Hi everyone,
I have a time series data set and I want to fill my line plot of this time
series with different colors
e.g
I want to fill portion related to 1995-1996 with blue , portion related to
1996-1997 with orange and then portion related to 1997-1998 with red
can anyone please help me.
Dan, google refine http://goo.gl/AeKml can actually transform zip codes
into longitude/latitude - http://goo.gl/1HDWb will show you how to do this
from street adresses, but it should also work from city names -- i think it
will allocate a default long/lat for a city, but not sure of the exact
mecha
Well, i believe writing correct date format would have served the purpose.
Suppose tfr contains Date as column and is a factor by class.
tft$Date <- as.Date(as.character(tfr$Date),"%d/%m%Y") should give you the
desired output.
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Dear Jean
Thanks a lot for your help.
The reason I did not provide producible code is that my work started with
reading in some large csv files, e.g. the data is not created by myself.
But the data is from the same data provider so I would expect to receive
data in exactly same data format.
I
lty =1 denotes the single continuous line
lty = 2 denotes the broken line
lty = 3 dotted line
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Hi, all
I am trying to use the label_wrap_gen function in this website.
https://github.com/hadley/ggplot2/wiki/labeller
I tried to make a long name like this
Light and heavy good vehicles (diesel) -\nGVX
f2 = facet_grid(vehicle ~ ., labeller=label_wrap_gen(width=15))
eventually, I got something
Hi,
How do I do a Duncan Multiple Range Test, with Nested ANOVA.
I am not sure how to write the nested variable within dmrt.
Any suggestions?
Archana
[[alternative HTML version deleted]]
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Hi,
Try this:
F<-function(x,type="local"){Y=3
x*Y}
F(3)
#[1] 9
Y<-4
F(3)
#[1] 9
Y<-5
F(3)
#[1] 9
A.K.
- Original Message -
From: "Schoenfeld, David Alan,Ph.D.,Biostatistics"
To: "'r-help@r-project.org'"
Cc:
Sent: Monday, August 6, 2012 5:07 PM
Subject: [R] Force evaluation of
Hello fellow R users,
I would need your help on GAM/GAMM models and interpolation on a marked
spatial point process (cases and controls).
I use the mgcv package to fit a GAMM model with a binary outcome, a
parametric part (var1+..+varn), a spline used for the spatial variation, and
a random
On Mon, Aug 6, 2012 at 9:03 PM, Schoenfeld, David
Alan,Ph.D.,Biostatistics wrote:
> Thank you both, this was very helpful. I need to study environments more. Do
> either of you know a good source?
Disclaimer: I really have no idea what I'm talking about.
They are a somewhat subtle, but excepti
I would like to find out how to apply commands found in the bayesm package, to
analyze data gathered via a choice-based conjoint study. Is there a web
resource where I can seek an R-Project consultant experienced in this, who I
could hire to walk me through the appropriate bayesm commands to use
Thank you both, this was very helpful. I need to study environments more. Do
either of you know a good source?
-Original Message-
From: Bert Gunter [mailto:gunter.ber...@gene.com]
Sent: Monday, August 06, 2012 6:03 PM
To: William Dunlap
Cc: Schoenfeld, David Alan,Ph.D.,Biostatistics;
Both of those approaches require the function to be created
at the same time that the environment containing some of its
bindings is created. You can also take an existing function and
assign a new environment to it. E.g.,
> f <- function(x) y * x
> ys <- c(2,3,5,7,11)
> fs <- lapply(ys, f
Thanks to both: Cute question, clever, informative answer.
However, Bill, I don't think you **quite** answered him, although the
modification needed is completely trivial. Of course, I could never
have figured it out without your response.
Anyway, I interpret the question as asking for the functi
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz wrote:
> is.letter <- function(x) grepl("[[:alpha:]]", x)
> is.number <- function(x) grepl("[[:digit:]]", x)
>
This does exactly what I wanted:
> x
[1] "a10" "b8" "c9" "d2" "e3" "f4" "g1" "h7" "i6" "j5" "k"
"l" "m" "n"
[15] "o" "p"
On Mon, Aug 6, 2012 at 4:30 PM, R. Michael Weylandt
wrote:
> On Sun, Aug 5, 2012 at 4:49 PM, Douglas Karabasz
> wrote:
>> I have a xts object made of daily closing prices I have acquired using
>> quantmod.
>>
>>
>>
>> Here is my code:
>>
>> library(xts)
>>
>> library(quantmod)
>>
>> library(lubri
On Sun, Aug 5, 2012 at 4:49 PM, Douglas Karabasz
wrote:
> I have a xts object made of daily closing prices I have acquired using
> quantmod.
>
>
>
> Here is my code:
>
> library(xts)
>
> library(quantmod)
>
> library(lubridate)
>
>
>
> # Gets SPY data
>
> getSymbols("SPY")
>
> # Subset Prices to j
You could use local(), as in
> F <- local({
+Y <- 3
+function(x) x * Y
+})
>F(7)
[1] 21
> Y <- 19
> F(5)
[1] 15
Look into 'environments' for more.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help
I am porting a program in matlab to R,
The problem is that Matlab has a feature where symbols that aren't arguments
are evaluated immediately.
That is:
Y=3
F=@(x) x*Y
Will yield a function such that F(2)=6.
If later say. Y=4 then F(2) will still equal 6.
R on the other hand has lazy evaluation.
Hi,
simpleRDA2 is still in the vegan package, but it is not exported.
I.e., the author only intends it for internal use and he doesn't make
it available to end users directly. If you need to get at it, you can
use
getAnywhere("simpleRDA2")
which will show it.
If you need to make it available to
On Aug 6, 2012, at 11:16 AM, hafida wrote:
I CANT FIND ANY ANSWER MR DAVID
When I suggested that you learn to use the shift key, I was hoping for
a sparing use of that key, such as at the beginning of sentences. The
caps-lock key is different than the shift key.
You are also posting to
Please find some reference online or textbook. This must be contained in
the model assessment part.
AIC, BIC, rolling prediction/forecasting error might be what you want.
Best wishes,
Jie
On Fri, Aug 3, 2012 at 4:07 AM, Soham wrote:
> Hi I am trying to fit a time series data.It gives a AR(2) mod
HI,
#These links
http://stackoverflow.com/questions/8545035/scatterplot-with-marginal-histograms-in-ggplot2
http://stackoverflow.com/questions/11022675/rotate-histogram-in-r-or-overlay-a-density-in-a-barplot
# might be helpful for you.
A.K.
- Original Message -
From: li li
To: r-he
HI,
You can subset the data
dados433<-subset(dados,dados[,3]=="433")
is.matrix(dados433)
#[1] TRUE
dados433
date value code
1 "2012-01-01" "0.56" "433"
2 "2012-02-01" "0.45" "433"
3 "2012-03-01" "0.21" "433"
4 "2012-04-01" "0.64" "433"
5 "2012-05-01" "0.36" "433"
6 "2012-06-01" "0.
Dear Rui and Arun,
Thanks a lot for your help. I will test all the proposed solutions ;-)
Best regards,
Henrique Andrade
2012/8/6 Rui Barradas :
> Hello,
>
> Try the following.
>
> split(data.frame(dados), dados[, "code"])
>
> Also, it's better to have data like 'dados' in a data.frame, like thi
I CANT FIND ANY ANSWER MR DAVID
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A OK I MAKE A MISTAKE
OK MR DAVID I WILL DO IT
THANK YOU
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R-help@r-project.org mail
You probably mean grepl('[a-zA-Z]', x)
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Mon, Aug 6, 2012 at 3:29 PM, Liviu Andronic wrote:
> On Mon, Aug 6, 2012 at 6:42 PM, Bert Gunter wrote:
See
example(layout)
for one idea. I think you might also want to look into rug plots.
Best,
Michael
On Mon, Aug 6, 2012 at 2:40 PM, li li wrote:
> Dear all,
> For two sets of random variables, say, x <- rnorm(1000, 10, 10) and y
> <- rnorm(1000. 3, 20).
> Is there any way to overlay the hi
Dear all,
For two sets of random variables, say, x <- rnorm(1000, 10, 10) and y
<- rnorm(1000. 3, 20).
Is there any way to overlay the histograms (and density curves) of x and y
on the plot of y vs. x?
The histogram of x is on the x axis and that of y is on the y axis.
The density curve here
On Mon, Aug 6, 2012 at 6:42 PM, Bert Gunter wrote:
> nzchar(x) & !is.na(x)
>
> No?
>
It doesn't work for what I need:
> x
[1] "a10" "b8" "c9" "d2" "e3" "f4" "g1" "h7" "i6" "j5" "k"
"l" "m" "n"
[15] "o" "p" "q" "r" "s" "t" "u" "v" "w" "x" "y"
"z" "1" "2"
[29]
Would just saving the results onto an object saving it work?
>From the help page example
summary(fm1 <- aov(breaks ~ wool + tension, data = warpbreaks))
myresults<- TukeyHSD(fm1, "tension", ordered = TRUE)
write.table (myresults, file = ksksk)
John Kane
Kingston ON Canada
> -Or
Well, I just posted the fourth copy to the list which I apologize
for. (I meant to delete the response I wrote.)
Re-re-posting an unclear message seems unwise on your part, 'hafida'.
You are not following the advice in the footer to all messages and you
are not following the advice in the
Hello,
Try the following.
split(data.frame(dados), dados[, "code"])
Also, it's better to have data like 'dados' in a data.frame, like this
you would have dates of class Date, and numbers of classes numeric or
integer:
dados2 <- data.frame(dados)
dados2$date <- as.Date(dados2$date)
dados2$v
On Aug 6, 2012, at 8:04 AM, hafida wrote:
Hi
can ANY body help me to programme this formula:
c[lj] and c[l'j] are matrix
A[j]^-1 is an invertible diagonal matrix
g[ll']=i[ll'] - sum *#from j=1 to k#* c[lj]c[l'j]A[j]^-1
WHERE
i[ll']= 1/n sum from i=1 to n z[il] z[il']
n,k,m are
HJ,
You don't provide any reproducible code, so I had to make up my own.
dat <- data.frame(a=letters[1:5], x=c(20110911001084, 20110911001084,
20110911001084, 20110911001084, 20110911001084),
y=c(2.10004e+12, 2.10004e+12, 2.10004e+12, 2.10004e+12,
2.10004e+12))
In my example,
On Aug 6, 2012, at 7:34 AM, Dominic Roye wrote:
Hello everyone,
I have a dataset with 5 colums (4 colums with thresholds of weather
stations and one with month - data of 5 years). Now I would like to
calculate the average for each month.
I tried this unsuccessfully:
lf.med <- sapply(LF[,1:4
Dominic,
It's great that you provided some example data, but a much smaller data
frame would have sufficed. For example, 10 randomly selected rows from
your data ...
LF <- structure(list(Serra.da.Foladoira = c(27.335652173913,
25.4632608695652,
24.464652173913, 22.550652173913, 22.217782
On 8/6/2012 11:54 AM, Achim Zeileis wrote:
On Mon, 6 Aug 2012, Michael Friendly wrote:
I have two versions of a bibtex database which have gotten badly out
of sync. I need to find find all the entries in bib2 which are not
contained in bib1, according to their bibtex keys. But I can't figure
Only an extra set of brackets:
is.letter <- function(x) grepl("[[:alpha:]]", x)
is.number <- function(x) grepl("[[:digit:]]", x)
Without them, the functions are fast, but wrong.
> x
[1] "a8" "b5" "c10" "d1" "e6" "f2" "g4" "h3" "i7" "j9" "k" "l"
[13] "m" "n" "o" "p" "q" "r
You would make it much easier for R-help readers to solve your problem if
you provided a small example data set with your code, so that we could
reproduce your results and troubleshoot the issues.
Jean
Naidraug wrote on 08/05/2012 09:08:25 AM:
>
> I've looked everywhere and tinkered for thre
On Aug 6, 2012, at 2:01 AM, tibr wrote:
I hope the original poster fixed this a long time ago, but I had the
same
problem and here is how I fixed it:
- go to the application Fontbook
- check if the Arial font has duplicates, and delete them, even if
they are
set to "Off"
- restart the com
- Forwarded Message -
From: arun
To: Liviu Andronic
Cc: R help
Sent: Monday, August 6, 2012 12:56 PM
Subject: Re: [R] test if elements of a character vector contain letters
Hi,
Not sure whether this is you wanted.
x<-letters
(x[1:10] <- paste(x[1:10], sample(1:10, 10), sep=''))
Dear R Community,
I'm trying to write a loop to split my data into different series. I
need to make a
new matrix (or series) according to the series code.
For instance, every time the "code" column assumes the value "433" I need to
save "date", "value", and "code" into the "dados433" matrix.
Ple
On Aug 6, 2012, at 12:06 PM, Marc Schwartz wrote:
> Perhaps I am missing something, but why use sapply() when grepl() is already
> vectorized?
>
> is.letter <- function(x) grepl("[:alpha:]", x)
> is.number <- function(x) grepl("[:digit:]", x)
Sorry, typos in the above from my C&P. Should be:
Dear R users
I read two csv data files into R and called them Tem1 and Tem5.
For the first column, data in Tem1 has 13 digits where in Tem5 there are 14
digits for each observation.
Originally there are 'numerical' as can be seen in my code below. But how
can I display/convert them using other
Hi,
Not sure whether this is you wanted.
x<-letters
(x[1:10] <- paste(x[1:10], sample(1:10, 10), sep=''))
x1<-c(x,1:26)
x1
[1] "a4" "b3" "c5" "d2" "e9" "f6" "g1" "h8" "i10" "j7" "k" "l"
[13] "m" "n" "o" "p" "q" "r" "s" "t" "u" "v" "w" "x"
[25] "y" "z"
Hi,
I am trying to run the command "forward.sel.par," however I receive
the error message: "Error: could not find function 'simpleRDA2'." I
have the vegan library loaded. The documentation on "varpart" has not
helped me to understand why I cannot call this function. Maybe I am
missing something ob
Hi
can ANY body help me to programme this formula:
c[lj] and c[l'j] are matrix
A[j]^-1 is an invertible diagonal matrix
g[ll']=i[ll'] - sum *#from j=1 to k#* c[lj]c[l'j]A[j]^-1
WHERE
i[ll']= 1/n sum from i=1 to n z[il] z[il']
n,k,m are given. j=1...k,l,l'=1...m,
it s co
Hi,
Here, the string with in the quotes are read exactly like that. So, you may
have to use the symbol instead of "friendly" or "numeric" from the link. Or
you have to convert those.
d1 <- data.frame(V1 = 1:4,
V2 = c("some text = 9", "some tèxt = 9", "some tèxt = 9", "some
tèxt = 9"))
Hi,
I run the second list of codes (is.between()) again from the sent mail. It
works fine for me. I am using R 2.15 on Ubuntu 12.04.
sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] L
Greetings,
I am using S4 classes and the code is organized by putting each class into a
separate source file
in a separate folder:
In Folder base/base.R:defines class "base" and does
setGeneric("showSelf",
function(this) standardGeneric("showSelf")
)
setMethod("showSelf",
signature(t
Dear all,
I would like to test the differences in dependent variable X depending on 2
grouping variables of each 10 levels.
I do this with a 2-way ANOVA, followed by a Tukey HSD test (TukeyHSD(x)).
However, since a lot of combinations are possible with 2 grouping variables,
each of 10 levels, the
It would be nice to be able to trigger NA returning NA with an argument to
the function, but you can easily get that result:
> ifelse(is.na(x), NA, nzchar(x))
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[13] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TR
Perhaps I am missing something, but why use sapply() when grepl() is already
vectorized?
is.letter <- function(x) grepl("[:alpha:]", x)
is.number <- function(x) grepl("[:digit:]", x)
x <- c(letters, 1:26)
x[1:10] <- paste(x[1:10], sample(1:10, 10), sep='')
x <- rep(x, 1e3)
> str(x)
chr [1:52
On 08/06/2012 09:51 AM, Rui Barradas wrote:
Hello,
Fun as an exercise in vectorization. 30 times faster. Don't look, guess.
> system.time(res0 <- grepl("[[:alpha:]]", x))
user system elapsed
0.060 0.000 0.061
> system.time(res1 <- has_letter(x))
user system elapsed
3.728 0.00
Hello,
Fun as an exercise in vectorization. 30 times faster. Don't look, guess.
Gave it up? Ok, here it is.
is_letter <- function(x, pattern=c(letters, LETTERS)){
sapply(x, function(y){
any(sapply(pattern, function(z) grepl(z, y, fixed=T)))
})
}
# test ascii codes, just one loo
On 08/06/2012 09:41 AM, Jie wrote:
After searching online, I found that clusterCall or foreach might be the
solution.
Re-write your outer loop as an lapply, then on non-Windows use
parallel::mclapply. Or on windows use makePSOCKcluster and parLapply. I
ended with
library(parallel)
library(M
Not that I've had a chance to really look at the problem, but I've removed
outer loops using parLapply from the parallel package. Works great.
On Mon, Aug 6, 2012 at 11:41 AM, Jie wrote:
> After searching online, I found that clusterCall or foreach might be the
> solution.
>
> Best wishes,
> Ji
nzchar(x) & !is.na(x)
No?
-- Bert
On Mon, Aug 6, 2012 at 9:25 AM, Liviu Andronic wrote:
> Dear all
> I'm pretty sure that I'm approaching the problem in a wrong way.
> Suppose the following character vector:
>> (x[1:10] <- paste(x[1:10], sample(1:10, 10), sep=''))
> [1] "a10" "b7" "c2" "d3"
After searching online, I found that clusterCall or foreach might be the
solution.
Best wishes,
Jie
On Sun, Aug 5, 2012 at 10:23 PM, Jie wrote:
> Dear All,
>
> Suppose I have a program as below: Outside is a loop for simulation (with
> random generated data), inside there are several sapply()'s
Sorry, forgot to Cc the list.
Em 06-08-2012 17:29, Rui Barradas escreveu:
Hello,
I'm glad it helped.
The result of function cut() is a factor variable so you can coerce it
to integer, giving more "normal" names, or, if you want to keep track
of the intervals the adjusted r2 belong to, got st
Dear all
I'm pretty sure that I'm approaching the problem in a wrong way.
Suppose the following character vector:
> (x[1:10] <- paste(x[1:10], sample(1:10, 10), sep=''))
[1] "a10" "b7" "c2" "d3" "e6" "f1" "g5" "h8" "i9" "j4"
> x
[1] "a10" "b7" "c2" "d3" "e6" "f1" "g5" "h8" "i9" "j
On Mon, 6 Aug 2012, Michael Friendly wrote:
I have two versions of a bibtex database which have gotten badly out of
sync. I need to find find all the entries in bib2 which are not
contained in bib1, according to their bibtex keys. But I can't figure
out how to extract a list of the bibentry ke
Thanks for the suggestion, got exactly what I needed from
library(splancs)
?pip
Alastair
Jeff Newmiller wrote
>
> This is off-topic (not about R), and a quick Web search of "test within
> polygon" yields many results, and adding "R" to the search when using
> Google provides hints about appl
Liviu:
Well, as usual, to a certain extent this is arbitrary and the only
issue is whether it is documented correctly.
To me, NA (of whatever mode) means ""indeterminate" or "unknown," so
since "" is known and of length 0, I would have expected NA as a
return. But the point is, not what our parti
This is off-topic (not about R), and a quick Web search of "test within
polygon" yields many results, and adding "R" to the search when using Google
provides hints about applying the algorithms in R.
---
Jeff Newmiller
I have two versions of a bibtex database which have gotten badly out of
sync. I need to find find all the entries in
bib2 which are not contained in bib1, according to their bibtex keys.
But I can't figure out how to extract a list of the bibentry keys in
these databases.
A minor question: Is
Hello,
With your example run and click 10 black points inside the area.
ploc <- locator(n=10)
points(ploc$x, ploc$y, pch = 19, col = "green", cex = 1)
Hope this helps,
Rui Barradas
Em 06-08-2012 16:05, Ally escreveu:
I have a complex 2D polygon with thousands of vertices, and I'd like to be
On Mon, Aug 6, 2012 at 9:53 AM, Liviu Andronic wrote:
> On Mon, Aug 6, 2012 at 4:48 PM, Liviu Andronic wrote:
>> string, something that I find strange. At best NA is the equivalent of
>> an empty string.
Certainly not to my mind, unless you think that zero and NA should be
the same for integers
Thanks Uwe but, actually, I did so.
Since “filetorun.exe” looks in the current folder for
“input.txt”, I tried moving all needed files to a newly created
temporary folder “tmp.id” (say, tmp.1) and running the executable.
This works fine by doing it directly from the windows command line but not
I have a complex 2D polygon with thousands of vertices, and I'd like to be
able to identify points from a large set contained within the polygon, and
was wondering if there might be an efficient way of doing this? Any advice
would be useful! Here is a small example of what I mean:
# make polygon
On Mon, Aug 6, 2012 at 4:48 PM, Liviu Andronic wrote:
> string, something that I find strange. At best NA is the equivalent of
> an empty string. In this sense, if you Hmisc::describe() the vector
> you get, as I would expect, that in the context of character vectors
> NA and '' values are conside
Dear all
I'm a bit surprised by the results output from nzchar(). The help page
says: "nzchar is a fast way to find out if elements of a character
vector are *non-empty strings*." (my emphasis. However, if you do
> x <- c(letters, NA, '')
> nzchar(x)
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE
Alan,
More RAM will definitely help. But if you have an object needing more than
2^31-1 ~ 2 billion elements, you'll hit a wall regardless. This could be
particularly limiting for matrices. It is less limiting for data.frame
objects (where each column could be 2 billion elements). But many R
a
Thanks Arun,
It works all right, I just found out that my problem was not with accents but
with the correct spelling of "some text".
Kind regards,
Luca
Il giorno 06/ago/2012, alle ore 15.01, arun ha scritto:
>
>
> Hi,
>
> Here, the string with in the quotes are read exactly like that.
Sorry but my previous email did not go through properly. Instead of the ? you
should really read an è or è according to
http://www.lookuptables.com/.
So there are extended ASCII characters I need to deal with.
I have tried
d1$V1[regexpr("some tèxt = 9",d1$V2)>0] <- 9
and
d1$V1[regexpr("some
On Sat, Aug 4, 2012 at 11:27 PM, Roberto wrote:
> Hi,
> I need to remove collinear variables to my Near-Infrared table of spectra.
>
> What package can I use?
>
> Something simple, because I am a novice about statistic.
>
There many methods of assessing multicollinearlity but to pick one
that ha
Hi, All,
I am using the supervised lda function (slda) from 'lda' package in R for
topic modeling (http://cran.r-project.org/web/packages/lda/index.html), my
data is a collection of documents, and within which each doc has a label.
There are about 97 different categories and 18K documents in tot
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http://r.7896
I hope the original poster fixed this a long time ago, but I had the same
problem and here is how I fixed it:
- go to the application Fontbook
- check if the Arial font has duplicates, and delete them, even if they are
set to "Off"
- restart the computer.
emmats wrote
>
> I was re-running som
Hi
It is better to use dput for presenting data for others. You probably want
?merge.
Something like
merge(datuak, datuak2, by = "calee_id", all.x=TRUE)
However calee_id seems to be a floating point number and it may be rounded
so you shall beware of it.
Regards
Petr
> Thank you very muc
Hi
>
> Hi,
>
> I appreciate your help with the segmented function. I am relatively new
to
> R. I followed the introduction of the 'segmented'-package by Vito
Muggeo,
> but still it does not work.
> Here are the lines I wrote:
>
> data_test<-data.frame(x=c(1:10),y=c(1,1,1,1,1,2,3,4,5,6))
> lr_t
On 2/08/2012 8:06 a.m., Thomas Steiner wrote:
Hi Ray,
2012/7/31 Ray Brownrigg :
On 07/28/12 23:46, Thomas Steiner wrote:
Hi,
I'd like to have a low resolution word map in R.
The "maps" package has this option, but if I use the argument, the map
looses sense: Russia and Australia get empty etc
On 06/08/2012 09:42, Uwe Ligges wrote:
On 06.08.2012 09:34, David Winsemius wrote:
On Aug 5, 2012, at 3:52 PM, alan.x.simp...@nab.com.au wrote:
Dear all
I have a Windows Server 2008 R2 Enterprise machine, with 64bit R
installed
running on 2 x Quad-core Intel Xeon 5500 processor with 24GB D
On 06.08.2012 09:34, David Winsemius wrote:
On Aug 5, 2012, at 3:52 PM, alan.x.simp...@nab.com.au wrote:
Dear all
I have a Windows Server 2008 R2 Enterprise machine, with 64bit R
installed
running on 2 x Quad-core Intel Xeon 5500 processor with 24GB DDR3 1066
Mhz
RAM. I am seeking to analy
Thanks arun and Rui; 3 fantastic suggestions.
The Season interval is not always a month so arun's suggestion works better
for this dataset. I couldn't get the as.between function to work on arun's
second suggestion, it only returned NAs.
However, arun's first suggestion worked a treat!
Many th
Thank you very much John, can you read it now?
mailto:jrkrid...@inbox.com]
Enviado el: 03 August 2012 20:17
Para: Nerea Lezama; r-help@r-project.org
Asunto: RE: [R] how to identify values from a column of a dataframe, and
insert them in other data.frame with the corresponding id?
Hi Nerea,
For
On Aug 5, 2012, at 3:52 PM, alan.x.simp...@nab.com.au wrote:
Dear all
I have a Windows Server 2008 R2 Enterprise machine, with 64bit R
installed
running on 2 x Quad-core Intel Xeon 5500 processor with 24GB DDR3
1066 Mhz
RAM. I am seeking to analyse very large data sets (perhaps as much a
Thank you.. It was very informative and helpful. It works
Sent from my iPhone
On Aug 5, 2012, at 10:21 PM, arun wrote:
> HI,
>
> Try this:
> dat1<-data.frame(x=c(NA,NA,rnorm(6,15),NA),y=c(NA,rnorm(8,15)),z=c(rnorm(7,15),NA,NA))
> dat1[which(colMeans(is.na(dat1))<=.15)]
> y
> 1
Thank you. It works great.
Sent from my iPhone
On Aug 5, 2012, at 9:08 PM, Jorge I Velez wrote:
> Hi Faz,
>
> Here is one way of doing it where "x" is your data frame:
>
> x[, colMeans(is.na(x)) <= .15]
>
> HTH,
> Jorge.-
>
>
> On Sun, Aug 5, 2012 at 9:04 PM, Faz Jones <> wrote:
> I have a
HI,
It works with me. I am using R 2.15 on Ubuntu 12.04.
d1 <- data.frame(V1 = 1:5, V2=c("some text = 9", "some téxt=9","sóme tèxt=9",
"söme text=9", "some têxt=9"))
d1
# V1 V2
#1 1 some text = 9
#2 2 some téxt=9
#3 3 sóme tèxt=9
#4 4 söme text=9
#5 5 some têxt=9
d
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