Am 28.07.2011 03:25, schrieb Fernando Andreacci:
I have a simple bar chart with annual precipitation (jan to dez).
I want to plot, above each bar (on a line), a square wich is color based on
a scale (0-100%). With 0 being white and 100 black, like a gradient. Is it
possible? How to?
Thanks
He
Dear all,
Can you pl. help me in plotting a matrix into 2D plot with more color
options. I tried to plot with myImagePlot with
RGB color ramp but it shows limited variation in color. There is a rainbow
plot option also. Can anyone help me.
Regards,
Hitendra
--
Hitendra Padalia, PhD
Scientist/Eng
Just a pointer, not a solution: if you use the method I described, it all
comes down to the initial correlation matrix that you specify for the
Gaussian case. If you had the "correct" initial correlation matrix, it would
lead to the desired linear correlation in your binomials. (But it is not
guar
Dear Bill,
For bivariate data, ordinary least product regression is equivalent to
geometric mean regression, or what Freedman/Pisani/Purves call the “SD Line”
(line passing through the centroid of the data, whose slope's magnitude is
the ratio of the standard deviations of the variables, and s
I have a simple bar chart with annual precipitation (jan to dez).
I want to plot, above each bar (on a line), a square wich is color based on
a scale (0-100%). With 0 being white and 100 black, like a gradient. Is it
possible? How to?
Thanks
--
Fernando Andreacci
[[alternative HTML vers
Some more skilled folks can help with the curve fitting, but the general
answer is yes -- R will handle this quite ably.
Consider the following code:
<<-->>
n = 1e5
length = 1e5
R = matrix(sample(c(-1,1),length*n,replace=T),nrow=n)
R = apply(R,1,cumsum) ## thi
Thank you both very much! The codes are pretty slick and should greatly help
me in my task.
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Hi:
Try this:
exampGood = lapply(2:4, function(x) matrix(rnorm(10 * x), ncol = x))
exampBad = lapply(1:3, function(x) matrix(rnorm(10 * x), ncol = x))
csfun <- function(m) {
if(ncol(m) == 1L) {return(m)} else {
t(as.matrix(apply(m, 1, cumsum)))
}
}
lapply(exampGood, csfun)
lapply(e
On Jul 27, 2011, at 9:42 PM, Dennis Murphy wrote:
Hi:
Is this more or less what you're after?
## Note: This is the preferred way to send your data by e-mail.
## I used dput(data-frame-name) to produce this,
## where data-frame-name = 'df' on my end.
df <- structure(list(V1 = c("chr1", "chr1",
Here's another question:
What's kind of difference does glht tests?
In other words,
If I get 4 intercepts and 4 slopes for all levels of GROUP,then glht tests
for the difference of the 4 intercepts or 4 slopes,or something else?
Thanks.
2011/7/27 Lao Meng
> Yes.
> According to your suggestio
You can replace the inner loop
for (i in 1:length) {
if (s[i] < 0 && s[i + 1] >= 0) {
z[i] = z[i] + 1
}
}
with the faster
z <- z + (diff(s>=0)==1)
(Using the latter forces you fix up the length of
z to be one less than you declared -- your loop never
touched the last entry in it
Hi:
Is this more or less what you're after?
## Note: This is the preferred way to send your data by e-mail.
## I used dput(data-frame-name) to produce this,
## where data-frame-name = 'df' on my end.
df <- structure(list(V1 = c("chr1", "chr1", "chr1", "chr1", "chr3",
"chr4", "chr4", "chr7", "chr7
On Jul 27, 2011, at 7:44 PM, Gene Leynes wrote:
> David,
>
> Thanks for the suggestion, but I think your answer only works
> because I was printing the wrong thing (because apply with margin=1
> transposes the results,
And if you want to change that, then the t() function is readily at
ha
Top posting cuz hotmail decided not to highlight...
Personally I would tend to use java or c++ for the inner loops
but you could of course later make an R package out of that.
This is especially true if your code will be used elsewhere
in a performance critical system. For example, I wrote som
Dear R folks,
Am Donnerstag, den 28.07.2011, 01:36 +0200 schrieb Paul Menzel:
> I need to simulate first passage times for iterated partial sums. The
> related papers are for example [1][2].
>
> As a start I want to simulate how long a simple random walk stays
> negative, which should result th
David,
Thanks for the suggestion, but I think your answer only works because I was
printing the wrong thing (because apply with margin=1 transposes the
results, something I always forget).
Check this to see what I mean:
str(answerGood)
str(answerBad)
Adding "as.matrix" is interesting and
Hi:
I don't get exactly the same results as you did in the second group
(how does temp.t[1] = -2.0 instead of -2.2?) but try this:
locality=c("USC00020958", "USC00020958", "USC00020958", "USC00020958",
"USC00020958", "USC00021001","USC00021001", "USC00021001", "USC00021001",
"USC00021001", "USC00
Dear R folks,
I need to simulate first passage times for iterated partial sums. The
related papers are for example [1][2].
As a start I want to simulate how long a simple random walk stays
negative, which should result that it behaves like n^(-½). My code looks
like this.
8< c
Yes, that is the general objective. I'll look-into aggregates in R and see if
anything helps.
Thanks,
a217
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_
On Jul 27, 2011, at 7:02 PM, a217 wrote:
Hello,
I have an input file:
http://r.789695.n4.nabble.com/file/n3700031/testOut.txt testOut.txt
where col 1 is chromosome, column2 is start of region, column 3 is
end of
region, column 4 and 5 is base position, column 6 is total reads,
column 7
i
Hi:
This is a linear model since you have no nonlinear term *in the
parameters*. The model you postulated is as far as you can go since
you have 1 df for error (unless you remove the intercept term). As the
plot below shows, a linear model does not fit the data well -
basically, it provides a quad
Hello,
I have an input file:
http://r.789695.n4.nabble.com/file/n3700031/testOut.txt testOut.txt
where col 1 is chromosome, column2 is start of region, column 3 is end of
region, column 4 and 5 is base position, column 6 is total reads, column 7
is methylation data, and column 8 is the strand.
I'm not sure what you're doing... but here are some tips about the parts I
can understand.
1) you don't need to use "which" as much. This works fine:
stnID <- stnid[!duplicated(stnid)]
2) "which" works within a for loop
3) Do you realize that "stnID" is shorter after you removed duplicates
Hello.
I am using R via PHP by making a system call, like this
$cmd = "echo source(\"$maindir\") | \"$Rterm\" --vanilla --slave --ess
--args ...
and I would like to know if there is any way of showing the identify()
interactivity in the browser on my website, save the selected points,
close
Daniel, thanks for the help. I finally made it, doing the merging separately.
Daniel Malter wrote:
>
> On a different note: how are you matching if AA has multiple matches in
> BB?
>
About that, all I have to do is check whether, for any of the BB which
matches with AA, the indicator equals 1.
On Jul 27, 2011, at 6:22 PM, Gene Leynes wrote:
I have tried a lot of ways around this, but I can't find a way to
make apply
work in a generalized way because it causes a failure whenever
reduces the
dimensions of its output.
The following example is easier to understand than the question.
Paul,
I agree completely. I didn't come at this with a programming background,
and I never noticed the other plot links, or understood their importance.
The "plot" help page was one of the first help pages I ever looked at and I
remember that It was totally confusing. In fact, help pages like th
Am Mittwoch, den 27.07.2011, 18:04 -0400 schrieb David Winsemius:
> On Jul 27, 2011, at 5:53 PM, Paul Menzel wrote:
> > as `plot.function` is not explicitly mentioned in `?plot`.
>
> Right, but the fact that it _is_ described as a "generic" function
> will tell the clueful that other methods ot
I have tried a lot of ways around this, but I can't find a way to make apply
work in a generalized way because it causes a failure whenever reduces the
dimensions of its output.
The following example is easier to understand than the question.
I wish it had a "drop=TRUE/FALSE" option like the "["
Dear all,
Does any one know if any R package or function can do Ordinary Least Products
regression? Many thanks!
Bill
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-proj
On Jul 27, 2011, at 5:53 PM, Paul Menzel wrote:
as `plot.function` is not explicitly mentioned in `?plot`.
Right, but the fact that it _is_ described as a "generic" function
will tell the clueful that other methods other than the default method
may exist, and that their availability may
Am Mittwoch, den 27.07.2011, 17:21 -0400 schrieb David Winsemius:
> On Jul 27, 2011, at 4:53 PM, Paul Menzel wrote:
> > Am Mittwoch, den 27.07.2011, 13:26 -0700 schrieb Bert Gunter:
> >> Paul:
> >> No such change is needed.
> >
> > Well the fact is, that I as a beginner was looking for who I could
Hi Cassie,
I'm not sure exactly what you are trying to get: i assume that each
station will have a different ppt value for each year-month
combination, but it looks like you're trying to get one ppt value for
each station.
Steve
On Thu, Jul 28, 2011 at 1:30 AM, Wilson, Cassandra J wrote:
> I a
Hi:
I prefer to use one of the summarization packages for this sort of
thing, but aggregate() works, too. Here are two versions of the same
idea:
# Uses ddply() in the plyr package:
index.date <- function(d) {
require('plyr')
out1 <- ddply(d, .(id, rcat), summarise, index = max(tdiff))
On Jul 27, 2011, at 4:53 PM, Paul Menzel wrote:
Dear Bert,
Am Mittwoch, den 27.07.2011, 13:26 -0700 schrieb Bert Gunter:
Paul:
No such change is needed.
Well the fact is, that I as a beginner was looking for who I could
plot
normal functions, so one more example would have helped me.
On Jul 27, 2011, at 8:40 AM, oaxacamatt wrote:
Greetings all,
I have two sets of data that I would like to investigate. The first is
gene/genome related data given different 'cell-states'. The second
set of
data is relates the genes to a biological pathway. /(I think in
pictures so
here go
Dear Bert,
Am Mittwoch, den 27.07.2011, 13:26 -0700 schrieb Bert Gunter:
> Paul:
> No such change is needed.
Well the fact is, that I as a beginner was looking for who I could plot
normal functions, so one more example would have helped me.
> You do not understand S3 methods.
That is probably
Hi all,
Is there any R package that fits binary regression models with power
(generalized) logistic link function?
Thanks a lot in advance,
Best,
Caio
[[alternative HTML version deleted]]
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https://stat.eth
Hi,
I am trying to replicate Engle and Manganelli's paper Conditional
Autoregressive Value at Risk (CAViaR) by Regression Quantiles. I have the
Matlab code which I cannot get to work as I have never used Matlab before,
does anyone know if there is the same code available to estimate the CAViaR
mod
Thanks, Enrico and Harold.
I have searched the archives for this topic, but all I can find is
about the multivariate normal or uniform generator, which doesn't help
in my case.
Harold's transformation is good for getting the correlation, but the
negative binomial distribution will be changed afte
Hi Daniel,
I am still failing to do that? May you look at my look. It is calculating the
estimates for one row. How do I incorporate the other data that has all other
columns?
Thanks
Ed
Sent from BlackBerry® wireless device
-Original Message-
From: "Daniel Malter [via R]"
Date: Sat,
Paul:
No such change is needed. You do not understand S3 methods. See
?plot.default and read about S3 methods (e.g. in the tutorial
Introduction to R or the Language manual).
-- Bert
On Wed, Jul 27, 2011 at 12:30 PM, Paul Menzel
wrote:
> Dear R folks,
>
>
> currently the section Examples contain
Hello I am trying to use predict from an arma-Garch model (arma(2, 2) +
garch(1, 1)) and I am getting the following error:
Error en arima(x = object@data, order = c(max(u, 1), 0, max(v, 1)), init =
c(ar, :
non-stationary AR part from CSS
Does anybody know what can be the reason of this error?
Dear R folks,
currently the section Examples contains the following as an example on
how to plot a “normal” function.
plot(sin, -pi, 2*pi)
Since it does not contain the argument for the function it would be
helpful to add for example the following.
## plots the graph of f(x) =
Am Mittwoch, den 27.07.2011, 14:36 -0400 schrieb David Winsemius:
> On Jul 27, 2011, at 10:12 AM, Paul Menzel wrote:
> > I am having problems getting good results when searching for R related
> > topics, that means I have not found out yet what keywords I should use
> > to get only relevant result
On Jul 27, 2011, at 10:12 AM, Paul Menzel wrote:
Dear R folks,
I am having problems getting good results when searching for R related
topics, that means I have not found out yet what keywords I should use
to get only relevant results. Most of the time I get also MATLAB
related
things and n
R Community -
I am attempting to fit a model as described in Hampton, Bossaerts, and
O'doherty (J. Neuroscience) 2006. They use a bayesian hidden markov model
to model the Reversal Learning data. I have tried using HMM and depmixS4
with no success. My data is a Reversal Learning Task in which t
On Jul 27, 2011, at 2:00 PM, Abraham Mathew wrote:
Lets say I have the following data frame.
df = data.frame(word = c("David", "James", "Sara", "Jamie", "Jon"))
df
I was trying to place brackets , [ ] , around each string.
I'll be exporting it with write.table and quotes=FALSE, so it will
ev
On 27/07/2011 10:12 AM, Paul Menzel wrote:
Dear R folks,
I am having problems getting good results when searching for R related
topics, that means I have not found out yet what keywords I should use
to get only relevant results. Most of the time I get also MATLAB related
things and nothing rela
On 27.07.2011 20:00, Abraham Mathew wrote:
Lets say I have the following data frame.
df = data.frame(word = c("David", "James", "Sara", "Jamie", "Jon"))
df
I was trying to place brackets , [ ] , around each string.
I'll be exporting it with write.table and quotes=FALSE, so it will
eventually
On 25.07.2011 19:42, eilunedpearce wrote:
Hi,
I'm trying to install CAIC directly into the newest version of R using the
code on the R-Forge CAIC website and I get an error message:
install.packages("CAIC", repos="http://R-Forge.R-project.org";)
Warning message:
In getDependencies(pkgs, depen
Use rseek.org.
Jeremy
On 27 July 2011 07:12, Paul Menzel wrote:
> Dear R folks,
>
>
> I am having problems getting good results when searching for R related
> topics, that means I have not found out yet what keywords I should use
> to get only relevant results. Most of the time I get also MATLAB
On 27/07/2011 9:21 AM, dean123 wrote:
Hi David,
I am trying to define t as a single numeric value and not as a vector.. I
want the function to realise that if I call test(t) and t is less than equal
10 return zero, if t greater than 10 return 2*t. Am i missing some code in
which to define t a nu
I spent all the day on this problem and I've just finally found a solution:
with options(digits=13), it now works.
If I hadn't found this, I would have used the solution of Renaud (but not
very convenient to me).
Nice end of day,
(Happy) Ptit Bleu.
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View this message in context:
http://r.789
Dear list,
I have a large dataset which is structured as follows:
locality=c("USC00020958", "USC00020958", "USC00020958", "USC00020958",
"USC00020958", "USC00021001","USC00021001", "USC00021001", "USC00021001",
"USC00021001", "USC00021001")
temp.a=c(-1.2, -1.2, -1.2, -1.2, -1.1, -2.2, -2.4, -2.6
Dear R-helpers
I have 7 data points that I want to fit a continuous curve to, that should look
similar to a sine wave
My data points would mark the local minima and maxima respectively.
This is what I’ve got so far. And I would keep doing so, but sadly nls() then
says that it has reached the ma
On Tue, Jul 26, 2011 at 10:12 AM, Ista Zahn
wrote:
> OK, easy enough:
>
> dat.tmp <- data.frame(id, month, value)
> my.plot <- function(dat) {print(dat[, c("id", "value")])}
> by(dat.tmp, id, my.plot)
Excellent. The output of that last line is:
* id value
1 110
2 112
3 111
4 1
Dear R folks,
I am having problems getting good results when searching for R related
topics, that means I have not found out yet what keywords I should use
to get only relevant results. Most of the time I get also MATLAB related
things and nothing related at all. The nature of this is of course t
Lets say I have the following data frame.
df = data.frame(word = c("David", "James", "Sara", "Jamie", "Jon"))
df
I was trying to place brackets , [ ] , around each string.
I'll be exporting it with write.table and quotes=FALSE, so it will
eventually look like:
[David]
[James]
[Sara]
Can
In fact, I always use a string for ID variables because you can get
some funky matching errors due to mismatches in precision.
Bill Gould at Stata Corp said this well in a recent blog post
"1. Use theory to check IDs if they are numeric
One way the id variable can become corrupted is if it is not
Dear
R users,
I
created a matrix that tells me the first day of use of a category by
id.
#Calculate
time difference
test$tdiff<-as.numeric(difftime(as.Date("2002-09-01"), test$ftime, units =
"days"))
#
obtain the index date per person and dcategory
Is anyone able to offer a reason that, when using weights in an lm fit, the
Cook's distances
shown in plot.lm(x, which=4) differ from those calculated by cooks.distance(x) ?
I can see that they _are_ different and the code tells me _how_ they differ
(essentially, one is using the weights and
Hi Marcus,
That's almost it! So ncol and column do the same thing in different
functions... gah!
Because my real use of xyplot I override the default plotting colors and
character, autokey doesn't 'know' this, so they don't match.
So the addition of the par.settings element is needed (from anot
I am having a hard time putting the below into a loop, where it pulls out ppt
from all he stations I have versus having to go through and hard code the data
to the specific stations. I tried
stnID <- stnid[which(duplicated(stnid)==FALSE)]
for(i in 1:length(stnID))
{
ppt[i] <- ppt[which(stnid==[i]
Jz, guys lay off.
I think you are being a little to hard on the poor soul. The manuals are
written in computer-eez and some of the tutorials are no better. /If you
can't help then don't answer./
In the meantime, Tia, I apologize. But I suggest you try and try a good
book or website that c
Greetings all,
I have two sets of data that I would like to investigate. The first is
gene/genome related data given different 'cell-states'. The second set of
data is relates the genes to a biological pathway. /(I think in pictures so
here goes.)/
*dataframe1*
gene, cell-state1, cell-state2
gene1,
Add this line of code to the end of your function, after the plot()
function.
points(B1[N], B2[N], pch=16, col=10)
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409 USA
Hi David,
I am trying to define t as a single numeric value and not as a vector.. I
want the function to realise that if I call test(t) and t is less than equal
10 return zero, if t greater than 10 return 2*t. Am i missing some code in
which to define t a numeric value instead of a vector?
Also,
On Jul 27, 2011, at 1:31 PM, Mark wrote:
Hi,
I want my xyplot legend to be flat, not tall, and there seems to be
no way
for xyplot's auto.key and key elements to do this: I tried many, many
permutations of what I could find in the archives and reading the
documentation. If there there's a
Don't really undertsand, are you looking for
points(B2[N], B1[N])
?
Uwe Ligges
On 27.07.2011 11:59, Rebecca Gray wrote:
Hello,
I would appreciate if someone could help me with this query. I would like to
plot a line chart of all of the points in a "for" loop. I would also like
to plot th
I ran the code I sent earlier and there were a few bugs. That's what I
get for using my phone to write code.
This works a little better.
#direction
angle<-seq(0,350,10)
#distance factor extends to
dist = sample.int(9000,size=length(angle))
#list of colors, order corrisponds to associated
On 25.07.2011 07:11, Vihan Pandey wrote:
Hi all,
I am an R newbie, and I have a question about scripting. I have the
following lines which I want to put int\
o a script which I can call from the shell of a Mac/Linux machine :
myrns<- read.csv(file="/Users/vihan/test.csv",sep="",header=FALSE)
On 24.07.2011 20:15, Madana_Babu wrote:
Hi Lei,
Thanks for your solution. It worked. Now I have another query.
After creating multiple DF[[i]]'s how do I aggregate them into one data frame
say DF (I want to bind all the data frames into one data frame). I have more
than 1000 DF[[i]]'s how ca
Hi Mark,
I believe the argument you are looking for is column not ncol.
Try the following:
library(lattice)
xyplot(1~1,
auto.key=list(
bty='n',
pch=rep(c(15,17),2),
col=c('brown','green','blue','red'),
columns=4,
text=c('How','to make','this into', 'a grob?'),
space="top"
))
Regards
Mar
On 27.07.2011 02:17, Jeremy Miles wrote:
This is clearly a message for the R-help mailing list, since it was
sent to the R help mailing list.
fisher.test(x)[1]
Since the test object is actually a list, your probably want
fisher.test(x)[[1]]
or more obvious
fisher.test(x)[["p.value"]]
Uwe
Hi,
I want my xyplot legend to be flat, not tall, and there seems to be no way
for xyplot's auto.key and key elements to do this: I tried many, many
permutations of what I could find in the archives and reading the
documentation. If there there's a way to make it flat, please tell me what
the mag
On 2011-07-27 05:35, Dieter Menne wrote:
Irene Prix wrote:
In a grouped Dotplot, is there any way to set the color of error bars to
be the same as the corresponding symbols?
...
require(lattice)
require(Hmisc)
data(barley)
Dotplot(variety~Cbind(yield, yield+2, yield-2)|year, groups=site,
dat
On 27/07/2011 12:50 PM, Megh Dal wrote:
Dear all, while executing some function, there are some custom messages popping
up onto the R console and I do not want to see them. While looking into the
corresponding codes of those function, I see that those are coming from
message() function.
Is th
Dear all, while executing some function, there are some custom messages popping
up onto the R console and I do not want to see them. While looking into the
corresponding codes of those function, I see that those are coming from
message() function.
Is there any way to stop those messages coming
Do you mean something like this?
> cors <- matrix(c(1, .9, .8, .8, .9, 1, .8, .8, .8, .8, 1, .9, .8, .8, .9, 1),
> 4)
> L <- chol(cors)
> N <- 1000
> dat <- cbind(rnbinom(N, mu = 4, size = 1), rnbinom(N, mu = 4, size = 1),
> rnbinom(N, mu = 4, size = 1), rnbinom(N, mu = 4, size = 1))
> result <-
Hi all,
I have written some functionalities in C# and want to call them in R, maybe
from another machine. I have found the following methods:
1. wrapper the C# code as Web Service: I noticed that there is a package named
SSOAP, but it seems not yet finished and might be not fully compatible wit
On Jul 27, 2011, at 10:52 AM, Marcus Mund wrote:
Hello everybody,
I hope this question is not too silly but I'm almost going crazy about
that and could not find a solution.
I have two variables, say A and B and I would like to combine them
in C.
In particular I want a C-value of B when B is
Hi,
use tableName[["columnName"]] without quotes, e.g.
tableName[[columnName]] # see ?"[["
and indexing is done via
x<-length(tableName[[columnName]][tableName[[columnName]] > 6] )
but you may consider
x<-sum(tableName[[columnName]]>6)
which saves some time and typing
Hth.
Am 27.07.2011 17:2
Thanks David,
Unfortunately, that is an option that I cannot use. The code that I provided
was a simplification of a more complex workflow that accommodates various
types of histograms (count, density or percentage). The actual panel
function of my histogram call contains several switches that all
On Jul 27, 2011, at 9:52 AM, Marcus Mund wrote:
> Hello everybody,
>
> I hope this question is not too silly but I'm almost going crazy about
> that and could not find a solution.
>
> I have two variables, say A and B and I would like to combine them in C.
> In particular I want a C-value of B w
Hi,
I'm using the tsDyn package to estimate a multivariate threshold VAR.
Currently, I'm in trouble with the selection of an external transition
variable. I cannot select it properly. My xls spreadsheet (rates3) contains
3 variables: I want to use the first and the second one as "data" and the
thir
Hi all,
I've been having trouble with something that seems like it should be
fairly straight forward. Any help at all from more experienced users is
appreciated!
I'd like to write a function that uses a column name as an argument.
However, I run into problems when I try to reference this colum
Hello everybody,
I hope this question is not too silly but I'm almost going crazy about
that and could not find a solution.
I have two variables, say A and B and I would like to combine them in C.
In particular I want a C-value of B when B is not NA and the A value in
case that B is NA:
A B
Hi,
I'm using heatmap.2 to cluster my data, using the centroid method for
clustering and the maximum method for calculating the distance matrix:
library("gplots")
library("RColorBrewer")
test <- matrix(c(0.96, 0.07, 0.97, 0.98, 0.50, 0.28, 0.29, 0.77,
0.08, 0.96, 0.51, 0.51, 0.
Tia:
R is a statistical/data analysis/graphics programming language. There
is a learning curve that you appear unwilling to climb. If I have
misjudged you, please work through the "Introduction to R" tutorial.
It will provide the information your need. If not, install
R-Commander or another R GUI
I'm not sure if there are any packages that do this, but I've created similar
plots in R. The easiest way I've found is to think in terms of a unit circle
in polar coordinates for drawing the plot.
I haven't tested the code below, but it will give you the idea.
dist=dist/9000
t = seq(0,360
On Jul 27, 2011, at 9:15 AM, Sébastien Bihorel wrote:
Hi
I would like to superimpose group-specific densityplots on top of an
overall
histogram using panel.histogram and panel.densityplot. Furthermore,
I would
like to automatically adjust the range of the y-axis to take into
account
the
Dear Jokel,
Unfortunately, this won't work. The derivation of the equation given in the
Handbook shows why this is so. First of all, note that
d = (m1 - m2) / sp,
where m1 and m2 are the means of the two groups and sp is the pooled SD. The
two independent samples t-test (assuming homoscedastic
Hi Yves,
Many thanks. Very straight forward, the way you put.
Best regards
Ogbos
On 27 July 2011 15:13, Yves REECHT wrote:
> **
> Hi,
>
> You may try something like:
>
> plot(rnorm(10), rnorm(10), main=expression("" %+-% 2*sigma))
>
> HTH,
> Yves
>
> Le 27/07/2011 14:57, ogbos okike a écrit :
>
Hi
I would like to superimpose group-specific densityplots on top of an overall
histogram using panel.histogram and panel.densityplot. Furthermore, I would
like to automatically adjust the range of the y-axis to take into account
the ranges of both histogram and densityplot. This last part is wher
On Jul 27, 2011, at 8:57 AM, ogbos okike wrote:
Dear List,
I am trying to label a plot with the symbol +/- sigma. Using
something like
- expression (2*sigma) gives me the symbol 2ó. However, adding +/-
to it
beats me.
The code I am using is: plot(x,y,type="l",main="
expression(paste("±"
Hi,
You may try something like:
plot(rnorm(10), rnorm(10), main=expression("" %+-% 2*sigma))
HTH,
Yves
Le 27/07/2011 14:57, ogbos okike a écrit :
> Dear List,
> I am trying to label a plot with the symbol +/- sigma. Using something like
> - expression (2*sigma) gives me the symbol 2ó. How
Dear List,
I am trying to label a plot with the symbol +/- sigma. Using something like
- expression (2*sigma) gives me the symbol 2ó. However, adding +/- to it
beats me.
The code I am using is: plot(x,y,type="l",main=" expression(paste("±",
plain(2*ó)),sep="").
Any suggestion will be appreciated.
> summary(n1)
Formula: LBM ~ c0 * time^gamma
Parameters:
Estimate Std. Error t value Pr(>|t|)
c0 2.584780.36767 7.030 8.93e-06 ***
gamma 0.292570.03781 7.738 3.21e-06 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.7124 on
Ram H. Sharma wrote:
>
> I want to overlay lattice scatter plot: I do not know why the following
> code
> is not plotting subscripts ! Sorry if this question is too simple:
>
> Working example shortened:
>
> .panel.xyplot(x, y, pch=16, col = "green4", ylim = c(0, 10))
>
>
Because they ar
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