I've installed R-2.14.0 from source on CentOS and on Kubuntu and in
both cases, I see something I've never seen before. Comments in
locally written functions disappear. I put comments there for a
purpose and I'd like to keep them. I can still use older versions of
R without that happening. Noth
Hi,
I want to perform Survival curves for case and control subjects in
the propensity score-matched cohort that
accounted for the clustering of matched pairs. How I can do it with R.
Thanks for your help,
Jan
[[alternative HTML version deleted]]
__
On Nov 21, 2011, at 05:50 , David Winsemius wrote:
>
> On Nov 20, 2011, at 6:34 PM, Paul Johnston wrote:
>> ...
>> I had intended to report logrank P values with the hazard ratio and CI
>> obtained from this function. In one case the P was 0.04 yet the CI
>> crossed one, which confused me, and
On 21/11/11 20:52, p_conno...@slingshot.co.nz wrote:
I've installed R-2.14.0 from source on CentOS and on Kubuntu and in
both cases, I see something I've never seen before. Comments in
locally written functions disappear. I put comments there for a
purpose and I'd like to keep them. I can stil
Hi to all
is it possible to build a legend with
plot(NA, type = "l",col="blue",lwd=2,ylim=c(-0.05,10),xlim=c(0,13),xaxt
= "n",xlab="",ylab="") # Grafikmodus mit erster Linie starten
Lcolors=c("blue","orchid","green")
lsymbols= c(16,17,15)
Ltext= c("text1","tex2","text3")
legend("topleft",cex=
On 11/21/2011 07:55 PM, Knut Krueger wrote:
Hi to all
is it possible to build a legend with
plot(NA, type = "l",col="blue",lwd=2,ylim=c(-0.05,10),xlim=c(0,13),xaxt
= "n",xlab="",ylab="") # Grafikmodus mit erster Linie starten
Lcolors=c("blue","orchid","green")
lsymbols= c(16,17,15)
Ltext= c("te
I am pleased to announce the first release of HiveR, a package to
draw 2D and 3D Hive Plots. It's currently on CRAN and on it's way to
a mirror near you.
Hive plots are a unique method of displaying networks of many types
in which node properties are mapped to axes using meaningful
properties rat
Hi to all,
I'm developping an user interface in TCL/TK but in some computers this error
occurs :
invalid command name "tk::MenuDup"
when the user runs :
tm <- tktoplevel(height=500,width=800)
topMenu <- tkmenu(tm)
tkconfigure(tm,menu=topMenu)
i don't understand why this code works on my compute
Hi fellow R users.
I have a problem when i try to change the Names of the columns on a dataset:
The names of the names dataset looks like this:
toPlot
> names(toPlot)
[1] "REPORT_20" "REPORT_21" "REPORT_22" "REPORT_23"
[5] "REPORT_24" "REPORT_25" "REPORT_QREP001" "
Hello,
I know there is plenty of people in this group who can give me a good answer :)
I have a 2^k model where k=4 like this:
Model 1) R~A*B*C*D
If I use the "*" in R among all elements it means to me to explore all
interactions and include them in the model i.e. I think this would be the so
On Fri, Nov 18, 2011 at 12:16 PM, Barry Rowlingson
wrote:
>
> On Fri, Nov 18, 2011 at 1:53 PM, Raphael Saldanha
> wrote:
>
> > I would like
> > to import this dataset direct into R, without saving the file on disk
> > (for using the most updated file),
>
> You don't really have much choice - I'
Hello,
I think I am not to far from a solution. I want to do lm regressions with
several variables which I define before in a list. What I've done so far is
like:
y <- c(1,5,6,2,5,10) # response
x1 <- c(2,12,8,1,16,17) # predictor
x2 <- c(2,14,5,1,17,17)
df <- data.frame(y,x1,x2)
predictorlist
Hello,
Couple of clarifications:
- A,B,C,D are factors and I am also interested in possible interactions but the
model that comes out from aov R~A*B*C*D violates the model assumptions
- My 2^k is unbalanced i.e. missing data and an additional level I also include
in one of the factors i.e. C
-
On 21.11.2011 11:00, habasque wrote:
Hi to all,
I'm developping an user interface in TCL/TK but in some computers this error
occurs :
invalid command name "tk::MenuDup"
when the user runs :
tm<- tktoplevel(height=500,width=800)
topMenu<- tkmenu(tm)
tkconfigure(tm,menu=topMenu)
i don't unde
My working example is in the package kml3d. The class 'clusterLongData' is
define by inehirance from class 'listClustering' and 'longitudinalData'. But
as you say, it is unusual, I will try an other way.
Thank you for your help.
Christophe
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On 11/20/2011 07:45 PM, sutada Mungpakdee wrote:
Hi,
I'm very beginner for R but I think it is a time to start as it is very useful.
I have a coverage read file (illusmp454merCbed) for whole genome ~ 450 Mbp.
This is head of this file.
Scaffoldsca_positioncoverage
Scaffold1 1
On 21.11.2011 13:34, Johannes Radinger wrote:
Hello,
I think I am not to far from a solution. I want to do lm regressions with
several variables which I define before in a list. What I've done so far is
like:
y<- c(1,5,6,2,5,10) # response
x1<- c(2,12,8,1,16,17) # predictor
x2<- c(2,14,5,1,
On 21.11.2011 11:40, Joel wrote:
Hi fellow R users.
I have a problem when i try to change the Names of the columns on a dataset:
The names of the names dataset looks like this:
toPlot
names(toPlot)
[1] "REPORT_20" "REPORT_21" "REPORT_22" "REPORT_23"
[5] "REPORT_24" "
On 21.11.2011 07:17, dilshan benaragama wrote:
Hi, R development team,
I am trying to use PCA in labdsv package.I need to build the ordination plot
from scratch. I used the following code (which is used in RDA) and I cannot get
the species (variable centroids) to the ordination plot, only
Hi,
I'm trying to select the best model for a particular problem. So far i
have managed to identify a set of variables that woudl explain my model
lm1 <- lm(Group ~ . , data=dataf))
> summary(lm1)
Df Sum Sq Mean Sq F valuePr(>F)
`A` 1 2.3963 2.3963 24.0390 7.328e-06
On 21.11.2011 01:32, Jonathan Greenberg wrote:
This is a follow-up to a question I asked a few years back. We have a pair
of computers that share a common home directory (and therefor a common
.Renviron) with identical hardware, but very different sets of libraries
such that using a "shared" R
Hi,
Original-Nachricht
> Datum: Mon, 21 Nov 2011 14:46:17 +0100
> Von: Uwe Ligges
> An: Johannes Radinger
> CC: r-help@r-project.org
> Betreff: Re: [R] lm and loop over variables
>
>
> On 21.11.2011 13:34, Johannes Radinger wrote:
> > Hello,
> >
> > I think I am not to far
Hi,
Thanks, your method does indeed work. Thank you.
Last night, I worked out something similar and found out about rowMeans as
well.
Kind wishes,
Ian
yy <- read.table( header = T, sep=",", text =
"Q20, Q21, Q22, Q23, Q24
0,1, 2,3,4
1,NA,2,3,4
2,1, 2,3,4
5,NA,3,NA,NA")
yy
Q20 Q21 Q22 Q23
Hi,
Thanks for the quick reply.
However, your suggestion doesn't solve the problem I'm afraid (i.e
dim(Matrix) still the same). What I want is to reduce the number of
rows (probesets) markedly based on their normalized intensity - thus
I've chosen a cut-off of 1.11
When I run your code (as well a
I could not install the tseries package.
Problem solved. I had an older version of R and it did not work well with
Mac OSX
I installed the 2.14 version and it works fine.
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Hello,
Using the technique suggested here (
https://stat.ethz.ch/pipermail/r-devel/2007-September/047028.html),
we can update a matrix in place (see example 1 below), with huge time and
memory savings.
Is it possible to tweak example 2 to update in place also? The only
difference is that the
hey guys
I have a panel data set that i want to perform some regressions on. I am
using the /plm/ package.
I defined a model in the following way:
PWBw.pool <- plm(*PWB* ~ log(*I_EQON*) + log(*RD*) + ... + *PAGRI*,
data = pfem, na.action=na.exclude, model="pooling")
When i run this it gives the
Inline.
On Mon, Nov 21, 2011 at 5:41 AM, Peter Davidsen wrote:
> Hi,
>
> Thanks for the quick reply.
> However, your suggestion doesn't solve the problem I'm afraid (i.e
> dim(Matrix) still the same). What I want is to reduce the number of
> rows (probesets) markedly based on their normalized int
Am 21.11.2011 10:06, schrieb Jim Lemon:
>
> Hi Knut,
> Have a look at the legendg function (plotrix).
>
Hi Jim
Thanks for the package it is very useful .. but I did not solve my problem.
As I have not much time I tried some basic configuration but the
characters are einther above the colored sqar
Giovanni:
1. Please read ?formula and/or An Introduction to R for how to specify
linear models in R.
2. Correct specification of what you want (if I understand correctly) is
log(R) ~ A*B + C + D
3. ... which presumably will also fail because some of your factors
have only one level, which means
I would guess you have factors instead of strings. Use as.character() or the
appropriate parameter to prevent generation of factors when you import the
names into R.
---
Jeff NewmillerThe .
Dear Michael,
First, thanks a lot for your suggestion. It seems to work very well.
However, trying to implement it into my code I realize the rest of my code
is simply running too slowly due to a lot of loops. The problem lies in the
step where I read in a text file showing the different communit
Remko,
Thank you for your reply. When I try your code, I'm getting the following
error:
acts10 <-read.table('RailAccident10.txt',sep=',',header=T)
> acts11 <- acts10[acts10$ACCDMG > 10^5,]
> h.test <- substr(acts11$CAUSE,1,1)
> h.data <- acts11[h.test == 'H']
Error in `[.data.frame`(acts11, h.tes
You need a statistician or at least someone who's take a stats course in the
last 10 years but it may be what the author was trying to get at.
At least the binomial is descrete as is
the z so it may be that the z was used as easier to calculate than a binomial?
How old is the paper. Before,
On Nov 21, 2011, at 10:00 AM, aclinard wrote:
Remko,
Thank you for your reply. When I try your code, I'm getting the
following
error:
acts10 <-read.table('RailAccident10.txt',sep=',',header=T)
acts11 <- acts10[acts10$ACCDMG > 10^5,]
h.test <- substr(acts11$CAUSE,1,1)
h.data <- acts11[h.te
Hello Bert,
Thank you for taking the time to try to answer.
1) I know this, however if one is interested in only interaction between two
specific factors then in R one uses I(A*B*C) meaning 3-way anova for that and
not the implicit 2-ways that would otherwise be computed.
2) True, but it fails
As I said before: please dput() some working data and I'll try to work
something up.
Without it, the only thing I can reasonably suggest is that perhaps
you are looking for the window() function to be applied before
min/max. Something like:
X <- ts(1:48, start = 1, frequency = 4)
Y <- ts(1:12, st
Giovanni,
Have you tried Bert suggestion 2)? Because his log(R) ~ A*B + C + D is NOT the
same as your log(R)~A+B+I(A*B)+C+D
Note that I(A * B) means: create a new variable that is the product of A and B.
Which is not meaningfull if A and B are factors (hence the warning you got).
So I(A * B) is
Can someone explain why the following happens?
---
:> quote(some.name)
some.name
:> bar <- structure(quote(some.name), class = "foo")
:> quote(some.name)
Error in print(some.name) : object 'some.name' not found
:> bar <- quote(some.name)
:> quote(some.name)
Error in print(some.name) : object 'some
the way I interpret the problem (and I may be wrong here, I don't think you
have been particularly clear with your question) is that you are trying to
make a factorial anova where you are trying to explain "R" as a result of
A,B,C and D, and their interaction terms. so using A*B*C*D.
what you sh
I'd appreciate it if you'd keep on list for the archives. That said, I
think this function does what you were hoping for.
Michael
powerset <- function(n, items = NULL){
if(!is.null(items)) {
if(n != length(items)) warning("Resetting n in preference to
length(items)")
n = lengt
This is a bug in R: it either should prohibit the
attaching of attributes to things of class "name"
or should make it so that attachment of an attribute
doesn't have such a global effect.
A more direct example of the problem is:
> attributes(quote(some.name))
NULL
> bar <- structure(quote(so
Many Thanks - Also for the link!
It works nice!
If i have further question, can i post them here or should i open a new
thread?
1)
If i want the following to make a function:
I do have to convert it, but i can´t get rid of these " " (brackets).
func<-function(y) {
library(quantmod)
getSymbols
Hello R users,
I'm trying to replace numerical values in a datamatrix with strings. R does
this except for numbers under 1 starting with a 9 (eg 98, 970, 9504
etc). This is really weird and I wondered whether someone had encountered
such a problem or knows the solution. I'm using the next scri
Dear R Users,
Do you know of an existing function that allows the production of
sensitivity and specificity forest plots?
See the following for an example:
http://www.google.co.uk/imgres?q=forest+plots+of+sensitivity+and+specificity&um=1&hl=en&authuser=0&biw=1920&bih=989&tbm=isch&tbnid=JLxXNU7iQ2N
Hello
I would appreciate your help on the followig. I want to generate random
binary matrices but I need to discard those with all-1 rows. That is, for a
10x10 matrix with five 1's
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]000000000 1
[2
1) It's sort of a cheap trick, but this works flexibly (i.e., you can
put in func(SPY) or func("SPY") and get the same output):
func <- function(y){
if(!require(quantmod)) stop("You need the quantmod package.")
chartSeries(get(getSymbols(as.character(substitute(y)), from =
"2011-11-01")))
This can't be reproduced without data -- kindly supply the result of
test_1 right after the first line using dput() if you would.
Michael
On Mon, Nov 21, 2011 at 10:42 AM, set wrote:
> Hello R users,
>
> I'm trying to replace numerical values in a datamatrix with strings. R does
> this except fo
On Nov 21, 2011, at 11:42 AM, R. Michael Weylandt wrote:
I'd appreciate it if you'd keep on list for the archives. That said, I
think this function does what you were hoping for.
Michael
powerset <- function(n, items = NULL){
if(!is.null(items)) {
if(n != length(items)) warning("Res
any()
Michael
On Mon, Nov 21, 2011 at 10:58 AM, Juan Antonio Balbuena wrote:
> Hello
> I would appreciate your help on the followig. I want to generate random
> binary matrices but I need to discard those with all-1 rows. That is, for a
> 10x10 matrix with five 1's
>
> [,1] [,2] [,3] [,4] [
Hi,
You write below:
> What I specifically need is a way to write "If any of the elements of the
> SUMROW vector == N.1s then flag <-true else flag<-false".
And in fact, ?any is just what you need, as you said.
any(SUMROW == N.1s)
should do the trick.
Sarah
On Mon, Nov 21, 2011 at 10:58 AM, J
1) "datamatrix" is not a defined term. I think you mean "data.frame".
2) you have not supplied any sample data, so your example is not reproducible.
3) All of the values in a vector (i.e. a column of a data.table must be of the
same type, be that character or numeric (or anything else, such as f
On Nov 21, 2011, at 10:58 AM, Juan Antonio Balbuena wrote:
Hello
I would appreciate your help on the followig. I want to generate
random
binary matrices but I need to discard those with all-1 rows. That
is, for a
10x10 matrix with five 1's
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
On Nov 21, 2011, at 11:58 AM, R. Michael Weylandt wrote:
1) It's sort of a cheap trick, but this works flexibly (i.e., you can
put in func(SPY) or func("SPY") and get the same output):
func <- function(y){
if(!require(quantmod)) stop("You need the quantmod package.")
chartSeries(get(getS
Dear Community,
I would like to know if the outliers from the outcome of aq.plot are the
ones with probability associated with D2 is < .001.
Thanks in advance, u...@host.com
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Hello!
I have a binary data from DGGE gel with 20 samples.
These samples are divided into 4 groups, so there are 4(groups) x
5(replicas) = 20.
I used ANOSIM method, and everything is ok but ... I have a problem with
boxplot...
like this:
distance <- vegdist(dgge2, method="jaccar
Dear list,
i'm a new R user, so I apologize if the topic is already being addressed
by some other user.
I'm trying to determine if the reproductive success of a species of bird
is related to a list of covariates.
These are the covariates:
§elev: elevation of nest (meters)
§seadist: d
You didn't dput() your data as I asked so I can't work with it, but,
to get you started, here's an example of how you could do
interpolation of the sort you seem interested in using xts objects.
(It also seems like you need to decide what to make of the repeated
measurements before trying to extrap
R-helpers:
Say I have two lists of arbitrary elements, e.g.:
list1=list(c(1:3),"R is fun!",c(3:6))
list2=list(c(10:5),c(5:3),c(13,5),"I am so confused")
I would like to produce a single new list that is composed of all
combinations of the "top level" of list1 and list2, e.g.:
listcombo=list(
Thanks Thierry:
I had missed that the OP's failure to read the formula docs and use of
I(A*B) was what caused the error. Mea Culpa.
However, I actually agree with Giovanni's remarks about the difference
between what is typically taught and what one faces in practice. Where
we disagree is that I t
Dear All,
I am very new to R - trying to teach myself it for some MSc coursework.
I am plotting temperature data for two different sites over the same time
period which I have downloaded from a university weather station data
archive.
I am using the following code to create the plot
plot ( x
Hello Rob,
Thank you for your suggestions. I tried glm too without success. Anyhow I
include all the information just in case someone with good knowledge can give
me a hand with this. I take log of the response variable because:
- its values span across multiple orders of magnitudes
- the diag
Hello!
I need to use Kruskal-Wallis test and post-hoc test (Dunn's test) for my
data. But when I searched around, I only found this function: kruskal.test. But
nothing for Dunn's test.
So I started to write one myself. But I do not know how to count ties in
the data frame. I can use
On Mon, 21 Nov 2011, William Dunlap wrote:
This is a bug in R: it either should prohibit the
attaching of attributes to things of class "name"
or should make it so that attachment of an attribute
doesn't have such a global effect.
A more direct example of the problem is:
> attributes(quote(som
On Mon, Nov 21, 2011 at 11:36 AM, Hao, Zhaozhe wrote:
> Hello!
>
> I need to use Kruskal-Wallis test and post-hoc test (Dunn's test) for my
> data. But when I searched around, I only found this function: kruskal.test.
> But nothing for Dunn's test.
>
> So I started to write one myself. Bu
On Nov 21, 2011, at 2:50 PM, Peter Langfelder wrote:
On Mon, Nov 21, 2011 at 11:36 AM, Hao, Zhaozhe
wrote:
Hello!
I need to use Kruskal-Wallis test and post-hoc test (Dunn's
test) for my data. But when I searched around, I only found this
function: kruskal.test. But nothing for Dunn'
On Nov 21, 2011, at 2:36 PM, Hao, Zhaozhe wrote:
Hello!
I need to use Kruskal-Wallis test and post-hoc test (Dunn's test)
for my data. But when I searched around,
"Searching around" is a bit vague as a search strategy:
Try :kruskal dunn" at the location that RSiteSearch gets you to:
h
On Nov 21, 2011, at 2:17 PM, SarahH wrote:
Dear All,
I am very new to R - trying to teach myself it for some MSc
coursework.
I am plotting temperature data for two different sites over the same
time
period which I have downloaded from a university weather station data
archive.
I am usi
On Nov 21, 2011, at 3:20 PM, Hao, Zhaozhe wrote:
Hi,
Thank you all for the the quick response. But there still some
questions.
1) nTies = length(x) - length(unique(x)) cannot distinguish
vector (1,2,2,2,3), and (1,2,2,3,3)
2) table(x)[table(x) >1] tells me the right number
I think the easiest way to do this is to set up a color vector with
ifelse and hand that off to the plot command: something like
col = ifelse(TEMP3[,"SITE"] == "BG1", "blue", "green") # Syntax is
ifelse(TEST, OUT_IF_TRUE, OUT_IF_FALSE)
For more complicated schemes, a set of nested ifelse()'s can
Hi, I am trying forestplot() and metaplot() from rmeta package to plot some
hazard ratios. Now foreatplot() can draw clipping with "<" or ">" at the ends
of the confidence line using "clip" argument, but can't use multiple colors for
different lines. On the other hand, metaplot() can use differe
Hi,
Thank you all for the the quick response. But there still some questions.
1) nTies = length(x) - length(unique(x)) cannot distinguish vector
(1,2,2,2,3), and (1,2,2,3,3)
2) table(x)[table(x) >1] tells me the right number, but how can I call
the numbers from the result o
Hi all
I hope you might help me with some aspects of producing a graph in lattice.
There are three things I have struggling with and that is: 1. to separate the
horizontal box rows from each other; 2. to change the colour of the horizontal
and vertical strips to white; and 3. to place the axes l
Hello everyone,
I'm trying to build R 2.14.0 from source on Win7 x64 using the
instructions from "R Installation and Administration" (version
2.13.2), but when I try to make the core files (as of step 3.1.3 of
the instructions), it fails with the following error:
...
x86_64-w64-mingw32-gcc -std=gn
On Nov 21, 2011, at 8:31 PM, Bert Gunter wrote:
> we disagree is that I think data analysts with limited statistical
> backgrounds should consult with local statisticians instead of trying
> to muddle through on their own thru lists like this. This is not meant
I think that people lacking reading
I just tried a litte bit with your dataset and hope to get a solution as
wanted ;-)
On 21.11.2011 21:55 (UTC+1), Andrew McFadden wrote:
Hi all
I hope you might help me with some aspects of producing a graph in lattice. There are three things
I have struggling with and that is: 1. to separate t
Hi:
Strictly a guess, but the following might be helpful. The call below
assumes that the referent data frame is test1, which consists of a
single column named x. Modify as appropriate.
test_lab <- with(test1, cut(x, c(0, 18954, 37791, 56951, 75944, 84885, 113835),
labels = c('I8
On Mon, Nov 21, 2011 at 7:42 AM, set wrote:
> Hello R users,
>
> I'm trying to replace numerical values in a datamatrix with strings. R does
> this except for numbers under 1 starting with a 9 (eg 98, 970, 9504
> etc). This is really weird and I wondered whether someone had encountered
> such
This sort of post seems to me to be completely unacceptable.
Is there a mechanism by which the list manager can unsubscribe
Mr. Azua and keep him unsubscribed until he learns some manners?
cheers,
Rolf Turner
On 22/11/11 10:28, Giovanni Azua wrote:
On Nov 21, 2011, at 8:31 PM, Ber
This is probably a very noobish question, but if I want to create a
function that allows an undetermined number of, say, numeric vectors to be
fed to it, I would use:
myfunction = function(...)
{
# Do something
}
Right? If so, how do I a) count the number of vectors "fed" to the
function, and b
You can do something like this:
test <- function(x,...){
print(x)
args = list(...)
if('y' %in% names(args))print(args$y)
if('z' %in% names(args))print(args$z)
}
Regards
Søren
Fra: r-help-boun...@r-project.org [r-help-boun...@r-project
Hi Jonathan,
On Mon, Nov 21, 2011 at 5:18 PM, Jonathan Greenberg wrote:
> This is probably a very noobish question, but if I want to create a
> function that allows an undetermined number of, say, numeric vectors to be
> fed to it, I would use:
>
> myfunction = function(...)
> {
> # Do something
Apologies for thickness - I'm sure that this operates as documented and with
good reason. However...
My understanding of arima.sim() is obviously imperfect. In the example below I
assume that x1 and x2 are similar white noise processes with a mean of 5 and a
standard deviation of 1. I thought x
On 22/11/11 13:04, Andy Bunn wrote:
Apologies for thickness - I'm sure that this operates as documented and with
good reason. However...
My understanding of arima.sim() is obviously imperfect. In the example below I
assume that x1 and x2 are similar white noise processes with a mean of 5 and a
Thank you everybody!
Eventually the Peter's trick did it!
Thanks!
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Hi,
I have two time series of NDVI data, one inside a protected area, and one
outside. I have decomposed the time series, to remove the seasonal component
and residual variation.
I want to compare the long-term trends inside and outside of the protected
area over time. So, at each time step (1 y
Hello, all,
I'm a new R user (new to any programming language in general, really), so I
apologize if this is easy/has already been answered (I've attempted
searching online but did not understand the pages I found).
My data is stored in text files with the headers LANE, RNA_NAME, SEQ, and
SEQCNT.
So basically I have made a HTML file with a table in it.
Column 3 contains a GenBank number and is always proceeded by "=GenBank">".
I want to read the file and return the number which comes directly after
this (the contents of column 3).
Ideally I would like to save this number as a string fo
I got the colour vector with ifelse to work, great! Thank you.
Is it possible to use the ifelse colour vector with other plot types? For
example with type=l ? I tried but the graphic came back with blue lines for
both sites and also a straight line connecting the start and end point of
the data?
Another approach would be to use ggplot2.
Code can look a bit daunting to begin with but ggplot2 is a
very versitile graphing package and well worth learning.
Simple example
=
library(ggplot2)
mydata <- data.frame(site=c("A","A","A", "
Hi
I am a newbie to R and I would like to ask a few questions about RServ Java
API...
My problem is, there will be many callers use my client web service (it uses
the rosuda Rengine API) to connect to RServ (running on linux), instead of
each caller establish a new connection, I would want to le
I don't think you can do different colors for a single line (not an
ifelse thing, just a what would that mean sort of thing), but a plot
type like "b" "o" or "h" will work the same way.
Michael
On Mon, Nov 21, 2011 at 4:23 PM, SarahH wrote:
> I got the colour vector with ifelse to work, great! T
Hi: You can make a list out of the ... by doing below.
myfunction = function(...)
{
params <- list(...)
print(params)
print(str(params))
}
On Mon, Nov 21, 2011 at 5:18 PM, Jonathan Greenberg wrote:
> This is probably a very noobish question, but if I want to create a
> function that allows an un
I have a scatter plot with 1 points. I would like to add a line that bins
every 50 points and connects the average of each bin. I'm looking for
something similar to line type "m" in Stata.
With this dataset of 1 points, I would also like to bin the data and make
boxplots at certain i
On Nov 21, 2011, at 10:18 PM, R. Michael Weylandt wrote:
I don't think you can do different colors for a single line (not an
ifelse thing, just a what would that mean sort of thing), but a plot
type like "b" "o" or "h" will work the same way.
I think Jim Lemon has a multicolored line function
On Nov 22, 2011, at 12:29 AM, Jeffrey Joh wrote:
I have a scatter plot with 1 points.
So you have numeric x and y values.
I would like to add a line that bins every 50 points and connects
the average of each bin.
What is the rule to be applied to form these bins? You may want to
l
Hi Jeffrey,
See ?factor ?rep and ?cut basically you just need to create another
variable that indicates what bin a point belongs to, and then you just
do regular plots. If you want to mix the scatterplot and the binned
points, you'll need to make sure the bins fall somehwere in the same
space.
A fast way to get **only** means of successive bins of width k of a
vector z of length nk is:
m <- colMeans(matrix(z,nrow = k))
(you wanted k =50)
This was instantaneous for a length 1e6 numeric vector on my laptop.
However, for anything else, you have to use something like cut() as
others have
On 11/22/2011 06:31 AM, Jonathan Greenberg wrote:
R-helpers:
Say I have two lists of arbitrary elements, e.g.:
list1=list(c(1:3),"R is fun!",c(3:6))
list2=list(c(10:5),c(5:3),c(13,5),"I am so confused")
I would like to produce a single new list that is composed of all
combinations of the "
On Mon, 21 Nov 2011, Frédéric Fournier wrote:
Hello everyone,
I'm trying to build R 2.14.0 from source on Win7 x64 using the
instructions from "R Installation and Administration" (version
2.13.2), but when I try to make the core files (as of step 3.1.3 of
the instructions), it fails with the fol
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