Well, code with 50+ ifelse statements (probably nested) seems to me quite
weird. I believe that such scheme could be solved differently but without
real code it is nothing to suggest.
r-help-boun...@r-project.org napsal dne 18.09.2009 13:51:45:
> On 9/18/2009 5:49 AM, premmad wrote:
> > I tried
Hi
r-help-boun...@r-project.org napsal dne 17.09.2009 04:14:24:
> Hi,
>
> I want to construct a data.frame 'y' by using x$x and x$y. I think
> that there might be better ways to do it (because, for example, we can
> use a_matrix[3:5,] to extract certain rows, where 'a_matrix' is a
> matrix). Can
Hi
I do not know rma but from help page boxplot requires as input a formula,
list (only some list of numerics), data frame or numeric vector. I am not
sure if your object is one of these. If not you need to convert it to
object which is acceptable for boxplot.
Regards
Petr
r-help-boun...@r-p
Dear all
I have a simple problem which I thought is easy to solve but what I tried
did not work. I want to change character variables to factor in data
frame. It goes easily from factor to character, but I am stuck in how to
do backwards conversion.
Here is an example
irisf<-iris
irisf[,2]<-f
Hi
r-help-boun...@r-project.org napsal dne 09.09.2009 10:07:49:
> Hi Henrique,
>
> Thanks for your reply.
> I tried you suggestion but it didn't work with the poLCA package.
>
> Maybe i didn't express myself good.
>
> The normal syntax is:
>
> f <- cbind(V1,V2,V3)~1
> poLCA(f,data)
You compl
Hi
what about reading each line by readLine and then split it to desired
portions?
x<-paste(letters, collapse="")
substring(x, c(1,3,5),c(2,4,15))
Regards
Petr
r-help-boun...@r-project.org napsal dne 08.09.2009 14:21:53:
> This data is from database and the maximum length of a field is
> def
Thank you.
hadley wickham napsal dne 07.09.2009 15:50:03:
> On Mon, Sep 7, 2009 at 8:36 AM, Petr PIKAL
wrote:
> > Dear all
> >
> > Colleague of mine ask me if R is capable of Andrews plot like
> > andrewsplot(x) in Matlab.
> >
> > Quick search did no
Dear all
Colleague of mine ask me if R is capable of Andrews plot like
andrewsplot(x) in Matlab.
Quick search did not reveal anything but before I start to write any
routine I would like to ask this ingenious audience if there is any
implementation of Andrews plots somewhere.
I know about p
the formats shown are more
scientific
> in focus. I still haven't been able to find a way of getting a "comma
> style".
AFAIK you can not format these in boxplot directly. You need to plot
without y axis and in axis you can use formating with prettyNum. I found
quite easily from s
Hi
r-help-boun...@r-project.org napsal dne 04.09.2009 13:53:24:
>
> Nordlund, Dan (DSHS/RDA) wrote:
> >
> >
> > You need to put the filename in quotes
> >
> > file = "C:/Documents and Settings/aelmore/Desktop/foo.pdf"
> >
> > Hope this is helpful,
> >
> > Dan
>
>
> Oh yes, this was very h
Hi
image<-matrix(0.1, 768,1024)
> image[1:10, 1:10]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
[2,] 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
[3,] 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
[4,] 0.1 0.1 0
Hi
it is rather difficult to understand what you mean by your
questions/answers without real reproducible code.
r-help-boun...@r-project.org napsal dne 03.09.2009 13:41:11:
>
> I'm posting answers to my own Q's here - as far as I have answers -
first so
> that people don't spend time on them,
Hi
use any of suitable selection ways that are in R.
E.g.
data[data$gender==1, ]
selects only female values
data$wage[(data$gender==1) & (data$race=1)] selects black female wages.
and see also ?subset
Regards
Petr
r-help-boun...@r-project.org napsal dne 03.09.2009 10:51:59:
> Dear all,
>
Hi
I had similar problem some time ago. I was advised to use eqscplot but it
did not suite my purpose so I used a little twist of biplot. (I added a
result)
fit<-princomp(some.data, cor=T)
biplot(fit, xlabs=rep("", no.of.poins)) #biplot without points
#some scaling of plot parameters
rrr<-a
Hi
that is a question which comes almost so often as "why R does not think
that my numbers are equal". So even I, non statistician, can deduct that
hist with probability =T can have any y axis range but the sum below curve
has to be below 1.
Regards
Petr
r-help-boun...@r-project.org napsal dn
Hi
r-help-boun...@r-project.org napsal dne 01.09.2009 10:39:42:
>
> HI, R user,
>
> I generate the vectors with the same length. I want to put each vector
into
> each column of data frame. Why it doesnt work`?
>
> rm<-data.frame()
>
> for(a in 1:6){
> rm[,a]<-getmeasure(p1,a,speech)
>
>
Hi
r-help-boun...@r-project.org napsal dne 29.08.2009 21:49:54:
> On Sat, Aug 29, 2009 at 8:14 PM, njhuang86 wrote:
> >
> > Hey guys,
> >
> > I was wondering how to create this sequence: 1, 2, 3, 1, 2, 3, 1, 2,
3, 1,
> > 2, 3... with the '1, 2, 3' repeated over 10 times.
>
> rep(1:3,10) # rep
Tinn-R
Petr
r-help-boun...@r-project.org napsal dne 28.08.2009 09:16:38:
> On Thu, 27 Aug 2009 12:43:41 -0700 Jonathan Greenberg
> wrote:
>
> JG> Quick informal poll: what is everyone's favorite text editor for
> JG> working with R? I'd like to hear from people who are using editors
> JG> tha
Hi
r-help-boun...@r-project.org napsal dne 25.08.2009 18:15:25:
>
> I want to draw lines using a matrix with the X-axe on the first column,
and
> the the Y-axe on the second column.
> These lines have to link the n points of a graph, so they are n-1, but
the
> resout using these two commands:
Hi
r-help-boun...@r-project.org napsal dne 26.08.2009 10:36:22:
> Hi,
>
> I'm trying to find an easy way to do this.
>
> I want to select the top three values of a specific column in a subset
> of rows in a data.frame. I'll demonstrate.
>
> ABC
> x21
> x41
> x3
Hi
r-help-boun...@r-project.org napsal dne 25.08.2009 12:07:11:
> Hi,
>
> the second step in my exercice is to calculate the sum of the amout for
each
> class et not the frequency
>
> i have this vector
>
> x y
> 1 100
> 2 1500
> 3 3250
> 4 625
> the goal of this exercice that to count the number of missing value,
number
> betwwen 0-1000 , 1000-3000, >3000.
>
> Thank you again for your help
>
>
>
>
>
> De : Petr PIKAL
>
> Cc : r-help@r-project.org
> Envoy
Hi
Do not use cbind as it results in matrix and it can have only data of one
type, in your case character. Use
MeanEst2000.Sz=data.frame(Sz,Pred)
instead. And you probably could go step further to put everything into one
data frame
DF <- data.frame(Sz,Pred, Obs)
then
with(DF, plot(Pred, O
Well
you shall consult (surprisingly :-)
?sample function
and for rating
?runif function
as was pointed out earlier
Regards
Petr
r-help-boun...@r-project.org napsal dne 25.08.2009 05:14:00:
> Dear All
> Â
> I know that you do not have to help me (as this is not a pure R
problem)Â but
> p
Hi
r-help-boun...@r-project.org napsal dne 25.08.2009 10:08:36:
> Hi Mark,
>
>
> Thank you for your answer !! it works but if i have "NA" in the vector z
what
> i shoud do to count its number in Z?
You do not have NA in z, you manage to convert it somehow to factor.
Please try to read about
Hi
r-help-boun...@r-project.org napsal dne 23.08.2009 05:00:11:
> Hi,
>
> It is easy to understand the types vector and frame.
>
> But I am wondering why the type factor is designed in R. What is the
> advantage of factor compare with other data types in R? Can somebody
> give an example in whi
Hi
r-help-boun...@r-project.org napsal dne 23.08.2009 17:29:48:
> On 8/23/2009 9:58 AM, David Winsemius wrote:
> > I still have problems with this statement. As I understand R, this
> should be impossible. I have looked at both you postings and neither of
> them clarify the issues. How can you ha
Hi
mapply("*", vect, mat, SIMPLIFY=F)
Regards
Petr
r-help-boun...@r-project.org napsal dne 21.08.2009 10:53:19:
>
> Suppose I have following list :
>
> mat <- vector("list")
> for (i in 1:4) mat[[i]] <- matrix(rnorm(25),5)
> mat
>
> > mat
> [[1]]
> [,1] [,2] [,3]
making your own
histogram function it is hard to say what do you want.
I am pretty sure you are inventing a wheel.
Regards
Petr
Conrad Addo napsal dne 20.08.2009 16:29:36:
> Petr
>
> Actually I wanted to find why the NUm vector was adding incrementally,
not
> about the CSV.
&
Hi
r-help-boun...@r-project.org napsal dne 20.08.2009 15:33:38:
> I'm trying to create a histogram from the following code, but my data
keeps
> adding the Num vector and plotting that. My data in the CSV file is
just
> one vector. Does anyone know why?
Well, I think that only you know why CS
ur procedures.
Regards
Petr
> http://paste.ubuntuusers.de/396117/ It even runs (though errors occur if
> I add th process r into the main loop, but I'm still not done with
it...).
>
> Petr PIKAL schrieb:
> > Hi
> >
> >
> > No. It is the problem of scoping. AFAIK function
ons for
manipulation with lists. E.g. structure lm(...) results in list and
summary(lm(...)) is again a list with quite complicated structure.
If I understand it correctly, during your computation you will have as a
result matrices with arbitrary dimensions.
I would make a list
lll<-vec
OIS
Thank you both for pointing me to it. I did not notice this as the
unscaled position of points was quite clear and strightforward according
to my knowledge of data. The scaled plot is slightly more distorted and
the relationships are not so obvious.
Thank you both
Petr Pikal
petr.pi
Ok
Thank you for your time.
Best regards
Petr Pikal
Duncan Murdoch napsal dne 19.08.2009 16:29:07:
> On 8/19/2009 10:14 AM, Petr PIKAL wrote:
> > Duncan Murdoch napsal dne 19.08.2009 15:25:00:
> >
> >> On 19/08/2009 9:02 AM, Petr PIKAL wrote:
> >> > Tha
Duncan Murdoch napsal dne 19.08.2009 15:25:00:
> On 19/08/2009 9:02 AM, Petr PIKAL wrote:
> > Thank you
> >
> > Duncan Murdoch napsal dne 19.08.2009 14:49:52:
> >
> >> On 19/08/2009 8:31 AM, Petr PIKAL wrote:
> >>> Dear all
> >>&g
Thank you
Duncan Murdoch napsal dne 19.08.2009 14:49:52:
> On 19/08/2009 8:31 AM, Petr PIKAL wrote:
> > Dear all
> >
>
> I would say the answer depends on the meaning of the variables. In the
> unusual case that they are measured in dimensionless units, it mi
Hi
r-help-boun...@r-project.org napsal dne 19.08.2009 13:04:39:
> Strange, it doesn't work for me:
>
> Error in database[4, 4][[1]][1, ] : incorrect number of dimensions
> Execution halted
>
> R version 2.9.0 (2009-04-17) on Arch Linux, no additional packages
> installed.
database[4,4][[1]][,
"ano" while highest group has low value of sio2.
But when I do the same with scale=T
fit<-prcomp(~iep+sio2+al2o3+p2o5+as.numeric(dus), data=rglp, factors=2,
scale=T)
biplot(fit, choices=2:3,xlabs=rglp$vzorek, cex=.8)
I get completely different picture which is not possible t
Hi
r-help-boun...@r-project.org napsal dne 18.08.2009 10:14:26:
> Hi Everbody
>
> Could somebody help me.?
>
> I need to remove the columns where the sum of it components is equal to
> zero.
>
> For example
>
> > a<-matrix(c(0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,1,0,0,1,0), ncol=4)
> > a
>
Hi
As matrix is vector with dim attribute
mymat<-ifelse(mymat==0, "A","P")
should be sufficient.
Even with data frame it works
mydf<-data.frame(mymat)
mydf<-ifelse(mydf==0, "A","P")
mydf
X1 X2 X3
[1,] "P" "P" "A"
[2,] "A" "A" "A"
[3,] "A" "A" "P"
[4,] "A" "A" "P"
[5,] "P" "P" "P"
R
Hi
r-help-boun...@r-project.org napsal dne 17.08.2009 17:41:59:
>
> Dear all,
>
> I'm trying to replace NA values with - in one column of a data
frame. I've
Well, Jim sent you a solution but I would rise a question, why do you want
to do that. R has many instruments how to smoothly handl
Hi
r-help-boun...@r-project.org napsal dne 15.08.2009 04:27:39:
> Say I have a csv file, each row contains several fields, one of them
> are whether the row is success.
>
> In history data, I have all the fields including the result of whether
> it is success. In future data, I only have fields
Hi
r-help-boun...@r-project.org napsal dne 18.08.2009 10:35:11:
> Dear all,
>
> I have a problem with the function read.xls from the gdata package,
error
> message see below. Two examples:
>
> First, I try to read my data, which does not work;
> Secondly, I tried the example code/data with th
Hi
r-help-boun...@r-project.org napsal dne 18.08.2009 06:13:09:
> Hi everyone,
> I want you all help me to give an idea, how to draw pentagon with
points?
> Maybe can use function, but I'm stuck.
Maybe. Maybe you can use a pencil, instead.
Basically this
plot(1:10, type="n")
polygon(c(2,3,5,6
r-help-boun...@r-project.org napsal dne 17.08.2009 16:18:43:
> Dear all,
> I would like to plot credible interval for a function estimate in R. I
> would like to plot the credible intervals as shaded region using polygon
> function. Does anyone ever used that? I tried several times but I could
> n
Hi
r-help-boun...@r-project.org napsal dne 18.08.2009 14:32:43:
> Hi,
>
> Actually, i use the function merge like this:
>
> (Data1 <- Data1[1:7,1:3])
> Policy.Number AXA.Entity Country
> 1106077BNL BNL
> 2 4001023 CH BNL
> 3106006 UK B
Hi
r-help-boun...@r-project.org napsal dne 17.08.2009 16:22:39:
> Hi,
> Can someone suggest an efficient way to substitute a vector/matrix
> which contains 1's and 0's to P's and A's (resp.)?
vec<-sample(0:1, 20, replace=T)
vec
[1] 1 1 0 1 0 1 0 1 0 0 1 1 0 0 0 0 0 0 1 0
vec<-ifelse(vec==0, "A",
Hi
r-help-boun...@r-project.org napsal dne 18.08.2009 14:13:12:
> I cannot understand why xyplot does not work within a simple for loop.
>
> This works up to the for loop; inside the for loop the png files are
> opened and closed, but nothing is plotted. No error messages are written
> to the co
Hi
r-help-boun...@r-project.org napsal dne 31.07.2009 08:27:24:
> The error message is: "Error in library(lattice) : there is no package
called 'lattice'"
>
> I have installed it twice because it did not work the first time, so I
was not
> sure if it worked.
> And YES, the folder lattice is in
Hi
r-help-boun...@r-project.org napsal dne 30.07.2009 10:19:21:
> I am attempting to replicate some of my experience from SAS in R and
assume
> there are best methods for using a combination of summary(), subset, and
> which() to produce a subset of mean values by categorical or ordinal
> factor
returns the position in the string x where that regular
> expression begins. Feed that to substring() and you get the desired
> result. Both substring() and regexpr() work on vectors of strings.
>
> Best regards,
> Chuck Taylor
> TIBCO Spotfire Seattle
>
>
> -Origi
trapply(s, "[0-9]*$", simplify = c)
>
>
>
> On Wed, Jul 29, 2009 at 9:10 AM, Petr PIKAL
wrote:
> > Hi all
> >
> > I have got something like that (actually those are column names)
> >
> > [51] "X19.2.300b...80" "X19.2.
> > x
> [1] "X19.2.400v...80"
> > strsplit(x,"\\...")[[1]][3]
> [1] "80"
>
> HTH,
>
> Jorge
>
> On Wed, Jul 29, 2009 at 9:10 AM, Petr PIKAL
wrote:
> Hi all
>
> I have got something like that (actually those are colum
Hi all
I have got something like that (actually those are column names)
[51] "X19.2.300b...80" "X19.2.400v...80" "X19.2.400b...80"
"X19.2.300v...90" "X19.2.300b...90"
[56] "X19.2.400v...90" "X19.2..400b..90" "X19.2.300v...100"
"X19.2.300b...100" "X19.2.400v
Hi
r-help-boun...@r-project.org napsal dne 28.07.2009 17:34:35:
> I am currently summarising a data set by collapsing data based on common
> identifiers in a column. I am using the 'aggregate' function to
summarise
> numeric columns, i.e. "aggregate(dat[,3], list(dat$gene), mean)". I
also w
rsion and unzip
it to library subdirectory.
> Then:
> library(TeachingDemos)
shall be executed without problem
Regards
Petr
>
>
> --- On Tue, 7/28/09, Jose Narillos de Santos
wrote:
>
> > From: Jose Narillos de Santos
> > Subject: Re: [R] pairs plot
Hi
If you want to use brute force without much knowledge about packages
make your own directory in librarary directory
select any package in your library directory and copy it to directory you
just made
leave only INDEX and DESCRIPTION files and R and Data directorires
change DESCRIPTION file ac
Hi
r-help-boun...@r-project.org napsal dne 28.07.2009 12:40:59:
> Dear Users
>
> This is my dataset called mydata4. I want to sort the dataframe on the
first
> column PxMid which is basically a column with dates.
>
> I've tried mydata4<-mydata4[order(mydata4$PxMid),] but it doesnt work.
Could
Hi
maybe outer?
Regards
Petr
r-help-boun...@r-project.org napsal dne 28.07.2009 11:36:10:
>
>
>
>
>
>
> How I can vary the parameters for a function?
>
> I have a function with 5 parameters I want to turn the function for a
range of
> numbers for one of these parameters!! i want to ha
Hi
r-help-boun...@r-project.org napsal dne 28.07.2009 09:55:11:
> Hi Greg I saw, read, the TeachingDemos you suggesttef but when run
pairs2
> function on my R module says "Can´t find function pairs2"
>
> How can I load the module or function pairs2?
Did you do
library(TeachingDemos)?
Regards
Hi
r-help-boun...@r-project.org napsal dne 28.07.2009 09:18:43:
>
>
> I tried searching but I couldn't quite find what I was looking for.
>
> Here's a dummy data matrix (with row and column labels):
> > y
>0 1 2 3 4
> 21 3 4 8 5 5
> 22 3 6 8 6 NA
> 23 4 5 11 4 3
> 24 4 2 1 4 6
> 25
Hi
r-help-boun...@r-project.org napsal dne 25.07.2009 23:15:04:
> Thanks for the answer Tal!
> But I can't get it to work correctly! :(
> Please bear with me this is the first time I am using R! and I am in a
rush
> to correct a paper
> in fact on the plane I am plotting a table
> > fullpointed=
Hi
I use saved R file stand.R which looks like this
"stand" <-
list( b110 = c(49.45000, 21.46, 11.468333, 24.33500, 28.112240),
b120 = c(49.77333, 19.386667, 7.736667, 20.87500, 21.753788),
b130 = c(49.60833, 18.365000, 5.708333, 19.23167, 17.265409),
b140 = c(5
Hi
maybe package akima with function interp can help you
Regards
Petr
r-help-boun...@r-project.org napsal dne 23.07.2009 11:04:25:
> That problem is that for every y I get a different x and thus cant
> create an array as in the help page examples.
> I was just wondering, whether there is a poss
Hi
Strange. Before starting with R about 10 years ago I used Excel for for
graphing too. As soon as I learned few tricks and read intro documents I
abandoned Excel for graphing almost completely. Now even for simple
visualisations of data I copy them ***from*** spreadsheet to R. But as you
mak
Hi
r-help-boun...@r-project.org napsal dne 21.07.2009 14:37:35:
>
>
> Graves, Gregory wrote:
> > I issued the following command to obtain the std dev for each month.
> >
> >
> >
> > psd<-numSummary(Sal, groups=month, statistics=c("sd"))
>
>
> numSummary is not in base R, is it? If not, whi
Hi
r-help-boun...@r-project.org napsal dne 21.07.2009 09:18:51:
>
> Hi Daniel,
> Thanks for the insight. My apologies for the unclearness of my original
> question.
> I have calculated the fit and se.fit values, see below
> predict(fm,newdata=test, se.fit=TRUE, type=c("response"))
> If I wasn't
Hi
r-help-boun...@r-project.org napsal dne 16.07.2009 11:31:25:
> I imagine I make a function whose results are a Matrix Z
>
> How cn I save in a txt or excel file the result of apply my function?
>
> something similar to save Z.txt.
write.table(tab, "clipboard", sep = "\t", row.names = F)
to
Hi
r-help-boun...@r-project.org napsal dne 15.07.2009 17:59:39:
>
> see ?ifelse
>
> you didn't specify what happens if a value is exactly zero in the
dataset
> and so i've just bundled it in with the negative case:
>
> x <- rnorm(20, 0, 1)
> y<-ifelse(x<=0, 10, 5)
For this simple case you ca
c graphics so it should not be
necessary.
> >>
> >> If you are getting blank plots when you shouldn't, that's a bug. If
you can
> >> put together a reproducible example that shows its a bug in R, rather
than a
> >> bug in your script, it will li
Hi
For this type of problems I do multipage pdf.
pdf("file", )
for (i in ...) {
do all stuff including plot
}
dev.off()
and then check the plots afterwards. Recently there was some post about
how to wait but you do not want only wait you want also to interactively
change plotting parameter
Hi
> str(read.table("test.txt", header=T))
'data.frame': 9 obs. of 12 variables:
$ snp : Factor w/ 9 levels
"rs1113188","rs1113397",..: 9 5 7 8 3 4 6 1 2
$ gene : Factor w/ 1 level "TRP2": 1 1 1 1 1 1 1 1 1
$ chromosome : int 3 3 3 3 3
Hi
matrix is virtually a vector so you can find index of values of x and add
those values to propper places in ax
ax <- a
idx <- which(!is.na(x))
ax[idx] <- x[idx]
ax
[,1] [,2] [,3]
[1,]321
[2,]422
[3,]552
Regards
Petr
r-help-boun...@r-project.org napsa
Hi
r-help-boun...@r-project.org napsal dne 13.07.2009 10:45:50:
> Dear R-users,
> I am using R(a package igraph) to calculate certain
> topological features of networks. When I try to draw a plot between
these
> features I get an error. Following is the code I am using :
>
Hi
r-help-boun...@r-project.org napsal dne 12.07.2009 22:24:29:
> On Sun, Jul 12, 2009 at 1:05 PM, David Winsemius
wrote:
> >
> > On Jul 12, 2009, at 3:35 PM, David Winsemius wrote:
> >
> >>
> >> On Jul 12, 2009, at 2:53 PM, Mark Knecht wrote:
> >>
> >>>
> >>> As a test I tried to print down to
Hi
r-help-boun...@r-project.org napsal dne 10.07.2009 12:07:03:
> I have a vector of values
>
>
>
> X = seq(1:10)
>
>
>
> I want to get another vector V of with sample (with replacement) of X
but
> with a constrain: V will have as much elements as those necessary to V
sum
> exactly 10.
>
Hi
you have to look to your data
when I used your function to some artificial data I got expected result
> myfun(visko,"konc")
Levels = 2
[[1]]
[1] NA
[[2]]
Welch Two Sample t-test
data: data[[nam[v]]] by data[[g]]
t = -1.7778, df = 4.541, p-value = 0.1415
alternative hypothesis:
Hi
maybe ?cumsum, or ?diff or ?aggregate
Regards
Petr
r-help-boun...@r-project.org napsal dne 10.07.2009 10:56:34:
> R users,
> I am a beginner in R. I have a time series data on various macro
variables
> like GDP, unemployment rate, consumer spending, imports, exports etc. I
am
> interested
Hi
Godmar Back napsal dne 09.07.2009 14:09:42:
> On Thu, Jul 9, 2009 at 4:50 AM, Petr PIKAL
wrote:
> Hi
>
> r-help-boun...@r-project.org napsal dne 09.07.2009 02:57:33:
>
> Not so weird. What do you expect from
>
> c(1:5, 10:20, 30:50)
>
> You mean wha
Hi
r-help-boun...@r-project.org napsal dne 09.07.2009 02:57:33:
> On Wed, Jul 8, 2009 at 8:34 PM, Jason Rupert
wrote:
> >
> > Maybe there is a great website out there or white paper that discusses
this
> but again my Google skills (or lack there of) let me down.
>
> Yeah, R is difficult to se
Hi
Godmar Back napsal dne 08.07.2009 15:28:50:
> On Wed, Jul 8, 2009 at 4:22 AM, Petr PIKAL
wrote:
> Hi
>
> r-help-boun...@r-project.org napsal dne 07.07.2009 19:06:17:
>
> > Hi,
> >
> > I am confused about how to select elements from a list.
> >
>
Hi
see
?lm and ?abline
Regards
Petr
r-help-boun...@r-project.org napsal dne 08.07.2009 11:31:19:
> Hi all,
> I am new to R. How does one go about fitting a trend-line to a
> scatter plot? Any help is appreciated.
>
> Thanks and regards,
>
> Anupam
>
>[[alternative HTML version de
Hi
r-help-boun...@r-project.org napsal dne 07.07.2009 19:06:17:
> Hi,
>
> I am confused about how to select elements from a list.
>
> I'm trying to select all rows of a table 'crossRsorted' such that the
> mean of a related vector is > 0. The related vector is accessible as
> a list element l[
ring to each alphabet
> where each alphabet will be in different column.
>
> thanks alot.
>
> regards,
> Hema.
> On Tue, Jul 7, 2009 at 4:12 PM, Petr PIKAL
wrote:
> Hi
>
> r-help-boun...@r-project.org napsal dne 07.07.2009 09:54:30:
>
> > Hi everyone,
>
Hi
r-help-boun...@r-project.org napsal dne 07.07.2009 10:05:09:
>
> Dear group:
> sorry for my beginners question, but I'm rather new to R and was
searching
> high and low without success:
>
> I have a data frame (df) with variables in the rows and observations in
the
> columns like (the ac
Hi
r-help-boun...@r-project.org napsal dne 07.07.2009 09:54:30:
> Hi everyone,
> Hi want to separate the string(column1) for example
Well, how did you get the data in R? Are they in separated columns of
data.frame? What do you mean by "separate"?
>
> column1 column2 column3 column4 column5 co
Hi
r-help-boun...@r-project.org napsal dne 06.07.2009 01:58:38:
> On Sun, Jul 5, 2009 at 1:44 PM, hadley wickham
wrote:
> >> I think the root cause of a number of my coding problems in R right
> >> now is my lack of skills in reading and grabbing portions of the data
> >> out of arrays. I'm ne
Quicvk and dirty
Insall R to PC and copy the appropriate directory to flash disk.
Regards
Petr
r-help-boun...@r-project.org napsal dne 03.07.2009 14:34:59:
> Hi!
>
> Is there some R version for flashdisk, pendrive etc ?
>
> --
> Regards,
>
> Raphael Saldanha
> saldanha.plan...@gmail.com
>
"%m")
[1] "02"
> as.numeric(format(strptime("20/2/06 11:16:16.683", "%d/%m/%y
%H:%M:%OS"), "%m"))
[1] 2
Regards
Petr
>
>
> Petr PIKAL wrote:
> Hi
>
> r-help-boun...@r-project.org napsal dne 02.07.2009 12:40:05:
>
>
Hi
r-help-boun...@r-project.org napsal dne 02.07.2009 12:40:05:
> I have a data frame (hf) that is all set up and the dates are working
> fine - however I need to extract the months and hours (2 separate
> columns) as numbers - however they are coming out as characters.
>
> I have tried both t
Hi
what about do inside some function a subset of your whole data frame
fff <- function( data, rows) {
data.1 <- data[1:rows,]
get all necessary stuf on data.1
return what you want
}
You can put a dimension check if you want the function to be more robust
Regards
Petr
r-help-boun...@r-pr
Hi
r-help-boun...@r-project.org napsal dne 29.06.2009 10:58:50:
>
> Hallo,
>
> I have a vector of several iterations and I have to remove /filter out
all
> the values < than 10,
> how can I do this?
vector[vector>=10]
does what you want but word "vector of iterations" frightens me a little
Petr Pikal
petr.pi...@precheza.cz
724008364, 581252140, 581252257
r-help-boun...@r-project.org napsal dne 26.06.2009 05:31:27:
> Hi all,
>
> This is a really basic question but I can't figure it out.
>
> I am trying to write a piece of code that will use two datasets, z
Hi
does
expand.grid(z=seq(0,6,2), y=seq(0,8,4), x=seq(0,10,2))
do what you want?
Petr
r-help-boun...@r-project.org napsal dne 26.06.2009 11:15:42:
> I would raise x,y and z in a loop but I won't raise of 1. I tried this
but it
> doesn't work
>
> mydata=matrix(nrow=1500,ncol=3)
> i=1
> for(
Hi
r-help-boun...@r-project.org napsal dne 24.06.2009 14:16:15:
> Hello
>
> I have a data frame d with columns "var1", "var2", "var3"
>
> Then I have two vectors:
> columns <- c("var2", "var3")
> values <- c(0, 1)
>
>
> Is there a compact way to subset the data frame
> using these two vectors
Hi
r-help-boun...@r-project.org napsal dne 23.06.2009 16:04:49:
> Try this:
>
> which(diff(is.na(inc)) < 0)
Shall it be
which(diff(is.na(inc)) < 0)+1
or
sum(is.na(inc))+1
or any other suitable construct with is.na
Regards
Petr
>
> On Tue, Jun 23, 2009 at 11:00 AM, Alfredo Alessandrini
omething like this.
> fff2<-function(x) length(unique(x))==1
> system.time(print(fff2(x)))
[1] FALSE
user system elapsed
0.390.080.47
Regards
Petr
> On Fri, Jun 19, 2009 at 8:18 AM, Petr PIKAL
wrote:
> Hi
>
> utkarshsinghal napsal dne
> 17.06.2009 15:29:
Hi
utkarshsinghal napsal dne
17.06.2009 15:29:34:
> I will wait for the next version-2.9.1 and presently using Petr's
suggestion, i.e.,
> (x[1]*length(x))==sum(x)
> which significantly reduced the run time.
>
> The problem is now there might be only small differences ,say, of the
order of
>
Hi
r-help-boun...@r-project.org napsal dne 19.06.2009 09:26:57:
> Ahoy. I'm trying to run a function for each country in a multinational
> dataset. Keeping it simple, an example is
>
>
> for(j in 11:14) {
> if(data$country_str == "j") {
>
> mu <- mean(data$ecdfs1)
> } else {
> mu <- 0
> }
> }
Hi
r-help-boun...@r-project.org napsal dne 17.06.2009 14:59:35:
>
> This is my function:
>
> zywnoscCalosc <- function( zywnosc, sklepik, sklslodycze) {
>
> b=as.vector(sklslodycze)
>
> fun=function(a,b){
> a=2+b
> }
> zywnosc=ifelse(is.na(sklepik),NA,fun(a,b))#Here I have a problem
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