Hi,
That's because the parser knows how to deal with that stuff. However,
when you type the same if/else at the command line, it will be parsed
line by line, and the evaluator will not wait for the else to evaluate
the if. Try to copy and paste your if/else to the command line.
Romain
Yihu
Hi,
I wonder whether there is any convenient function (or package) to
extract tables from a HTML page? e.g. from
http://www.google.com/finance/historical?q=SHE:002251
I know we can readLines('URL'), gsub('...', '...', source), ...
and at last get the numbers; I'm writing to ask whether someone ha
Hi list members,
?"else" tells us
In particular, you should not have a newline between '}' and
'else' to avoid a syntax error in entering a 'if ... else'
construct at the keyboard or via 'source'.
but there's no syntax error when you break the line between "}" and
"else" in a func
Agree, it is ambiguous. Will try again
Suppose I have the following
X<-matrix(c(1:20),nrow=4,byrow=T), hence this is a 4 by 5 matrix
Y<-matrix(c(2,4,6,8),ncol=1), and this is a 4 by 1 vector, each value denote
y1, y2,y3, y4
Step 1 - write each row of X as a vector, so I have vectors t(x_
Hi, since bigglm doesn't have update, I was wondering how to achieve
something like (similar to the example in ff package manual using biglm):
first <- TRUE
ffrowapply ({
if (first) {
first <- FALSE
fit <- bigglm(eqn, as.data.frame(bigdata[i1:i2,,drop=FALSE]), chunksize =
1, family =
Hello,
Can anyone in this group translate what the following C Code is doing? Any
help appreciated.
Jim:
#include
#include
#include
double rtrun(double mu, double sigma,double trunpt, int above)
{
/* function to draw truncated normal
above=1 means from above b=trunpt, a
On 3/04/2009, at 4:36 PM, Gundala Viswanath wrote:
Dear all,
With this snippet:
xlim=c(-126000,50)
in the current plot, the x-axis appear in the following tick marks:
-10, 10, 30, ...etc
How can we force a point where x=0 to appear in the plot?
Yielding:
-10, 0, 10
hi,
Previously my email very simple and less information. Let say I have raw
data as below:-
user_id website time
20google0930
21yahoo0935
20facebook1000
25facebook1015
61google0940
60yaho
Does load() work?
2009/4/3 purpleshirt :
>
> I login Linux machine on my windows desktop via Exceed, then run R.
>
> Somehow my desktop got rebooted by accident, when I re-login, I can see the
> R session is still listed there (run >top). My question is, is there any way
> I can bring back that R
Dear all,
With this snippet:
xlim=c(-126000,50)
in the current plot, the x-axis appear in the following tick marks:
-10, 10, 30, ...etc
How can we force a point where x=0 to appear in the plot?
Yielding:
-10, 0, 10, 30, ...etc
^
in the x-axis tick m
Thanks for the bug report. In the interim these two (and
variations work):
plot(z, screens=c(1,2,2), ylim = c(1, 40))
plot(z, screens=c(1,2,2), ylim = list(c(1, 40), c(1, 40), c(1, 40)))
On Thu, Apr 2, 2009 at 3:41 PM, Scott Tennican wrote:
>
> I am plotting multiple graphs per window with multi
On 3/04/2009, at 3:47 PM, Gundala Viswanath wrote:
Dear all,
I have a plot which contain the following parameter:
xlim=c(-126000,50)
In the actual plot it self the x-axis appear as:
-1e+05, 0e+00, 1e+05, 2e+05 ..etc...
Is there a way we can disable such representation
of x-axis to plai
Dear all,
I have a plot which contain the following parameter:
xlim=c(-126000,50)
In the actual plot it self the x-axis appear as:
-1e+05, 0e+00, 1e+05, 2e+05 ..etc...
Is there a way we can disable such representation
of x-axis to plain number:
-10, 0 , 10, 20, ...etc
- Gunda
> X1 X2
> 1 11 0
> 2 11 0
> 3 11 0
> 4 11 1
> 5 12 0
> 6 12 0
> 7 12 0
> 8 13 0
> 9 13 1
> 10 13 1
>
>
> and I want to select all rows pertaining to factor levels of X1 for
> which exists at least one "1" for X2. To be clear, I want rows 1:4
> (since there exists at least one o
Generally unambiguous questions get answered quickly. Perhaps other
readers are having the same difficulties I am experiencing. So let's
try to parse what you are asking:
On Apr 2, 2009, at 8:26 AM, Kutlwano Ramaboa wrote:
Hello
This may have been answered elsewhere, and I have looked on t
I think the unique function is superfluous:
> x[x$X1 %in% x$X1[x$X2==1], ]
X1 X2
1 11 0
2 11 0
3 11 0
4 11 1
8 13 0
9 13 1
10 13 1
--
David Winsemius
On Apr 2, 2009, at 12:43 PM, Patrizio Frederic wrote:
or the exactly equivalent form:
x[x$X1 %in% unique(x[x$X2>0,"X1"]),
Hello,
I've succeeded in running an ordered probit model, and I am currently trying to
perform some diagnostics to determine the appropriateness of my model for the
data. I cannot figure out how to get residuals for the ordered probit so that I
can plot them against the variables in my mode
I login Linux machine on my windows desktop via Exceed, then run R.
Somehow my desktop got rebooted by accident, when I re-login, I can see the
R session is still listed there (run >top). My question is, is there any way
I can bring back that R session?
Thanks,
--
View this message in context:
> df <- read.table(textConnection('Experiment_id Treament_type Value
+ 12345"control"3
+ 12345"full treatment"4
+ 12345"full treatment"5
+ 12345"partial treatment"4
+ 1"control"1'), header = TRUE)
> df
Experiment_id Treament_type Value
1 12
Hello,
I'm accessing a MySQL database from inside R and had a problem with a
function. In the database, there is data from years 1985 to 2007 that I would
like to retrieve. I can easily get the data from one year by the following:
info1985=dbGetQuery(con, statement='the SQL query')
Inside the
We're offering advanced course in USA this april.
Regards - Sue
---
XLSolutions Corporation (www.xlsolutions-corp.com) is proud to announce
our*** R Advanced Programming***courses at USA locations for March -
April 2009.
http:
I am sure this is not the most elegant method, but it will "work".
> new <- merge(nn,uu, by = c("a","b"), all.x=T)
> new
a bx.xy.x x.y y.y
1 1 1 0.03648491 0.69644753 100 -100
2 1 2 0.76826365 0.28821953 NA NA
3 1 3 0.62744363 0.08373154 NA NA
4 2 1 0.27503185 0.557
Dear Michael,
How about the following?
y <- model.response(model.frame(base.mod))
TSS <- sum((y - mean(y))^2)
1 - drop.mod$RSS/TSS
I hope this helps,
John
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Michael Friendly
>
Folks,
Updating values in a table is simple in SAS or SQL, but I've wracked my
brain looking for an elegant solution in R to no avail thus far. Certainly
this is a common need that's been solved in dozens of different ways.
Given an initial dataframe nn and a smaller dataframe of updates uu, I'
On Thu, Apr 2, 2009 at 3:16 PM, Judith Flores wrote:
>
> Hello dear R-community,
>
> I have been trying to figure out a way to generate histograms of a numeric
> variable observed in different entities ('individual' below). Each one of
> this entities is classified as "A" or "B" (according to
Hello dear R-community,
I have been trying to figure out a way to generate histograms of a numeric
variable observed in different entities ('individual' below). Each one of this
entities is classified as "A" or "B" (according to the pseudo-code below):
variable<-sample(rep(1:10,10))
indiv
This is probably simple, but I just can't see it...
I want to calculate the R^2s for a series of linear models where each
term is dropped in turn. I can get the
RSS from drop1(), and the r.squared from summary() for a given model,
but don't know how to use the
result of drop1() to get the r.s
just for curiosity,
`%ni%` <- Negate(`%in%`)
> 1 %ni% c(2,1)
[1] FALSE
d1[id %ni% c(1,4), ]
baptiste
On 2 Apr 2009, at 22:17, gina patel wrote:
I have another question, if I now want to remove multiple id's e.g.
id=1 or 4 is there a simple OR command I can use?
I tried d2<-(d1[id != 1 |
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of gina patel
> Sent: Thursday, April 02, 2009 2:17 PM
> To: Brian Lee Yung (Portfolio Analytics)Rowe; hadley wickham
> Cc: R-help@r-project.org
> Subject: Re: [R] Deleting rows based
On 3/04/2009, at 9:30 AM, gina patel wrote:
I have created this data frame to help illustrate my problem.
id<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
x<-rep((seq(1:5)),5)
y<-c(0, 0.1, 0.5, 0.4, 0.2, 0, 0.1, 0.5, 0.4, 0.12, 0, 0.1, 0.5,
0.55, 0.2, 0, 0.1, 0.5, 0.3, 0.2, 0, 0.1, 0.6, 0
I have another question, if I now want to remove multiple id's e.g. id=1 or 4
is there a simple OR command I can use?
I tried d2<-(d1[id != 1 | 4, ])
however this does not delete anything
PS d2<-(d1[id != 4, ]) worked to remove id=4
Thanks
Gina
--- On Thu, 4/2/09, hadley wickham wrote:
F
On Thu, 2009-04-02 at 14:37 +0200, nboe...@uni-potsdam.de wrote:
> Dear List-Members,
>
> does anyone know a simple way to automatically install all R packages
> on a unix system to the default library path? Not from inside R, it
> should rather work as a shell script - job at startup.
>
> So
On Thu, Apr 2, 2009 at 3:37 PM, Rowe, Brian Lee Yung (Portfolio
Analytics) wrote:
> Is this what you want:
>> d1[which(id != 4),]
Or just
d1[id != 4, ]
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/li
Is this what you want:
> d1[which(id != 4),]
id xy
[1,] 1 1 0.00
[2,] 1 2 0.10
[3,] 1 3 0.50
[4,] 1 4 0.40
[5,] 1 5 0.20
[6,] 2 1 0.00
[7,] 2 2 0.10
[8,] 2 3 0.50
[9,] 2 4 0.40
[10,] 2 5 0.12
[11,] 3 1 0.00
[12,] 3 2 0.10
[13,] 3 3 0.50
[14,] 3 4 0.55
[15,] 3 5 0.2
I am plotting multiple graphs per window with multiple series on each graph.
When I try to set ylim I get the error below:
Error in ylim[[idx]] : subscript out of bounds
Am I incorrectly specifying my ylim list or is this a bug?
Here is a simple reproduction:
z <- zoo(cbind(a = 1:10, b = 11:20,
I have created this data frame to help illustrate my problem.
id<-c(rep(1,5),rep(2,5),rep(3,5),rep(4,5),rep(5,5))
x<-rep((seq(1:5)),5)
y<-c(0, 0.1, 0.5, 0.4, 0.2, 0, 0.1, 0.5, 0.4, 0.12, 0, 0.1, 0.5, 0.55, 0.2, 0,
0.1, 0.5, 0.3, 0.2, 0, 0.1, 0.6, 0.4, 0.1)
d1<-cbind(id,x,y)
I would like to delet
nboe...@uni-potsdam.de schreef:
Dear List-Members,
does anyone know a simple way to automatically install all R packages
on a unix system to the default library path? Not from inside R, it
should rather work as a shell script - job at startup.
Something like R cmd install -l pkgs ### where p
Stacey Burrows schreef:
I have a data.frame, data, with 30 factor variables. I would like to tabulate
the frequencies of each variable and output to a tex file using Sweave. Here is
my code chunk:
<>=
library(xtable)
for(j in 1:30){
cap <- paste("Frequency counts for Q",j,".", sep="")
xtab
Hi List,
I am looking for an R advanced programming course in USA for April. Need to
attend the course before April 31st.
Regards - Steve
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/l
I have a data.frame, data, with 30 factor variables. I would like to tabulate
the frequencies of each variable and output to a tex file using Sweave. Here is
my code chunk:
<>=
library(xtable)
for(j in 1:30){
cap <- paste("Frequency counts for Q",j,".", sep="")
xtable(t(table(data[,j])),capt
On 2/04/2009, at 11:27 PM, Uwe Ligges wrote:
Not many of us are clairvoyants
Hey! Wish I'd said that! :-)
cheers,
Rolf
P. S. Great minds think alike.
R.
###
Ravi,
You are right. I did not specified the type of problem:
large problems with smooth functions and not large-scale discrete
problems. I am sorry for confusion. For non-smooth
functions, I can suggest subplex.
http://cran.r-project.org/web/packages/subplex/index.html
Florin
On Thu, 2 A
Hello Wesely,
The appropriate way to address irregular time series depends on what want to
use the estimates for. If your objective is to estimate the times that you
don't observe (interpolate) then a natural cubic spline is a good method to
provide such an estimate. If your objective is to obta
I think this works in general, although it's
a little bit clunky:
id_y <- array(1:10,dim=c(2,1,5))
id_yt <- aperm(id_y,c(2,1,3))
m_id <- array(dim=c(dim(id_y)[1],dim(id_y)[1],dim(id_y)[3]))
for (i in 1:dim(id_y)[3]){
m1 <- array(id_y[,,i],dim=dim(id_y)[1:2])
m2 <- array(id_yt[,,i],dim=di
Would your blood still circulate if you take
off your heart?
That is, the answer is, "No, that's where the
work is done in the function." If you want to
understand the computations that function is
doing, then you need to examine the C code
that it is calling.
Patrick Burns
patr...@burns-stat.
You need to put in calls to 'as.matrix'. It's
a bit tricky though -- what you want depends
on whether it is the first or second subscript
that has length 1.
as.matrix(id_y[,,i])
if the second dimension has length 1.
t(as.matrix(id_y[,,i]))
if the first dimension has length 1.
Patrick Burns
Can someone recommend a more sophisticated way to annotate heatmaps than the
ColSideColors argument of heatmap and heatmap.2? In particular, I would
like to be able to annotate columns with more than one piece of information,
like in Figure 1 of the article at
http://www3.interscience.wiley.com/cg
On 4/2/09, In Hee Park wrote:
> Thanks for your comment. Beside axis location, I have another
> question for you.
>
> If I wanted to change the matrix column names shown on the lattice
> plot, which argument should be used to handle it? (I'd like to know
> the overall lattice plot level sche
I have a data frame that looks something like...
Column 1 is an experiment_id, Column 2 is the type of treatment ("control",
"full treatment", or "partial treatment"), and Column 3 is a value.
Experiment_id Treament_type Value
12345"control"3
12345"full treatment"4
12345"full
Thanks for your comment. Beside axis location, I have another
question for you.
If I wanted to change the matrix column names shown on the lattice
plot, which argument should be used to handle it? (I'd like to know
the overall lattice plot level scheme, for example, which argument
controls which
Hello,
I'm trying to use the function svyglm in the library survey.
I create a data survey object:
data_svy<- svydesign(id=~PSU, strata=~sample_domain,
weights=~sample_weight, data=data, nest=TRUE)
and I try to use svyglm() with little success:
R<-svyglm(data_svy[,4]~(data_svy[,iCol]==listMo
Thanks, that helped! I didn't realize there was another version of the book.
On Wed, Apr 1, 2009 at 2:51 PM, Tobias Verbeke <
tobias.verb...@openanalytics.be> wrote:
> Hi,
>
> I'm trying to follow the ggplot introduction here:
>> http://had.co.nz/ggplot/ggplot-introduction.pdf
>>
>> I've install
I tried the example in ggplot as well and rec'd the same error message.
ggpoint is not found in the ggplot2 library.
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034
steve_fried...@nps.gov
Office (3
or the exactly equivalent form:
x[x$X1 %in% unique(x[x$X2>0,"X1"]), ]
Patrizio
2009/4/2 Nutter, Benjamin :
> x <-
> data.frame(matrix(c(rep(11,4),rep(12,3),rep(13,3),rep(0,3),1,rep(0,4),re
> p(1,2)),ncol=2))
>
> id.keep <- unique(subset(x,X2>0)$X1)
>
> x2 <- subset(x,X1 %in% id.keep)
>
> x2
>
>
Hello all,
I am looking at the one program of the marketing textbook. I understand what
the following function is doing especially the .C command
Can someone help?
If I take off the .C comand will it still run??
drawwc = function(w, mu, y, sigi) {
.C("draww", w = as.double(w), as.double
Hello,
Is there a function in r to find the best fitting model for a set of data?
I would like to know if my data are related exponentially,linearly or if there
is a logarithmic correlation between my x and y values.
To get a better imagination I've added the graphics at the end of this mail
as an
Harald von Waldow wrote:
>
> For the purpose of model selection I am looking for a way to
> exhaustively (and efficiently) search for best subsets of predictor
> variables for a logistic regression model.
>
Of all the dangerous ways of doing this and getting confusing results, gl1ce
in lasso2
On 3/29/09, In Hee Park wrote:
> Dear R users:
>
>
> I am having difficulty to place x-axis location alternatively
> top/bottom side in Lattice plot, which is composed of seven-column as
> following:
>
> E1 E2 E3 E4 E5 E6 E7
> Case1 -505.85 -75.32
On Thu, Apr 2, 2009 at 3:49 PM, wrote:
> Sorry I sent a description of the function I was trying to minimize but I
> must not have sent it to this group (and you). Hopefully with this clearer
> description of my problem you might have some suggestions.
>
> It is basically a warehouse placement
Hi again,
I understood what you guys explained...
But, there isn't a way to do a multiplication of matrix with a FOR command
or otherelse where one o my dimension is ONE...
Well, as my data file is small, I did the procedure at the excel... But,
this is not the good procedure...
Thanks,
Marcio
U
Florin,
Please allow me to clarify some issues:
Many, if not most, problems in science involve optimization in one form or
another. Consequently, "optimization" is a vast area. There are many
different types of optimization. Here is a one way to classify optimzation
problems (neither mutually
Just a quick thought: if you have the locations
of the items along a line segment (the "warehouse"),
can you find a way to formulate the problem so that
the average distances represent a matrix calculation
(i.e., multiply some huge matrix by the locations)
then it's a linear problem ...
It's t
x <-
data.frame(matrix(c(rep(11,4),rep(12,3),rep(13,3),rep(0,3),1,rep(0,4),re
p(1,2)),ncol=2))
id.keep <- unique(subset(x,X2>0)$X1)
x2 <- subset(x,X1 %in% id.keep)
x2
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Stephan Lindner
I tried many optimizers in R on my large scale optimization problems.
I am not satisfied with their speed on large op problems. But you may
try in this order
nlminb
ucminfucminf package
spq BB package
optim
Is here someone that try to port Ipopt in R?
https://projects.co
Hi all,
I was wondering how to construct a seasonal differenced time series
variable.
I used the following code to construct a 12 span seasonal difference
seasonal<-diff(V2, lag=12, differences=1)
is this correct?
thank you in advance
joe
[[alternative HTML version deleted]]
__
Dear all,
I'm trying to select from a dataframe all rows which correspond to a
factor (the id variable) for which there exists at least one positive
value of a certain variable. As an example:
x <-
data.frame(matrix(c(rep(11,4),rep(12,3),rep(13,3),rep(0,3),1,rep(0,4),rep(1,2)),ncol=2))
> x
Dear R-users,
For the purpose of model selection I am looking for a way to
exhaustively (and efficiently) search for best subsets of predictor
variables for a logistic regression model.
I am looking for something like leaps() but that works with glm.
Any feedback highly appreciated.
--
Harald
Dear List-Members,
does anyone know a simple way to automatically install all R packages
on a unix system to the default library path? Not from inside R, it
should rather work as a shell script - job at startup.
Something like R cmd install -l pkgs ### where pkgs should mean all
packages
Dear friends,
this time I have a problem with using waba function. Firstly, I'll explain you
my situation.
In the survey a gruop of supervisors judge the dipendents of a company.
One supervisor reported on more than one subordinate.
Thus, I need to show that lack of independence is not a problem
Thanks for your mail. I guess that the constant row sum on X would create
problems in a simulation framework because you might end up with linearly
dependent columns or even with columns of zeros (which I believe do not make
much sense).
First of all, I think there is a problem with your examp
Sorry I sent a description of the function I was trying to minimize but I must
not have sent it to this group (and you). Hopefully with this clearer
description of my problem you might have some suggestions.
It is basically a warehouse placement problem. You have a warehouse that has
many item
Did you load the package after installing it?
install.packages("ggplot2")
library(ggplot2)
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
--- On Wed, 4/1/09, haettulegur wrote:
> From: haettulegur
> Subject: [R]
While not wholly directed towards OO programming, 'R Programming for
Bioinformatics' does cover both the S3 and S4 systems (and how to make
them play nicely together) at an accessible level.
http://www.bioconductor.org/pub/RBioinf/
Best,
Jim
Krzysztof Sakrejda-Leavitt wrote:
Tom Backer Jo
Tom Backer Johnsen wrote:
This is interesting. The fact that there are so few texts on the
subject probably means that very few are using OO programming methods in R.
That's overstating the problem. I frequently run into OO R packages and
many of them are very nicely done. Some good example
Your R version seems to be out of date. This would happen in older
versions of R, prior to 2.8.0 I believe, due to a bug in socketSelect
that was fixed last August.
luke
On Wed, 1 Apr 2009, Ubuntu Diego wrote:
I'm trying to use snow in my dual-core (hopefully later this is going to
run in a c
Wacek Kusnierczyk wrote:
Tom Backer Johnsen wrote:
This is interesting. The fact that there are so few texts on the
subject probably means that very few are using OO programming methods
in R.
I will probably look in the Python direction. Pity. I would have
preferred R.
if you insist on oo
Tom Backer Johnsen writes:
> This is interesting. The fact that there are so few texts on the
> subject probably means that very few are using OO programming methods
> in R.
Using python is great; your deduction about use of OO in R harder to
support. The Bioconductor project (300+ packages, no
Tom Backer Johnsen wrote:
> This is interesting. The fact that there are so few texts on the
> subject probably means that very few are using OO programming methods
> in R.
>
> I will probably look in the Python direction. Pity. I would have
> preferred R.
>
if you insist on oo, python might be
I have a data.frame y as below, and I want to calculate the "weighted modal
value", where v... are the values, and w... are the weights.
At tha moment I am doing it as shown below, but my data.frame has 165900 and
more rows, and it takes ages to do the calculations. I am sure there is a
much better
This is interesting. The fact that there are so few texts on the
subject probably means that very few are using OO programming methods in
R.
I will probably look in the Python direction. Pity. I would have
preferred R.
Tom
krzysztof.sakre...@gmail.com wrote:
I have been using Chambers f
May be:
plot(c(-1, 1) , c(-1, 1), type = "n")
n = 4
a = rep(0, n)
b = 1:n/n
fun = function(i, a, b, col = 1 , ...) {
abline(a[i], b[i], col = col[i], ...)
}
lapply(1:n, fun, a=a, b=b, col = 1:n)
Andrea
Thomas Levine wrote:
I really want to do this:
abline(
a=tan(-kT*pi/180),
b=kY-tan(-
Hello
This may have been answered elsewhere, and I have looked on the web, but
nothing helps. I am trying to do the following:
X<-matrix(c(1:15),nrow=3,byrow=T)
Y<-matrix(c(2,4,6,8,10),ncol=1)
I need to sum the product of each row of X by the remaining j rows multiplied
by j y values (i.e s
What you are trying to do is make R behave like a macro language, which is not
a good idea at all. Better to use R as it was designed to be used rather than
make it behave like something it wasn't.
It also follows that trying to do this is going to be tricky. Here is one way:
RENAME <- functi
Dear R Users:
Greetings!
Finally, I got what I required...! Thanks a lot for all of your valuable
support, especially to Jim and Duncan Murdoch. Great help guys...
By the way, herewith I am furnishing the code that I used for adjusting the
axis interval unit... Hope it will be useful to others, a
Ioannis,
Here's an illustrative example. Note that: glm also objects to X4; X1,..,X4
are defined as factors.
I've looked (albeit in a crude way) at various examples using the perturb
package and it seems to confirm that X4 is the source of multicollinearity.
As I say, I think the constant row-su
Mark, Ted, Gabor,
Thanks for all your input.
José
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: 01 April 2009 18:12
To: Jose Iparraguirre D'Elia
Cc: r-help@r-project.org
Subject: Re: [R] A query about na.omit
First input the data frame:
> Lines <- "
Try this where we read in a zoo series and then merge it with
a zero width regularly spaced series to create the result.
Lines <- "V1,V2
2008-10-14 08:45:00,94411.08
2008-10-14 08:50:00,90745.45
2008-10-14 08:55:00,82963.35
2008-10-14 09:00:00,75684.38
2008-10-14 09:05:00,78931.82
2008-10-14 09:20
I am running on 64-bit Ubuntu, R version 2.8.1
If I do anything Tcl/Tk related like:
> library( Rcmdr )
or
> available.packages()
I get the error:
Error in structure(.External("dotTclObjv", objv, PACKAGE="tcltk"), class =
"tclObj") :
[tcl] unknown color name "red"
--
Example:
see_context <- function(word, data, context) {
dat <- data$strings
temp <- grep(paste("^", word, "$", sep=""), dat)
if(length(temp))
index <- lapply(temp, function(x) seq(max(0, x - context),
min(x + context, length(dat
sapply(index, function(x) paste(dat[x]
>> rename(x,"C","Z")
> x <- rename(x,"C","Z")
> x
A B Z D
1 2 3 4
--
Regards,
Hans-Peter
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
an
j.k wrote:
Hello alltogheter,
I have the following problem and maybe someone can help me with it.
I have a list of values with times. They look like that:
V1 V2
1 2008-10-14 08:45:00 94411.08
2 2008-10-14 08:50:00 90745.45
3 2008-10-14 08:55:00 82963.35
RJCallahan wrote:
Hi all,
I have a (hopefully quick) question. I've got a fascinating set of fitted
surfaces in three dimensions corresponding to local linear multiple
regressions. I'd like to add rugs to the X and Y axes (corresponding to my
independent variables) in order to get a sense for
Tammy Ma wrote:
Hi, All.
I have scatterplot in 2-D
I'd like to see which proportion of data goes into a range. How shall I figure
it out?
Perhaps some accumulative graph or something?
See ?hist and it argument breaks as well as ?cut
Uwe Ligges
Kind regards,
Tammy
___
MarcioRibeiro wrote:
Hi listers,
I am having some trouble in a matrix multiplication...
I have already checked some posts, but I didn't find my problem...
I have the following code...
But I am not getting the right multiplication...
I checked the dimension and they are fine...
id_y <- array(1:
The forecast package has a number of forecasting methods
you could try.
On Thu, Apr 2, 2009 at 3:17 AM, minben wrote:
> I want to forecaste the call number everyday for a call-center. Now I
> have removed the influence of the fluctuation with some method, so
> only thing left is to analyze the tr
You have so many strange special characters (including different quotes
at strange places) below, that it is impossible to find out anything.
If you are using the same editor you are writing your e-mails with, I am
sure you will have diverse problems with your code
Uwe Ligges
markle..
Hi
r-help-boun...@r-project.org napsal dne 02.04.2009 11:14:34:
> HI,
> I need help..
> How to sort and group the data below:-
> USER NAME
> 12 admin
> 12 admin
> 10 admin
> 10 advertising
> 61 process
> 17snapshot
> 61ticket
> 61ticket
> 30snapshot
> 10adv
I have answered almost the same question a few minutes ago.
Why do you send questions more than once? It won't make responses
quicker but will confuse people who spend their time in answering
messages who do not see that the questions is already answered in
another thread
Uwe Ligges
a) This is the *R-help* mailing list. If you want others to make your
data analyses, there are consultants around ...
b) If you still want help from the list, follow the posting guide and
provide relevant information such as some insight to your data. Not many
of us are clairvoyants and hence
Dear R users,
I have a table with say three columns and 70 rows.
The second column has dates in format dd.mm.yy (i.e. 01.10.07 indicating
record for October 1, 2007).
I read the text file, containing the records. Now, I want to find out the
weekday for each date.
However, I cannot do so.
1)
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