Dear R helpers,
I have one trivial problem while writing an output file in csv format.
I have two dataframes say df1 and df2 which I am reading from two different csv
files.
df1 has column names as date, r1, r2, r3 while the dataframe df2 has column
names as date, 1w, 2w.
(the dates in both
On Wed, Jul 18, 2012 at 06:02:27PM -0700, bilelsan wrote:
> Leave the Taylor expansion aside, how is it possible to compute with [R]:
> f(e) = e1 + e2 #for r = 1
> + 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 #for r = 2, excluding e2*e1
> + 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 #for r = 3,
Great! Thank you ever so much Rui.
Santi
>
> From: Rui Barradas
>To: Santiago Guallar
>Cc: r-help@r-project.org
>Sent: Wednesday, July 18, 2012 8:29 PM
>Subject: Re: [R] Imposing more than one condition to if
>
>Hello,
>
>You're right. I had thought of this,
Hello every one
can any one tell me how to draw contour with this data set
c zshock
1 0.45450237 0
2 0.02663337 0
3 -2.08444556 0
4 -0.12715275 0
5 0.67066360 0
6 -0.73540081 0
I want to draw contour for shock i.e my z matrix is shock
Hi all,
I am running into a problem using forecast with ARIMA models, hope you can
help shed some light onto this.
I am fitting several ARIMA models using the auto.arima() function onto
several time series, which are basically the residuals from a linear model
fit. There are 40 such data points i
Maybe I'm missing something too but from your example seems like you are
looking for
xyplot(rnorm(12) ~ 1:12 , type="l",
scales=list(x=list(at=seq(2,12,2),labels=c(1, ' ', 3 , ' ' , 5 , ' ' ))),
par.settings=list(axis.components=list(bottom=list(tck=c(0,1)
See "scales" in ?xyplot and str(trel
Leave the Taylor expansion aside, how is it possible to compute with [R]:
f(e) = e1 + e2 #for r = 1
+ 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 #for r = 2, excluding e2*e1
+ 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 #for r = 3, excluding
e2*e1^2 and e1*e2^2
+ ... #for r = k
In other words, I
Dear R community,
I'm having hard time to understand the kde function in "ks" package. Let me
use a 3-dimensional kernel smooth example to explain my question using the
elevation data in geoR.
### here is what I did ###
library(ks)
require(geoR)
data(elevation)
elev.df <- data.frame(x = elevatio
Hi,
I'm looking for an "easy" way to setup a decision tree.
This is *not* any kind of regression, but a very simple DAG with probability on
each edge and a value at each node. I just need a way to input the graph and
calculate the expected value at each end node. The whole thing could be done
There are obviously a large variety of non-smooth problems;
for CVAR problems, if by this you mean conditional value at
risk portfolio problems, you can use modern interior point
linear programming methods. Further details are here:
http://www.econ.uiuc.edu/~roger/research/risk/risk.html
type ?merge in R
-
Yasir Kaheil
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in biplot you could set the limits xlim, ylim of the axes to zoom in on the
plot.
-
Yasir Kaheil
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type ?read.table in R
-
Yasir Kaheil
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Hello,
Not sure whether I understand it well.
If you want your output to include only Patient A &B,
this should work:
dat1<-read.table(text="
Patient Cycle Variable1 Variable2
A 1 4 5
A 2 3 3
A 3 4 NA
B 1 6 6
B 2 NA 6
C 1 6 5
C 3 2 2
",sep="",header=TRUE)
subset(dat1,!dat1$Patient=="C")
Pa
You can do this using only xtabs. Using Rui's data.frame, d:
x <- xtabs(Count~Live+Age, d)
barplot(x, beside=T, legend.text=TRUE, args.legend=list(x="topleft"))
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station
Cren bancaakros.it> writes:
>
The most robust solver for non-smooth functions I know of in R is Nelder-Mead
in the 'dfoptim' package (that also allows for box constraints).
First throw out the equality constraint by using c(w1, w1, 1-w1-w2) as input.
This will enlarge the domain a bit, but com
First. Your example does not make sense.
for (n in 2:10) {s[n]=1+s[n]+rnorm(1,mean=0,sd=1)}
adds 1 + 1 + rnorm(1) so it will result in values ranging from 0 - 4 with a
few exceeding that range. It will never result in 9 values greater than 4.
For example I get
> s <- rep(1,10)
> set.seed(42)
>
I've looked at the lattice book and the 'R Graphics Cookbook' without
seeing how to change the labels along the x axis for groups in a box plot,
specifically cenbox().
The attached example has a main and axes labels with default group labels.
Please point me to a reference on how I can change
Did you follow the directions?
There's a special README for Ubuntu to help you get the current version.
http://cran.r-project.org/bin/linux/ubuntu/README
But you also don't tell us what version of Ubuntu you're running.
Sarah
On Wed, Jul 18, 2012 at 5:27 PM, Chet Seligman wrote:
> This doesn't
This doesn't work, what should I do?
sudo apt-get install r-base
[sudo] password for cseligman:
Reading package lists... Done
Building dependency tree
Reading state information... Done
r-base is already the newest version.
0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.
Also:
> .
For reading data into R you should start with
http://cran.r-project.org/doc/manuals/R-data.html (or the local copy
that was installed with R on your machine).
For the example above the read.table function should be fine. If you
want to change the shape of the resulting data frame then look at the
Hello,
Sorry, I forgot the barplot.
dmelt <- melt(d, id.vars=c("Live", "Age"), measure.vars="Count")
wide <- dcast(Live ~ Age, data = dmelt)
col <- rainbow(4)
barplot(as.matrix(wide[, 2:5]), beside=T, col=col)
legend("topleft", wide[, 1], fill=col)
Hope this helps,
Rui Barradas
Em 18-07-20
Hello,
Install package 'reshape2' and try the following.
#install.packages('reshape2')
library(reshape2)
d <- read.table(text="
LiveAgeCount
ParentsAge19324
AnotherAge1937
OwnPlaceAge19116
GroupAge1958
OtherAge195
ParentsAge20378
Another
On Wed, Jul 18, 2012 at 9:47 AM, bilelsan wrote:
> Dear list,
>
> I have a big deal concerning the development of a Taylor expansion.
>
> require(Matrix)
> e1 <- as.vector(1:5)
> e2 <- as.vector(6:10)
>
> in order to obtain all the combinations between these two vectors following
> a Taylor expans
Tena koe Lib
In case you have receive a reply to this (I didn't notice one), here is one
option:
> lib
A X1 X4 X5
1 A 2 3 3
2 A 3 4 NA
3 B 1 6 6
4 B 2 NA 6
5 C 1 6 5
6 C 3 2 2
> str(lib)
'data.frame': 6 obs. of 4 variables:
$ A : chr "A" "A" "B" "B" ...
$ X1: num 2 3 1 2
You have only saved the last run of the for() loop (when i is 4 and j is
7) to the object called model. If you want to save all of the lme fits,
you need to set give some dimensionality to the object model. For
example, you could define model as a list, then store the lme fits as
separate ele
> land<-c(0,0,0,0,1,1,1,0,0,0,1,1,0,0,0,0)
> expect <- c(1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3)
> cumsum(c(TRUE, diff(land)==1))
[1] 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3
> all.equal(.Last.value, expect)
[1] TRUE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-
Thanks. Also helpful.
DAV
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Wednesday, July 18, 2012 4:20 PM
To: David A Vavra; 'jim holtman'
Cc: r-help@r-project.org
Subject: RE: [R] Trapping option settings
Try using trace(), as in
trace(options, quote(print
Hi,
Entering data from a given file is the hardest part of learning R for me.
For example, I have this datafile from Moore's text:
LiveAge Count
Parents Age19 324
Another Age19 37
OwnPlaceAge19 116
Group Age19 58
Other Age19 5
Parents Age20 378
Another Age20 47
O
Hi Jean,
Is there something missing in the function?
ids <- a$id
for(i in 2:4){
for(j in 5:7){
y <- a[, j]
x <- a[, i]
model<-lme(y ~ x , random= ~1|ids, na.action="na.exclude")
}}
summary(model)
Linear mixed-effects model fit by REML
Data: NULL
AIC
On Wed, Jul 18, 2012 at 7:41 PM, Gary Dong wrote:
> For your second question:
> 1) by relative long and lat, I mean distances to the city center from
> south-north and east-west directions.
> 2) I know the coordinate of the reference point.
The problem here is that going 10 miles north along a
Nice! It works. Thank you, Rui
There's something that takes me back to the original question, though. The code
takes the first value that meets the critical condition (light ==0); however,
this value can appear more than once a day because these animals nest in dark
caves, and once they are in
HI,
Check this link:
http://stackoverflow.com/questions/1444306/how-to-use-outlier-tests-in-r-code
Hope it would be helpful.
A.K.
- Original Message -
From: Sajeeka Nanayakkara
To: "r-help@r-project.org"
Cc:
Sent: Wednesday, July 18, 2012 9:27 AM
Subject: [R] R code for to check out
Dear list,
I have a big deal concerning the development of a Taylor expansion.
require(Matrix)
e1 <- as.vector(1:5)
e2 <- as.vector(6:10)
in order to obtain all the combinations between these two vectors following
a Taylor expansion (or more simply through a Maclaurin series) for real
numbe
Thanks a lot, it works fine with me !!
François Maurice
De : "Nutter, Benjamin"
À : Francois Maurice ; r-help@r-project.org
Envoyé le : mercredi 18 juillet 2012 8h21
Objet : RE: [R] Variable labels
I have my own function for doing this that is similar to the one presented
below. Others may
got it... another merge did the trick
narrow6<-merge(narrow2,narrow5,by=c("gene","gender"))
Thanks for the help Rui
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I am trying to run the cspade function in the arulesSequences package in R.
After I successfully read in my transactions using read_baskets, I try to
execute the cspade function against the transactions object I read in.
However, when I execute the command, I obtain an error: system invocation
fa
Hi,
You could use "merge", "join" etc.
merge(DF3,DF2)
# station lat lon data
#1 ST002 41 2 3
#2 ST003 42 3 7
library(plyr)
join(DF3,DF2,type="inner")
Joining by: station
# station lat lon data
#1 ST002 41 2 3
#2 ST003 42 3 7
#or
join(DF3,DF2,type="right")
Ho
Hi
Try this:
mydat <-read.table(text="
ABC XYZ
12 6
6 50
90 100
55 85
100 25
",sep="",header=TRUE)
apply(mydat,2,quantile,probs=0.2)
ABC XYZ
10.8 21.2
A.K.
- Original Message -
From: Rantony
To: r-help@r-project.org
Cc:
Sent: Wednesda
Thanks,
in a way this has worked... with a slight modification to this:
narrow3<-aggregate(narrow2$value~narrow2$gene+narrow2$gender,data=narrow2,mean)
narrow4<-aggregate(narrow2$value~narrow2$gene+narrow2$gender,data=narrow2,sd)
which gives a table of the 24000 gene&gender means (narro
# Whoops! I have just seen there's a little mistake
# in the 'sharpe' function, because I had to use
# 'w' array instead of 'ead' in the cm.CVaR function!
# This does not change the main features of my,
# but you should be aware of it
---
# The function to be minimized
sharpe <- function(w) {
# Hi all,
# consider the following code (please, run it:
# it's fully working and requires just few minutes
# to finish):
require(CreditMetrics)
require(clusterGeneration)
install.packages("Rdonlp2", repos= c("http://R-Forge.R-project.org";,
getOption("repos")))
install.packages("Rsolnp2", repos=
Yes,
I would be interested in both the ggplot2 and lattice ways of doing
this. Unfortunately, I am not interested in creating a panel for each
chromosome. Actually, I would like to create a Manhattan plot using
xyplot. Thus I would need to alternate tick marks and tick labels.
Thanks!
On Mon,
Anyone knows how to convert a deldir$delsgs to a X3D IndexedTriangleSet?
Are there already any functions/packages?
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PLEASE do r
Hello, I need to subset my data to only look at the parts that have "holes"
in it. I already have a formula to get rid of inconsistencies, but now I
need to look only at the problem data to reconfigure it. In my data set
where there are multiple "cycles" per "patient," and I want to highlight
the p
Thank you! I had to make some modifications since my data is between a data
set and a subset not one data set to itself, but I was able to use
essentially your method to get it working.
I did have some trouble with the matrix command on the last line - it kept
returning lots of NAs. So I just
I am trying to assign increasing trip numbers to a binary variable ("land";
1=home and 0=away) where a string of 1's shouldn't increment the trip_no
more than once.
For example; based on
land<-c(0,0,0,0,1,1,1,0,0,0,1,1,0,0,0,0)
the "trip_no" sequence produced should be 1,1,1,1,2,2,2,2,2,2,3,3,
Hi nice people,
i am trying to made a double "for" cycle, i wish that for cycle on k
activates t times the for cycle on s
the first cycle is this:
s<-rep(1,10)
s
[1] 1 1 1 1 1 1 1 1 1 1
> for (n in 2:10) {
+ s[n]=1+s[n]+rnorm(1,mean=0,sd=1)
+ }
> s
[1] 1 4.75 4.86 4.05 4.09 4.56 4.63 4.65 4
hi,
i am new to using R, so if u can help me in merging my csv file having
different sheets .if u can help me with the commands for it.
regards
karan
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Oh, I clearly misunderstood what you were doing there. I don't know anything
about Manhattan plots but a quick google for "manhattan plot r package" turns
up a number of items so the type of plot you want may already exist.
Sorry to not be of more help.
John Kane
Kingston ON Canada
> -
On 2012-07-18 04:27, Rui Barradas wrote:
Helo,
All problems should be easy.
d <- read.table(text="
gene variable value gender line rep
1 CG1 X208.F1.30456 4.758010 Female 208 1
2 CG1 X365.F2.30478 4.915395 Female 365 2
3 CG1 X799.F2.30509 4.641636 Female 799 2
4 CG1 X306.M2.326
Try using trace(), as in
trace(options, quote(print(as.list(sys.calls()
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of David A Vavra
> Sent: Wednesday, July 18, 20
Jim,
Thanks.
It wasn't sure if merely overriding the options function by placing one in
the global environment would guarantee it would be the one actually called.
In any case, I didn't know how to identify the caller. This is quite helpful
and looks promising. I'll give it a try.
DAV
-
Hello,
Inline
Em 18-07-2012 18:44, Nordlund, Dan (DSHS/RDA) escreveu:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Sajeeka Nanayakkara
Sent: Wednesday, July 18, 2012 6:28 AM
To: r-help@r-project.org
Subject: [R] R code for to
Thanks, Sara.
For your first question:
Geographic Coordinate System: GCS_North_American_1983_HARN
Projected Coordinate System:
NAD_1983_HARN_StatePlane_Oregon_North_FIPS_3601_Feet_Intl
I have imported the shp file in R by read.shaplefiles().
For your second question:
1) by relative long and la
Hi,
On Wed, Jul 18, 2012 at 2:11 PM, Gary Dong wrote:
> Dear R users,
>
> I have a city map in shape file (polygon). I also have some points that I
> hope to plot them to the city map.
What projection is the shape file in? Have you successfully imported
it into R yet?
This page has a nice overv
Hello,
You're right. I had thought of this, and I believe there's a day when it
happens to have a zero in the middle of the day. R doesn't allow
conditions like the one you've written but it does allow multiple
conditions, combined using the logical connectives, 'not' - '!', 'or' -
'|' and 'a
Hi all,
I am also confused about the manual:
a. The input arguments:
wt.method are the weights case weights (giving the relative importance of
case, so a weight of 2 means there are two of these) or the inverse of the
variances, so a weight of two means this error is half as variable
Hi Terry, Greg, and Marc,
Thanks for your advice about this. I think I have a pretty good starting point
now for the analysis.
Appreciate your help.
Paul
--- On Wed, 7/18/12, Terry Therneau wrote:
From: Terry Therneau
Subject: Re: [R] Power analysis for Cox regression with a time-varyin
Thank you Rui. Very helpful. Yolande.
On Wed, Jul 18, 2012 at 1:37 PM, Rui Barradas wrote:
> Hello,
>
> Try the following.
>
>
> for(i in 1:dim(diveData_2008)[1]){
> # Which dive is this observation from
> thisIndex <- as.character(index(diveData_**2008[i,]))
> thisIndex <- as.PO
Dear R users,
I have a city map in shape file (polygon). I also have some points that I
hope to plot them to the city map. The only information I have about those
points are their relative longitude and latitude to the city center by
miles. Is there a way that R can help me to do this? Thanks.
Ga
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Sajeeka Nanayakkara
> Sent: Wednesday, July 18, 2012 6:28 AM
> To: r-help@r-project.org
> Subject: [R] R code for to check outliers
>
>
>
>
>
> What is the R code to check w
Hello,
Try the following.
for(i in 1:dim(diveData_2008)[1]){
# Which dive is this observation from
thisIndex <- as.character(index(diveData_2008[i,]))
thisIndex <- as.POSIXct(thisIndex)
dive_id <- diveCond_all$dive_id[diveCond_all$timestamp == thisIndex]
#
it is a function from library(xts)
On Wed, Jul 18, 2012 at 1:15 PM, Jean V Adams wrote:
> What is the function index() that you use in this line of code?
> thisIndex <- as.character(index(diveData_2008[i,]))
>
> Is it from some package? Or a function you wrote yourself?
> I'm trying to run the
What is the function index() that you use in this line of code?
thisIndex <- as.character(index(diveData_2008[i,]))
Is it from some package? Or a function you wrote yourself?
I'm trying to run the code you submitted, but I don't have a function
called index().
Jean
Yolande Tra wrote on 07/18
At 00:10 18/07/2012, eeste...@ualg.pt wrote:
Dear All,
I'm having trouble tweaking a forest plot made using the R
meta-analysis package metafor. I did the analysis based upon the
correlation coeff from studies and plotted the corresponding forest
plot easily
q2<-rma(yi,vi,mods=cbind(grupo),dat
David, thanks a lot!
I tried x[-1] myself but forgot to delete 'group' from the keyby
statement - this explains why it did not work for me.
This is amazing - just 2 lines instead of my many-many.
Great learning!
Dimitri
On Tue, Jul 17, 2012 at 10:49 PM, David Freedman wrote:
> Honestly, I wasn't
I'm not sure why, but lme() doesn't seem to like the variables to be
referenced as part of a list using [ or $.
Here's an easy workaround ...
ids <- a$id
for(i in 2:4){
for(j in 5:7){
y <- a[, j]
x <- a[, i]
lme(y ~ x , random= ~1|ids, na.action="na.exclude")
}}
J
> Bert Gunter
> on Wed, 18 Jul 2012 07:14:31 -0700 writes:
> checkforoutliers <- function(series) NULL
> Cheers, Bert
> *Explanation: There is no such thing as a statistical
> outlier -- or, rather,"outlier" is a fraudulent
> statistical concept, defined arbitrari
On 18/07/2012 10:14 AM, Bert Gunter wrote:
checkforoutliers <- function(series)NULL
Cheers,
Bert
*Explanation: There is no such thing as a statistical outlier -- or,
rather,"outlier" is a fraudulent statistical concept, defined arbitrarily
and without scientific legitimacy. The typical unstated
> >> What is the R code to check whether data series have
>>> outliers or not?
In case noone else has pointed you there, you could try the 'outliers' package.
That contains some of the 'standard' methods of outlier testing for univariate
data.
What you do with them when you find them is a
To further what Bert says:
You would almost certainly prefer to use robust statistics than
"outlier detection".
I believe Greg Snow's TeachingDemos package has a data set "outliers"
suggesting some of the perils of doing things the outlier-removal way.
Best,
Michael
On Wed, Jul 18, 2012 at 9:14
On 18.07.2012 18:03, Mati Bielawski wrote:
I would like to know how to use the keep parameter in the step function. Or
the way to always keep one of my Xs, and the other Xs to be keep or not by
step forward method.
See ?step and in particular the "lower" part of its scope argument:
step(
I would like to know how to use the keep parameter in the step function. Or
the way to always keep one of my Xs, and the other Xs to be keep or not by
step forward method.
Thank you all
Mati Bielawski
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R-he
abhisarihan,
Please don't crosspost!
http://stackoverflow.com/questions/11530800/installing-packages-from-rprofile-site-file
Thanks,
Garrett
On Tue, Jul 17, 2012 at 2:34 PM, abhisarihan wrote:
> I am trying to install custom packages upon starting R. A lot of the code
> that is written by us ri
On 17.07.2012 21:34, abhisarihan wrote:
I am trying to install custom packages upon starting R. A lot of the code
that is written by us right now is available for editing to the users. To
try and protect the code, I am packaging the production level code and
having the users install it on their
On 14.07.2012 00:41, Julien Salanie wrote:
I have a read a lot about the benefits of vectorization in R. I have a
program that takes "almost forever" to run. A good way to see if I have
learned something ... My problem can be summarized like this : I have a
nonlinear function of several variables
I think you'll need to roll your own using tryCatch() around open().
Michael
On Jul 18, 2012, at 5:09 AM, Berry Boessenkool
wrote:
>
>
> Hi all,
>
> I'm working on a function that reads online data that is only available to
> certain IPs. It then writes a subset of the data into a file.
>
checkforoutliers <- function(series)NULL
Cheers,
Bert
*Explanation: There is no such thing as a statistical outlier -- or,
rather,"outlier" is a fraudulent statistical concept, defined arbitrarily
and without scientific legitimacy. The typical unstated purpose of such
identification is to remove
On 18/07/2012 8:27 AM, bunnylove...@optonline.net wrote:
What i have is a for loop to name files. I want it to skip over the file if it
doesn't exist. Is there a way to do that?
s=seq(from = chron("03/15/2012"), to = chron("06/15/2012"))
day=format(as.Date(s), "%Y%m%d")
for (k in 1:length(day))
Hello.
I have an ordered dependent variable (scale 0 - 3), 5 measurement points. I
am afraid I have strong ceiling effects in my data, and would like to plot
the data (trajectories).
However, you know that plotting ordered variables isn't really feasible.
My question: would you think it appropri
Dear R-list,
I have a data set (in the following example called "a") which have:
one "subject indicator" variable (called "id")
three dependent variables (varD, varE, var F)
three independent variables (varA, varB, varC)
I want to fit 9 lme models, one per posible combination (DA, DB, DC, E
Hello,
I have the following loop for two data sets: diveData_2008 and
diveData_2009. It uses two other data: diveCond_all and fishTable. The
problem is at the point to identify the dive_id for the given index (index
is timestamp). It keeps on saying
for the1st loop
Error in fishReport$dive_id[i] <
Ok, thank you Dan!
I was already afraid that I would get this answer. I will solve it by
defining date/time both as a date/time object and as a character object ( in
another column)
Regards,
Sandy
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What is the R code to check whether data series have outliers or not?
Thanks,
Sajeeka Nanayakkara
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PLEASE do read the pos
Marc gave the referencer for Schoenfeld's article. It's actually quite
simple.
Sample size for a Cox model has two parts:
1. Easy part: how many deaths to I need
d = (za + zb)^2 / [var(x) * coef^2]
za = cutoff for your alpah, usually 1.96 (.05 two-sided)
zb = cutoff for pow
Here is an example of how to do it.
> library(survival)
> vfit <- coxph(Surv(time, status) ~ celltype + trt, data=veteran)
> userinput <- data.frame(celltype="smallcell", trt = 1)
> usercurve <- survfit(vfit, newdata=userinput) #the entire predicted
survival curve
> user2 <- summary(usercurve,
Dear R users,
I am new to lattice but am trying to get to grips with it.
I need to add contour lines for the topography to a plot I created with spplot.
The topography is available as a
SpatialPixelsDataFrame object Z_sfc. If I plot it as spatial lines, after
converting to a SpatialLines objec
What i have is a for loop to name files. I want it to skip over the file if it
doesn't exist. Is there a way to do that?
s=seq(from = chron("03/15/2012"), to = chron("06/15/2012"))
day=format(as.Date(s), "%Y%m%d")
for (k in 1:length(day)){
B1=read.csv(paste("S:/file_", day[k], ".csv", sep=""))
Da
I have my own function for doing this that is similar to the one presented
below. Others may have other ideas that work better. As a general rule, I
would caution against writing out just the label without the variable name.
The only reason I see to separate the labels and names is if you are
On 07/18/2012 10:11 PM, Jim Lemon wrote:
On 07/18/2012 09:42 PM, Sarah Goslee wrote:
On Wed, Jul 18, 2012 at 6:42 AM, Jim Lemon wrote:
On 07/18/2012 08:02 PM, Stuart Leask wrote:
Hi there.
I have been plotting some circles using 'symbols', with radii
representing
my data, but the radii looke
# MCT ordered by the second column of MCT, from smallest to largest values
MCT[order(MCT[, 2]), ]
Jean
Trying To learn again wrote on 07/17/2012
04:42:24 PM:
> Hi all,
>
> I want to order a series that is included on the second column in
MCT.csv.
>
> I do but R doesn´t order, could be beca
On 07/18/2012 09:42 PM, Sarah Goslee wrote:
On Wed, Jul 18, 2012 at 6:42 AM, Jim Lemon wrote:
On 07/18/2012 08:02 PM, Stuart Leask wrote:
Hi there.
I have been plotting some circles using 'symbols', with radii representing
my data, but the radii looked incorrect.
It seems to happen with a s
On Wed, Jul 18, 2012 at 6:42 AM, Jim Lemon wrote:
> On 07/18/2012 08:02 PM, Stuart Leask wrote:
>>
>> Hi there.
>>
>> I have been plotting some circles using 'symbols', with radii representing
>> my data, but the radii looked incorrect.
>>
>> It seems to happen with a single circle too:
>>
>> Symb
Helo,
All problems should be easy.
d <- read.table(text="
gene variable value gender line rep
1 CG1 X208.F1.30456 4.758010 Female 208 1
2 CG1 X365.F2.30478 4.915395 Female 365 2
3 CG1 X799.F2.30509 4.641636 Female 799 2
4 CG1 X306.M2.32650 4.550676 Male 306 2
5 CG1 X712.M2.3
Hello,
Try the following.
# This is your code
df_sorted = df[order(as.Date(df$date, "%m/%d/%Y"), decreasing = TRUE),]
# This is my code
nams <- as.character(unique(dat1$name))
nums <- sapply(nams, function(nm) which(names(df_sorted) %in% nm))
df_sorted[, sort(nums)] <- df_sorted[, nams]
names(d
Dear Mr Rui Barradas,
Thanks a lot for your wonderful suggestion. It worked and will help me
immensely in future too. Really heartfelt thanks once again.
Vincy
--- On Wed, 7/18/12, Rui Barradas wrote:
From: Rui Barradas
Subject: Re: [R] How to have original (name) order after melt and cast c
Hi all,
I'm working on a function that reads online data that is only available to
certain IPs. It then writes a subset of the data into a file.
So whenever I'm logged in elsewhere or am not connected to the internet, I get
an error, and the function is terminated.
I want it to rather print a
Hi
I think/hope there will be a simple solution to this but google-ing has
provided no answers (probably not using the right words)
I have a long data frame of >2 000 000 rows, and 6 columns. Across this
there are 24 000 combinations of gene in a column (n=12000) and gender in a
column (n=2... obv
Hello,
Now I don't understand. Inline
Em 18-07-2012 11:56, Akhil dua escreveu:
why are you writing ncol=2 ?
I have levels=100 for x1 and the x1 is my z matrix
In your contour instruction x3 is the z matrix, not x1. And your dataset
shows a 2x3 grid, hence ncol=2, for (x1 times x2) 0:1x1:3.
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