That worked!
download.file(myurl, destfile=myfile, mode="wb")
Thanks a lot,
paolo
On 01/21/2011 02:53 PM, William Dunlap wrote:
Try mode="wb" ('b' for binary mode) in the
call to download.file(). It should make a
difference on Windows (& Mac?) and be innocuous on
Unix.
Bill Dunlap
Spotfire,
(1) I have a master data frame that reads
ClientID |date |value
(2) I also have a control data frame that reads
Client ID| Min date| Max date| control parameters
The control data set may not have all client IDs .
I want to use the control data frame on the master data frame to
remove client ID
Hi:
To spare the OP from tearing his hair out over a nuance of regular
expressions, here's a toy example:
df <- data.frame(key = paste('POSTOFF', 1:3, sep = '_'),
as.data.frame(matrix(rpois(180, 10), ncol = 60)))
nms <- scan(what = "")
"key""AMR.pa1.M" "AMR.pa2.M" "AMR.
Hi all,
I am trying to download, decompress and read a csv file. My code:
myurl <-
"ftp://ftp.ncbi.nih.gov/pub/geo/DATA/supplementary/series/GSE24729/GSE24729_MitoNuclear_suppl_male_stats.csv.gz";
#
myfile <- "GSE24729_MitoNuclear_suppl_male_stats.csv.gz"
#
download.file(myurl, destfile=myfi
Hi:
See inline, please.
On Thu, Jan 20, 2011 at 4:22 PM, Nathan Miller wrote:
> Hello,
>
> I am trying to generate a set of boxplots using ggplot with the following
> data with 4 columns (Day, Site, VO2, Cruise)
>
> AllCorbulaMR
>Day Site VO2 Cruise
> 1 11 148.43632670
h_profiles_Flick/",Filename,".pdf",sep=""))
caption = paste(aLocation," (", aYear,".",aMonth,".",aDay,")",sep="")
credits = paste("splineWriter.R / hel...@ucsd.edu / 20110120")
xrng = range(x)
yrng = range(y)
pred = qplot
Bill & Duncan
Thanks for your quick reply. I would still be looking for days.
Now I have to figure out how the bad data got into cts since I generate this
file each month.
--
View this message in context:
http://r.789695.n4.nabble.com/Unexpected-Gap-in-simple-line-plot-tp3228853p3228920.htm
Perhaps the easiest way to incorporate the heteroskedasticity
consistent SE's and output them in a familiar and easy to interpret
format is to use coeftest() in the lmtest package.
coeftest(myModel, vcov=vcovHC(myModel))
Andrew Miles
On Jan 20, 2011, at 4:42 PM, Achim Zeileis wrote:
On Th
You do have missing values. Setting xlim does not subset the data.
How about
link <- "http://processtrends.com/files/RClimate_CTS_latest.csv";
cts <- read.csv(link, header = TRUE)
scts <- subset(cts, !is.na(GISS) & !is.na(cts)) ## remove defectives
plot(GISS ~ yr_frac, scts, type =
On 20/01/2011 8:12 PM, D Kelly O'Day wrote:
I am getting an unexpected gap in a simple plot of monthly data series.
I have created a csv file of monthly climate observations that I store
on-line. When I download the csv file and plot one of the series, I get a
gap even though there is data for
I am getting an unexpected gap in a simple plot of monthly data series.
I have created a csv file of monthly climate observations that I store
on-line. When I download the csv file and plot one of the series, I get a
gap even though there is data for the missing point.
Here is a snippet to show
Dear Bulent,
It's not clear to me what you want to do, but the following may be relevant:
Graphs -> Plot of means; Graphs -> Boxplot; and, after fitting the model,
Models -> Graphs -> Effects plots.
I hope this helps,
John
> -Original Message-
> From: r-help-boun...@r-project.org [mailt
Hello,
I am trying to generate a set of boxplots using ggplot with the following
data with 4 columns (Day, Site, VO2, Cruise)
AllCorbulaMR
Day Site VO2 Cruise
1 11 148.43632670 1
2 11 61.73864969 1
3 11 92.64536096 1
4 11 73.35434957
Hi, R users,
I'm running a Tobit model but convergence can not be reached within 30
iterations. Is there anyway I can change the max number of iterations?
Thanks.
Gary
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
h
expression(Z[i][i])
also works as expected, though without the comma.
S Ellison
>>> Gerrit Eichner 01/20/11 12:09 PM >>>
Try
> plot (1, 1, ylab = expression (Z[list(i,i)]))
Hth -- Gerrit
> Original Message
> Subject: puzzled with plotmath
> Date: Thu, 20 Jan 2011 12:48:1
ROCR
On Thu, Jan 20, 2011 at 3:04 PM, He, Yulei wrote:
> Hi, there.
>
> Suppose I already have sensitivities and specificities. What is the quick
> R-function to calculate AUC for the ROC plot? There seem to be many R
> functions to calculate AUC.
>
> Thanks.
>
> Yulei
>
>
>[[alternati
Hi, there.
Suppose I already have sensitivities and specificities. What is the quick
R-function to calculate AUC for the ROC plot? There seem to be many R functions
to calculate AUC.
Thanks.
Yulei
[[alternative HTML version deleted]]
__
R-
On Jan 20, 2011, at 5:53 PM, Lars Bishop wrote:
Hi,
I'm running R 2.11.1 on Windows 32-bit. I fitted a glm to a "large"
data set
with no problems (i.e., nothing is wrong with the created glm object).
However, when I try to use the "effects" package to plot the effects
I get
the error mes
Hi,
I'm running R 2.11.1 on Windows 32-bit. I fitted a glm to a "large" data set
with no problems (i.e., nothing is wrong with the created glm object).
However, when I try to use the "effects" package to plot the effects I get
the error message shown below. I get the same message if I try to plot
I recently started using R and I have a simple question. I am running R
(v.
2.12.1) and Rcmdr (v.1-6.3) on Mac (Snow Leopard).
I am using a data set I used before for practicing ANOVA with R, so I
know
what the results should look like. I can get ANOVA table using both Rcmdr
and GUI. However,
It looks like you might be asking for this. This gets the "lower right"
diagonal.
> mat <- matrix(c(1,4,3,4,1,2,3,2,1), nrow = 3, ncol=3, byrow=TRUE)
> mat
[,1] [,2] [,3]
[1,]143
[2,]412
[3,]321
> diag(mat[rev(2:nrow(mat)),-1])
[1] 2 2
>
This gets the uppe
Haseeb - thank you for republishing your solution.
Prof Brian Ripley - thanks for pointing to my error.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.
On Thu, 20 Jan 2011, Mojo wrote:
I'm new to R and some what new to the world of stats. I got frustrated with
excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file that I load and store in a dataframe.
ds <- read.csv("boo
I'm not sure what a "real ANOVA diagram" is supposed to look like, nor
do I know what your data look like.
But this might get you started:
fakedata <- runif(100)
fakegroups <- sample(rep(letters[1:5], each=20))
boxplot(fakedata ~ fakegroups)
If that isn't what you're after, a clearer explanation
Joe P King wrote:
So many matrices are square symmetrical (i.e. variance-covariance matrices),
is there any way to get R to split the matrix on its diagonal and just
return one diagonal?
mat<-matrix(c(1,4,3,4,1,2,3,2,1), nrow = 3, ncol=3, byrow=TRUE)
is there anyway to get the lower rig
Ok, that's a known bug:
https://github.com/hadley/ggplot2/issues/labels/facet#issue/96
Thanks for the reproducible example though!
Hadley
On Thu, Jan 20, 2011 at 3:46 AM, Sandy Small wrote:
>
> Thank you.
> That seems to work - also on my much larger data set.
> I'm not sure I understand
On Thu, 20-Jan-2011 at 10:34AM +0200, E Hofstadler wrote:
|> Dear all,
|>
|> Being a newbie to R, I've trawled through many old posts on this list
|> looking for a solution to my problem, but unfortunately couldn't quite
|> figure it out myself. I'd be very grateful if someone here on this
|> list
So many matrices are square symmetrical (i.e. variance-covariance matrices),
is there any way to get R to split the matrix on its diagonal and just
return one diagonal?
So if I have
mat<-matrix(c(1,4,3,4,1,2,3,2,1), nrow = 3, ncol=3, byrow=TRUE)
is there anyway to get the lower right di
Dear,
I am trying to visualise a time-progressing line (it's supposed to
represent spread patterns) using brew package and Google Earth.
The idea is to have a function which takes start and end point
geographic coordinates, as well as number of intervals to chop the path
up, and returns the coll
>> I think I should be able to do this using the "reshape" function, but
>> I cannot get it to work. I think I need some help to understand
>> this...
>>
>>
>> (If I could split the "variable" into three separate columns splitting
>> by ".", that would be even better.)
>
> Use strsplit and "["
Or
On Jan 20, 2011, at 2:37 PM, > wrote:
Hello, I have fit a simple spline model to the following data.
Data
x y
0 1.298
2 0.605
3 0.507
4 0.399
5 0.281
6 0.203
7 0.150
8 0.101
Model
Sp.1=lm(y~bs(x,df=4))
Now I wish to inverse predict the
On 1/20/2011 3:37 PM, David Winsemius wrote:
On Jan 20, 2011, at 2:08 PM, Mojo wrote:
I'm new to R and some what new to the world of stats. I got
frustrated with excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file that
On Jan 20, 2011, at 2:08 PM, Mojo wrote:
I'm new to R and some what new to the world of stats. I got
frustrated with excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file that I load and store in a dataframe.
> ds <- re
Hi again
thank you to those who replied. I actually figured out a simple way
(which does entail transforming the data, oh well...) and I thought I
throw it out there just in case someone needs to solve a similar
problem. Apologies if it is in fact obvious.
Once the right levels for the contouring
The same way you interpret densities less than one?
On Jan 20, 2011, at 2:28 PM, Paul Ramer wrote:
How does one then interpret kernel density distributions with values
greater
than one?
The same way you interpret densities less than one?
density != probability
My output from the densit
combine them and then use acf:
x <- ts(rnorm(10))
y <- ts(x+ rnorm(10))
u <- ts.union(x, y)
(acf(u, na.action=na.pass))
I don't know if it's correct, but it gives an answer... I'm too afraid to
check if it's correct.
--
View this message in context:
http://r.789695.n4.nabble.com/CCF-and-mis
Dear List,
I recently started using R and I have a simple question. I am running R (v.
2.12.1) and Rcmdr (v.1-6.3) on Mac (Snow Leopard).
I am using a data set I used before for practicing ANOVA with R, so I know
what the results should look like. I can get ANOVA table using both Rcmdr
and GUI. H
I'm new to R and some what new to the world of stats. I got frustrated
with excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file that I load and store in a dataframe.
> ds <- read.csv("book2.csv")
> df <- data.frame(ds)
I
How does one then interpret kernel density distributions with values greater
than one?
My output from the density function.
---
density(delt.m[[1]][,6], na.rm=TRUE)
Call:
density.default(x = delt.m[[1]][, 6], na.rm = TRUE)
Data: delt.m[[1]][, 6] (171 obs.); Bandwidth 'bw' = 0.0
Hello, I have fit a simple spline model to the following data.
Data
x y
0 1.298
2 0.605
3 0.507
4 0.399
5 0.281
6 0.203
7 0.150
8 0.101
Model
Sp.1=lm(y~bs(x,df=4))
Now I wish to inverse predict the x for y=.75, say. Optimize works fine
for a
This did the trick:
panel.circ3 <- function(...)
{
args <- list(...)
circ1 <- ellipse(diag(rep(1, 2)), t = 1)
panel.xyplot(circ1[,1], circ1[,2],
type = "l",
lty = trellis.par.get("reference.line")$lty,
col = trellis.par.get("referenc
iling list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__ Information from ESET NOD32 Antivirus, version of virus signature
databas
I getting "Error in matrix(0, n, n) : too many elements specified"
while building randomForest model, which looks like memory allocation
error.
Software versions are: randomForest 4.5-25, R version 2.7.1
Dataset is big (~90K rows, ~200 columns), but this is on a big machine (
~120G RAM)
and I call
Take a look in prod function:
prod(1:4)
On Thu, Jan 20, 2011 at 3:18 PM, Jannis wrote:
> Dear list,
>
>
> is there a function in R that returns the product of a vector?
>
> E.g. if the vector is c(1,2,3,4) it should return 1*2*3*4=24
>
>
> Cheers
> Jannis
>
>
>
>
>
>
> _
Is this what you want:
> prod(c(1,2,3,4))
[1] 24
>
On Thu, Jan 20, 2011 at 12:18 PM, Jannis wrote:
> Dear list,
>
>
> is there a function in R that returns the product of a vector?
>
> E.g. if the vector is c(1,2,3,4) it should return 1*2*3*4=24
>
>
> Cheers
> Jannis
>
>
>
>
>
>
> _
Jeremy,
I don't seem to remember signing up to have R-help be my personal e-mail :-)
It is common to use phi to represent the standard normal distribution (with
mean 0, and variance 1), the "script/small/lower case" version is for the
distribution (height of the bell curve) and the "block/capit
Dear list,
is there a function in R that returns the product of a vector?
E.g. if the vector is c(1,2,3,4) it should return 1*2*3*4=24
Cheers
Jannis
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do re
On Thu, Jan 20, 2011 at 10:12 AM, Ortiz, John wrote:
> Hi everybody.
>
> I want to identify duplicate numbers and to increase a value of 0.01 for each
> time that it is duplicated.
>
> Example:
> x=c(1,2,3,5,6,2,8,9,2,2)
>
> I want to do this:
>
> 1
> 2 + 0.01
> 3
> 5
> 6
> 2 + 0.02
> 8
> 9
> 2 +
Hi Andy,
this is my way of doing it:
#first data:
wetMeans<-c(9.904762,6.344828,6.346154,6.855769,9.074324,9.953988,13.482966,
14.546053,10.841584,9.752033,6.739336,8.955056)
burnMeans<-c(0.06214286,0.05396552,0.04096154,0.05302885,0.05831081,0.073926
38,0.29969940,0.25596217,0.09316832,0.14441
I've gotten good results using the sine function to map colors.
For example, when plotting x, map the range(x) to (-pi/2,pi/2)
which the sine will transform to (-1,1), then add 1 and multiply by
half the desired number of colors. Now the integer values will
pick the colors and give a pleasing
Hi everyone,
I'm trying to perform a linear regression y = b1x1 + b2x2 + b3x3 + b4x4 +
b5x5 while constraining the coefficients such that -3 <= bi <= 3, and the
sum of bi =1. I've searched R-help and have found solutions for
constrained regression using quadratic programming (solve.QP) where th
Stoker, Ben wrote:
> A reproducible example would be very useful. In general
> I find that I can just index an mcmc object as though it
> were a matrix, e.g.:
>
> z <- cbind(a=runif(20),b=runif(20),c=runif(20))
> library(coda)
> m <- mcmc(data=z)
> t(apply(m,2,quantile,c(0.025,0.975)))
>
> If
Dear all,
This may be a simple question, but I cannot find the solution. I have a
point shapefile containing several hundreds of points, and a grid file
with a certain pixels size. Certain grid cells have several points
falling into them. I am looking for a way to randomly select one point
pe
Hello John,
If many numbers are duplicated, then one way is to coerce to a factor
and use the levels() function. For instance:
x <- c(1,1,2,2,2,3,3,4,1,1,2,4)
X <- factor(x)
for (i in levels(X))
{
loc <- (X==i); len = length(loc)
x[loc] <- x[loc] + 0.01 * (1:len)
}
x
[1] 1.01 1.
Dear list,
My query follows on from a question I posted a few days ago. I have the
following 2 sets of data:
wetMeans[1] 9.904762[2] 6.344828[3] 6.346154[4] 6.855769[5] 9.074324[6]
9.953988[7] 13.482966[8] 14.546053[9] 10.841584[10] 9.752033[11] 6.739336[12]
8.955056burnMeans[1] 0.06214286[2] 0
The following worked for me recently:
library(RMySQL)
MySQL. <- MySQL()
MySQLcon <- dbConnect(MySQL., user='thisuser', password='thispassword',
dbname='desiredDB')
I have the following suggestions and questions for you:
1. Have you tried supplyi
Try this:
replace(x + ave(x, x, FUN = seq) * .01, !(duplicated(x) | duplicated(x,
fromLast = TRUE)), x)
On Thu, Jan 20, 2011 at 1:12 PM, Ortiz, John wrote:
> Hi everybody.
>
> I want to identify duplicate numbers and to increase a value of 0.01 for
> each time that it is duplicated.
>
> Example
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ortiz, John
> Sent: Thursday, January 20, 2011 7:13 AM
> To: r-help@r-project.org
> Subject: [R] Identify duplicate numbers and to increase a value
>
> Hi everybody.
>
> I want
On Jan 20, 2011, at 10:51 AM, Fredrik Karlsson wrote:
Dear list,
I need to convert this data.frame
names(codesM)
[1] "key""AMR.pa1.M" "AMR.pa2.M" "AMR.pa3.M" "AMR.pa4.M"
[6] "AMR.pa5.M" "AMR.pa6.M" "AMR.pa7.M" "AMR.pa8.M" "AMR.pa9.M"
[11] "AMR.pa10.M" "AMR.ta1.M" "AMR.ta2.M"
On 2011-01-19 20:15, Max Kuhn wrote:
Hello everyone,
I'm stumped. I'd like to create a scatterplot matrix with circular
reference lines. Here is an example in 2d:
library(ellipse)
set.seed(1)
dat<- matrix(rnorm(300), ncol = 3)
colnames(dat)<- c("X1", "X2", "X3")
dat<- as.data.frame(dat)
grps<-
As for your second question, you could certainly do
newcodesM = transform(newcodesM, variable1 =
unlist(strsplit(variable,'\\.'))[1], variable2 = unlist(strsplit(variable,
'\\.'))[2], variable3 = unlist(strsplit(variable,'\\.'))[3])
though I'm sure there is a more efficient use of strsplit in t
I run a multinomial regression on a data set with an outcome that
has three values. First, I build an initial model, b.mod. Then I
run a loop to bootstrap the coefficients. For the initial model,
using "predict()", I can print the wrong/false predictions table.
But how do I get this table (that
I would think that the following code should work:
newcodesM = reshape(codesM, id=1)
If other variables in the data.frame are factors, reshape thinks all of them
are ID variables and tries to use all of them as "keys". Specifying the id
variable you want to keep (I used id=1 since "key" is in t
Hi John,
If you only have one duplicated number (e.g., just 2), then this will work:
x <- c(1,2,3,5,6,2,8,9,2,2)
xd <- duplicated(x)
x[xd] <- x[xd] + seq(sum(xd))/100
x
otherwise, I think a different framework than duplicated() will be
necessary, because it will matter not just if the number is
Dear list,
I need to convert this data.frame
> names(codesM)
[1] "key""AMR.pa1.M" "AMR.pa2.M" "AMR.pa3.M" "AMR.pa4.M"
[6] "AMR.pa5.M" "AMR.pa6.M" "AMR.pa7.M" "AMR.pa8.M" "AMR.pa9.M"
[11] "AMR.pa10.M" "AMR.ta1.M" "AMR.ta2.M" "AMR.ta3.M" "AMR.ta4.M"
[16] "AMR.ta5.M" "AMR.ta6.M"
If you haven't got so much data a loop should do:
while(sum(duplicated(x))>0) #if this condition is TRUE then there are still
duplicates in x
{
x[duplicated(x)] <- x[duplicated(x)]+0.01 #using duplicated(x) to
index the x vector
}
Hope this helps,
Regards
Moritz
Thanks Josh and David
Barth
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This transmittal and any attachments may contain PRIVILEGED AND
CONFIDENTIAL information and is intended only for the use of the
addressee. If you are not the designated recipient, or an employe
Hi,
On Thu, Jan 20, 2011 at 6:59 AM, Maas James Dr (MED) wrote:
> I'm attempting to generalise a function that reads individual list
> components, in this case they are matrices, and converts them into 3
> dimensional array. I can input each matrix individually, but want to do it
> for about
Many thanks for the correction David.
Josh
On Thu, Jan 20, 2011 at 7:17 AM, David Scott wrote:
> I think you need poly(X, 2, raw = TRUE) to interpret the coefficients in
> the manner described below.
>
> poly uses orthogonal polynomials by default:
>
> poly package:stats
I think you need poly(X, 2, raw = TRUE) to interpret the coefficients
in the manner described below.
poly uses orthogonal polynomials by default:
poly package:stats R Documentation
Compute Orthogonal Polynomials
Description:
Returns or evaluates orth
Hi, my name is Guilherme and I'm trying to solve an optimization problem in
R, regarding reliability and survival time of equipments. I Have to write
the function in the image attached to this email, please take a look, where
the f(t) is weibull's distribution density function, Ca=1000 and Cb=100 a
Hi Ben Bolker,
Just to say thank you VERY much for the reply below and for taking the time to
go through my code. I think you're absolutely right and I have been using the
wrong formula completely. I have been trying out several more examples (am
still struggling with this) and will submit any f
Hello,
I used to use RMySQL but as there is no more package for windows, I decided
to move to RODBC.
I installed ODBC driver for MySQL (downloaded on the MySQL website) and then
the RODBC package.
I finally discovered that it was not needed to "register" your database with
ODBC before using it.
Hi everybody.
I want to identify duplicate numbers and to increase a value of 0.01 for each
time that it is duplicated.
Example:
x=c(1,2,3,5,6,2,8,9,2,2)
I want to do this:
1
2 + 0.01
3
5
6
2 + 0.02
8
9
2 + 0.03
2 + 0.04
I am trying to get something like this:
1
2.01
3
5
6
2.02
8
9
2.03
2
I'm attempting to generalise a function that reads individual list components,
in this case they are matrices, and converts them into 3 dimensional array. I
can input each matrix individually, but want to do it for about 1,000 of them
...
This works
array2 <- abind(list1[[1]],list1[[2]],list1[
Hi Barth,
Here is an option fitting a linear model toa second order polynomial
and extracting the coefficients. The Intercept corresponds to "c" in
your email, then poly(...)1 to "b" and poly(...)2 to "a".
dat <- read.table(textConnection("
Y X
0.159529 0
0.5 0.773019
1 1"), he
Dear R helpers
I am having recovery rates as given below and I am trying to estimate the Loss
Given Default (LGD) and for this I am using Kernel Density estimation method.
recovery_rates =
c(0.61,0.12,0.10,0.68,0.87,0.19,0.84,0.81,0.87,0.54,0.08,0.65,0.91,
0.56,0.52,0.30,0.41,0.24,0.66,0.35,0.3
Hello
I would like to obtain the coefficients for a quadratic function (ax^2 + bx +
c) given three sets of points on the quadratic curve. For instance:
Y X
0.1595290
0.5 0.773019
1 1
Is there a function in R to obta
Peter,
Look for 'comma-separated list' on the help page!!
Yes, seeing the solution I also understand why list is the solution.
The special meaning of list () in plotmath was only in my passive vocabulary -
and after this discussion I think it is upgraded to active ;-)
I have to admit that my
On 2011-01-19 08:48, Paul Ossenbruggen wrote:
Hi,
I have missing values in my time series. "na.action = na.pass" works
for acf and pacf. Why do I get the following error for the ccf?
ts(matrix(c(dev$u[1:10],dev$q[1:10]),ncol=2),start=1,freq=1)
Time Series:
Start = 1
End = 10
Frequen
Hi,
On Tue, Jan 18, 2011 at 3:48 PM, ADias wrote:
>
> I have done like this to get the result I need more directly
>
> A<-c("Tell me how many different letter this vector has?")
> prop.table(table(strsplit(A,"")))
>
> ? a c d e f h i l m n o r s
> t
>
Robinson, David G sandia.gov> writes:
>
> I have an mcmc object and I'm trying to plot the quantiles of the
> variables - and not as a function of the
> iterations as in cumuplot.
>
> I cannot seem to find the right combination of indexing to
> access the variables; after which I'm sure I can
On 2011-01-20 02:05, Taras Zakharko wrote:
Hello Den,
your problem is not as it may seem so Ivan's suggestion is only a partial
answer. I see that each patient can have
more then one diagnosis and I take that you want to isolate patients based on
particular conditions.
Thus, simply looking for
Claudia,
Mittelhessen! ;-)
thanks & viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do
with an index list)
Something is missing; this last sentence of yours appears to be not
complete, doe
Hi
I want to do correlation of genes but I need to know p-values to
construct netwrok based on correlation. But I,m getting too many zero
for some reasons. I read one paper and they did analysed their data
by ::
1- R values
2-P-values
3-Spearman p values - FDR corrected (Bejamini-Hochberg)
Do yo
On Thu, Jan 20, 2011 at 10:53:01AM +0200, Den wrote:
> Dear R people
> Could you please help.
>
> Basically, there are two variables in my data set. Each patient ('id')
> may have one or more diseases ('diagnosis'). It looks like
>
> iddiagnosis
> 1 ah
> 2 ah
> 2 ihd
> 2 im
>
Prof. Brian Ripley and Tal Galili,
Thanks for your replies.
I cannot make the
*install.packages("ei",repos="http://r.iq.harvard.edu";, type='source')*
command work.
Alternatively, I can find the way (thanks to stackoverflow guys) I can
install that file from a local repo. For the sake of recor
Hi Taras,
Indeed, I've overlooked the problem. Anyway, I'm not sure I would have
been able to give a complete answer like you did!
Ivan
Le 1/20/2011 11:05, Taras Zakharko a écrit :
Hello Den,
your problem is not as it may seem so Ivan's suggestion is only a partial
answer. I see that each
I did try it. It gave me
[[1]]
id diagnosis
1 1ah
5 3ah
7 4ah
8 4 ihd
10 5 ihd
[[2]]
id diagnosis
1 1ah
2 2ah
5 3ah
7 4ah
[[3]]
id diagnosis
3 2 ihd
8 4 ihd
10 5 ihd
Which isn't what t
On 01/20/2011 02:11 PM, Uwe Ligges wrote:
On 20.01.2011 14:08, Claudia Beleites wrote:
Gerrit,
thanks & viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do
with an index list)
Works for me
Dear all,
I think I have a rather strange question, but I'd like to give it a try:
I want to perform a simulation numerous times, thats why I can't do it by
hand. I sample a small dataset from a very large one, and use backward
selection to select significant predictors for some arbitrary outcome
Hello Den,
your problem is not as it may seem so Ivan's suggestion is only a partial
answer. I see that each patient can have
more then one diagnosis and I take that you want to isolate patients based on
particular conditions.
Thus, simply looking for "ah" or "idh" as Ivan suggests will yield p
Hi to all members of this group,
how can i download a webpage using R (html), which is generated using
javasript when trying to open that webpage in the browser?
When i try to save the page as text file i get only some java code inside,
not the actual data.
When opening website in the browser and
You have to tell R that a column has date/time content. Here is an example:
test<-"3/23/2010 20:55" #this is your format
as.POSIXct(test, format="%m/%d/%Y %H:%M") #with the format parameter you
tell R how to interpret your data
___
Moritz Grenke
http://www.360mix.de
On 2011-01-20 04:09, Gerrit Eichner wrote:
Try
plot (1, 1, ylab = expression (Z[list(i,i)]))
Hth -- Gerrit
Original Message
Subject: puzzled with plotmath
Date: Thu, 20 Jan 2011 12:48:18 +0100
From: Claudia Beleites
To: R Help
Dear all,
I'm puzzled with matrix indices
Is this what you are trying to do:
> x <- read.table(textConnection("3/23/2010 20:55, -0.10655,
> -0.121454561, -0.12032, -0.111680001,-0.122429997
+ 3/23/2010 21:25, -0.099166997, -0.114189997, -0.11287,
-0.104647976,-0.114720002"), sep = ',', as.is = TRUE)
> closeAllConnectio
On 20.01.2011 14:08, Claudia Beleites wrote:
Gerrit,
thanks & viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do
with an index list)
Works for me with a recent R version.
for future sea
Gerrit,
thanks & viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do with an
index list)
for future searches: probably the easiest cheat is, of course,
plot (1, 1, ylab = expression (Z["i, i"])
Dear,
I am trying to visualise a time-progressing line (it's supposed to
represent spread patterns) using brew package and Google Earth.
The idea is to have a function which takes start and end point
geographic coordinates, as well as number of intervals to chop the path
up, and returns the coll
On 18.01.2011 07:42, fayazvf wrote:
I ve installed wordnet 2.1 and R 2.11.0 on windows 7.
Whenver i try to load wordnet in R, i get an error
initDict()
FALSE
cannot find wordnet 'dict' directory, Please set the WNHOME variable to its
parent.
I have tried setting WNHOME to "C:\Program Files
(x
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