hi, guys,
I am just a beginner to the excellent R package, quantmod. I quite don't
know how to change the y-axis name in the chartSeries function.
Actually, I want to write some sort of the following function, by which
I could use just one code sentence to complete the financial analysis.
Hi R-helpers,
Is it possible to change the color of the boxes when plotting decision
trees using 'fancyRpartPlot()' from rpart.plot package ?
--
Regards,
Abhinaba Roy
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
h
On Jul 9, 2014, at 7:47 PM, Sébastien Bihorel wrote:
> Hi,
>
> I have this set of R scripts which are ran on a linux box and create plots
> with the lattice package. I do not specify any custom font family, so I
> believe that whatever is the default font on my system is used in the plot.
> 1- h
Hi,
I have this set of R scripts which are ran on a linux box and create plots
with the lattice package. I do not specify any custom font family, so I
believe that whatever is the default font on my system is used in the plot.
1- how can I know which is the default font used in my plots?
2- is thi
Dear John
There is my data set.
Thanks.
On Wed, Jul 9, 2014 at 8:12 PM, John Fox wrote:
> Dear Judith,
>
> I take it from your reply that you have *more* observations than there are
> response variables in the multivariate linear model, but since you still
> haven't provided access to the dat
Dear John,
I am including abundance values ââin my data set so obviously I have zero
abundances.
The problem is that if plot only the factors (biomasa, altdosel, altsoto,
cobertura, riqarb, elevacion, temperatura, precipitacion) I get the
graphic, the same happen when I included only the famil
Dear R Help,
I'm trying to install the rjags and rgdal packages on a linux cluster
running R 3.0.3. However, I'm having problems installing them
successfully. Both packages require external programs (JAGS and GDAL,
respectively), which have been successfully installed.
For rjags, the error
Dear Judith,
I take it from your reply that you have *more* observations than there are
response variables in the multivariate linear model, but since you still
haven't provided access to the data, it's still impossible to tell what the
problem is.
I don't follow your application, possibly be
I think you are mistaken. Please provide an example of how you used this
function in any version of R that behaved as you describe.
Also, please post in plain text to avoid the what-you-see-is-not-what-we-see
feature that HTML email provides.
--
Grumpy today, Jeff?
For the concrete issue, I'd conjecture that the base problem is that there are
way too many columns in the data and that the nature of the method is not
properly understood. It is not obvious that k-means clustering based on
Euclidean distance makes sense in 1426-dimensional
Hi R community,
i created a function (mkdate) as follows:
mkdate = function(x) {
x$date = as.Date(paste(x$year, x$month, x$day, sep="-"))
x$wy = ifelse(x$month >=10, x$year+1, x$year)
x$yd = as.integer(format(as.Date(x$date), format="%j"))
x$wyd = cal.wyd(x)
x
}
the function results in adding the
After reading the metals data frame, I would do this:
metals$result <- as.numeric(gsub('<','',metals$Cedar.Creek))
metals$flag <- ifelse(grepl('<',metals$Cedar.Creek),'<','h')
Also, assuming you got your data into R using read.table(),
read.csv(), or similar, I would include
stringsAsFactors=T
Dear Maria Judith Carmona Higuita,
Since you didn't include enough information (such as your access to your data)
to reproduce the error, one can only guess. My guess: you have fewer
observations in your data set than response variables on the LHS of the
multivariate linear model.
I hope this
On 10/07/14 04:24, Jeff Newmiller wrote:
Grumpy today, Bert?
Bert is ***always*** grumpy! :-) If he weren't, I'd get worried.
But then someone else, not more than a million miles from this email,
has a strong tendency to be grumpy (acerbic?) as well.
Of course ***I*** am ***never*** gr
Oh dear,
I seem to have suffered a case of reversed arguments.
This explains my surprise why R didn't have this in a function already -
as it does!
I was following the pattern of search.vector %in% pattern, but match()
arguments are opposite this.
Thanks to both Davids.
Michael
-Original Me
On Jul 9, 2014, at 1:13 PM, Folkes, Michael wrote:
> So nice!
> Apply wins again.
I doubt that `sapply( ..., which(,) )` would win a foot race with `match`:
> match(Tset, pop.df$pop)
[1] 5 4 2
--
David.
> Thanks David.
> Michael
>
> -Original Message-
> From: David L Carlson [mailto
So nice!
Apply wins again.
Thanks David.
Michael
-Original Message-
From: David L Carlson [mailto:dcarl...@tamu.edu]
Sent: July-09-14 1:11 PM
To: Folkes, Michael; r-help@r-project.org
Subject: RE: using match to obtain non-sorted index values from
non-sortedvector
There may be a faster
Hi Sam,
> But this may not be the important issue here at all. If " value is left censored at k -- i.e. we know it's less than k but not
> how much less -- than Sarah's proposal is not what you want to do.
> Exactly what you do want to do depends on context, and as it concerns
> statistical method
There may be a faster way, but
> sapply(Tset, function(x) which(pop.df$pop==x))
[1] 5 4 2
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-
Hello all,
I've been struggling with the best way to find index values from a large
vector with elements that will match elements of a subset vector [the
table argument in match()].
BUT the index values can't come out sorted (as we'd get in which(X %in%
Y) ).
My 'population' vector can't be so
Hi,
I have a problem using the function Candisc from Candisc Package.
bosques1<-read.csv("bosques1.csv",header=TRUE,encoding="latin1")
bosques1<-na.exclude(bosques1)
attach(bosques1)
#Modelo de regresión
mod <-
lm(cbind(biomasa,altdosel,altsoto,cobertura,riqarb,elevacion,temperatura,precipitaci
Actually, this doesn't _quite_ do what I want;
I want different R colors (1, 2, 3, &c.) to select
different pens in HPGL ("SP1", "SP2", "SP3", &c.),
but the HPGL file I get selects only pen 1.
A hacky way to do this would be to generate
a few different postscript files for the different
colors on
Thanks for all the responses. It sometimes difficult to outline
exactly what you need. These response were helpful to get there.
Speaking to Bert's point a bit, I needed a column to identify where
the < symbol was used. If I knew more about R I think I might be
embarrassed to post my solution to th
@ Sarah
Thanks a lot, paste does the job perfectly!
On 09/07/2014 17:46, Sarah Goslee wrote:
How about:
plot(dd$Sepal.Length, dd$Petal.Length, main=paste("These are the
levels:", paste(levels(dd$Species), collapse=", ")))
Thanks for the actual reproducible example!
Sarah
On Wed, Jul 9, 20
Well, ?grep and ?regex are clearly apropos here -- dealing with
character data is an essential skill for handling input from diverse
sources with various formatting conventions. I suggest you go through
one of the many regular expression tutorials on the web to learn more.
But this may not be the
On Jul 9, 2014, at 12:19 PM, Sam Albers wrote:
> Hello,
>
> I have recently received a dataset from a metal analysis company. The
> dataset is filled with less than symbols. What I am looking for is a
> efficient way to subset for any whole numbers from the dataset. The column
> is automatically
Hi Sam,
I'd take the similar tack of removing the < instead. Note that if you
import the data frame using the stringsAsFactors=FALSE argument, you
don't need the first step.
metals$Cedar.Creek <- as.character(metals$Cedar.Creek)
metals$Cedar.Creek <- gsub("<", "", metals$Cedar.Creek)
metals$Cedar
Hello,
I have recently received a dataset from a metal analysis company. The
dataset is filled with less than symbols. What I am looking for is a
efficient way to subset for any whole numbers from the dataset. The column
is automatically formatted as a factor because of the "<" symbols making it
d
Thanks for the suggestion.
I found that I get the result I wanted with this simple command:
split(unlist(A2),unlist(A1))
$`1`
[1] 2.718282
$`2`
[1] 7.389056 7.389056
$`3`
[1] 20.08554
$`4`
[1] 54.59815 54.59815 54.59815
$`5`
[1] 148.4132 148.4132
$`13`
[1] 442413.4
$`23`
[1] 9744803446
whi
Is the following 'g' what you want? A better example might be with
A2a <- lapply(A1, function(x)x+seq_along(x)/(100*length(x)))
g <- function (x, y) {
xLengths <- vapply(x, FUN = length, FUN.VALUE = 0L)
yLengths <- vapply(y, FUN = length, FUN.VALUE = 0L)
stopifnot(identical(xLength
The code is actually available at the websites you provide. Try "View page
source" in your browser. The most cryptic code isn't needed because the math
functions (e.g, incomplete gamma function) are available in R.
-Original Message-
From: Paul Miller [mailto:pjmiller...@yahoo.com]
S
Grumpy today, Bert?
While it is a fact that RStudio is a separate tool from R, it is clear from the
question that the OP is interested in capabilities that R is providing and he
simply cannot tell the difference.
OP:
1) "Better" is a word that leads to pointless arguments. You will have to be
Hi Alvaro,
this was a tricky problem. Under Windows R uses the trio library
(different from the package Trio which creates very similar error
messages) for printf support. arules currently contains a bug that
results in an invalid format string for printf when an error message is
created. For
How about:
plot(dd$Sepal.Length, dd$Petal.Length, main=paste("These are the
levels:", paste(levels(dd$Species), collapse=", ")))
Thanks for the actual reproducible example!
Sarah
On Wed, Jul 9, 2014 at 11:24 AM, Bea GD wrote:
> Hi all,
>
> I'd like to include the levels of one of my variable
To cut the tree, the clustering algorithm must produce consistently increasing
height values with no reversals. You used one of the two options in hclust that
does not do this. Note the following from the hclust manual page:
"Note however, that methods "median" and "centroid" are not leading to
Hi all,
I'd like to include the levels of one of my variables in the title of a
plot. I'd like these factor levels to be concatenated. E.g. 'These are
the levels: setosa, versicolor, virginica'.
I've been working with this code but I don't get the desired results.
Any suggestions would be a
Revolution Analytics staff and guests write about R every weekday at
the Revolutions blog:
http://blog.revolutionanalytics.com
and every month I post a summary of articles from the previous month
of particular interest to readers of r-help.
In case you missed them, here are some articles related
Thanks Michael, everything going perfect right now! I didn't expect such an
extensive list of itemsets given the insights on the data that I have. And the
error message didn't gave me the right clue. Now is working fine! Thanks for
your time!
Best Regards.
Alvaro.
-Mensaje original--
RStudio is a separate product with its own support. Post there, not here.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Tue, Jul 8, 2014 at 7:34 PM, Phan
Oh it was easier than I thought.
postscript('project-contracts.ps')
hist(log(projects$n.contracts))
dev.off()
Then run this from the shell.
pstoedit -f plot-hpgl project-contracts.ps project-contracts.hpgl
And send it to the plotter.
On 09 Jul 13:10, Thomas Levine wrote:
> Hi,
>
> I w
Hi,
I want to print plots on a Roland DXY-1100 plotter.
How can I do this from R? I think the easiest thing
would be a graphics device for Printer Command
Language or Hewlett-Packard Graphics Language, but
I haven't managed to find any of those.
Thanks
Tom
__
I have a data.frame with 300 observations of 36 numerical, categorical, and
NA variables. I am trying to evaluate the partitioning around medoids
clustering algorithm for a marketing segmentation study. My original
dataset has over 130,000 observations, but I took a sample for easy
reproducibility
Hi R'er,
I have a dataset which has a matrix of 7502 x 1426 (rows x columns).
The data is in a CSV format which has a size around 68Mb. This dataset is less
than 10% of our dataset.
I have been adopting the Anomaly detection method as described by
http://www.mattpeeples.net/kmeans.html .
It has
http://cran.r-project.org/web/packages/h2o/
Please try h2o,
H2O is fast scalable open source R package for Generalized Linear Modeling,
Deep Learning, Gradient Boosting, RandomForest, k-means for large tera-byte
datasets. This package allows you to scale R over Hadoop like datasets
in-memory on mu
Dear R people,
The new packge 'icd9' provides a range of tools for working with ICD-9-CM codes.
http://cran.r-project.org/web/packages/icd9/index.html
https://github.com/jackwasey/icd9
ICD-9 (clinical modification) is primarily used for categorizing
diseases in the USA for hospital administratio
Hi,
I'd like to cut a hierachical cluster tree calculated with hclust at a
specific height.
However ever get following error message:
"Error in cutree(hc, h = 60) :
the 'height' component of 'tree' is not sorted (increasingly)"
Here is a working example to show that when specifing a height in
Em 05-07-2014 00:43, John McKown escreveu:
I messed up my original response by not including r-help in the
distribution. And now I won't look as bad because, after a short nap,
I have new, much shorted (but more difficult, for me, to understand)
answer.
#
# The original data is in the variable "
On 9 Jul 2014 17:19, "Takatsugu Kobayashi" wrote:
>
> Hi R-users,
>
> This should be a simple question: How can I delay each loop process in
some
> minutes? The reason for this is I need to avoid too much traffic to get
> longitudes and latitudes of 2000 addresses using google API.
>
R is too fas
Dear R-users,
three weeks ago I sent the mail below, but I didn't receive any
response. Maybe it was overlooked.
Thanks anyway for all the help you gave us by this mailing-list,
Giancarlo Camarda
[...]
I have a matrix with a series of "not-overlapping in a row dimension"
vectors in a given
Thanks Bill and the other guys for the variety of useful replies!
In fact I'm working with pretty big lists (with ~35000 sublists) and Bill's
solution is the fastest one in terms of computing time.
Now comes the second part of the question... :-)
I've my usual list of values and time indices to so
On 09/07/2014 08:17, Takatsugu Kobayashi wrote:
Hi R-users,
This should be a simple question: How can I delay each loop process in some
minutes? The reason for this is I need to avoid too much traffic to get
longitudes and latitudes of 2000 addresses using google API.
I am searching for solutio
Hi R-users,
This should be a simple question: How can I delay each loop process in some
minutes? The reason for this is I need to avoid too much traffic to get
longitudes and latitudes of 2000 addresses using google API.
I am searching for solutions with keywords like interval, minutes, delay,
bu
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