Hi All,
I have the following problem (read the commented bit below):
a<-matrix(1:9,nrow=3)
a
[,1] [,2] [,3]
[1,]147
[2,]258
[3,]369
div<-1:3
apply(a,2,function(x)x/div) ##want to divide each column by div-
instead each row is divided##
[,1] [,
I just looked more carefully at your code.
You are computing the unequal-variance (Welch) version of the t-test, so
that's why there isn't a problem. Compare it with the equal-variance
t-test, using the pooled variance estimate, which does have a problem, as
below
-thomas
tstat4 <- function
In help(gc) I read, "...the primary purpose of calling 'gc' is for the
report on memory usage".
What memory usage does gc() report? And more importantly, which memory
uses does it NOT report? Because I see one answer from gc():
used (Mb) gc trigger (Mb) max used (Mb)
Ncells 148759
Simon,
Many thanks as always for your help.
I see and appreciate the example that you cited, but I'm having a hard
time generalizing it to a multivariate case. A bit about my context --
my dataset is response ratios; the log of a treatment over a control.
One of my explanatory variables is t
OK,although the variance ratio was already 2.25 to 1, tried sigma1=10,
sigma2=25, which makes the ratios of the variances 6.25 to 1.
Still no change. See:
http://msemac.redwoods.edu/~darnold/math15/R/chapter11/DistributionForTwoIndependentSamplesPartII.html
D.
--
View this message in context
Dear David,
On Wed, Apr 17, 2013 at 6:24 PM, David Arnold wrote:
> Hi,
[snip]
>
> D.
Before posting to StackExchange, check out the Wikipedia entry for
"Behrens-Fisher problem".
Cheers,
Jay
--
G. Jay Kerns, Ph.D.
Youngstown State University
http://people.ysu.edu/~gkerns/
_
On 04/17/2013 06:24 PM, David Arnold wrote:
Hi,
Typical things you read when new to stats are cautions about using a
t-statistic when comparing independent samples. You are steered toward a
pooled test or welch's approximation of the degrees of freedom in order to
make the distribution a t-dist
Hi,
Typical things you read when new to stats are cautions about using a
t-statistic when comparing independent samples. You are steered toward a
pooled test or welch's approximation of the degrees of freedom in order to
make the distribution a t-distribution. However, most texts give no
informati
Dear list,
I am trying to do an automated model selection of a glmm (function glmer;
package: lme4) containing a large number of predictors. As far as i
understand, glmulti is able to devide the process into chuncks and proceed
by parallel processing on on multiple cores. Unfortunately this does n
Dear R experts/ users
Full Information Maximum Likelihood (FIML) estimation approach is
considered robust over Seemingly Unrelated Regression (SUR) approach
for analysing data of multivariate sample selection problem. The zero
cases in my dependent variables are resulted from three sources:
Irreve
tapply(t(data.matrix(sim_sub)),rep( rep(1:4, each=5), each=10),mean)
# 1 2 3 4
#0.0086 0.0074 0.0082 0.0108
unlist(lapply(split(sim_sub,((seq_len(nrow(sim_sub))-1)%/%5)+1),function(x)
mean(unlist(x
# 1 2 3 4
#0.0086 0.0074 0.0082 0.0108
A.K.
On Apr 17, 2013, at 9:54 AM, Keith S Weintraub wrote:
> Folks,
> I recently was given a simulated data set like the following subset:
>
> sim_sub<-structure(list(V11 = c(0.01, 0, 0, 0.01, 0, 0.01, 0, 0, 0, 0,
> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), V12 = c(0, 0, 0, 0.01, 0.03, 0,
> 0, 0, 0, 0, 0, 0.
On 17/04/2013 11:33 AM, Ivan Alves wrote:
Dear Duncan and A.K.
Many thanks for your super quick help. The modified lapply did the trick, mapply died
with a error "Error in dots[[2L]][[1L]] : object of type 'builtin' is not
subsettable".
That's due to a typo: I should have said
mapply(plot,
On Apr 16, 2013, at 11:02 PM, Pascal Oettli wrote:
> Hello,
>
> Don't reply only to me.
>
> 1) Filter the unwanted mails,
There is also an option to get Rhelp postings in digest-format.
--
David.
> 2) It takes few days to unsubscribe you.
>
> Regards,
> Pascal
>
>
> On 04/17/2013 02:59 P
Dear R experts,
I am trying to simulate a list containing data matrices. Unfortunately, I don't
manage to get it to work.
A small example:
n=5
nbootstrap=2
subsets<-list()
for (i in 1:n){
subsets[[i]] <- rnorm(5, mean=80, sd=1)
for (j in 1:nbootstrap){
test<-list()
On Wed, Apr 17, 2013 at 10:36 AM, R. Michael Weylandt
>
> It sounds like you want .Deprecate
>
> ?.Deprecate
>
Perhaps you meant Deprecated?
?Deprecated
Best,
Peter
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PL
Jan,
What mgcv version are you using, please? (Older versions have a poor
p-value approximation when select=TRUE, but of course it's possible that
you've managed to break the newer approximation as well)
The 'select=TRUE' option adds a penalty to each smooth, to allow it to
be penalized out
On Apr 17, 2013, at 10:17 AM, Jannis wrote:
> Dear R community,
>
>
> what would be the proper R way to handle obsolete function names? I have
> created several packages with functions and sometimes would like to change
> the name of a function but would like to create a mechanism that othe
No, you don't have to use four dots.
Please check these links for further details:
http://stackoverflow.com/questions/5890576/usage-of-three-dots-or-dot-dot-dot-in-functions
http://cran.r-project.org/doc/manuals/R-lang.pdf
A.K.
From: Janesh Devkota
To: arun
Hello Arun,
Thank you so much for the prompt reply. I have one simple question here.
DOes three dots (...) in the reduce function means we are applying for
three dataframes here ? So, if we were to combine four would that dots be
four dots ?
Thanks.
On Wed, Apr 17, 2013 at 12:16 PM, arun wrote
I have 11 possible predictor variables and use them to model quite a few
target variables.
In search for a consistent manner and possibly non-manual manner to identify
the significant predictor vars out of the eleven I thought the option
"select=T" might do.
Example: (here only 4 pedictors)
firs
HI Janesh,
YOu can use:
library(plyr)
?join_all()
#From the help page:
dfs <- list(
a = data.frame(x = 1:10, a = runif(10)),
b = data.frame(x = 1:10, b = runif(10)),
c = data.frame(x = 1:10, c = runif(10))
)
join_all(dfs)
join_all(dfs, "x")
join_all(dfs,
On Apr 17, 2013, at 16:47 , Rémi Lesmerises wrote:
> Dear all,
>
> I'm trying to compare two sets of variables, the first set is composed
> exclusively of numerical variables and the second regroups factors and
> numerical variables. I can't use a Manova because of this inclusion of
> numeric
Hello,
Try the following.
blocks <- rep(1:(1 + nrow(sim_sub) %/% 5), each = 5)[seq_len(nrow(sim_sub))]
aggregate(sim_sub, list(blocks), FUN = mean)
Hope this helps,
Rui Barradas
Em 17-04-2013 18:04, arun escreveu:
do.call(rbind,lapply(split(sim_sub,((seq_len(nrow(sim_sub))-1)%/%
5)+1),col
Also,
do.call(rbind,lapply(split(sim_sub,rep(1:(1+nrow(sim_sub)/5),each=5)[seq_len(nrow(sim_sub))]),colMeans))
# V11 V12 V13 V14 V15 V16 V17 V18 V19 V20
#1 0.004 0.008 0.002 0.008 0.004 0.01 0.004 0.014 0.022 0.010
#2 0.002 0.000 0.002 0.006 0.006 0.01 0.018 0.006 0.010 0.014
#3
Hi, I have a quick question here. Lets say he has three data frames and he
needs to combine those three data frame using merge. Can we simply use
merge to join three data frames ? I remember I had some problem using merge
for more than two dataframes.
Thanks.
On Wed, Apr 17, 2013 at 1:05 AM, Far
do.call(rbind,lapply(split(sim_sub,((seq_len(nrow(sim_sub))-1)%/%
5)+1),colMeans))
# V11 V12 V13 V14 V15 V16 V17 V18 V19 V20
#1 0.004 0.008 0.002 0.008 0.004 0.01 0.004 0.014 0.022 0.010
#2 0.002 0.000 0.002 0.006 0.006 0.01 0.018 0.006 0.010 0.014
#3 0.000 0.004 0.002 0.012 0.
Dear Ivan,
No problem.
If you want it in a single plot:
matplot(do.call(cbind,g),ylab="value",pch=1:2,main="Some
plot",col=c("red","orange"),type="o")
legend("topleft",inset=.01,lty=c(1,1),title="Plot",col=c("red","orange"),names(g),horiz=TRUE)
A.K.
From: Iv
Folks,
I recently was given a simulated data set like the following subset:
sim_sub<-structure(list(V11 = c(0.01, 0, 0, 0.01, 0, 0.01, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0), V12 = c(0, 0, 0, 0.01, 0.03, 0,
0, 0, 0, 0, 0, 0.01, 0, 0.01, 0, 0, 0, 0, 0, 0.04), V13 = c(0,
0, 0, 0.01, 0, 0, 0,
Le mardi 16 avril 2013 à 10:15 -0700, Paul Miller a écrit :
> Hi Milan and Max,
>
> Thanks to each of you for your reply to my post. Thus far, I've
> managed to find answers to some of the questions I asked initially.
>
> I am now able to control the justification of the leftmost column in
> my t
Hi Levi,
Thank you very much for your reply and concern.
There was a package till 2011...or even 2012...entitled "Generalized Path
Seeking" Regression in "R" which was using a combination of all of these
methods...as a way to regularize regression.
However, I presently can't find it. Thus, I was
Dear Claudia,
Your question has been posed on many previous occasions.
The (short) answer has always been the same: have a look at the Anova function
in the car package but before doing that, get a copy of John Fox's "Applied
Regression Analysis and Generalized Linear Models" book.
Best,
José
Dear Remi,
On Wed, 17 Apr 2013 08:23:07 -0700 (PDT)
Rémi Lesmerises wrote:
> Dear John,
>
> Thanks for your comments! But when I tried your suggestion, the output was as
> the following:
>
>
> Response Dist_arbre :
> Df Sum Sq Mean Sq F value Pr(>F)
> Poids
Dear Duncan and A.K.
Many thanks for your super quick help. The modified lapply did the trick,
mapply died with a error "Error in dots[[2L]][[1L]] : object of type 'builtin'
is not subsettable".
Kind regards,
Ivan
On 17 Apr 2013, at 17:12, Duncan Murdoch wrote:
> On 17/04/2013 11:04 AM, Ivan Al
Dear John,
Thanks for your comments! But when I tried your suggestion, the output was as
the following:
 Response Dist_arbre :
      Df   Sum Sq   Mean Sq F value   Pr(>F)  Â
Poids     1 0.00010398 0.00010398  6.2910 0.0364733 * Â
Age      1 0.520
Hi.
No problem.
cc:ing to Rhelp.
A.K.
From: GUANGUAN LUO
Sent: Wednesday, April 17, 2013 10:25 AM
Subject: Re: how to change the date into an interval of date?
Thank you so much . That is exactly the things i want.
GG
Hi,
>Try this:
>library(mondate)
>
Dear R community,
what would be the proper R way to handle obsolete function names? I have
created several packages with functions and sometimes would like to
change the name of a function but would like to create a mechanism that
other scripts of functions using the old name still work.
C
Hi,
Try:
set.seed(25)
g<- list(sample(1:40,20,replace=TRUE),sample(40:60,20,replace=TRUE))
names(g)<- c("2009-10-07","2012-02-29")
pdf("Trialnew.pdf")
lapply(seq_along(g),function(i) plot(g[[i]],main=names(g)[i]))
dev.off()
A.K.
- Original Message -
From: Ivan Alves
To: "R-help@r-proj
On 17/04/2013 11:04 AM, Ivan Alves wrote:
Dear all,
List g has 2 elements
> names(g)
[1] "2009-10-07" "2012-02-29"
and the list plot
lapply(g, plot, main=names(g))
results in equal plot titles with both list names, whereas distinct titles
names(g[1]) and names(g[2]) are sought. Clearly, lap
Dear Remi,
Take a look at the Anova() function in the car package. In your case, you could
use
Anova(lm(as.matrix(Y) ~ x + z))
or, for more detail,
summary(Anova(lm(as.matrix(Y) ~ x + z)))
I hope this helps,
John
John Fox
Sen. William McMas
Dear all,
List g has 2 elements
> names(g)
[1] "2009-10-07" "2012-02-29"
and the list plot
lapply(g, plot, main=names(g))
results in equal plot titles with both list names, whereas distinct titles
names(g[1]) and names(g[2]) are sought. Clearly, lapply is passing 'g' in stead
of consecutivel
Dear r-helpers,
I have a graph created using the library 'spatgraphs'.
library(spatstat)
library(spatgraphs)
xy <- rbind(c(28.39, -16.27), c(30.62, -20.13), c(32.25, -28.7), c(22.43,
-27.22), c(27.5, -21.17), c(31.22, -24.52), c(17.93, -26.92), c(18.72,
-17.95), c(24.15, -17.82), c(29.23, -22.85))
Exel is hardly the epitome of good graphing. There are a couple of ways to do
what you want and Jim has shown you one , but ...
What about using two panels to present the data as in
opar <- par(mfrow = c(2, 1))
plot(dat1$Date, dat1$Weight, col = "red", xlab = "", ylab = "Weight")
plot(dat1$Da
Dear all,
I'm trying to compare two sets of variables, the first set is composed
exclusively of numerical variables and the second regroups factors and
numerical variables. I can't use a Manova because of this inclusion of
numerical variables in the second set. The solution should be to perform
Hi
Your files have only one column, so melted data is virtualy the same.
When I read them as test1 and test2 i can do
plot(ecdf(test1$Down))
plot(ecdf(test2$Up), add=T, col=2)
Or using previously ustated ggplot2 package
test<-rbind(melt(test1),melt(test2))
p<-ggplot(test, aes(x=value, colour
On 17-04-2013, at 10:08, "S'dumo Masango" wrote:
> I have installed the arm package and its dependents (e.g MATRIX, etc), but
> cannot use the functions "invlogit" and "bayesglm" because it gives me the
> error message "Error: could not find function "invlogit" or Error: could not
> find functio
Hi Carrington,
You also need the boot package (see
http://stat.ethz.ch/R-manual/R-patched/library/boot/html/inv.logit.html )
As for the other function, please load the arm package, e.g.,
require(arm)
require(boot)
and then you will be able to use the functions mentioned below.
HTH,
Jorge.-
O
Have you loaded it?
library(arm)
John Kane
Kingston ON Canada
> -Original Message-
> From: masan...@uniswa.sz
> Sent: Wed, 17 Apr 2013 10:08:39 +0200
> To: r-help@r-project.org
> Subject: [R] Error: could not find function "invlogit" and "bayesglm"
>
> I have installed the arm package
Dear R-enthusiats,
While using the regression fonction 'brunch' of CAPER (with R v2.15.4),
in a simple case (binary variable Yes/No vs. a continuous variable)
I ended with an unexplained error:
Error in if (any(stRes > robust)) { :
missing value where TRUE/FALSE needed
I simplified my code
Hello everybody,
I have got a data set with about 400 companies. Each company has a
score for its enviroment comportment between 0 and 100. These companies
belong to about 15 different countries. I have e.g. 70 companies from
UK and 5 from Luxembourg,- so the data set is pretty unbalanced and
Dear Sir,
Thanks a lot for your valuable suggestions and help.
Regards
Katherine
--- On Wed, 17/4/13, Jim Lemon wrote:
From: Jim Lemon
Subject: Re: [R] Creating a vector with repeating dates
To: "Katherine Gobin"
Cc: r-help@r-project.org
Date: Wednesday, 17 April, 2013, 10:35 AM
On 04/17
I have installed the arm package and its dependents (e.g MATRIX, etc), but
cannot use the functions "invlogit" and "bayesglm" because it gives me the
error message "Error: could not find function "invlogit" or Error: could not
find function "invlogit". What could be the problem.
Regards
Car
Dear,
When i multiply the y of a regression by 10, I would expect that the
coefficient would be multiply by 10 and the z value to stay constant. Here
some reproducible code to support the case.
*Ex 1*
library(mvtnorm)
library(VGAM)
set.seed(1)
x=rmvnorm(1000,sigma=matrix(c(1,0.75,0.75,1),2,2))
s
Hi,
Try this:
library(mondate)
mutate(evt_c.1,t=ave(round(as.numeric(mondate(paste(evt_c.1[,2],"-01",sep="",patient_id,FUN=function(x)
c(0,cumsum(diff(x)
# patient_id responsed_at t
#1 1 2010-5 0
#2 1 2010-7 2
#3 1 2010-8 3
#4 1
Paul,
#1: I've never tried but you might be able to escape the required tags in
your text (e.g. in html you could write out the in your text).
#3: Which output? Is this in text?
#2: I may be possible and maybe easy to implement. So if you want to dig
into it, have at it. For me, I'm completely
Perhaps I am wrong, but I think there are only a few packages supporting
Elastic Net, and none of them also perform Best Subsets.
On Wed, Apr 17, 2013 at 8:46 AM, Christos Giannoulis wrote:
> Merhaba, Hello to you too Mehmet (Yasu ki sena)
>
> Thank you for your email and especially for sharing
Hi
Could you perhaps possibly help me. I would like to use the package
VIF but cannot get results
I attach the .csv file and my R code. What do I have to do ? Any help
is greatly appreciated.
library(VIF)
coal <- read.csv("e:/freekvif/cqa1.csv",header=TRUE)
y <- as.numeric(coal$AI)
x <-
Hi,
I hope this is what you are looking for:
library(plyr)
mutate(evt_c.1,t=ave(as.numeric(gsub(".*\\-","",responsed_at)),patient_id,gsub("-.*","",responsed_at),FUN=function(x)
c(0,cumsum(diff(x)
# patient_id responsed_at t
#1 1 2010-5 0
#2 1 2010-7 2
#3
Thank you,
I see this was due to me using %y instead of %Y,
See ?strptime
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: 17 April 2013 12:31
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Problem with DateVisit-gives wrong year?
HI,
test.date<- rep(c("14/05/2012","
On Tue, Apr 16, 2013 at 1:25 PM, Sarah Goslee wrote:
> Given that we occasionally run into problems with comparing Excel
> results to R results, and other spreadsheet-induced errors, I thought
> this might be of interest.
>
> http://www.nextnewdeal.net/rortybomb/researchers-finally-replicated-rein
On 17-04-2013, at 14:39, Christofer Bogaso wrote:
> Hello again,
>
> Let say I have a matrix:
>
> Mat <- matrix(1:12, 4, 3)
>
> And a vector:
>
> Vec <- 5:8
>
> Now I want to do following:
>
> Each element of row-i in 'Mat' will be divided by i-th element of Vec
>
> Is there any direct wa
Hi,
Try:
evt_c.1<- read.table(text="
patient_id responsed_at
1 2010-5
1 2010-7
1 2010-8
1 2010-9
2 2010-5
2 2010-6
2 2010-7
",sep="",header=TRUE,stringsAsFactors=FALSE)
lst1<-split(evt_c.1,evt_c.1$patient_id)
res<-do.
Hi,
Try:
sweep(Mat,1,Vec,"/")
# [,1] [,2] [,3]
#[1,] 0.200 1 1.80
#[2,] 0.333 1 1.67
#[3,] 0.4285714 1 1.571429
#[4,] 0.500 1 1.50
do.call(rbind,lapply(seq_len(nrow(Mat)),function(i) Mat[i,]/Vec[i]))
# [,1] [,2] [,3]
#[1,] 0.200
Merhaba, Hello to you too Mehmet (Yasu ki sena)
Thank you for your email and especially for sharing this package. I
appreciate it.
However, my feeling is that this package does not have the third component
of Best Subsets (pls correct me if I am wrong). It uses only a combination
of Ridge and Las
Hello again,
Let say I have a matrix:
Mat <- matrix(1:12, 4, 3)
And a vector:
Vec <- 5:8
Now I want to do following:
Each element of row-i in 'Mat' will be divided by i-th element of Vec
Is there any direct way to doing that?
Thanks for your help
___
On Wed, Apr 17, 2013 at 2:33 PM, Marc Schwartz wrote:
> If you only want up to say second order interactions:
>
>> summary(lm(mpg ~ (cyl + disp + hp + drat) ^ 2, data = mtcars))
>
This is what I was looking for. Thank you so much.
> This is covered in ?formula
>
Indeed. I tried to parse ?formula
Hi,
dat1<- read.table(text="
V1 V2
A 1
B 2
C 1
D 3
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- read.table(text="
V3 V2
AAA 1
BBB 2
CCC 3
",sep="",header=TRUE,stringsAsFactors=FALSE)
library(plyr)
join(dat1,dat2,by="V2",type="full")
# V1 V2
On Apr 17, 2013, at 7:23 AM, Liviu Andronic wrote:
> Dear all,
> Consider the model below:
>
>> x <- lm(mpg ~ cyl * disp * hp * drat, mtcars)
>> summary(x)
>
> Call:
> lm(formula = mpg ~ cyl * disp * hp * drat, data = mtcars)
>
> Residuals:
>Min 1Q Median 3Q Max
> -3.5725 -
Dear all,
Consider the model below:
> x <- lm(mpg ~ cyl * disp * hp * drat, mtcars)
> summary(x)
Call:
lm(formula = mpg ~ cyl * disp * hp * drat, data = mtcars)
Residuals:
Min 1Q Median 3Q Max
-3.5725 -0.6603 0.0108 1.1017 2.6956
Coefficients:
Estimate S
On Tue, Apr 16, 2013 at 4:33 PM, Jim Lemon wrote:
> On 04/17/2013 03:25 AM, Sarah Goslee wrote:
The final point does relate to Excel and any application that hides what is
> going on to the casual observer. I will treasure this URL to give to anyone
> who chastises my moaning when I have to perf
Hi,
You may also try:
dfab<-data.frame(A,B)
library(plyr)
dfcd<-subset(mutate(dfab,C=A*2,D=B*3),select=-c(A,B))
#or
dfcd1<-subset(within(dfab,{D<-B*3;C<-A*2}),select=-c(A,B))
dfcd$C
#[1] 2 4 6
dfcd$D
#[1] 12 18 21
A.K.
- Original Message -
From: jpm miao
To: r-help
Cc:
Sent: Wedn
HI,
test.date<- rep(c("14/05/2012","01/10/2012","28/09/2012"),each=6)
as.Date(test.date,"%d/%m/%Y")
# [1] "2012-05-14" "2012-05-14" "2012-05-14" "2012-05-14" "2012-05-14"
#[6] "2012-05-14" "2012-10-01" "2012-10-01" "2012-10-01" "2012-10-01"
#[11] "2012-10-01" "2012-10-01" "2012-09-28" "2012-09-28
On 04/17/2013 09:18 PM, Pancho Mulongeni wrote:
Hi I have the following factor of dates that I want to converted to Date class
so I can extract the month
test.date
[1] 14/05/2012 14/05/2012 14/05/2012 14/05/2012 14/05/2012 14/05/2012
[7] 14/05/2012 14/05/2012 14/05/2012 14/05/2012
201 Leve
Hi all,
I would greatly appreciate if someone was so kind and share with us a
package or method that uses a regularized regression approach that balances
a regression model performance and model complexity.
That said I would be most grateful is there is an R-package that combines
Ridge (sum of sq
Hi I have the following factor of dates that I want to converted to Date class
so I can extract the month
> test.date
[1] 14/05/2012 14/05/2012 14/05/2012 14/05/2012 14/05/2012 14/05/2012
[7] 14/05/2012 14/05/2012 14/05/2012 14/05/2012
201 Levels: 01/10/2012 01/11/2012 01/12/2012 02/07/2012 ...
Molo uni-hohenheim.de> writes:
>
...
> *As there are some regions without neighbours in my data I use the following
> code to create the Weights Matrix:*
>
> > W_Matrix<- nb2listw(location_nbq, style="W", zero.policy=TRUE)
> >W_Matrix
>
> *And get this Output:*
>
...
> /("Error in print.listw
On Apr 17, 2013, at 10:41 , Patrick Burns wrote:
> There is a blog post about this:
>
> http://www.portfolioprobe.com/2012/07/26/r-inferno-ism-order-is-not-rank/
>
> And proof that it is possible to confuse them
> even when you know the difference.
It usually helps to remember that x[order(x)]
On 04/17/2013 07:11 PM, Katherine Gobin wrote:
Dear R forum
I have a data.frame
df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013", "4/12/2013"),
values = c(47, 38, 56, 92))
I need to to create a vector by repeating the dates as
"Current_date", 4/15/2013, 4/14/2013, 4/13/2013, 4
Hi.
Here are some examples that can maybe help you:
a <- "Current date"
b <- Sys.Date()-1:5
a
b
class(a)
class(b)
c(a,b)
mode(b)
as.numeric(b)
class(c(a,b))
c(a, as.character(b))
class(c(a,b))
class(c(a,as.character(b)))
Hope this helps.
On Wed, Apr 17, 2013 at 11:21 AM, Katherine Gobin wr
Hello,
Try the following.
rep(c("Current_date", as.character(df$dates)), 3)
Hope this helps,
Rui Barradas
Em 17-04-2013 10:11, Katherine Gobin escreveu:
Dear R forum
I have a data.frame
df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013", "4/12/2013"),
values = c(47, 38, 56,
Dear Andrija Djurovic,
Thanks for the suggestion. Ia m aware of "rep". However, here I need to repeat
not only dates, but a string "Current_date". Thus, I need to create a vector (
to be included in some other data.frame) with the name say "dt" which will
contain
dt
Current_date
4/15/2013
4/
hi Andrew.
gam does suppress the intercept, it's just that this doesn't force the
smooth through the intercept in the way that you would like. Basically
for the parameteric component of the model '-1' behaves exactly like it
does in 'lm' (it's using the same code). The smooths are 'added on' t
?rep
On Wed, Apr 17, 2013 at 11:11 AM, Katherine Gobin wrote:
> Dear R forum
>
> I have a data.frame
>
> df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013",
> "4/12/2013"), values = c(47, 38, 56, 92))
>
> I need to to create a vector by repeating the dates as
>
> "Current_date", 4/
Dear R forum
I have a data.frame
df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013", "4/12/2013"),
values = c(47, 38, 56, 92))
I need to to create a vector by repeating the dates as
"Current_date", 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, "Current_date",
4/15/2013, 4/14/2013
On Apr 17, 2013, at 10:16 , Shane Carey wrote:
> Can you resend this link please?
>
Psst:
https://stat.ethz.ch/pipermail/r-help/2013-April/351669.html
--
Peter Dalgaard, Professor
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)381
There is a blog post about this:
http://www.portfolioprobe.com/2012/07/26/r-inferno-ism-order-is-not-rank/
And proof that it is possible to confuse them
even when you know the difference.
Pat
On 16/04/2013 19:10, Julio Sergio wrote:
Julio Sergio gmail.com> writes:
I thought I've understoo
Can you resend this link please?
Thanks
On Tue, Apr 16, 2013 at 10:33 PM, Jim Lemon wrote:
> On 04/17/2013 03:25 AM, Sarah Goslee wrote:
>
>> ...
>> Ouch.
>>
>> (Note: I know nothing about the site, the author of the article, or
>> the study in question. I was pointed to it by someone else. Bu
Thanks a lot:)
Sent from my iPad
On Apr 16, 2013, at 10:15 PM, arun wrote:
> Hi Farnoosh,
> YOu can use either ?merge() or ?join()
> DataA<- read.table(text="
> ID v1
> 1 10
> 2 1
> 3 22
> 4 15
> 5 3
> 6 6
> 7 8
> ",sep="",header=TRUE)
>
> DataB<- read.
88 matches
Mail list logo
