1. That you think order() or sort() "did not work " is preposterous.
2. I suggest that you carefully read both ?sort and ?order, as you
obviously do not understand what these functions do. Here is a hint to get
you started:
> 3:1
[1] 3 2 1
> order(3:1)
[1] 3 2 1
> sort(3:1)
[1] 1 2 3
-- Bert
O
Dr. Bolker thank you for the help. I have figured out how to make this
do what I wanted - re:inline. I wanted to use the pch variable as the
number for the R plotting symbol. I hope that makes more sense.
kindest regards,
Stephen
On Tue 17 Jul 2012 10:31:07 PM CDT, Ben Bolker wrote:
Stephe
Dear Uwe,
I will enter the correct user::password values and try again. As per your
suggestion I will contact the local sysadmin
if required.
Thanks for your help.
Regards,
Kishor
On Wed, Jul 18, 2012 at 12:46 AM, Uwe Ligges <
lig...@statistik.tu-dortmund.de> wrote:
>
>
> On 16.07.2012 14:45,
Hi,
You should be able to reach your goal using functions "which" and "apply"
On 7/17/2012 3:03 PM, Christopher Desjardins wrote:
Hi,
Let's say I have the following data:
a=matrix(c(1,2,4,4,2,1,1,2,4),nrow=3,byrow=T)
a
[,1] [,2] [,3]
[1,]124
[2,]421
[3,]1
Hi all,
I want to order a series that is included on the second column in MCT.csv.
I do but R doesn´t order, could be because is a csv?
I have prove
MCT<-read.csv("MCT.csv")
a<-order(MCTor[,2],2,decreasing = FALSE)
a<-order(MCTor[,2],1,decreasing = FALSE)
or the same with sort but didn´t wo
Hi Yolande,
For your first error regaring factor, I am getting a single number
thisIndex <- as.character(index(diveData_2008[2,]))
> dive_id <- diveCond_all$dive_id[diveCond_all$timestamp==thisIndex]
> dive_id
[1] 1
#because there is only one TRUE value below and that refers to dive_id 1.
diveC
Hi,
Try,
apply(a,1,which.max)
[1] 3 1 3
A.K.
- Original Message -
From: Christopher Desjardins
To: r-help@r-project.org
Cc:
Sent: Tuesday, July 17, 2012 6:03 PM
Subject: [R] Finding the column with the maximum value by row
Hi,
Let's say I have the following data:
> a=matrix(c(1,2,4,
Stephen Sefick auburn.edu> writes:
> I'm sorry I didn't give the correct code, or explaination; the
> shape=pch doesn't work anymore. For example this used to work, but no
> longer can accept continuous values:
>
> #example code
> a <- 1:10
> b <- 1:10
> pch <- 1:10
>
> q <- data.frame(a,b
Dear Emily,
The lavaan package is typically used to fit models with latent
variables, and these models are typically fit to the covariance matrix
(and not necessarily to the raw data). Thus, it is usually not
straightforward to get data residuals from the fitted models. In your
case, it appe
Hi,
I see a lot of folks verify the regression identity SST = SSE + SSR
numerically, but I cannot seem to find a proof. I wonder if any folks on
this list could guide me to a mathematical proof of this fact.
Thanks.
David.
--
View this message in context:
http://r.789695.n4.nabble.com/Regressi
Honestly, I wasn't sure what you wanted to do with 'group'. Here it is with
the 'group' variable deleted
library(data.table)
dt=data.table(x[,-1])
dt[,lapply(.SD, function(x)weighted.mean(x,myweight)), keyby='myweek']
--
View this message in context:
http://r.789695.n4.nabble.com/weighted
This is applicable to either using optim or the differential equation
solver or any similar solver
Suppose I want to use the differential equation solver and this is my code
d<-y[2]
vdpol<-function(t,y)
{
list(c(1,
d,
3,
4
)
}
stiff<-ode(y=rep(0,4),times=c(0,1),func
David, many thanks.
Did something get ommitted from your line?:
ddply(x ,.(myweek), summarize, m1=weighted.mean(var1,myweight),
m2=weighted.mean(var2,myweight))
Because it just reproduces x - in a somewhat different order...
Thank you!
Dimitri
On Tue, Jul 17, 2012 at 9:22 PM, David Freedman wr
If there are many variables, I'd then suggest the data.table package:
library(data.table)
dt=data.table(x)
dt[,lapply(.SD, function(x)weighted.mean(x,myweight)), keyby=c('group',
'myweek')]
The '.SD' is an abbreviation for all the variables in the data table
(excluding the grouping variables). T
Thanks a lot, David.
Indeed, it's much shorter.
Unfortunately, in my real task I am dozens and dozens of variables
like var1 and var2 so that manually specifying things like in
"m1=weighted.mean(var1,myweight)" would take a lot of code and a very
long time.
Dimitri
On Tue, Jul 17, 2012 at 6:34 PM,
I'm sorry I didn't give the correct code, or explaination; the
shape=pch doesn't work anymore. For example this used to work, but no
longer can accept continuous values:
#example code
a <- 1:10
b <- 1:10
pch <- 1:10
q <- data.frame(a,b,pch)
qplot(a,b, shape=pch)
many thanks,
Stephen Sefic
On 2012-07-17 11:15, chester123 wrote:
Thank you in advance.
Now I want to make comparison of the different bandwidth h in a normal
distribution graph.
This is the table of bandwidth h: thumb rule (normal)--0.00205; thumb
rule(Epanech.)--0.00452; Plug-in (normal)--0.0009;
Plug-in(Epanech.)--0.0
Is there a way to use a continuous variable to pch in qplot? I believe
this worked in previous version. I need to specify certain values of a
shape for particular points so that multiple graphs all show the same
shapes for the same streams. I have gone to the original data and added
a pch co
Inline
On 2012-07-17 04:16, MK wrote:
Hi,
How do I use the subset function without losing the attributes?
You don't. At least not without modifying subset.data.frame().
The key sentence on the ?Extract help page is this:
"Subsetting (except by an empty index) will drop all
attributes exc
Dear R users,
I'm trying to determine local population centers in a region through kernel
smoothing. What I have is population density in each neighborhood, which
can be presented as point density. So in my data set pop, I have three
columns for about 1000 neighborhoods: x (x coordinate), y (y c
Dear All,
I'm having trouble tweaking a forest plot made using the R
meta-analysis package metafor. I did the analysis based upon the
correlation coeff from studies and plotted the corresponding forest
plot easily
q2<-rma(yi,vi,mods=cbind(grupo),data=qim)
q2
forest (q2,transf=transf.ztor
Amusing that someone named RICCI is asking about tensors
(sorry!)
Kjetil
On Tue, Jul 17, 2012 at 6:31 AM, Peppe Ricci wrote:
> Hi guys,
>
> I need some help to analyzing my data.
> I start to describe my data: I have 21 matrices, every matrix on the
> rows has users and on columns has items
The plyr package is very helpful for this:
library(plyr)
ddply(x ,.(myweek), summarize, m1=weighted.mean(var1,myweight),
m2=weighted.mean(var2,myweight))
--
View this message in context:
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Sent from the R help mailing list
Venables & Ripley:
"Modern Applied Statistics with S (fourth Edition)"
(known as MASS)
Kjetil
On Sat, Jul 14, 2012 at 4:01 PM, Larry White wrote:
> I'm looking for a single book that provides a deep, yet readable
> introduction to applied data analysis for general readers.
>
> I'm looking for c
Hello!
I wrote a code that works, but it looks ugly to me - it's full of loops.
I am sure there is a much more elegant and shorter way to do it.
Thanks a lot for any hints!
Dimitri
# I have a data frame:
x<-data.frame(group=c("group1","group2","group1","group2"),
myweight=c(0.4,0.6,0.4,0.6),
On 18/07/12 07:58, Courtney Saltonstall Couch wrote:
Could you please remove my email from the listserv?
You have to do that yourself. See:
https://stat.ethz.ch/mailman/listinfo/r-help
cheers,
Rolf Turner
__
R-help@r-project.org maili
apply(a, 1, which.max)
Apply the which.max() function to the matrix a row-wise (1)
Michael
On Tue, Jul 17, 2012 at 5:03 PM, Christopher Desjardins
wrote:
> Hi,
> Let's say I have the following data:
>
>> a=matrix(c(1,2,4,4,2,1,1,2,4),nrow=3,byrow=T)
>
>> a
>
> [,1] [,2] [,3]
>
> [1,]1
Hi,
Let's say I have the following data:
> a=matrix(c(1,2,4,4,2,1,1,2,4),nrow=3,byrow=T)
> a
[,1] [,2] [,3]
[1,]124
[2,]421
[3,]124
What syntax should I use to get R to tell me the column that corresponds to
the maximum value for each row?
For my exa
By the way, here is a related function I wrote in response to an R-help question
a while back that finds sequences that start when a signal rises about a certain
threshold and end when the signal drops below a lower threshold. This avoids
hysteresis-like problems.
f1 <- function(x, startThreshold
Dear list,
I have a problem with defining a function (see below) to read my testfile
(see testfile). My function only returns mydata I wish to work with
attr(mydata, 'fc') as well (for labelling a plot). Principally it works if
I do not insist on this function but it would be much easer if it is
p
Hi,
I'm using write.table() to export dataframe to Excel. All works fine except
that I want to export the variable labels instead of variable names.
I can see the labels in the R consol using attr(), but I just don't know how
to use the labels instead of the names.
Thanks,
François Mauric
Hi,
I'm using write.table() to export dataframe to Excel. All works fine except
that I want to export the variable labels instead of variable names.
I can see the labels in the R consol using attr(), but I just don't know how
to use the labels instead of the names.
Thanks,
François Mauri
Could you please remove my email from the listserv?
Thanks,
Courtney
--
--
Courtney Saltonstall Couch
Ph.D. Candidate
Corson Hall E323
Department of Ecology and Evolutionary Biology
Cornell University
Ithaca, NY 14853
Office: 607-254-4296
[[alternative HTML version deleted]]
__
Hi.
You should check function addtable2plot from plotrix package.
Also, here is an example of modified addtable2plot function that I
sometimes use.
This function is probably to far from perfect but, maybe, it will give you
some ideas how to solve your problem:
t2p <- function(x=par("usr")[1],y=
There are functions that allow for the "plotting" of text and R
objects that could be used to plot to a bitmap. Look at the
'addtable2plot' function in the plotrix package and the textplot
function in the gplots package (look for alternative spellings if you
don't find them based on those exact sp
Look at ?as.numeric.difftime
> z <- f(d, d$Score > 150)
> elapsedTime <- z$endDate - z$startDate
> units(elapsedTime)
[1] "secs"
> as.numeric(elapsedTime, units="hours")
[1] 7.616667 0.00 192.70 4.37 62.97 0.00
12.50 0.00
> as.numeric(elapsed
Homework? -- we don't do homework here.
-- Bert
On Tue, Jul 17, 2012 at 11:13 AM, Chandramouli Banerjee <
banerjee.chandramo...@gmail.com> wrote:
> Dear Community,
>
> I have a problem I am trying to code in R.
>
> Suppose there are 7 products and each have a ROI value attached to it.
> There ar
thanks for the link! I should read it through. that said, I didn't find any
good general solution to the problem so here I post some attempts for
general input. maybe someone knows how to speed this up. both my solutions
are theoretically O(n) for creating a list of n elements. The function to
impr
Simon, in your recent reply to me, you noted that EDF=1 corresponds to a
straight line fit. Does the edf always indicate the polynomial exponent
directly? We ask because Hothorn and Everitt said "Roughly, the
complexity of a cubic spline is about the same as a polynomial of degree
one less than
Yes! That does exactly what I want it to. Thank you so much.
One question, though, is it possible that the time difference be in hours
instead of seconds (other than dividing by 3600)? Looking at the code, I
don't know what I would change.
Does this do what you want?
> firstInRun <- functio
I am trying to install custom packages upon starting R. A lot of the code
that is written by us right now is available for editing to the users. To
try and protect the code, I am packaging the production level code and
having the users install it on their machine during start up.
However, when I t
Thank you in advance.
Now I want to make comparison of the different bandwidth h in a normal
distribution graph.
This is the table of bandwidth h: thumb rule (normal)--0.00205; thumb
rule(Epanech.)--0.00452; Plug-in (normal)--0.0009;
Plug-in(Epanech.)--0.002.
this is the condition: N=1010 data
I would like to output a nicely formatted data frame to a bitmap.
Is this possible in R?
--
View this message in context:
http://r.789695.n4.nabble.com/Table-Frame-output-tp4636790.html
Sent from the R help mailing list archive at Nabble.com.
__
R-
Dear Community,
I have a problem I am trying to code in R.
Suppose there are 7 products and each have a ROI value attached to it.
There are 400 representatives who calls 150,000 customers with these
product detailing and achieve sales through the calls.
There is a cost per call and revenue earned
Thank for your time, Rui.
Now, I get this error message:
Error en rbind(deparse.level, ...) :
numbers of columns of arguments do not match
Apparently, some columns have missing values and rbind doesn't work. I tried:
require(plyr)
do.call(rbind.fill, by(z, z$date, f))
Then the code runs t
Hi, The following distribution is known as Kumaraswamy binomial Distribution.
http://r.789695.n4.nabble.com/file/n4636782/kb.png
For a given data I need to estimate the paramters (alpha and beta) of this
distribution(Known as Kumaraswamy binomial Distribution, A Binomial Like
Distribution). For t
Hello,
Print DateTime, and see what's wrong.
When you paste(), you use the order ymd HM with space as separator but
when you try to convert to POSIXct you pass a wrong format, not the one
created by paste().
Solution: as.POSIXct(DAteTime, format="%Y %m %d %H %M")
Hope this helps,
Rui Barra
On Tue, Jul 17, 2012 at 2:00 AM, Christos Michalopoulos
wrote:
>
>
>
>
> I am working on a two stage threshold quantile regression model in R, and my
> aim is to estimate the threshold of the reduced-form equation (call it
> rhohat), and the threshold of the structural equation (call it qhat), i
Dear Uwe, Dear List,
This is exactly it. I knew there had to be an easier way. And I had
even looked up in par for it! Knowledge is a beautiful thing; thank
you so much.
> On Tue, Jul 17, 2012 at 4:28 PM, Uwe Ligges
> wrote:
>
> Insert
>
> par(yaxs="i")
>
> after you are finsihed with the image.
Hi Greg and Paul,
I had initially contemplated a solution similar to Greg's, which is simulation.
However, I might just throw out, that if based upon Terry's comments, time
varying covariates do not impact the power/sample size considerations for the
Cox model, then Schoenfeld's 1983 article in
Insert
par(yaxs="i")
after you are finsihed with the image.
See ?par for details what is does.
Uwe Ligges
On 17.07.2012 13:13, Tiago R M wrote:
Dear Mailing list,
I want to plot a matrix using image() and on its side I want to give
information to every line of the matrix using colors. I tr
On 12.07.2012 17:38, Saint wrote:
Hi!
Would be grateful if somebody helped me understand this error message after
trying to run a panel data:
PanelData <- read.xls("/Users/Bahman/Desktop/Taylor
rule/Data/PanelData2.xls")
attach(PanelData)
# Panel Data regresson for all countrys.
Answer <-
On 16.07.2012 14:45, Kishor Tappita wrote:
Dear R- Users,
I am unable to install R packages over the internet. Below is my session
information. My guess is that this may be
a proxy issue. I have set the http_proxy_user : my user id(network id) :
password (machine password) .
Maybe not corre
Thank for your time, Rui.
Now, I get this error message:
Error en rbind(deparse.level, ...) :
numbers of columns of arguments do not match
Apparently, some columns have missing values and rbind doesn't work. I tried:
require(plyr)
do.call(rbind.fill, by(z, z$date, f))
Then the code runs thro
On 13.07.2012 14:00, Gabor Grothendieck wrote:
On Thu, Jul 12, 2012 at 8:17 AM, Bharat Warule wrote:
Hello,
I am using read.csv.sql first time for reading the large data file.If I am
ran this code that showns warning “closing unused connection”.
Is it I am missing any argument from my comma
On Tue, Jul 17, 2012 at 2:23 PM, R. Michael Weylandt
wrote:
> On Tue, Jul 17, 2012 at 10:55 AM, Sarah Goslee wrote:
>> In general, you use assign() to do that, but having object names
>> containing spaces should be avoided.
>>
>>> mytext <- "Experiment name : CONTROL DB AD_1"
>>> mytext
>> [1] "E
On 17.07.2012 19:58, Michael Weylandt wrote:
Reproducible code? The error seems to be rather clear.
Also, this seems more of a BUGS question than an R question.
Indeed, the reproducible code is somthing for a BUGS mailing list. just
a guess: The typical row vs. column confusion in BUGS.
b
On Tue, Jul 17, 2012 at 10:55 AM, Sarah Goslee wrote:
> In general, you use assign() to do that, but having object names
> containing spaces should be avoided.
>
>> mytext <- "Experiment name : CONTROL DB AD_1"
>> mytext
> [1] "Experiment name : CONTROL DB AD_1"
>> mytext <- sub("Experiment name :
Reproducible code? The error seems to be rather clear.
Also, this seems more of a BUGS question than an R question.
Michael
On Jul 17, 2012, at 11:06 AM, PRAGYA SUR wrote:
> Dear R users,
>Can anyone tell me why I might get the error message
> "the array index is greater t
I will repost on R-devel.
-- Bert
On Tue, Jul 17, 2012 at 10:32 AM, Prof Brian Ripley
wrote:
> On 17/07/2012 18:20, Bert Gunter wrote:
>
>> Folks:
>>
>> sessionInfo()
>> R version 2.15.0 (2012-03-30)
>> Platform: i386-pc-mingw32/i386 (32-bit)
>>
>> locale:
>> [1] LC_COLLATE=English_United States
Hello,
I think there is some mistake in the format.
Try this:
dat1<-data.frame(Year=rep(2009,2),Month=rep(10,2),Day=rep(5,2),hour=rep(0,2),minute=c(0,15),second=c(11.288,11.258))
###Your code
DateTime<-with(dat1,paste(Year,Month,Day,hour,minute))
DateTime
#[1] "2009 10 5 0 0" "2009 10 5 0
Hi,
Try this:
dat1<-data.frame(a,b,c)
dat1[a<=c& b>c,"abc_check"]<-TRUE
> dat1
a b c abc_check
1 10 100 50 TRUE
2 20 200 60 TRUE
3 30 300 70 TRUE
#or
dat1<-within(dat1,{abc_check<-ifelse(a<=c & b>c,"TRUE","FALSE")})
dat1
a b c abc_check
1 10 100 50 TRUE
2 20 200
I am hoping to run a non-parametric correlation with a covariate.
Unfortunately, for the two correlation I am running, one is not normally
distributed and the other is categorical data. Therefore, I cannot use
partial correlation for these two analysis. Any help would be greatly
appreciated!
Thank
On 17/07/2012 18:20, Bert Gunter wrote:
Folks:
sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=En
One quick (though probably not canned) approach to get a feel for what an
analysis might be like is to analyze a sample data set (from the survival
package, a textbook, or a past analysis). Choose something that has some
similarity to the planned study. Now look at the widths of the confidence
in
I think that you need to supply some data with your question.
Use dput() to output your data or a subset of it and paste the output into the
email
See ?dput for more information.
Probably a subset of the data would be sufficient. Something like
dput(head(localRaw, 50) should be fine.
John
Folks:
sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base pa
Does this do what you want?
> firstInRun <- function(x) c(TRUE, x[-1] != x[-length(x)])
> lastInRun <- function(x) c(x[-1] != x[-length(x)], TRUE)
> f <- function(data, condition) {
+ with(data, data.frame(startDate = Date[firstInRun(condition)],
+ endDate =
On 17/07/2012, Rui Barradas wrote:
> Hello,
>
> That's not the error I've had. You must be aware that read.table creates
> a data.frame and therefore the object 'timestamps' is NOT holding time
> stamps, it's holding a vector, 'V1', of time stamps.
>
Was not aware of the significance of the data
I would STRONGLY recommend that you talk to your local statistician.
Further:
1. R-help is not a statistical consulting forum
2. Remote statistical consulting is very risky due to the inherent
difficulty in communicating all essential context of the problem and data.
You have been warned!
-- Be
Dear R users,
Can anyone tell me why I might get the error message
"the array index is greater than the upper bound for t" in WinBUGS ? t is a
vector I have defined. I have checked repeatedly that the array index for t
does match with the upper bound till which the loop is run.
Dear Bert and Sarah,
Thank you very much for your clarifications on this matter. I will
have to study more closely the way extracting subsets of data
structures is performed, and I will change my programming habits
accordingly.
Best regards,
Paulo Barata
---
I have the following data (a subset of the real data set):
structure(list(Date = structure(c(1209625080, 1209641460, 1209652500,
1209676800, 1209682860, 1209692100, 1209706980, 1209722580, 1209726300,
1209739620, 1209762780, 1209765720, 1209770520, 1209791040, 1209812580,
1209829920, 1209837180
Dear Mailing list,
I want to plot a matrix using image() and on its side I want to give
information to every line of the matrix using colors. I tried to use
barplot, but cannot align the two plots. Code below. I succeed in doing
what I want using another image() plot, but was wondering whether the
Hello,
I have a question concerning the "bestglm" procedure (function that finds
the best subset of explanatory variables for the variable we want to model
by checking all possible models), available in the package "bestglm".
When I use the function for a database that has over 45 variables (that
Hello!
I am optimizing my code in R and for this I need to know a bit more about
the internals. It would help tremendously if someone could link me to a
page with O()-complexities of all the operations.
In this particular case, I need something like a linked list with O(1)
insertLast/First ability
Hi,
I'm hoping that someone will be able to help. I would like to compare how
covariates associate with the risk of a binary outcome during two periods.
Period 1 will be non-exposure to a treatment and period 2 will be exposure
to a treatment. The same individuals will be examined in each group
I am working on a two stage threshold quantile regression model in R, and my
aim is to estimate the threshold of the reduced-form equation (call it rhohat),
and the threshold of the structural equation (call it qhat), in two stages. On
the first stage, i estimate rhohat by quantile regressio
Have you found a solution for this? I am also trying to find a way to
retrieve the confidence intervals for the predictions.
Best regards,
João Oliveirinha
On Sunday, October 30, 2011 7:04:03 PM UTC, Muhammed Akbulut wrote:
>
> Hi,
>
> Is it possible to calculate confidence intervals for suppor
What's pos?
What's o?
What is your loop looping over?
Where's your reproducible example provided using dput()?
And more generally, have you read R FAQ 7.31 and the posting guide?
Most likely either o is wrong, or there are NA values in pos, but we
have no way to diagnose that.
Sarah
On Tue, Jul
In general, you use assign() to do that, but having object names
containing spaces should be avoided.
> mytext <- "Experiment name : CONTROL DB AD_1"
> mytext
[1] "Experiment name : CONTROL DB AD_1"
> mytext <- sub("Experiment name : ", "", mytext)
> mytext
[1] "CONTROL DB AD_1"
>
> assign(mytext,
Hi,
I'd like to simulate survival times according to continuous covariate
values. Suppose we have a continuos covariate whose value of zero
corresponds to a mean survival time of 180 days and whose value of 0.5
corresponds to a mean survival time of 240 days. How can I use this
information to sim
Hello all,
I am a medical student and as a capstone for my summer research project I am
going to create a simple online web "calculator" for users to input their
relevant data, and a probability of relapse within 5 years will be computed
and returned based on the Cox PH model I have developed.
The
Yepp, that makes much more sense, look at Eiks post for an answer.
I'm terribly sorry, i confused Rd with RData.
On 17.07.2012, at 15:18, Jessica Streicher wrote:
> Oh, i think i get it now, you're talking about formatting a table in a
> documentation file?
>
> On 17.07.2012, at 14:45, purus
Thanks in advance here.
I use dpik() function to calculate the bandwidth h. Following is the related
code:
h<-dpik(x,scalest="minim",level=2L,kernel="normal",canonical=FALSE,gridsize=401L,range.x=range(x),truncate=TRUE)
But there is warning messages:
1: In bkfe(gcounts, 6L, alpha, range.x = c(sa
Hi. Try this:
a<=c & c wrote:
> Hi
>
> I 've a data
>
> a=c(10,20,30)
> b=c(100,200,300)
> c=c(50,60,70)
>
> I want to compare a[1]<=c[1]
> How to compare for all the records
>
> -
> Thanks in Advance
> Arun
> --
> View this message in context:
> http://r.789695.n4.nabble.com/comparin
Any reason I'd get an error like this?
Error in if (round(pos[o + 1]) == (pos[o + 1] - 0.4)) { :
missing value where TRUE/FALSE needed
but when i do it individually, out of the for loop,
> (round(pos[o+1])==(pos[o+1]-.4) )
65
TRUE
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Hello,
>titletool<-read.csv("TotalCSVData.csv",header=FALSE,sep=",")
> class(titletool)
[1] "data.frame"
>titletool[1,1]
[1] Experiment name : CONTROL DB AD_1
>t<-titletootl[1,1]
>t
[1] Experiment name : CONTROL DB AD_1
>class(t)
[1] "character"
now i want to create an object (vector) with th
Hi
I 've a data
a=c(10,20,30)
b=c(100,200,300)
c=c(50,60,70)
I want to compare a[1]<=c[1]http://r.789695.n4.nabble.com/comparing-three-vectors-tp4636728.html
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R-help@r-project.org mailing lis
Inline below.
-- Bert
On Tue, Jul 17, 2012 at 7:40 AM, Paulo Barata
wrote:
>
> Dear Frans and Peter,
>
> Yes, the notation df[,'var'] is able to catch a non-existent
> variable var inside a data frame df. But the notation df$var
> isn't.
>
> So we have this situation, where two different notatio
Hi,
On Tue, Jul 17, 2012 at 10:40 AM, Paulo Barata
wrote:
>
> Dear Frans and Peter,
>
> Yes, the notation df[,'var'] is able to catch a non-existent
> variable var inside a data frame df. But the notation df$var
> isn't.
>
> So we have this situation, where two different notations, which
> (as fa
I was wondering why I get instead of the timestamp in the following.
Thanks.
> dataDir <- file.path(wd)
> localRaw <- read.csv(file.path(dataDir,"LOCAL.csv"), as.is=T,stringsAsFactors
= FALSE)
> localRaw[1:2,]
Year Month Day hour minute second Temp1mab Temp7mab Temp14mab Salinity1mab
1 2009
On Jul 16, 2012, at 6:55 PM, Noah Silverman wrote:
> Hello,
>
> I'm tasked with putting together a cost effectiveness analysis on a proposed
> medical treatment.
>
> The "standard" suggested by someone is an expensive commercial package names
> "TreeAge Pro" which looks like its just a fancy
Dear Frans and Peter,
Yes, the notation df[,'var'] is able to catch a non-existent
variable var inside a data frame df. But the notation df$var
isn't.
So we have this situation, where two different notations, which
(as far as I understand) perform the same action, have different
kinds of respon
Yes your right , pls tell how do format table in Rd files
Thanks
B.Purushothama
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Oh, i think i get it now, you're talking about formatting a table in a
documentation file?
On 17.07.2012, at 14:45, purushothaman wrote:
> Hi,
>
> i need to create table like this
>
> -
Try again please,
so you have a table/dataframe or some such with 4(?) columns:
module_name, class_name, function_name and function_description
I guess under the second line shall be an example of data in the table, but i
cannot see what belongs to what column.
If you have an example table alre
Hi,
I think \tabular does what you want, see
see http://cran.r-project.org/doc/manuals/R-exts.html#Lists-and-tables
hth.
Am 17.07.2012 14:45, schrieb purushothaman:
> Hi,
>
> i need to create table like this
>
> ---
Hi,
i need to create table like this
---
module name class namefunction name
I think you need to elaborate a bit. At least I get no idea of what you want
from that one sentence.
On 17.07.2012, at 13:44, purushothaman wrote:
> Hi
>
> how to create table with file link in Rd File
>
> Thanks
> B.Purushothaman
>
> --
> View this message in context:
> http://r.789695.n4
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