Perhaps:
G10 <- predict(wei, list(Group =10 , UsefulLife =10), type="response")
You should get a single km-type object and will need to extract a time-specific
value.
(You really _should_ post what you have tried.)
--
David
Sent from my iPhone
Sent from my iPhone
> On Mar 14, 2014, at 9:11
Perhaps:
G10 <- predict(wei, list(Group =10 , UsefulLife =10), type="response")
You should get a single km-type object and will need to extract a time-specific
value.
(You really _should_ post what you have tried.)
--
David
Sent from my iPhone
> On Mar 14, 2014, at 7:11 PM, shilpi harpavat
Perhaps:
G10 <- predict(wei, list(Group =10 , UsefulLife =10), type="response")
You should get a single km-type object and will need to extract a time-specific
value.
(You really _should_ post what you have tried.)
--
David
Sent from my iPhone
> On Mar 14, 2014, at 7:11 PM, shilpi harpavat
Hello,
I am using the function survfit in the 'survival' package. Calling the function
produces the median survival
time automatically, as below.
sleepfit <- survfit(Surv(timeb, death)~1)
> sleepfit
Call: survfit(formula = Surv(timeb, death) ~ 1)
records n.max n.start events median 0.95LCL 0
On Fri, 14 Mar 2014, Duncan Murdoch wrote:
On 14-03-14 8:59 PM, Mike Miller wrote:
What I'm using:
R version 3.0.1 (2013-05-16) -- "Good Sport"
Copyright (C) 2013 The R Foundation for Statistical Computing
Platform: x86_64-unknown-linux-gnu (64-bit)
That's not current, but it's not very old.
On 14-03-14 8:59 PM, Mike Miller wrote:
What I'm using:
R version 3.0.1 (2013-05-16) -- "Good Sport"
Copyright (C) 2013 The R Foundation for Statistical Computing
Platform: x86_64-unknown-linux-gnu (64-bit)
That's not current, but it's not very old...
According to some docs, options(digits)
What I'm using:
R version 3.0.1 (2013-05-16) -- "Good Sport"
Copyright (C) 2013 The R Foundation for Statistical Computing
Platform: x86_64-unknown-linux-gnu (64-bit)
According to some docs, options(digits) controls numerical precision in
output of write.table(). I'm using the default value f
David Carlson's solution is obviously better! (sorry, I just saw it)
David Stevens
On 3/14/2014 4:57 PM, David Stevens wrote:
Renalda - your figure got stripped out. I'll assume you mean to have
depth on the vertical axis with a zero at the top and increasing (+)
depth as you go down. This is
Renalda - your figure got stripped out. I'll assume you mean to have
depth on the vertical axis with a zero at the top and increasing (+)
depth as you go down. This is pretty simple to do.
1) use the (-) of depth on the y-axis
2) make the plot without the y-axis labels (yaxt='n') using
ylim=c
Make your years numeric, not character.
Then
plot(Year, AnnualInflation, type='l')
(don't need the data frame for this plot)
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 3/14/14 11:15 AM, "Miguel A. Buitrago"
wrote:
Dear R-helpers,
here is a novice question. I want to plot a two variable graph: year and
Inflation Rate.
I created two vectors which I combined into a data frame:
Year <- c("1990", "1991", "1992", "1993", "1994", "1995", "1996", "1997",
"1998", "1999", "2000", "2001", "2002", "2003", "2004", "20
Greetings,
I'm running a series of Chi-square tests to examine differences across
categorical variables. The situation is this:
I have three variables: sex (M/F), habitat (5 levels), season
(W,Sp,Su,F). A Cochran-Mantel-Haenzel test detects non-indepedence
across my sex strata. I then sub
On 03/14/2014 06:55 PM, Renalda Munubi-Misinzo wrote:
I would like to have a plot with values in a reversed order (as shown in
fig 2). I tried to plot the graph in R using the following codes
plot(density,depth,pch=as.numeric(species),xlab="fish density
(#/m2)",ylab="water depth (m)")
legend(
Often, there is a mix of information available from the various studies that
needs to be used to compute the effect sizes or outcomes to be used for the
meta-analysis. Then you have to 'build up' your dataset in multiple steps and
you cannot bypass first using escalc().
As a very basic example,
Dear all,
I want to describe the time interval between mammography screenings as a
function of several clinical and demographic variables. For each woman (ID), I
have multiple events and each event corresponds to the time interval between
two consecutive mammography screenings. Also, women's da
R-help strips attachments so your post has three problems:
1. No plot to look at
2. No data to replicate your code
3. Email is in html format instead of plain text
Try again using plain text mail, copy the results of
dput(your.data.frame) into your email and provide code that
generates the figure
Hi,
Please use ?dput() to show the data.
dat <- read.table(text="id name seq
101 as 1
102 ld 2
103 eg 3
101 as 4
104 bs 5
103 eg 6
105 ka 7
104 bs 8",sep="",header=TRUE,stringsAsFactors=FALSE)
res1 <- aggregate(seq~.,data=dat,FUN=I)
res1 <- res1[order(res1$id),] ##creates the column 'seq'
Ummm...
totalfunc(0)
totalfunc(100)
totalfunc() always gives the same result, because its argument is not
used in the function body. You probably want something like:
totalfunc <- function (y,z) {Reduce(func1, x = y,
accumulate = TRUE, init = z)
}
Also note that yo
Hi,
You could use ?strsplit()
dat <- read.table(text="'Column 1'
'AAA--> BBB'
'CCC--> DDD'",sep="",header=TRUE,check.names=FALSE,stringsAsFactors=FALSE)
dat1 <- setNames(as.data.frame(do.call(rbind,strsplit(da
Hi,
Try:
res2 <- res1[!is.na(res1$Amount),]
A.K.
On Friday, March 14, 2014 3:41 AM, dila radi wrote:
Hi all,
Regarding the previous post, here is part of my data.
structure(list(Year = c(1949L, 1949L, 1949L, 1949L, 1949L, 1949L,
1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L,
If possible I would like to apply a function, that contains a "reduce"
statement, across the rows of 2d array. This code works for me right up
until the last. Can it be done or is there a better way? My next challenge
is to pass it a vector of starting values, i.e. each row will have a
different in
Dear readers,
I am currently trying to estimate some panel data models in R using PLM
package. This includes the estimation of basic pooled, fixed effects and
random effects models. Therefore I make use of this code:
Now here's the problem:
Now here's the problem: I can without any problem esti
Hi,
I am fitting a weibull model as follows
my models is
s <- Surv(DFBR$Time,DFBR$Censor)
wei <- survreg(s~Group+UsefulLife,data = DFBR,dist="weibull")
How can i predict the probabilty of failure in next 10 days, for a new data
with group =10 and usefuleLife =100
Thx
shilpi
Something like:
tmp <- boxplot(V ~ date, data=pippo, plot=FALSE)
bxp(tmp, at=sort(unique(pippo$date))
You may need to adjust the x-axis limits and the box widths.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 3/13/1
I would like to have a plot with values in a reversed order (as shown in
fig 2). I tried to plot the graph in R using the following codes
plot(density,depth,pch=as.numeric(species),xlab="fish density
(#/m2)",ylab="water depth (m)")
legend(locator(n=1),legend=c("Tropheus brichardi","Petrochromis
I am getting odd results from the bayesglm function in the ARM package.
Running R 3.0.2 and downloaded the most resent ARM package. Every dataset I
try runs, but the output is incorrect - I get a null deviance that is less than
the residual deviance. I have tested this with datasets I have us
I would like to have a plot with values in a reversed order (as shown in
fig 2). I tried to plot the graph in R using the following codes
plot(density,depth,pch=as.numeric(species),xlab="fish density
(#/m2)",ylab="water depth (m)")
legend(locator(n=1),legend=c("Tropheus brichardi","Petrochromis
Thank you so much for all your answers my problem was solved.
I really appreciate all your emails and really fast reaction.
All the best,
Malgorzata Gazda
Date sent: Thu, 13 Mar 2014 22:38:54 +0100
From: Arunkumar Srinivasan
To: arun , Dennis Murphy
Copies to: R help
Subject:
Hi all,
Regarding the previous post, here is part of my data.
structure(list(Year = c(1949L, 1949L, 1949L, 1949L, 1949L, 1949L,
1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L,
1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1949L,
1949L, 1949L, 1949L, 1949L, 1949L, 1949L, 1
Dear All
As you can specify the data directly to rma.uni via n1i, m1i, sd1i, etc in
Metafor, why would you ever want to use escalc to calculate yi and vi? Aren't
these just intermediate steps to the final pooled effect size which is
calculated by rma.uni; or is there some advantage to calcul
Perhaps this blog post will be of use:
http://www.r-bloggers.com/generating-balanced-incomplete-block-designs-bibd/
HTH,
-- David
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of n.hub...@ncmls.ru.nl
Sent: Friday, March 14, 2014 4
You could fit a linear model to original/predicted y values and get rsquared
from that.
Chris
On Mar 13, 2014 5:26 PM, Greg Snow <538...@gmail.com> wrote:
>
> Well if I had it and you asked nicely, then I would be happy to give
> it to you. Oh, you mean the gls function, not GLS as my initials
Hi,
using data.table in package code i.e
setkey(aligtable,transition_group_id,align_origfilename)
aligtable = aligtable[CJ(unique(transition_group_id),
unique(align_origfilename))]
When I do run R package check I got warnings such as:
convert2msExperiment: no visible binding for global variable
On Thu, 13 Mar 2014, Tim Marcella wrote:
Hi,
I am working with hurdle models in the pscl package to model zero
inflated overdispersed count data and want to incorporate censored
observations into the equation. 33% of the observed positive count data
is right censored, i.e. subject lost to fo
On 03/14/2014 08:19 PM, al Vel wrote:
Hi,
Exactly the matrix transformation is the one that pasted in my first
mail. That is here:
new_point <- function(x1, x2, y1, y2, grad=1.73206){
b1 <- y1-(grad*x1)
b2 <- y2-(-grad*x2)
M <- matrix(c(grad, -grad, -1,-1), ncol=2)
intercepts <- as.matrix(c(b1,b
Hi,
Exactly the matrix transformation is the one that pasted in my first mail.
That is here:
new_point <- function(x1, x2, y1, y2, grad=1.73206){
b1 <- y1-(grad*x1)
b2 <- y2-(-grad*x2)
M <- matrix(c(grad, -grad, -1,-1), ncol=2)
intercepts <- as.matrix(c(b1,b2))
t_mat <- -solve(M) %*% intercep
I am wondering whether anyone could explain what'd be the difference between
running a 'generalized additive regression' versus 'generalized linear
regression' with splines.
The smooth terms in mgcv::gam are represented using *penalized*
regression splines, with the degree of penalization sel
Dear all,
This is something for the combinatorics freaks amongst you (which I am
certainly not :-).
I do the following problem to solve and was wondering if this can be easily
done using R:
I have a set of, let's say, 10 variables: n(varibale) = 10
e.g. var <- c(a,b,c,d,e,f,g,h,i,j)
On 13 Mar 2014, at 22:26 , Greg Snow <538...@gmail.com> wrote:
> Oh, you mean the gls function, not GLS as my initials (my
> parents are OLS and WLS, perhaps I was destined to regress), sorry.
Fortune candidate?
> The gls function in the nlme package (is that the one that you are
> asking about
On 03/14/2014 07:05 PM, al Vel wrote:
Hi,
I was very clear in the mail that i want to create a four edged
Barycentric diagram like Ternary plot (like the rhombus in the piper
diagram (http://en.wikipedia.org/wiki/Piper_diagram)). The idea is to
graphically depicts the ratios of the four variabl
Thanks for very quick reply
Cheers
Frederik Borup
On Fri, Mar 14, 2014 at 6:09 AM, Jim Lemon wrote:
> On 03/14/2014 06:52 AM, Frederik Borup wrote:
>
>> Dear all,
>>
>> I want to create a boxplot with whiskers. I want to compare several
>> studies. For each study I have
>>
>> 1. mean
>
Hi,
I was very clear in the mail that i want to create a four edged Barycentric
diagram like Ternary plot (like the rhombus in the piper diagram (
http://en.wikipedia.org/wiki/Piper_diagram)). The idea is to graphically
depicts the ratios of the four variables as positions in an rhombus/diamond
s
42 matches
Mail list logo
