This might not seem helpful, but there really is a communication gap here. You
need to read the Posting Guide, post in plain text so your example code is not
mangled by the HTML, and include a bit of data that is representative of the
data you are working with. You may find some helpful advice h
Not much. Columns are not normally measured in bytes.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live:
Hi Yuan,
On 01/10/2014 06:49 PM, Yuan Luo wrote:
How to find the package of a class given classname?
For example, there is a class called GAlignments, I want to do something
like
attr("GAlignments", "package") that gives you the package where the class
is defined? But of course, attr("GAlignment
This probably doesn't answer your question, but, amazingly enough, try
googling the class name.
It worked for GAlignmnets.
Incidentally, there's no reason class names have to be unique among R
packages.. And they can be hidden in namespaces.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biosta
On 11/01/14 16:54, srinivasa babu wrote:
Hi,
I deleted my account. I need to delete all of messages started by me in
here
https://stat.ethz.ch/pipermail/r-help/2009-December/221628.html
http://comments.gmane.org/gmane.comp.lang.r.general/172827
Can you please guide how to do so?
You can't.
Hi,
I deleted my account. I need to delete all of messages started by me in
here
https://stat.ethz.ch/pipermail/r-help/2009-December/221628.html
http://comments.gmane.org/gmane.comp.lang.r.general/172827
Can you please guide how to do so?
thanks,
Sri.
[[alternative HTML version delet
How to find the package of a class given classname?
For example, there is a class called GAlignments, I want to do something
like
attr("GAlignments", "package") that gives you the package where the class
is defined? But of course, attr("GAlignments", "package") won't work...
Thanks for any help!
Y
Hi
I’m using bar plot function
And I have in my data some zero, but my data are numerics and when i try to use
bar plot i have a warning message of « NA was introduced during the automatic
conversion » and i can’t see my bar plot.
So if somebody have an idea for that?
Thanks
clémence
Hi Adel,
You may have a look at the *showtext *package (on CRAN and
https://github.com/yixuan/showtext), which could convert text into lines
and curves in the pdf file, so it doesn't rely on those fonts after the
file is created.
Best,
Yixuan
2014/1/10 Adel ESSAFI
> Hello list
> I generated p
ok, I think I got it. Here is the code I used. Thanks for the help guys..
fake <- cbind(c(1,2,5,8,12,19), c(2,5,8,12,19,20), c(1,1,1,1,1,1),
c(-10,-10,-10,-10,-10,-10), c(10,10,10,10,10,10), runif(6, 0, 2), runif(6, 0,
2))
datalim <- c(-0,10)
matrix3plus <- fake[,-(1:2)]
dimnames(fake)[[2]]
Hi,
I'm relatively new to R and need to some help in converting some of my data
into a ts object that can then be used to run an STL analysis.
My initial input file is a .csv output from some atmospheric measurement
instruments. The first column is the year fraction and extends over 2 years
with
HI,
I couldn't reproduce the error.
complaints <- data.frame(text=c("Do your homework", "Provide reproducible
example","Read the R-help manual"))
library(tm)
myCorpus <- Corpus(VectorSource(as.character(complaints$text)))
myCorpus
#A corpus with 3 text documents
A.K.
On Friday, January 10, 201
Sorry for bugging you again. I was wondering whether it is possible to include
multiple different colors instead of just two.
In the example below, the color.scale() function goes from yellow to red, but
never passes through plain white. Would it be possible to tweak the function or
would I hav
It seems like emdist does not like to compare matrices with all 0 values. I
ended up removing those from my 3D array and have ~8000 matrices instead of
13000.
I am using res2 <- unlist(mclapply(seq_len(ncol(indx)),function(i) {x1 <-
indx[,i]; emd2d(results[x1[1],,],results[x1[2],,]) }) )
But even
Hi all,
I am attempting to do SDM in R, mostly I have used the 'dismo' package. My
problem is that all the methods call for a raster stack as the input
representing the environmental layers. However, my processing steps leave
me with a table of extracted values for each occurrence (much like sec
?as.POSIXct for time-formatting. This function makes a structured list of
time data, where you specify an input time and format i.e.
as.POSIXct('2014-01-09 01:30:00', format='%Y-%m-%d %H:%M:%S')
On 1/10/14, 15:24 , "Santosh" wrote:
I don't think apropos or indexing would help. I am open to your
Hi,
Maybe it is not directly related to R but sine many are statistical experts so
I post it here for help:
I have two variables (say x and y) of length n. Now the cor(x,y) is close to 0.
I need to find the subset in {1,.. n} so that the correlation between x and y
using the subset data is maxi
Hi Silvano
To save you typing try paste0 (...) instead of paste(, sep = "")
Just checking quickly
In
### code chunk number 3: Serie1
for (i in 1:7){
aux <- paste("Q", i, sep="")}
will always end up with "Q7"
Is the } to go elsewhere?
OR
Is it superfluous?
Duncan
-Original Message
I agree with Don... focus on identifying the names of the columns and then use
column name indexing to extract the columns you want. You will probably want to
rename them to a standard set of names once you have gone to all this
trouble... just assign the new vector of names to the names functio
At the risk of being annoying ...
Your original question was,
"Is there a way to dynamically include columns in a dataframe?"
The answer is yes. One way, and I think the simplest, is to calculate the
names of the columns you want to keep, and then use an expression like I
suggested, that is, one
I don't think apropos or indexing would help. I am open to your
suggestions/tips.
I usually get multiple versions of a dataset (even with the same column
names). In the source data, I occasionally notice inconsistencies...
formatting issues, column naming issues etc..
As shown In the "a1" example
Don's response seems apropos to me. Do you understanding indexing,
i.e. the "[" operator? If not, you should read An Introduction to R or
other tutorial (there are many good ones on the web). If that is not
the issue, you need to explain more clearly why his answer does not
suffice.
Cheers,
Bert
Dear users,
An eBook (2. Ed.) of Tinn-R was published today.
Dowload (free):
- http://www.uesc.br/editora/index.php?item=conteudo_livros_digitais.php
- http://www.uesc.br/editora/livrosdigitais2/tredit.pdf
- Http://nbcgib.uesc.br/lec/software/editores/tinn-r/en # h10-ebook
All the best,
--
///\
My intention is to include certain columns if they meet certain criteria.
For example, if "b" is one of the columns of a1, then keep otherwise don't.
HTH..
santosh
On Fri, Jan 10, 2014 at 1:01 PM, MacQueen, Don wrote:
> Apologies, but all that ifelse() stuff is too hard to follow.
>
> What I w
... In fact, on my computer, the pmax construction took longer. I
would guess that's due to the data.frame construction. Nevertheless,
the comment is still germane.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge.
It probably doesn't make much difference in this small example, but
maybe it's worth noting that **if possible** (it typically is not),
things can be speeded up by generating all the random numbers at once
then applying a vectorized operation to the entire ensemble. In this
case, this becomes:
ran
Apologies, but all that ifelse() stuff is too hard to follow.
What I would do is compute a character vector of column names to keep,
then do
a1[ , names.to.keep]
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/10/1
Dear Rxperts...
I would like to conditionally include an element (as a column) in a
dataframe. Please see the sample code below:
There is a correction to the earlier post.. my apologies...
a1 <- data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],
R=rep(c("A","B"),each=5))
lc1 <- list(C1 =
Dear Rxperts...
I would like to conditionally include an element (as a column) in a
dataframe. Please see the sample code below:
a1 <-
data.frame(P=rep(1,10),Qr=LETTERS[1:10],b=letters[1:10],R=rep(c("A","B"),each=5))
lc1 <- list(C1 = "P",C2 =
ifelse(is.element("Q",names(a1)),"Q",ifelse(is.elem
In R package "psychometrics" an estimate of SE of R squared of /sersq <-
sqrt((4*rsq*(1-rsq)^2*(n-k-1)^2)/((n^2-1)*(n+3))) with n sample size,
and k number of parameters if sample size greater than 60 is found.
Does anyone have a formula for smaller sample size or an exact formula?
I have been
Hi,
May be this helps:
set.seed(42)
vec1 <- sapply(1:1000,function(x) max(rnorm(1000,0,1)))
#or
set.seed(42)
vec2 <- replicate(1000,max(rnorm(1000,0,1)))
identical(vec1,vec2)
#[1] TRUE
set.seed(598)
res <- sample(vec1,100,replace=FALSE)
max(res)
#[1] 4.408794
A.K.
Hello I need some help in p
On 10/01/2014 19:29, Adel ESSAFI wrote:
Hello list
I generated pdf files with R that I integrated
in .tex file and compiled with pdflatex.
however, all the fonts of my R figure not embadded. so the file is rejected
from EDAS web site.
Could you help please.
If you have suitable licenses for t
Hello list
I generated pdf files with R that I integrated
in .tex file and compiled with pdflatex.
however, all the fonts of my R figure not embadded. so the file is rejected
from EDAS web site.
Could you help please.
[adel@localhost hcw]$ pdffonts hcw.pdf
name
library(ggplot2)
a <- data.frame(x=rnorm(10), y=rnorm(10))
#conversion of mm to points 2.83464567
plot.demo <- qplot(x, y, data=a, xlab="test.x",
ylab="test.y")+theme(panel.border=element_rect(fill=NA, size=1),
axis.title.x = element_text(size=10*2.83464567, family="Arial"),
axis.title.y = element
The column length is 4000 bytes long if that helps.
--
View this message in context:
http://r.789695.n4.nabble.com/How-do-you-transform-a-dataframe-to-a-corpus-tp4683396p4683402.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@
What would you like to get from the following?
0 1 0 0 0
1 1 1 0 0
1 1 1 1 0
0 0 1 1 0
0 0 1 1 1
0 0 0 1 1
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of
There are many ways, but this is not a statistical theory support forum for
helping you choose among them. Your question becomes more appropriate when you
ask something along the lines of "how do I regress data using AR model with R?"
(Though reading the Time Series Task View on CRAN would be ev
Hi;
I have a data frame complains w/ dimensions 11335291 ( 1.13m obs 1 col)&
I am trying to transform it into a corpus
using the following code: myCorpus <-Corpus(VectorSource(complaints$text))
Error in .Source(readPlain, encoding, length(x), FALSE, names(x), 0, TRUE,
:
vectorized sources
On Jan 10, 2014, at 10:11 AM, ramoss wrote:
Hi,
I am trying to use the package TM on a dataframe & get the following
error:
complaints <- tm_map(complaints, tolower)
Error in UseMethod("tm_map", x) :
no applicable method for 'tm_map' applied to an object of class
"data.frame"
Tm doesn't
Hi,
I am trying to use the package TM on a dataframe & get the following error:
complaints <- tm_map(complaints, tolower)
Error in UseMethod("tm_map", x) :
no applicable method for 'tm_map' applied to an object of class
"data.frame"
Tm doesn't work on dataframes? My data frame consists of 1
Hi,
Use ?reshape()
dat1<- read.table(text="V1 V2 V3 V4 V5
1 210505 ARS B A
1 210505 BFGL A B
1 210505 NGS B B
1 210506 ARS B B
1 210506 BFGL A A
1 210506 NGS B B
1 210507 ARS B B
1 210507 BFGL A B
1 210507 NGS A B",sep="",header=TRUE,stringsAsFact
Dear list,
I'm using MuMIn for model averaging and I had a question about the
z-test that MuMIn performs when using the summary call. e.g.:
> model.1 <- subset(model.sel(model.1.glmm.list, rank = AICc), delta<3)
> summary(model.avg(model.1))
>
> ...
>
>
On 01/10/2014 09:46 PM, Adel ESSAFI wrote:
Hello list
I have this data frame which represent values grouped by algorithm.
remark that for each algorithm, we have the same value in V3 but differnt
values in V4.
I want to make a plot with algorithms in axes, and for each algorithm I
draw a bar (fo
I have monthly claims data from 2009 to end of 2013 and would like to
forecast for 2014. How is this done in R.
Charles.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLE
Hello list
I have this data frame which represent values grouped by algorithm.
remark that for each algorithm, we have the same value in V3 but differnt
values in V4.
I want to make a plot with algorithms in axes, and for each algorithm I
draw a bar (for V3) and a bwplot for V4.
is thart possible.
Hello list
I have this data frame which represent values grouped by algorithm.
remark that for each algorithm, we have the same value in V3 but differnt
values in V4.
I want to make a plot with algorithms in axes, and for each algorithm I
draw a bar (for V3) and a bwplot for V4.
is thart possible.
On Jan 10, 2014, at 2:04 AM, Long Vo wrote:
Hi R users,
I need to apply a function on a list of vectors. This is simple when
I use
functions that returns only one numerical value such as 'mean' or
'variance'. Things get complex when I use functions returning a list
of
value, such as 'acf'
you can find it by use of indexing with setting 1=T and 0=False in your
global environment of R.
On Fri, Jan 10, 2014 at 12:32 PM, email wrote:
> Dear all,
>
> I have a binary matrix
>
> 0 0 0 0 0
> 1 1 1 0 0
> 1 1 1 0 0
> 0 0 0 0 0
> 0 0 0 1 1
> 0 0 0 1 1
>
> I want to find the location of all
On 01/10/2014 06:56 AM, email wrote:
Hi:
I am trying to analyze an yeast gene expression data
http://arep.med.harvard.edu/biclustering/yeast.matrix
I need to convert the real-valued data matrix to a binary (0,1)
matrix. Is there any package available? How can it be done?
Hi John,
It looks li
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