Dear all,
I generate data under IRT mdel. I conducted 1000 replications. When I run, some
replication was not fit with my model. So, replications were fit model were
less than 1000 replication. If I want its run until 1000 replicaions. How
shouldI write functin.
Thank you,
Kamontip
On 30-11-2012, at 07:18, C W wrote:
thanks, Berend. Both of your code works great. Is there a function that can
do it?
Something like this:
x - matrix(NA, nrow=15, ncol=2)
for(i in 1:15){
x[i,] - sample(c(NA, 20, 77), 2, prob=c(0.2, 0.3, 0.4))
}
x
[,1] [,2]
[1,]
Just a lame question: is there any chance to generate SVG maps with
googleVis and import that in LaTeX?
Best,
Gergely
On Fri, Nov 30, 2012 at 2:02 AM, Yihui Xie x...@yihui.name wrote:
Then you can take a look at the ggmap (on CRAN) or snippets package
(on Simon's RFroge:
If you mean Little's Chi Sq test in particular, go to the BaylorEdPsych package.
Regards,
José
José Iparraguirre
Chief Economist
Age UK
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Sileshi Melesse
Sent: 30 November 2012 05:57
kernlab package version 0.9-15 has been posted on CRAN,
it contains several improvements and fixes along with new features e.g.
variance estimation for Gaussian Process regression.
cheers
Alexandros
___
R-packages mailing list
anna.fechner at de.pwc.com writes:
Dear R community,
I'm trying to graphically illustrate my data with a worlmap.
Unfortunately, my data is partly on country basis and partly on regional
basis (e.g. certain African countries are aggregated to one region). I am
using the package
Dear R users developers,
I coming across the following issue since R 2.15.2 32-bit (running on
Windows XP 32.bit; some output left out for conciseness):
setInternet2(TRUE)
require(rJava)
.jinit()
getCRANmirrors()
system(ls , intern = TRUE)
Error in system(ls, intern = TRUE) : CreateThread
Dear All,
#I have the following code
Dose-1000
Tinf -0.5
INTERVAL -8
TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL)))
TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL)))
TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL)))
TIME5 -matrix(c((0*INTERVAL):(4*INTERVAL)))
TIME4 -matrix(c((0*INTERVAL):(5*INTERVAL)))
TIME3
Hello,
You don't have to exchange 'object' by the name of your model, you call
the function with the name of your model:
x - 1:20
y - x + rnorm(20)
fit - lm(y ~ x)
ci_lm(fit)
beta lowerupper
(Intercept) 0.6741130 -0.9834827 2.331709
x 0.9575906 0.8192171
Hello R-Users, Hello R-help-team.
I found a nice way to create a xy-plot with a single log-axis (e.g. the
y-axis). This is often needed to show biological data.
First attach the desired dataset you want to plot.
Now use the following commands:
plot(x,y,log=y, yaxt=n, ylim=c(0.1,10), ... )
Hi
I see a changed behaviour in xts indexed on class Date in the latest
versions, versus 2.
It seems to be related to changes to/from daylight savings time,
happens those weekends.
Is it not intended that class Date be used like this, or is this new
behaviour incorrect?
Giles
Example:
Hello
I have dataframe
101 2008-07 0.2898966
102 2008-08 0.3101667
103 2008-09 0.3730476
104 2008-10 0.2717037
105 2008-11 0.1344286
106 2008-12 0.1375000
107 2009-01 0.1781000
108 2009-02 0.2146667
109 2009-03 0.2808235
110 2009-04 0.4326250
111 2009-05 0.3420741
112 2009-06 0.2675238
113 2009-07
Hi R Users!
I am having some difficulties in using the TSCov function from RTAQ package
that should calculate the two time scale realized volatility (Zhang et al,
2005).
Let's suppose I have tick by tick data, let's say aaa and bbb.
If I write in R:
/stock1=aaa$PRICE
stock2=bbb$PRICE
Dear R users,
given the patient sample with their Gender and Age
GENDER Age
[1,] 2 45
[2,] 1 58
[3,] 1 54
[4,] 2 71
[5,] 2 64
...
I would like to create an another column which groups the patients wrt
Gender specific Age quantiles, following
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__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Hi,
This is sasi.
I installed R 64 bit and oracle client 64 bit in fedora 17. After R is
working fine. Oracle database also working fine. but problem is, If i try
to connect oracle from R its not connecting.
So, pls guide me how to connect from R to oracle.
how, i tried:--
odbcinst.ini file
Hi,
this is likely very easy to do but I don't find the right terms to search
for the solution.
I want to show in a bar chart how much different categories deviate from the
average percentage change.
Thus, in the following example I would like to set the zero line or base or
whatever it is
Hello, Everyone.
Does anyone know how to create a Nightingales Rose chart by using R?
Hopefully, the graph could be displayed like this:
http://mbostock.github.com/protovis/ex/crimea-rose.html
Thanks a lot.
Kind regards,
Henry
[[alternative HTML version deleted]]
Hello,
I have the starting date in UNIX_TIME LONG number (number of milliseconds
since 1970 jan 1st).
Is there a way to generete the sequence of dates with monthly interval in
POSIXct format like
2000-01-01
2000-02-01
2000-03-01
etc
Thanks
[[alternative HTML version deleted]]
HI,
Use,
?na.omit()
na.omit(data)
A.K.
- Original Message -
From: Vasilchenko Aleksander vasilchenko@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, November 30, 2012 7:07 AM
Subject: [R] missed values
Hello
I have dataframe
101 2008-07 0.2898966
102 2008-08 0.3101667
103
Hi,
I was wondering if it's possible to separate elements in multiple rows
that actually should appear in different columns. I have a file where
in certain lines there are elements not separated, and they certainly
should appear in different columns (an example of the file is
attached).
Hello,
Take a look at the graph in
http://gallery.r-enthusiasts.com/graph/Rose_diagram,97
Hope this helps,
Rui Barradas
Em 30-11-2012 12:24, Henry Smith escreveu:
Hello, Everyone.
Does anyone know how to create a Nightingales Rose chart by using R?
Hopefully, the graph could be displayed
Hello,
Try the following.
Y - 2000:2001
M - 1:12
D - 1
x - expand.grid(Y, M, D)
Dates - sort(as.Date(apply(x, 1, paste, collapse = -)))
Hope this helps,
Rui Barradas
Em 30-11-2012 12:26, Vasilchenko Aleksander escreveu:
Hello,
I have the starting date in UNIX_TIME LONG number (number of
Hello,
Try the following.
x - scan(text=
-100 -100-3456-3456-3456-100 -100
23 -3456-3456-189 34 56 78
-100 34 56 21 44 65 78
, what = )
fun - function(x){
f - function(.x){
if(grepl(-[[:digit:]]+, .x)){
g - gregexpr(-[[:digit:]]+, .x)
.y -
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Stodola, Kirk
Sent: Thursday, November 29, 2012 5:32 PM
To: r-help@r-project.org
Subject: [R] Deleting certain observations (and their imprint?)
I'm manipulating a large
Hi
try to make your example easier and reproducible.
I get
CONCF -matrix(c(CONC1+CONC2+CONC3+CONC4+CONC5+CONC6+CONC7+CONC8))
Error in CONC1 + CONC2 : non-conformable arrays
so I cannot understand what shall the final plot look like.
Regards
Petr
-Original Message-
From:
On 30-11-2012, at 13:55, Jaime Otero Villar wrote:
Hi,
I was wondering if it's possible to separate elements in multiple rows that
actually should appear in different columns. I have a file where in certain
lines there are elements not separated, and they certainly should appear in
Thanks Eric. It would be good to show your entire script next time as stated
in the posting guidance.
Regarding matching with SPSS please describe the bootstrapping algorithm
used there. In rms I do the unconditional bootstrap, i.e., I sample with
replacement from the rows of the raw data. And
I think your request is internally inconsistent: you want units of percent but
you want to show the average(I assume you intend mean)?
You can subtract the mean from each value and then divide by the mean, then
subtract 1, then multiply by 100. You inherently lose visibility to what the
mean
Hello,
I have a variable in a data frame that contains NA values. I just want to
subset so that I get the obs where that variable is missing.
In SAS I would do:
data missing;
set test;
if myvalue=' ';
run;
How can I perform this simple task in R?
Thanks in advance for your help.
--
On Nov 30, 2012, at 9:27 AM, ramoss ramine.mossad...@finra.org wrote:
Hello,
I have a variable in a data frame that contains NA values. I just want to
subset so that I get the obs where that variable is missing.
In SAS I would do:
data missing;
set test;
if myvalue=' ';
run;
How
Hello user,
I have large data containing subject id, time and response where
subjects are measured repeatedly. However some time are duplicates. I
only want data with unique time points per id. I mean if time is
repeated, then take only one.
Here is a sample data.
id timeres
1 2
Hi Andras
Now it works but I suppose you shall simplify or explain it a bit more.
CONC7-c(rep(0,3),CONC7)
CONC8-c(rep(0,6),CONC8)
CONC6+CONC7+CONC8
[,1]
[1,] 0.00
[2,] 2.330056
[3,] 2.133774
[4,] 1.954027
[5,] 4.119477
[6,] 3.772456
[7,] 3.454667
[8,] 5.493705
[9,] 5.030920
[10,]
Hi
see ?is.na
x -sample(c(1:3, NA), 20, replace=T)
x
[1] 2 NA 2 3 3 3 3 2 3 NA 3 2 1 2 NA 3 3 3 2 2
y-rnorm(20)
y[is.na(x)]
[1] 0.1600417 1.3264063 -0.6175832
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
Le vendredi 30 novembre 2012 à 07:27 -0800, ramoss a écrit :
Hello,
I have a variable in a data frame that contains NA values. I just want to
subset so that I get the obs where that variable is missing.
In SAS I would do:
data missing;
set test;
if myvalue=' ';
run;
How can I
?seq.Date
seq(as.Date(2000-01-01), len=5, by=1 mon)
[1] 2000-01-01 2000-02-01 2000-03-01 2000-04-01 2000-05-01
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Vasilchenko Aleksander
Sent: Friday, November 30, 2012 6:27 AM
Hello Bikek,
please use dput() next time to provide the data, its easier to use that.
also: looking at the data provided, how would you want to decide which value of
the non-unique times to retain? Just take the first one? They aren't all the
same.
On 30.11.2012, at 16:59, bibek sharma wrote:
Hi,
Try this:
Lines--100 -100-3456-3456-3456-100 -100
23 -3456-3456-189 34 56 78
-100 34 56 21 44 65 78
res-unlist(strsplit(gsub(\\-, -,Lines),\n))
res1-do.call(rbind,lapply(lapply(split(res,seq_along(res)),function(x)
unlist(strsplit(x, ))),function(x) as.numeric(x[x!=])))
res1
## [,1] [,2]
Hi,
Try this:
dat1-read.table(text=
id time res
1 2 0.64
1 3 0.78
1 3 6.5
1 3 4.5
1 4 4
1 5 3.4
2 10 5.7
2 11 5.8
2 11 9.3
2 11 3.4
2 12 3.4
2 13 6.7
3 3 5.6
3 3 3.4
3 4 2.3
3 5 5.6
3 12
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of bibek sharma
Sent: Friday, November 30, 2012 5:00 PM
To: R-help@r-project.org
Subject: [R] For loop
Hello user,
I have large data containing subject id, time and
I found the answer;
Its mymissing - subset(mydata,is.na(myvar))
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Dear all,
I am using the zeroinfl() function from the pscl package to develop a
zero-inflated Poisson GLM. I would like to calculate the standard errors of
predicted values. I've tried code posted in a previous discussion on this topic
On 11/29/2012 2:53 PM, Filoche wrote:
Dear R users.
I'm currently making a report with knitr (RStudio) where I would like to
plot a googleVis map. However, the map generated is an HTML file which I
don't know how to integrate it in my report.
So my question is how to include a map generated
You could use the duplicated function maybe:
mtest
id time value
[1,] 13 1
[2,] 13 2
[3,] 12 3
[4,] 11 4
[5,] 21 5
[6,] 23 6
[7,] 23 7
[8,] 23 8
duplicated(mtest[,1:2])
[1] FALSE TRUE FALSE FALSE FALSE FALSE TRUE
Hi,
You could also try this:
Lines--100 -100-3456-3456-3456-100 -100
23 -3456-3456-189 34 56 78
-100 34 56 21 44 65 78
Lines1-readLines(textConnection(Lines))
res1-unlist(strsplit(gsub([-], -,Lines2), ))
matrix(as.numeric(res1[res1!=]),nrow=length(Lines1),byrow=TRUE)
[,1] [,2] [,3] [,4]
Hi
ifelse shall work just fine.
x-1:10
a-rnorm(10)
b-rnorm(10)+100
tinf-5
ifelse(xtinf, a,b)
[1] -0.32157732 -0.11065857 0.07127818 0.35928359 100.06200507
[6] 98.77752570 100.92743376 99.73918386 99.79837539 101.32640626
It is only your value of Tinf which is 0.5 compared to TIME
Hi,
May be this helps:
dat1-read.table(text=
---data---
,header=TRUE,stringsAsFactors=FALSE)
library(zoo)
dat1$date-as.yearmon(dat1$date,format=%Y-%m)
lm1-lm(value~date,dat1)
dat2-data.frame(date=dat1[,1])
dat1$fit-predict(lm1,newdata=dat2)
Hello R-experts,
I would like to know if there is a solution to read files with extension
.gsheet directly into R - see http://www.fileinfo.com/extension/gsheet for more
info on this file format.
Thank you,
Luca
Mr. Luca Meyer
www.lucameyer.com
R 2.15.1
Mac OS X 10.8.2
Suppose I have the following square, non-negative matrices
g=matrix(c(0,2,4,0.25,0,0,0,0.6,0),3,3,byrow=T);
I want to create a list where this matrix is repeated multiple times. if I
do this brute force (manually), using
env - list(g,g,g)
works fine. Yields
[[1]]
[,1] [,2] [,3]
[1,]
You are so close:
rep(list(g), 3)
[[1]]
[,1] [,2] [,3]
[1,] 0.00 2.04
[2,] 0.25 0.00
[3,] 0.00 0.60
[[2]]
[,1] [,2] [,3]
[1,] 0.00 2.04
[2,] 0.25 0.00
[3,] 0.00 0.60
[[3]]
[,1] [,2] [,3]
[1,] 0.00 2.04
[2,] 0.25 0.00
[3,] 0.00 0.60
Hi,
Try this:
lapply(1:3,function(x) g)
A.K.
- Original Message -
From: Anser Chen anser.c...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, November 30, 2012 12:50 PM
Subject: [R] repeating matrices in a list
Suppose I have the following square, non-negative matrices
Hi,
I have a question regarding the cv.glm function in the package boot. What is
exactly the cost? Is it the threshold value for an estimated value to be
classified as either 0 or 1? I have troubles understanding the explanation
in R. Lets say I want all estimations 0.65 to be classified as 1s
Petr,
sorry, here is a better example that should work:
Dose-200
Tinf -0.5
INTERVAL -3
TIME8 -matrix(c((0*INTERVAL):(1*INTERVAL)))
TIME7 -matrix(c((0*INTERVAL):(2*INTERVAL)))
TIME6 -matrix(c((0*INTERVAL):(3*INTERVAL)))
CDURINF6 -((Dose/Tinf)*(1/(0.088*76.9)))*(1-exp(-0.088*TIME6))
CAFTINF6
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1 2 3 4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
for(j in 1:ia) {
H = function(x) {sen(x) + a[j]}
for(i in 1:ib) {
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1,2,3,4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
int1 = numeric(ib)
b = numeric(ib)
for(j in 1:ia) {
H = function(x)
Hi,
I would like to create calendarheat map with Hours of day on Y axis and
Date on X axis.
Is there any function to do so? Can someone help me with reference.
Thanks
Sharan
[[alternative HTML version deleted]]
__
R-help@r-project.org
Hi,
YOu could also use:
set.seed(5)
dat1-data.frame(col1=sample(c(1:4,NA),10,replace=TRUE),col2=runif(10,0,1))
dat1[!complete.cases(dat1),]
# col1 col2
#3 NA 0.3184040
#8 NA 0.8878698
#9 NA 0.5549226
A.K.
- Original Message -
From: ramoss ramine.mossad...@finra.org
To:
Hi, thanks a lot.
The package from your link seems could plot only one type of variable.
If I have following three variables:If I have following three
variables:
a b c
1 3 7 8
2 5 4 1
3 5 1 7
4 3 1 5
5 1 7 3
6 7 1 1
7 1 1 1
8 3 3 2
Do you have any idea how to plot it by using variable
Petr,
thanks for the solution, putting the zeros in there makes it work just how I
want it. Also, thanks for picking up on the error in my code below... I did not
notice that it only changes the 1st value, and yes, your assumption is right:
there is a chance where more values should be
When plotting a heatmap with heatmap.2 the shortest euclidean distance
cluster is plotted in the top left corner of the heatmap
http://r.789695.n4.nabble.com/file/n4651430/heatmap.2.jpeg
when plotting the same data with pheatmap the very same cluster is in the
bottom right corner
Hello everybody,
I have written a script with two inline cfunctions. The script crashes from
time to time with:
*** caught segfault ***
address 0x10, cause 'memory not mapped'
The crashs happen within R code after the cfunctions were executed.
Nevertheless I think
that the pointers in my
Dear Dennis,
Many thanks!
Yes, ggplot2 could be used to illustrate a simple rose chart (category +
one type variable)
If I have a data frame like this:
DF - data.frame(month = factor(month.abb, levels = month.abb),
freq1 = rpois(12, 80),freq2=rpois(12,
80),freq3=rpois(12, 80))
Thank you Michael.
I'll try that. Maybe I can find a solution to convert the HTML file into a
let's say SVG file.
Regards,
Phil
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Sent from the R help mailing list archive at
Thank everyone for the answers.
I looked into ggmap but could not find how to plot a map that would like
this:
http://www.google.ca/trends/explore#q=flu
I looked into get_map but it can only be of this type:
maptype = c(terrain, satellite, roadmap, hybrid, toner,
watercolor)
Any suggestions
Now i've managed to do this:
funcs - list()
funcs[]
# loop through to define functions
for(i in 1:ib-1){
# Make function name
funcName - paste( 'func', i, sep = '' )
# make function
func = paste('function(x){sin(x + a[', i,'])))}',sep = '')
funcs[[funcName]] =
a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9)
b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1)
cor(a,b)= -1
a'=qbinom(a, 1, 0.5)
b'=qbinom(b, 1, 0.5)
why cor(a',b') becomes -0.5 ?
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Sent from the R help
Hello,
If you want Age quantiles by gender, you have to split the data by
gender, apply the same code then recombine the result.
fun - function(x){
Age_group - cut(x[, Age], labels=c(1:10),
breaks=quantile(x[, Age], seq(0,1,.1)),
include.lowest = TRUE)
cbind(x,
HI,
You could also use:
res1-sapply(1:3,function(x) g,simplify=FALSE)
#or
res2-replicate(3,g,simplify=FALSE)
identical(res1,res2)
#[1] TRUE
A.K.
- Original Message -
From: Anser Chen anser.c...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Friday, November 30, 2012 12:50 PM
Subject:
Hello,
Your code doesn't run without initializing 'ss' to something. And I've
made some changes, but I don't understand what you are trying to do. See
comments inline.
a = c(1,2,3,4)
ia = length(a)
x = seq(1, 100, by=0.1) # It was 'length = 0.1' (!)
ib = length(x)
ss - numeric(ia) #
On 30-11-2012, at 16:08, faeriewhisper wrote:
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1,2,3,4)
ia = length(a)
x = seq(1,100,length=0.1)
ib = length(x)
int1 =
Thats not a very precise question. I'll try anyway..
- if you use c, you need to separate the values by commas
- i think you mean seq(1,100,0.1), otherwise x only has one value
- function sen is not defined
- If you call int(1), upper will be 1, not x[i]
- why are you making a function and
On 30-11-2012, at 19:34, Berend Hasselman wrote:
On 30-11-2012, at 16:08, faeriewhisper wrote:
Hi guys!
I have to compute something and i don't know what i'm doing wrong. my code
is a bit complex, but imagine that is something like this:
a = c(1,2,3,4)
ia = length(a)
x =
On 30/11/2012 12:02 PM, Donatella Quagli wrote:
Hello everybody,
I have written a script with two inline cfunctions. The script crashes from
time to time with:
*** caught segfault ***
address 0x10, cause 'memory not mapped'
The crashs happen within R code after the cfunctions were
On Fri, Nov 30, 2012 at 9:47 AM, jaybell stephe...@yahoo.com.tw wrote:
a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9)
b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1)
cor(a,b)= -1
a'=qbinom(a, 1, 0.5)
b'=qbinom(b, 1, 0.5)
why cor(a',b') becomes -0.5 ?
On my computer the correlation is
Hi,
You could try this:
Sys.setenv(TZ=GMT)
a-as.Date(15423:15426)
x-xts(seq_along(a),a)
print(index(x))
#[1] 2012-03-24 2012-03-25 2012-03-26 2012-03-27
print(as.numeric(index(x)))
#[1] 15423 15424 15425 15426
A.K.
- Original Message -
From: Giles giles.heyw...@cantab.net
To:
Hello,
Arun's code is much better.
Rui Barradas
Em 30-11-2012 18:50, arun escreveu:
Hi,
You could also try ?ave()
dat$Age_group-ave(dat$Age,dat$GENDER,FUN=function(x){cut(x,labels=1:10,breaks=quantile(x,seq(0,1,.1)),include.lowest=TRUE)})
dat
# GENDER Age Age_group
#1 2 45 1
I think this does it.
require(reshape2)
df1 - melt(DF, id=month)
ggplot(df1, aes(x = month, y = value, fill = variable)) +
theme_bw() +
geom_bar(stat = identity) +
coord_polar()
John Kane
Kingston ON Canada
-Original Message-
From: henry.helsi...@gmail.com
Hi,
You could also try ?ave()
dat$Age_group-ave(dat$Age,dat$GENDER,FUN=function(x){cut(x,labels=1:10,breaks=quantile(x,seq(0,1,.1)),include.lowest=TRUE)})
dat
# GENDER Age Age_group
#1 2 45 1
#2 1 58 10
#3 1 54 1
#4 2 71 10
#5 2 64
Hi Frank,
My apologies for not posting the entire script - I have repasted it below.
library(rms)
library(foreign)
temp=read.spss('coxdata.sav', to.data.frame=T)
formula=Surv(months, recidivate) ~ fac1 + fac2 + fac3 + fac4 + fac5 + fac6 +
fac7 + fac8
fit=cph(formula, data=temp, x=T, y=T)
hello guys!
thank u for the help, but u didnt understood what i need.
1st, it is a[i] cuz i want to sum 1 + x[i], for all i's not j.
but i've solved it! :)
like i said, my code is more complex, but, if you need to integrate several
functions in a loop, thats what you should do:
w2 =
m - matrix(nrow=5, ncol=5)
m - ifelse(row(m)==col(m), 1, 0.2)
c - chol(m)# Choleski decomposition
u - matrix(rnorm(2000*5), ncol=5)
uc - u %*% c
cr - pnorm(uc)
cr - qbinom(cr,1,0.5)
cor(cr)
I expected that the cor(cr) to be 0.2
On Nov 30, 2012, at 10:34 AM, jaybell wrote:
m - matrix(nrow=5, ncol=5)
m - ifelse(row(m)==col(m), 1, 0.2)
c - chol(m)# Choleski decomposition
u - matrix(rnorm(2000*5), ncol=5)
uc - u %*% c
cr - pnorm(uc)
cr - qbinom(cr,1,0.5)
cor(cr)
It will be crucial to know the details of the test statistic and P-value
calculations from SPSS. It's also running anova on both the bootcov and the
original fits to see if SPSS is ignoring the bootstrap when computing the
covariance matrix.
Frank
Eric Claus wrote
Hi Frank,
My apologies for
Is it possible to get a line number with an error report?
I have a long script and an error:
Error in `[.xts`(x, xsubset) : subscript out of bounds
It would be very helpful, and save a lot of time, if there was some
indication in the error message which line the error was.
I can find it using
One thing that I do when I have a long script is to put progress
report messages. These have some comments so I can chart the progress
and also print out the current CPU and memory usage so I can also
isolate where potential problems might be. This will help narrow down
the section of code where
On Nov 30, 2012, at 1:22 PM, Worik R wrote:
Is it possible to get a line number with an error report?
I have a long script and an error:
Error in `[.xts`(x, xsubset) : subscript out of bounds
It would be very helpful, and save a lot of time, if there was some
indication in the error
Hello R usuer,
The code given below superimposes a pie diagram on another plot
containing some points. However, I would like to center the pie
diagram on the xy location on the plot, but not on the center. is
there any way to re-center pic diagram.
Any suggestion or better alternative are highly
Hello,
Package sos does a good job at finding things available for R.
library(sos)
findFn('gsheet')
found 0 matches
x has zero rows; nothing to display.
Warning message:
In findFn(gsheet) : HIT not found in HTML; processing one page only.
So I guess there's nothing yet.
Hope this helps,
Try subplot() from the TeachingDemos package. E.g.,
plot(c(1,2,3), c(1,3,2), xlim=c(0,5), ylim=c(0,10))
subplot(pie(1:5), x=4, y=6)
abline(h=6, v=4)
subplot(pie(1:5), x=1, y=8, size=c(0.5, 0.5))
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
Hi,
I don't know if the information is still current, but there were some
discussions a while back pertaining to reading Google Docs documents in R.
There is a blog post here from David Smith that might be helpful:
Dear list,
// IN SHORT //
What are possible workarounds to consolidate documentation for S4
methods that are scattered across different packages (generic and some
custom methods in one package, additional custom methods in another
package) in a *single* Rd help file while using package
sorry, I repost the question again
a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9)
b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1)
cor(a,b)= -1
pa=qbinom(a, 1, 0.5)
pb=qbinom(b, 1, 0.5)
cor(pa,pb)=-0.8
but when
pa=qbinom(a,10,0.5)
pb=qbinom(b,10,0.5)
cor(pa,pb) becomes -1 again
Hi guy,
I have recently encountered a problem while I was just trying to generate
some random numbers with the function rnorm, the problem is shown below:
case 1
rnorm(20*0.2)
[1] -1.2765922 -0.5732654 -1.2246126 -0.4734006
case 2###
* rnorm(20*(1-0.8))
[1]
I recently lost the partitions on my hard drive (second time in 6 months)
so I had to have our IT folks image all my files over to a new drive. I
completely reinstalled R (now 2.15.2) and all my libraries to my computer
(Dell Latitude running Windows 7). A few of my previous workspaces
For this it would be best to insert an intermediate group container, as its
add method allows you to add the RGtk2 widgets. Something like:
library(gWidgets)
options(guiToolkit=RGtk2)
library(RGtk2)
w - gwindow()
lyt - glayout(cont=w)
lyt[1,1] - (g - ggroup(cont=lyt))
widget - gtkButton(click
Dear Arun and Rui, thank you for replying, your commands are very helpful.
Before you replied, Ive solved the issue with the following approach (for
quartiles):
myFun - function(x, GENDER)
{
x.male - cut(x, labels=c(1:4), breaks=quantile(split(x,GENDER) $ '1',
seq(0,1,.25),
Just noticed that I get a similar error about object 'kronecker' in
Matrix package when trying to load lme4. So this is a more pervasive
problem.
Brian
Brian S. Cade, PhD
U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO 80526-8818
email:
Hi all,
please, can someone give me an example of LDA with MASS package with
training, validation and test sets?
I never used it, so I need and example to avoid to make errors.
Thank you very much.
Best,
Roberto
--
View this message in context:
Potentially serious? Only to you: this looks like R FAQ 7.31 striking again.
Here's a hint:
identical(20*(1-0.8), 4)
and another
rnorm(round(20*(1-0.8)))
Sarah
On Fri, Nov 30, 2012 at 4:05 PM, liuxf li...@math.mcmaster.ca wrote:
Hi guy,
I have recently encountered a problem while I was
[See in-line below]
On 30-Nov-2012 21:05:23 liuxf wrote:
Hi guy,
I have recently encountered a problem while I was just trying to generate
some random numbers with the function rnorm, the problem is shown below:
case 1
rnorm(20*0.2)
[1] -1.2765922 -0.5732654 -1.2246126
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