}
thank you guys for your reply.
i was sure that it was related to changes in the generation of ps
files with the latest R release.
now the question is: how can i restore the old behavior in R2.13?
thanks
gaetano
On 7/21/11, Ted Harding wrote:
> Yes, Peter, your suggestion does the trick (at an
Context added back in:
SamiC wrote:
Hi,
So I am trying to plot my results of a model. what i have is the
majority
of the data between the values of 0 and 30, then one outlier at 80
and
another at 130. the model plots a nice line through the data
between 0 to
30, however given the outl
On Jul 21, 2011, at 3:38 PM, zlu wrote:
Hi Peter,
I'm not sure how many people still have 9 month old postings on their
mail client and will know that Peter Dalgaard is the intended recipient.
Do you have any idea or codes of construct a score test based
confidence
interval for coeffic
Hi, couldn't find much in the archives. I had checked before posting.
Anyway the plotrix package was a good hint. So anyone who reads this in
future, i ended up using the gap.plot function in the plotrix package, which
does the trick. Thanks
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Hi Peter,
Do you have any idea or codes of construct a score test based confidence
interval for coefficients in logistic regression?
James
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Hi Sean,
Here is one option that I believe does what you want. The logic is simple,
1) instatiate a test value that does not meet your criteria
2) use a while loop that will keep looping until your criteria is met
(be sure you set a meetable criteria or this will continue ad
nauseum!!!),
3) rand
On 2011-07-21 08:56, marcel wrote:
I am using lattice for a bar plot, having a little trouble removing ticks,
tick labels from x-axis, but keeping them on the y-axis. I looked around
quite a bit (http://tolstoy.newcastle.edu.au/R/e7/help/09/06/1733.html, help
pages, etc), tried variations of "sca
Hi all,
I have a dataframe of behavioral observations from 360 fish, each with 241
observation points(rows), which looks like this:
> head(d)
fishtreatment tank trial video tid pid ang.chgabs.ac
t lenvel d2p x y
1 1
Yes, Peter, your suggestion does the trick (at any rate with
Gaetano's files).
I edited his volc2.13.eps (the slow one) as follows (the original
commented out with "%%##") making just the following change:
%%## /setrgb { srgb setcolor } def
/setrgb { setrgbcolor } def
%%## End of editing
(at lin
On 2011-07-21 09:59, cassie jones wrote:
Hi all,
I have a matrix which looks like the following,
x=matrix(c(1,2,3,4,3,5,4,6,4,9),5,2,byrow=T)
Now, I want to replace the 2nd columns of those 4 rows which have duplicate
entries in the 1st column. For example, x[2,1]=x[3,1]=3 and I want to
replac
Hi all,
Currently i am trying to this on R which is running on multicore processor.
I am not sure how to use mclapply() function on this task. Can anyone help
me.
# Setting up directory
setwd("/XXX////2011/07/20")
library(sqldf)
# Data is available in the form of multiple struct
This is due to the introduction of sRGB. Since this actually does something
(Google for sRGB and you will be approximately as wise as me...), I don't think
it is likely to be taken out. You can, however, always edit .ps.prolog. (I
would expect that the line
/setrgb { setrgbcolor } def
instead
Replace the soon-to-be Design with rms.
Specify surv=TRUE to cph so that approximate rather than fully correct
standard errors will be computed by survplot/survest.
Frank
christopher.a.hane wrote:
>
> Hi,
>
> I have a Cox PH model that's large for my server, 120K rows, ~300 factors
> with 3 le
Hi,
I have a Cox PH model that's large for my server, 120K rows, ~300 factors
with 3 levels each, so about 1000 columns. The 300 factors all pass a
preliminary test of association with the outcome. Solving this with cph from
Design takes about 3 hours. I have created the fit with x=T, y=T to save
On Jul 21, 2011, at 1:42 PM, Benjamin Caldwell wrote:
That is, run all possible combinations of the two vectors through the
equation.
*Ben *
For "all combinations" the usual route is data preparation with either
expand.grid() or outer()
On Thu, Jul 21, 2011 at 10:04 AM, Benjamin Caldwel
[...]
I do not understand why you expect to get comparable results with SAS
"discrete" and coxph "exact". They are two different approaches to
handling ties (as Terry explained; of course, some comparability
should be expected in "normal" cases).
[...]
I didn't know precisely the specifities of
On Thu, Jul 21, 2011 at 11:29 AM, Dimitri Liakhovitski
wrote:
> Hello!
> It's a shoot in the dark, but I'll try. If one has a total of 100
> (e.g., %), and three components of the total, e.g.,
> mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with
> 3 sectors representing x, y, a
That is, run all possible combinations of the two vectors through the
equation.
*Ben *
On Thu, Jul 21, 2011 at 10:04 AM, Benjamin Caldwell wrote:
> Hi,
> I'm trying to run a basic simulation and sensitivity test by running an
> equation with two variables and then plotting the results against
Achim, this is great! Thanks a lot!
Dimitri
On Thu, Jul 21, 2011 at 1:09 PM, Achim Zeileis wrote:
> On Thu, 21 Jul 2011, Dimitri Liakhovitski wrote:
>
>> Hello!
>>
>> I am trying to build a mosaic plot that has different colors for each
>> entry (cell). My data that goes into the plot is NOT real
On Thu, 21 Jul 2011, Dimitri Liakhovitski wrote:
Hello!
I am trying to build a mosaic plot that has different colors for each
entry (cell). My data that goes into the plot is NOT really a
contingency table and I would like each shape to have its own
color.Looks like mosaic and mosaicplot treat
On Jul 21, 2011, at 9:36 AM, Daniel Ponce wrote:
> Dear Sr. of R-project:
>
> In the document "R: Regulatory Compliance and Validation Issues. A
> guidance Document for the Use of R in Regulated Clinical Trial
> Environments", in the point 6.3 say that exist a set of validation
> test, that are
Hi,
I'm trying to run a basic simulation and sensitivity test by running an
equation with two variables and then plotting the results against each of
the vectors. R is running the vectors like this : 0 with 0, 1 with 1, etc. I
would like it to run them like 0 for 1:100, 1 for 1:100, and then the
re
On 11-07-21 10:36 AM, Daniel Ponce wrote:
Dear Sr. of R-project:
In the document "R: Regulatory Compliance and Validation Issues. A
guidance Document for the Use of R in Regulated Clinical Trial
Environments", in the point 6.3 say that exist a set of validation
test, that are localized in the te
On Thu, Jul 21, 2011 at 12:33 PM, Dimitri Liakhovitski
wrote:
> Thanks a lot, Sarah.
> I am definitely going to explore.
> A quick question about
> mosaicplot(matrix(c(50, 30, 20, 0), 2, 2))
> mosaicplot(matrix(c(0, 50, 20, 30), 2, 2))
>
> What are the numbers that appear on the graphs (2.1, 2.2,1
Hi all,
I have a matrix which looks like the following,
x=matrix(c(1,2,3,4,3,5,4,6,4,9),5,2,byrow=T)
Now, I want to replace the 2nd columns of those 4 rows which have duplicate
entries in the 1st column. For example, x[2,1]=x[3,1]=3 and I want to
replace x[3,2] by x[2,2]. Similarly, I want to re
Hello, everyone.
I am trying to show the correlation matrix using
color2D.matplot(). And I want to map (0,1) to (dark red to dark green).
But by default color2D.matplot() will map(min element of given matrix,
max element of given matrix) to (dark red to dark green).
Can anyone help me with thi
Hello!
I am trying to build a mosaic plot that has different colors for each
entry (cell). My data that goes into the plot is NOT really a
contingency table and I would like each shape to have its own
color.Looks like mosaic and mosaicplot treat the data as a contingency
table and seem to allow di
Hello all-
I am doing a survival analysis for two species of invasive plants I
outplanted to edges and interiors of island and mainland sites in a local
reservoir. I am using the KM estimate and had no problem doing survdiff for
my data using the following code:
S4<-Surv(outplant$SurvTime, outpl
I am using lattice for a bar plot, having a little trouble removing ticks,
tick labels from x-axis, but keeping them on the y-axis. I looked around
quite a bit (http://tolstoy.newcastle.edu.au/R/e7/help/09/06/1733.html, help
pages, etc), tried variations of "scales = list(alternating = c(0,0)",
"sc
Hi everyone,
I have been working some days in a nested loop in R but I can't find the
solution.
I have a data.frame with an unique ID for individuals and unique ID for
different stands, for each indiviadual I have a dbh record and a SBA (stand
basal area) field.
Pma<-rep (1:40)
P<-seq(1,4,
Dear Sr. of R-project:
In the document "R: Regulatory Compliance and Validation Issues. A
guidance Document for the Use of R in Regulated Clinical Trial
Environments", in the point 6.3 say that exist a set of validation
test, that are localized in the test subdirectory. I search the place
in the p
Thanks a lot, Sarah.
I am definitely going to explore.
A quick question about
mosaicplot(matrix(c(50, 30, 20, 0), 2, 2))
mosaicplot(matrix(c(0, 50, 20, 30), 2, 2))
What are the numbers that appear on the graphs (2.1, 2.2,1.1,1.2)?
Dimitri
On Thu, Jul 21, 2011 at 12:20 PM, Sarah Goslee wrote:
>
On Thu, Jul 21, 2011 at 12:06 PM, Dimitri Liakhovitski
wrote:
> I tried the mosaic chart:
>
> mytotal=data.frame(x=50,y=30,z=20)
> require(stats)
> mosaicplot(mytotal1)
>
> It's good, but it only creates rectangles that are stacked on top of
> each other, which is not exactly what I was looking fo
Duncan,
Makes perfect sense, thank you very much.
Gene
On Thu, Jul 21, 2011 at 10:46 AM, Duncan Murdoch
wrote:
> On 21/07/2011 11:37 AM, Gene Leynes wrote:
>
>> It occurred to me that it would be nice to be able to save a library that
>> I
>> have installed an loaded into my workspace for easy
I tried the mosaic chart:
mytotal=data.frame(x=50,y=30,z=20)
require(stats)
mosaicplot(mytotal1)
It's good, but it only creates rectangles that are stacked on top of
each other, which is not exactly what I was looking for.
Is there a R package for waffle chart?
I tried help.search("waffle") and
On 21/07/2011 11:37 AM, Gene Leynes wrote:
It occurred to me that it would be nice to be able to save a library that I
have installed an loaded into my workspace for easy sharing. I suppose it
might not work if someone's on a different version of R, but usually since
the library is just a collec
It occurred to me that it would be nice to be able to save a library that I
have installed an loaded into my workspace for easy sharing. I suppose it
might not work if someone's on a different version of R, but usually since
the library is just a collection of functions, it seems plausible that yo
This is called a squarified pie chart or a waffle chart (if you want
to keep the food metaphor going):
http://eagereyes.org/communication/Engaging-readers-with-square-pie-waffle-charts.html
Hadley
On Thu, Jul 21, 2011 at 10:29 AM, Dimitri Liakhovitski
wrote:
> Hello!
> It's a shoot in the dark,
It's more complex than what you describe, but what about a
mosaic plot?
http://conprogram.weebly.com/program-schedule.html
They're very useful, and much better than pie charts because they don't
rely on the visual estimation of angles, something people aren't very good
at.
Sarah
On Thu, Jul 21,
Hello!
It's a shoot in the dark, but I'll try. If one has a total of 100
(e.g., %), and three components of the total, e.g.,
mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with
3 sectors representing x, y, and z according to their proportions in
the total.
I am wondering if it's
On 21-Jul-11 13:24:32, Duncan Murdoch wrote:
> On 11-07-21 3:23 AM, pilchat wrote:
>> Dear R users,
>>
>> I have a desktop computer and a laptop, both of them
>> with Ubuntu Lucid. The former has R2.10 installed from
>> Ubuntu repositories (this is the most recent version
>> in the repositories), w
Hello all useRs,
I am tring make a simple surface plot ( 2 by 2 terms of a MARS model
(with earth package)
but I get the follow error message:
> plotmo( mars )
Error: bad index (missing column in x?)
I don't no how to workround this... :-(
I thanks in advanced by some help!
Thanks.
Cleber
##
In order to apply the bootstrap, you must resample, uniformly at random
from the independent units of measurement in your data. Assuming that
these represent the rows of 'data', consider the following:
est <- function(y, x, obeta = c(1,1), verbose=FALSE) {
n <- length(x)
X <- cbind(rep(1,
On Jul 19, 2011, at 7:19 PM, J. wrote:
@Dimitri: I tried to enter it as numeric and still got the same
outcome. I
still wonder if there is any way to get the same result from both
programs.
@David, Bert: Yes, I found that the gender coefficient is R is
exactly twice
that of the one from S
On 2011-07-20 23:50, Jim Silverton wrote:
Hello everyone,
Peter (see my earlier post) recommended the following script for finding the
means and standard deviations and putting them in table form.
However, I would like the standard deviations under the means in brackets.
Can anyone check this co
On Jul 21, 2011, at 9:42 AM, SamiC wrote:
Hi,
So I am trying to plot my results of a model. what i have is the
majority
of the data between the values of 0 and 30, then one outlier at 80 and
another at 130. the model plots a nice line through the data
between 0 to
30, however given the
On Jul 21, 2011, at 12:22 AM, B. Jonathan B. Jonathan wrote:
Thanks David for your pointer. Here my original matrix is VCV matrix
which is the utmost important matrix in finance. However in reality
what happens is that due to incomplete data. lot of missing values
(or some other problems)
Hi,
So I am trying to plot my results of a model. what i have is the majority
of the data between the values of 0 and 30, then one outlier at 80 and
another at 130. the model plots a nice line through the data between 0 to
30, however given the outliers you cant seen this unless you change the y
On 11-07-21 3:23 AM, pilchat wrote:
Dear R users,
I have a desktop computer and a laptop, both of them with Ubuntu Lucid. The
former has R2.10 installed from Ubuntu repositories (this is the most recent
version in the repositories), while the latter has R2.13 from the CRAN
repositories.
I notic
Hi Ted,
the two ps files are generated using exactly the same script. For example:
--
setEPS()
postscript (file="volc.eps",width=5,height=4)
image(volcano)
dev.off()
--
With R2.10 I get a file with size 182K, while R2.13 gives a file of 186K. I
am
Hi, all
Here is what I try to do.
> strptime('20110101 0900',"%Y%m%d %H%M")
[1] "2011-01-01 09:00:00"
> strptime('20110101 900',"%Y%m%d %H%M")
[1] NA
If I have a 3-digit hour like '900', please show me how to convert it to
9:00:00.
Thank you in advance.
Wonjae
[[alternative HTML version d
On 21-Jul-11 07:23:54, pilchat wrote:
> Dear R users,
>
> I have a desktop computer and a laptop, both of them with
> Ubuntu Lucid. The former has R2.10 installed from Ubuntu
> repositories (this is the most recent version in the
> repositories), while the latter has R2.13 from the CRAN
> reposito
Just to say, I got this solved.
On Wed, Jul 20, 2011 at 7:55 PM, Roland Sookias wrote:
> Hi
>
> Is it possible to use grofit to get the AIC of several (e.g. two) growth
> models and compare both these and model parameters? All I can get it to do
> so far is return parameters for a single model.
if using the GUI, turn off buffering, or use 'flush.console()' after 'cat'.
Sent from my iPad
On Jul 21, 2011, at 7:34, "Bogaso Christofer"
wrote:
> Hi all, I have a lengthy 'for' loop and for each loop I want to track the
> iteration number that is currently going on. For this, I have tried
>
Hi all, I have a lengthy 'for' loop and for each loop I want to track the
iteration number that is currently going on. For this, I have tried
following:
> for (i in 1:10) {
+ DumDat <- rnorm(1000)
+ cat("iteration:", i, " \n")
+ }
iteration: 1
iteration: 2
iteration: 3
iteration: 4
Hi
Slides of my talk on Time Series Analysis and Mining with R at
Canberra R Users Group on 18 July are available at
http://www.rdatamining.com/docs. It presents time series
decomposition, forecasting, clustering and classification with R code
examples.
Regards
--
Yanchang Zhao
PhD, Data Miner
E
Layout did it! Thanks guys :)
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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On Wed, 2011-07-20 at 23:38 +1200, Rolf Turner wrote:
> On 20/07/11 11:07, DrCJones wrote:
> > Hi,
> >
> > par(mfrow = c(2,2))
> >
> > will create a 2x2 window that I can use to plot 4 diferent figures in:
> > [plot1 plot2]
> > [plot3 plot4]
> >
> > But how can do 3 so that the bottom spans the wid
Dear all,
Thank you so much for your help.
Best,
Sumona
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-c
On Thu, Jul 21, 2011 at 1:44 AM, Madana_Babu wrote:
> Hi all,
>
> I have R installed on a box, which is running on a machine with 16 core and
> Redhat - Linux. I am handling huge (size of dataset will be 5 GB) dataset.
> Lets assume that my data is in the form of structured (multiple) logs. I
> ac
cran.r-project.org/doc/manuals/R-intro.pdf
cran.r-project.org/doc/contrib/Paradis-rdebuts_en.pdf
cran.r-project.org/doc/contrib/Verzani-SimpleR.pdf
cran.r-project.org/doc/contrib/Lemon-kickstart/kr_intro.html
https://stat.ethz.ch/mailman/listinfo/r-help
On Thursday 21 July 2011 10:46:43 Varsha Ag
On Jul 20, 2011, at 13:33 , Rolf Turner wrote:
> On 20/07/11 21:08, Simon Knapp wrote:
>> Hi All,
>>
>> This is not really an R question but a statistical one. If someone could
>> either give me the brief explanation or point me to a reference that might
>> help, I'd appreciate it.
>>
>> I want
I have good experiences with the foreach package, available on cran. It
includes some tutorials which might help you.
cheers,
Paul
On 07/20/2011 11:44 PM, Madana_Babu wrote:
> Hi all,
>
> I have R installed on a box, which is running on a machine with 16 core and
> Redhat - Linux. I am handling
Hi Filipe and David,
Thanks both of you for the suggestion. It works perfect.
best,
- Vickie
Subject: RES: [R] Question about converting list items in matrix
Date: Wed, 20 Jul 2011 18:39:45 -0300
From: filipe.bote...@vpar.com.br
To: is...@live.com; r-help@r-project.org
"This message and its
On 07/21/2011 05:16 AM, Varsha Agrawal wrote:
> I am a new user and want to learn R from the most basic level.
>
> Suggest me a reading or a link.
Googling 'R introduction' lead to a great number of interesting links.
Also take a look at CRAN, primarily the documentation sections. Books
related t
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