http://finzi.psych.upenn.edu/R/library/datasets/html/morley.html >
"The classical data of Michaelson and Morley on the speed of light"
Can you provide more information about the data? How were they obtained,
etc.? I do not have the book "Genstat Primer" and the nearest location where
it is av
Thanks to David and Michael
Michael:
That works, however with a glitch. Each of my 24 files has got two columns:
"Name", and "Rank/score".
file_list <- list.files()
list_of_files <- lapply(file_list, read.csv) # Read in each file
# I can see the 2-columns at this stage. However, the following l
On Apr 11, 2012, at 12:17 AM, R. Michael Weylandt wrote:
Your problem is that you can't merge the file names, but you need to
load them into R and merge the resulting objects.
This should be straightforward enough to do:
file_list <- list.files()
list_of_files <- lapply(file_list, read.csv) #
On Apr 10, 2012, at 11:48 PM, Christopher Kelvin wrote:
Hello,
I wish to censor 10% of my sample units of 50 from a Weibull
distribution. Below is the code for it.
I will need to know whether what i have done is correct and if not,
can i have any suggestion to improve it?
Thank you
p=2;b
Your problem is that you can't merge the file names, but you need to
load them into R and merge the resulting objects.
This should be straightforward enough to do:
file_list <- list.files()
list_of_files <- lapply(file_list, read.csv) # Read in each file
merge_all(list_of_files, by = "Name")
Mic
You don't need to return() from the for loop -- just put your outputs
in a variable:
set.seed(1) # For reproducibility
x <- numeric(20)
for(i in 1:20) x[i] <- bob(0.5, sqrt)
or (more elegant but basically the same thing)
set.seed(1)
x1 <- replicate(20, bob(0.5, sqrt)) # Same calculation done 20x
Hello,
I wish to censor 10% of my sample units of 50 from a Weibull distribution.
Below is the code for it.
I will need to know whether what i have done is correct and if not, can i have
any suggestion to improve it?
Thank you
p=2;b=120
n=50
r=45
t<-rweibull(r,shape=p,scale=b)
meantrue<-gamma(
Hi all,
I wish to merge 24 .csv files, each having a common identifier-column
("Name") and do two things:
1. Retrieve the common one's. [Analogy: while merging 2-dataframes, similar
to using: merge ( ,by="Name", all=FALSE) ]
2. Retrieve all, i.e., the union of the rows of 24 files. [again,
You were told before this isn't a Mac question so please don't cc R-SIG-Mac.
I'm not sure what this bit of your reply means "My question is to find
any command to plot the data I got from the field;" but your reply
later suggests that your problem is that you are overriding previous
plots on a giv
Thanks.
So, I want the function to return results from the if statement.
bob <- function(var1, func)
{
#func: a simple function
num1 <- var1
num2 <- func(var1)
if(ruinf(1) wrote:
> The error message should make it pretty clear -- you aren't inside a
> function so you can't return()
Dear Michael (and Davis), Your answer is not what I want to know. My question
is to find any command to plot the data I got from the field; such as a set of
(x,y) data ( I actually have these data) and together withe the derived ones .
I brought these data to plot on x-y plane, getting a graph
Two ways around this:
I = Easy) Just use zoo/xts objects. ts objects a real pain in the
proverbial donkey because of things like this.
Something like:
library(xts)
PI1.yq <- as.xts(PI1) # Specialty class for quarterly data (or regular
zoo works)
lag(PI1.yq)
II = Hard) lag on a ts actually chang
This is the same malformatted message you posted on R-SIG-Mac even
after David specifically asked for clarification not to reward bad
behavior, but perhaps this will enlighten:
# Minimal reproducible data!
x <- runif(15, 0, 5)
y <- 3*x - 2 + runif(15)
dat <- data.frame(x = x, y = y)
rm(list =
The error message should make it pretty clear -- you aren't inside a
function so you can't return() a value which is, by definition, what
functions (and only functions) do.** Not sure there's a great
reference for this though beyond the error message
Incidentally, what are you trying to do her
Hello,
I am writing codes for time series computation but encountering some
problems
Given the quarterly data from 1983Q1 to 1984Q2
PI1<-ts(c(2.747365190,2.791594762, -0.009953715, -0.015059485,
-1.190061246, -0.553031799, 0.686874720, 0.953911035),
start=c(1983,1), frequency=4)
> PI1
hi, I'm doing some data on least square curve fitting. What I like to have is
to compare the scatter plot whilst having the fitting curve on the same
coordinates. Any suggestting command besides plot(x,y). TaweeMac OSX 10.7.3
[[alternative HTM
Dear all, I get an error using the return function. The following is a
simpler version.
for (j in 1:10)
{
samples = 5*j
return(samples)
}
Error: no function to return from, jumping to top level
Similar warning happens with if statement.
Why do I get an error? print() works fine. I d
On Apr 10, 2012, at 7:57 PM, Josh O'Brien wrote:
Many thanks for your help.
I mistakenly deleted my original message (not even knowing that was
possible). Apologies for that.
Heh. I suspect it disappeared from your mail-client but not from the
Archive (or for that matter the multiple rhelp
The Milwaukee Chapter of the ASA (MILWASA) in cooperation
with The Medical College of Wisconsin,
Marquette University,
The Children's Research Institute,
The Clinical and Translational Science Institute (CTSI)
and Quantitative Health Sciences
are proud to announce
R Programming Workshops with Bill
Many thanks for your help.
I mistakenly deleted my original message (not even knowing that was
possible). Apologies for that.
For future reference, when a section of an R manual (like the bit from the
'Installing R Under Windows Section' of the R-admin manual) no longer
applies to the current ve
On Apr 10, 2012, at 8:01 PM, Worik R wrote:
> Thank you.
>
> That was exactly what I need.
>
> Looking at '?[' I see...
>
> drop: For matrices and arrays. If TRUE the result is coerced to
> the lowest possible dimension (see the examples). This only
> works for extract
Thank you.
That was exactly what I need.
Looking at '?[' I see...
drop: For matrices and arrays. If TRUE the result is coerced to
the lowest possible dimension (see the examples). This only
works for extracting elements, not for the replacement. See
drop
On Apr 10, 2012, at 7:33 PM, Worik R wrote:
Friends
I am extracting sub-sets of the rows of a matrix. Generally the
result is
a matrix. But there is a special case. When the result returned is a
single row it is returned as a vector (in the example below an integer
vector). If there are
Friends
I am extracting sub-sets of the rows of a matrix. Generally the result is
a matrix. But there is a special case. When the result returned is a
single row it is returned as a vector (in the example below an integer
vector). If there are 0, or more than 1 rows returned the result is a
ma
Sorry this is so late:
But you could try a "nerge" (from the 'caroline' package)
nerge(list(a,b,c))
Just have to make sure that the rows for each dataframe are renamed with
the date columns.
On 1/30/2012 11:44 PM, Massimo Bressan wrote:
thanks don
I have here enough to study for a while...
On Apr 10, 2012, at 22:03 , nerak13 wrote:
> Hi,
>
> I've got the following data:
>
> x<-c(1,3,5,7)
> y<-c(37.98,11.68,3.65,3.93)
> penetrationks28<-dataframe(x=x,y=y)
>
> now I need to fit a non linear function so I did:
>
> fit <- nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, star
hello everyone,
i wanted help to build a* user defined function in R for multiple sequence
alignment*.
i tried using pairwisealignment () for the multiple sequences, but the
results seem to be wrong.
i have 10 sequences which i have to align ( by building profiles).
is there a function to build a
On 10.04.2012 20:24, Josh O'Brien wrote:
I am attempting to build a customized R installer on Windows, using the Inno
Setup installer.
I am following the instructions in Section 3.1.8 of the R Installation and
Administration Manual ("Building the Inno Setup installer"), which includes
the foll
On 10.04.2012 23:03, Katharine Miller wrote:
Hi,
I have been using some code for multivariate random forests. The output
from this code is a list object with all the same values as from
randomForest, but the model object is, of course, not of the class
randomForest. So, I was hoping to modif
Hi,
I have been using some code for multivariate random forests. The output
from this code is a list object with all the same values as from
randomForest, but the model object is, of course, not of the class
randomForest. So, I was hoping to modify the code for predict.randomForest
to work for p
On Apr 10, 2012, at 4:03 PM, nerak13 wrote:
Hi,
I've got the following data:
x<-c(1,3,5,7)
y<-c(37.98,11.68,3.65,3.93)
penetrationks28<-dataframe(x=x,y=y)
now I need to fit a non linear function so I did:
fit <- nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start =
list(a=0,b = 1,
Hi,
I've got the following data:
x<-c(1,3,5,7)
y<-c(37.98,11.68,3.65,3.93)
penetrationks28<-dataframe(x=x,y=y)
now I need to fit a non linear function so I did:
fit <- nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start =
list(a=0,b = 1,c=1), trace = T)
The error message I get is:
Er
you need to re-install it.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playi
Hi All,
I have a self-cooked package and save it to a zip file after
running make, say named xxx.zip. After installing it to R by running "Install
packages from local zip files" under Packages menu in R (Windows), I realized I
needed to change some source codes and re-make it. My q
If you brought the data in from Excel, it is probably a dataframe.
You probably need to read the "Intro to R" on how to access
information in a dataframe.
You should at least show what you did. most likely VIQ is a column in
your Excel file and depending on how you read it in, you would
probably
How many levels can the the nlme function in nlme library handle when a
multilevel dataset is fitted? It will be greatly appreciated if any examples
with 3 or 4 levels can be shared. Thanks for your attention.
Hua
[[alternative HTML version deleted]]
_
I am attempting to build a customized R installer on Windows, using the Inno
Setup installer.
I am following the instructions in Section 3.1.8 of the R Installation and
Administration Manual ("Building the Inno Setup installer"), which includes
the following passage:
An alternative way to c
Hi,
I am very new to R and the stats world. I have enjoyed working with R so far
but I have come across an error message in a very simple command that I am
unable to resolve.
I bring data in through excel .csv files and check them to be sure R reads
them correctly and has everything assigned as
So it's a machine/OS issue: if you really want to trace it down, take
a look here: http://svn.r-project.org/R/trunk/src/main/seq.c
seq.int() in R goes to do_seq() in C, but at this point it's probably
best to identify it as floating-point gremlins and to work around.
Michael
On Tue, Apr 10, 2012
On Apr 10, 2012, at 1:08 PM, Steve Lavrenz wrote:
I definitely need a loop - the example I gave was only a simple
one. Say I
want to do more complex calculations in each step, such that the
numeric
difference between consecutive terms is not constant.
I will try out some of the methods th
Hi everyone,
Thank you so much for your help. I see that one can use tricks, such as double
brackets, and sparing use of gc() to help with memory usage in R, the fact is
that a new copy of the object is made every time a named object is assigned,
and R's garbage collection should take care of
On 04/10/2012 09:11 AM, stella wrote:
Hi,
Sorry, I am bad with regular expression and a beginner with R. How do I get
only the numbers 0009987 from the following entry?
GO:0009987~cellular process
sub(".*:(.*)~.*", "\\1", "GO:009987~cellular process")
but you might also be interested in
s
On Apr 10, 2012, at 1:08 PM, Steve Lavrenz wrote:
I definitely need a loop - the example I gave was only a simple one.
Say I
want to do more complex calculations in each step, such that the
numeric
difference between consecutive terms is not constant.
You can always use:
for( i in seq(0,
Hello,
>
> In Reply To
> compare two matrices
> Apr 10, 2012; 9:26am — by Kehl Dániel Kehl Dániel
> Dear Members,
>
> I have two estimated transition matrices and I want to compare them.
> In fact I want to check the hypothesis if they come from the same process.
> I tried to look for some test
Is mgcv and particularly its gam available for Splus? I've been using it
happily in R and need to implement something in Splus for which the
automatic smoothing parameter selection is needed.
Thanks for any guidance,
David Katz
da...@davidkatzconsulting.com
--
View this message in context:
htt
While you can build up a vector like this in a for loop, this is exactly
the sort of construction that leads to excessive memory growth because on
each iteration of the loop R creates a new copy of the vector x - old
copies have no references to them, but are not deallocated until the next
automati
Hi,
Sorry, I am bad with regular expression and a beginner with R. How do I get
only the numbers 0009987 from the following entry?
GO:0009987~cellular process
Thanks a lot,
Stella
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View this message in context:
http://r.789695.n4.nabble.com/Get-part-of-a-GO-term-tp4546125p4546125.html
Sent f
Thank you Michael
It is indeed the OS
identical(seq.int(0,1,length.out = 11), seq.int(0,1, by = 0.1)) # FALSE
Michael Weylandt wrote
>
> What difference is it you are worried about:?
>
> identical(seq.int(0,1,length.out = 11), seq.int(0,1, by = 0.1)) # TRUE
>
> Though that may be OS dependen
I definitely need a loop - the example I gave was only a simple one. Say I
want to do more complex calculations in each step, such that the numeric
difference between consecutive terms is not constant.
I will try out some of the methods that have been shared so far. Thank you!
-Steve
Fr
Do you need a loop at all?
Will this do the trick?
seq(from=0, to=100, by=5)
Jean
Steve Lavrenz wrote on 04/10/2012 09:48:34 AM:
> Everyone,
>
> I'm very new to R, especially when it comes to loops and functions, so
> please bear with me if this is an elementary question. I cannot seem to
>
What's wrong with
num <- rep(0,10)
done <- FALSE
i <- 2
while(!done){
num[i] <- num[i-1] + 5
if(num[i] > 20) done <- TRUE
i <- i + 1
}
num <- num[1:(i-1)]
You can delete the unused tail when you finish. It
On Apr 10, 2012, at 12:19 PM, David Winsemius wrote:
On Apr 10, 2012, at 11:58 AM, Rainer Schuermann wrote:
cbind() works as well, but only if c is attached to the existing
test variable:
tst <- cbind( test, c )
tst
ab c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5
On Apr 10, 2012, at 11:53 AM, Steve Lavrenz wrote:
Albyn,
Thanks for your help. This however, still assumes that I have to
define an
array of length 10. Is there a way that I can construct this so that
my
array is exactly as long as the number of spots I need to reach my
threshold
value
On Apr 10, 2012, at 11:58 AM, Rainer Schuermann wrote:
cbind() works as well, but only if c is attached to the existing
test variable:
tst <- cbind( test, c )
tst
ab c
1 1 0.3 y1
2 2 0.4 y2
3 3 0.5 y3
4 4 0.6 y4
5 5 0.7 y5
str( tst )
'data.frame': 5 obs. of 3 va
I don't know the multicore package, but if possible, it might be
easier to upgrade to 2.15 and use the new built-in parallel package
that was introduced in R 2.14.
Then your syntax would be something like
mclapply(files, illumqc)
Michael
On Tue, Apr 10, 2012 at 11:33 AM, Wyatt McMahon wrote:
>
Sorry, I missed that the OP's real question was in character/factor,
not in the "why are these all factors" bit...good catch.
Rant about cbind() still stands though. :-) [Your way with cbind()
would give him all characters, not some characters and some numerics
since cbind() gives a matrix by de
On Tue, Apr 10, 2012 at 5:32 PM, R. Michael Weylandt
wrote:
> You might try the Defaults package.
>
Thanks for the hint. Unfortunately
> library(Defaults)
> setDefaults('create_progress_bar', name = "text")
doesn't do the trick. But one can set defaults for each *ply()
function individually:
> s
Hi,
?sd
?cor.test
...
Regards,
Pascal
De : anaraster
À : r-help@r-project.org
Envoyé le : Mardi 10 avril 2012 22h50
Objet : [R] taylor.diagram from plotrix package
Is there a way to access the numeric results (standard deviation and
correlation) obtained with
As the error message suggests, see ?memory.size, and you'll find that the
problem is arising because R is running out of memory. If you were able to
run this analysis before, then one possible reason why it now fails is that
the workspace has increased in size in the interim - more objects and
resu
The key to using cv.glm is that you have to have a fitted model object with
all the data to validate. In your case, that would appear to be a model
like this:
lm(base[,ncol(base)]~NDI)
where NDI is calculated from two bands in the dataframe base. However, if
the ground truth data is independently c
Hello everyone,
I'm trying to parallelize an R script I have written. To do this, I am
first trying to use the multicore package, because I've had some previous
success with that.
The function I'm trying to parallelize is illumqc. I'd like to create a
separate process for each of 8 files,
> x<-numeric(1)
> x
[1] 0
> x[2]<-2
> x
[1] 0 2
you don't really need to define the length?
Am 10.04.2012 um 17:45 schrieb Albyn Jones:
> Here are a couple of constructions that work.
>
> albyn
> ===
>
> num <- rep(0,10)
> for (i in 2:10) {
>
cbind() works as well, but only if c is attached to the existing test variable:
> tst <- cbind( test, c )
>
> tst
>
ab
Still didn't work for me without cbind , although you really don't need it ;)
worked after i set options(stringsAsFactors=F).
> options(stringsAsFactors=F)
> df<-data.frame(intVec,chaVec)
> df
intVec chaVec
1 1 a
2 2 b
3 3 c
> df$chaVec
[1] "a" "b" "c"
documentat
Albyn,
Thanks for your help. This however, still assumes that I have to define an
array of length 10. Is there a way that I can construct this so that my
array is exactly as long as the number of spots I need to reach my threshold
value?
Thanks,
-Steve
-Original Message-
From: Albyn Jon
Incoherent. Please read the posting guide . Also, no homework.
Bert
Sent from my iPhone -- please excuse typos.
On Apr 10, 2012, at 7:12 AM, Mariam wrote:
> People, help me please!
> How to use lm() function to defind a cofficient for 7-polinom, and what
> expression should I put in /formula/
What difference is it you are worried about:?
identical(seq.int(0,1,length.out = 11), seq.int(0,1, by = 0.1)) # TRUE
Though that may be OS dependent.
M
On Tue, Apr 10, 2012 at 10:51 AM, Alexander wrote:
>
> Berend Hasselman wrote
>>
>> On 10-04-2012, at 15:54, Alexander wrote:
>>
>>> I am work
Here are a couple of constructions that work.
albyn
===
num <- rep(0,10)
for (i in 2:10) {
num[i] <- num[i-1] + 5
if(num[i] > 20) break
}
> num
[1] 0 5 10 15 20 25 0 0 0 0
or
num <- rep(0,10)
done <- FALSE
i <- 2
while(!done){
You probably have more objects in your workspace than you did
previously. Clean them out (or just use a new R session) and things
should go back to normal.
You might also want to follow up on the help(memory.size) hint though
-- doesn't Windows impose a memory limit unless you ask it for more?
Mi
http://cran.r-project.org/doc/manuals/R-lang.html#while
i<-2
while(value <=100){
num[i] <- num[i-1] +5
value <- num[i]
i <- i+1
}
something like this?
greetings Jessi
Am 10.04.2012 um 16:48 schrieb Steve Lavrenz:
> Everyone,
>
> I'm very new to R, especially when it c
Don't use cbind() -- it forces everything into a single type, here
string, which in turn becomes factor.
Simply,
data.frame(a, b, c)
Like David mentioned a few days ago, I have no idea who is promoting
this data.frame(cbind(...)) idiom, but it's a terrible idea (albeit
one that seems to be very
Hi Mariam,
Check out the ?poly function.
Best,
Jorge.-
On Tue, Apr 10, 2012 at 10:12 AM, Mariam <> wrote:
> People, help me please!
> How to use lm() function to defind a cofficient for 7-polinom, and what
> expression should I put in /formula/
>
> --
> View this message in context:
> http://r
You might want to use a while loop instead, something like:
while(TRUE){
# Do things
# Test: if your condition has occured
if(conditionHappened) break # break will end loop.
}
Michael
On Tue, Apr 10, 2012 at 10:48 AM, Steve Lavrenz
wrote:
> Everyone,
>
> I'm very new to R, especially when
You might try the Defaults package.
Michael
On Tue, Apr 10, 2012 at 10:54 AM, Liviu Andronic wrote:
> Dear all
> Is it possible to set globally the option .progress = "text" to all
> the apply functions in 'plyr'. For example, current default is
> daply(..., .progress = "none"). I would like to
Berend Hasselman wrote
>
> On 10-04-2012, at 15:54, Alexander wrote:
>
>> I am working under R2.11.1 Windows and I was wondering why there is a
>> difference between
>>
>> seq.int(0,1,by=0.1)[4]-0.3
>> seq.int(0,1,length.out=11)[4]-0.3
>>
>> there is also the fact that
>>
>> seq(0,1,by=0.1)[
P.S. : "\xb2" works but must be used with perl = TRUE.
# 'y' defined as above.
grep("\xb2", y, perl = TRUE)
gsub("\xb2", "HERE", y, perl = TRUE)
Rui Barradas
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A few days ago I responded to Ramiro with a suggestion that turns out to be
incorrect.
> Ramiro
> >
> > I think the problem is the loop - R doesn't release memory allocated
> inside
> > an expression until the expression completes. A for loop is an
> expression,
> > so it duplicates fit and datase
Hello:
While running R doing the analysis of my data I (using packages such as
BIOMOD or e1071) get the following error as a result of several of my
analysis:
Error: cannot allocate vector of size 998.5 Mb
In addition: Warning messages:
1: In array(c(rep.int(c(1, numeric(n)), n - 1L), 1),
Hello,
To use the octal code works with me.
# I've created a file with that byte only.
x <- readLines("ascii0178")
y <- c(as.character(1:4), x, as.character(6:10))
y
grep("\262", y) # should return 5
gsub("\262", "HERE", y)
Hope this helps,
Rui Barradas
--
View this message in context:
htt
Greetings:
rockchalk is a collection of functions to facilitate presentation of regression
models.
It includes some functions that I have been circulating for quite some time
(such as
"outreg") as well as several others. The main aim is to allow people who do not
understand very much R to surviv
Complete newbie to R -- struggling with something which should be pretty
basic. Trying to create a simple data set (which I gather R refers to as a
data.frame). So
> a <- c(1,2,3,4,5);
> b <- c(0.3,0.4,0.5,0,6,0.7);
Stick the two together into a data frame (call test) using cbind
> test <- dat
Dear All,
I have just uploaded a new version of the QCA (Qualitative Comparative
Analysis) on CRAN, and it will be propagated in a couple of days.
This is version 1.0-0 ("Easter edition") of the package, straight from
the previous version 0.6-5, and it represent a major re-write of the
package in
People, help me please!
How to use lm() function to defind a cofficient for 7-polinom, and what
expression should I put in /formula/
--
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Sent from the R help mailing list archive at Nabble.com.
_
Everyone,
I'm very new to R, especially when it comes to loops and functions, so
please bear with me if this is an elementary question. I cannot seem to
figure out how to construct a loop which runs a function until a certain
value is computed. For example, say I have the following:
num = numer
A slightly easier formulation of the second proposal from Jessica:
plot(c(0,0) , xlim = range(x1, x2, x3), ylim = range(y), type = "n")
will set the canvas correctly.
On Tue, Apr 10, 2012 at 10:10 AM, Jessica Streicher
wrote:
> Hello Arunkamar!
>
> Basically:
>
> plot(x1,y)
> lines(x2,y)
> line
I've got the strange problem with clock24.plot that only the first data
point (phase = 23.38, size = 0.44) from the phases/sizes numeric vectors
is plotted.
Does anyone have an idea why this could be?
library(plotrix)
phases <- c(23.38, 22.29, 22.71)
sizes <- c(0.44, 0.30, 0.30)
clock24.plot(sizes
Dear all
Is it possible to set globally the option .progress = "text" to all
the apply functions in 'plyr'. For example, current default is
daply(..., .progress = "none"). I would like to set it to daply(...,
.progress = "text"), so as to avoid writing the argument every time I
call such a function
On Apr 10, 2012, at 10:21 AM, Dennis Fisher wrote:
R 2.14.1
OS X
Colleagues,
I am making a graphic with two y-axes. I create the label for the
right-side axis with:
mtext(side=4, line=1, "Some text")
The label is rotated 90° counterclockwise. I would prefer that it
be rotated 9
On 10-04-2012, at 15:54, Alexander wrote:
> I am working under R2.11.1 Windows and I was wondering why there is a
> difference between
>
> seq.int(0,1,by=0.1)[4]-0.3
> seq.int(0,1,length.out=11)[4]-0.3
>
> there is also the fact that
>
> seq(0,1,by=0.1)[4]-0.3
> seq(0,1,length.out=11)[4]-0.3
R 2.14.1
OS X
Colleagues,
I am making a graphic with two y-axes. I create the label for the right-side
axis with:
mtext(side=4, line=1, "Some text")
The label is rotated 90° counterclockwise. I would prefer that it be rotated
90° clockwise. However, srt is not supported for mtext and
Hi,
I am wondering if anybody has experience with
scatterplot matrices where some (but NOT all) axis
are transformed and the labels are nicely plotted.
So far I looked into
1) pairs()
2) scatterplotMatrix() from package 'car'
3) splom() from packagae 'lattice'
4) plotmatrix() from 'ggplot2'
I ca
Hello Arunkamar!
Basically:
plot(x1,y)
lines(x2,y)
lines(x3,y)
You might need to adjust the first plot so all data is shown. For that you
could use something like
plot(c(min(x),max(x)) , c(min(y),max(y)),type="n")
x is all data from x1,x2,x3. type="n" says that these points won't be shown in
t
Il giorno mar, 10/04/2012 alle 09.54 -0400, David Winsemius ha scritto:
> Perhaps, modulo encoding issues I'm not expert in, one more
> backslash
> than you tried:
>
> > gsub( "\\\xb2" , "2",
> "(MPa)\t(mm3)\t(nM)\t(mm3/g)\t(mm3/g)\t(%)\t(m
> \xb2/g)\t")
> [1] "(MPa)\t(mm3)\t(nM)\t(mm3/g)\t(mm
I am working under R2.11.1 Windows and I was wondering why there is a
difference between
seq.int(0,1,by=0.1)[4]-0.3
seq.int(0,1,length.out=11)[4]-0.3
there is also the fact that
seq(0,1,by=0.1)[4]-0.3
seq(0,1,length.out=11)[4]-0.3
but I think this can be explained by floating precision...
A
Is there a way to access the numeric results (standard deviation and
correlation) obtained with the taylor.diagram ?
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Sent from the R help mailing list archive at Nabble.com.
Hi
I have four sets of datas
x1, x2, x3,y
I want to sactter plot between (x1,y) and line chart between (x2 ,y) and
(x3,y)
all these should come in a single graph
Can anyone help
-
Thanks in Advance
Arun
--
View this message in context:
http://r.789695.n4.nabble.com/plotting-mul
On Apr 10, 2012, at 9:44 AM, ottorino wrote:
Deae R helpers,
the problem I'm facing today is related to the manipulation of a
string.
The string is coming from a a porosimeter, whose control is under a
complicate set-up of two computers
One (running on DOS) is controlling directly the hard
Hi,
I run R on a unix server and login from a Mac with ssh -X. When I want to run a
graphics function like hist, I get the following x11 message:
Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma, d$colortype,
:
unable to start device X11cairo
In addition: Warning message:
In f
Deae R helpers,
the problem I'm facing today is related to the manipulation of a string.
The string is coming from a a porosimeter, whose control is under a
complicate set-up of two computers
One (running on DOS) is controlling directly the hardware, while the
other (running on win XP) which proc
On Apr 10, 2012, at 8:59 AM, David Winsemius wrote:
On Apr 10, 2012, at 3:16 AM, aajit75 wrote:
I have got solution using within function as below
dd$Seg <- 1
dd <- within(dd, Seg[x2> 0 & x3> 200] <- 1)
In this instance the first of your assignments appears superfluous.
dd <- within(dd,
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