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
> > 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
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.
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
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
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 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 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 14:32:43:
> 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
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 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
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 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 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
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 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
>
"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 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]][,
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
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
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
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
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
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
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
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
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]
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
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
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
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
> 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
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
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 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:
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 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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
see ?format
output of format can be transferred to numeric and you can use logical
operators to select only desired data.
x[as.numeric(format(x, "%Y")) > 2008,]
another option is to compare it to some date like
x > as.Date("2008-1-1")
this assumes that x has class "Date". If not you can ch
Hi
did you try
?complete.cases or ?na.omit?
nadata[complete.cases(nadata),]
Regards
Petr
r-help-boun...@r-project.org napsal dne 21.09.2009 08:16:04:
>
> I have to remove missing data both in character and numeric datatype.I
tried
> using NA condition but it is not working ,please help me t
Hi
I believe you are quite near.
r-help-boun...@r-project.org napsal dne 21.09.2009 11:38:29:
>
> Thank you so much for trying to help me.
> Thus, I still can't get it to work.
> I will clearify a bit. If you somehow have time helping me I would much
appreciate it.
> NAD and Prot.amount are
Hi
I tried but
> describe(c(5,3,76,4/0))
Error: could not find function "describe"
>
What shall I do? Quick search did not find function on CRAN and I do not
have enough time to dig it from somewhere.
Regards
Petr
r-help-boun...@r-project.org napsal dne 21.09.2009 12:06:09:
> Hi,
>
> I hav
Hi
r-help-boun...@r-project.org napsal dne 22.09.2009 11:51:18:
>
> Hi,everyone i need to calculate quartile values of a variable grouped by
the
> other variable .
> same as in aggregate function(only median,mean or functions is
possible-i
> think so)
> Could you please help me to achieve the
Hi
r-help-boun...@r-project.org napsal dne 22.09.2009 14:57:59:
> Hi all, I have been trying to solve this problem and have had no luck so
far.
>
> I have numeric vectors VAR1, VAR2, and VAR3 which I am trying to cbind.
I also
> have a character vector "VAR1,VAR2,VAR3". How do I manipulate th
Hi
I am not sure if I understand what you want but if your matrix is not so
big you can try
> x[,1]*x[,2]
[1] 5 12 21 32
> cumsum(x[,1]*x[,2])
[1] 5 17 38 70
>
and than check value of cumsum according to your condition.
Regards
Petr
r-help-boun...@r-project.org napsal dne 23.09.2009 05:25
Hi
r-help-boun...@r-project.org napsal dne 23.09.2009 14:49:38:
> On Wed, Sep 23, 2009 at 07:29:30AM -0500, Peng Yu wrote:
> > On Wed, Sep 23, 2009 at 1:24 AM, Peter Dalgaard
> > wrote:
> > > Peng Yu wrote:
> >
> > Is there an operation on a factor to get a subset and keep only the
> > correspo
Hi
You can use outer. If your data are in data frame test then
DIFF <- as.vector(t(outer(test$val, test$val, "-")))
returns a vector, You just need to add suitable names to rows.
CASE <- as.vector(t(outer(test$ID, test$ID, paste, sep="-")))
data.frame(CASE, DIFF)
will put it together.
Regard
Hi
r-help-boun...@r-project.org napsal dne 23.09.2009 21:17:10:
>
> Hi,
>
> Is there a way I can plot the median as well as the quantiles in the
actual
> boxplot using the "boxplot" command?
AFAIK boxplot produces box which marks upper and lower quartile and
median. So you shall be more prec
Hi
r-help-boun...@r-project.org napsal dne 25.09.2009 05:31:58:
>
> I want to know, how do I retain the current plot and axes properties
such
> that subsequent graphing commands add to the existing graph.
Are you looking for
?lines, ?points
Regards
Petr
>
> Thank you very much!!
>
> --
Hi
r-help-boun...@r-project.org napsal dne 16.06.2009 12:45:04:
> Stefan Uhmann wrote:
> > Dear List,
> >
> > why does this not work?
> >
> > df <- data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7),
> > fac = c('A', 'A', 'B'))
> > tapply(cbind(df$var1, df$var2, df$var3), df$fac, me
Hi
r-help-boun...@r-project.org napsal dne 16.06.2009 14:31:21:
> Hi Jim,
>
> What you are saying is correct. Although, my computer might not have
> same speed and I am getting the following for 10M entries:
>
>user system elapsed
> 0.559 0.038 0.607
With numbers you may speed it a
Hi
r-help-boun...@r-project.org napsal dne 17.06.2009 07:51:27:
>
> 2 - is.na(a) - it's superb! but I need call a function: wy[i]<-
> ifelse(((is.na(a))), call_fun1(x), call_fun2(x)
You did not grasp how to use ifelse. It goes not cycle through logical
vector is.na(a).
from help page
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