Hi,
Is there any function available to do stepwise selection of variables in
Conditional(matched) logistic regression( clogit)? step, stepwise etc are
failing in case of conditional logistic regression. Please help.
Thanks
P.T. Subha
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On 02/17/2012 06:42 AM, Jordan Patricia Sinclair wrote:
Hello all.
I need to graph multiple lines of different lengths on the same graph. When I
try to add lines I get an error due to different lengths. The only thing I
could find when reading up on it was that 'if the data are inputted separ
Hi Alex,
I cannot say how to implement such constraints with fPortfolio, but in
general you can use heuristics to solve such problems. An example for
selecting a number of assets from a larger universe is given in a
vignette of the NMOF package (of which I am the author) and in the code
exam
Hi
My data looks like this
data is a vector
data=var1 var2 var3
100 120 130
i want to put it in an XML
xmlOutput=NULL
xmlOutput<- newXMLNode("results")
for( i in 1 : length(data))
{
newXMLNode("variable",attrs=c(name =names(data)[i] ), value =
data[i]), parent =
On Thu, 16-Feb-2012 at 07:30AM +, Prof Brian Ripley wrote:
|> Try read.S in package foreign. It was written back in the days of
|> that S-PLUS format.
Great. Knows all about big-endian and does exactly what I needed very
simply. All that time I've known about the foreign package and never
On 17/02/2012 05:39, R. Michael Weylandt wrote:
Looks like its in the CRAN archives here --
http://cran.r-project.org/src/contrib/Archive/rimage/ -- then the
steps for doing a source install are well documented (and easily
googleable) but (unstated) OS specific.
Yes, but the package was archive
On 02/16/2012 10:39 PM, Hey, Kirsty wrote:
Dear all,
I am wanting to plot points that have a defined absolute size relative
to the scale of the axis.
For example, the following gives me a plot with sizes relative to the
defined "size" variable but are not of the specified absolute size
(relative
Hi all,
I am fairly new to mixed effects models and lmer, so bear with me.
Here is a subset of my data, which includes a binary variable (lake (TOM or
JAN)), one other fixed factor (Age) and a random factor (Year).
lake FishID Age Increment Year
1 TOM 1 1 0.304 2007
2 TOM
On 17-02-2012, at 05:58, chuck.01 wrote:
> Hello,
> I have recently had issues with read.csv where i get the following warning,
> and this happens on both my OSX and Linux machines. Here is the warning
> and an example CSV file is attached:
>
> Warning message:
> In read.table(file = file, he
Looks like its in the CRAN archives here --
http://cran.r-project.org/src/contrib/Archive/rimage/ -- then the
steps for doing a source install are well documented (and easily
googleable) but (unstated) OS specific.
Michael
On Thu, Feb 16, 2012 at 5:41 PM, chuan_zl wrote:
> Dear all:
>
> Kindly t
Simply:
The R2 value we obtain at an optimized lambda using glmnet: how do we state
whether that's significant or not?
Using the standard lm() function, we are able to run an ANOVA and test for
significance. We have no such output with glmnet.
Thanks!
--
View this message in context:
http://r.
A kind respondent mentioned that I gave no data for this data set. Here is a
subset in CSV format:
"","levels","m101","m10"
"1",0.15,0.00166736248903737,0.00525782746279257
"2",0.16,0.000671421429082792,0.00263263808833129
"3",0.17,0.00024891450838,0.00125478603676572
"4",0.18,8.627005347
Hello,
I have recently had issues with read.csv where i get the following warning,
and this happens on both my OSX and Linux machines. Here is the warning
and an example CSV file is attached:
Warning message:
In read.table(file = file, header = header, sep = sep, quote = quote, :
incomplete
Dear all:
Kindly to ask from R users about the package "rimage" is not available in R
version 2.14.1. May I know how I can install this package? Thanks al ot
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Sent from the R help ma
What are you expecting to get?
At the moment it appears that you are just ploting one data vector. table() is
giving you a one row table.
BTW you probably should not use data as a data.frame name. It is a reserved
word.
John Kane
Kingston ON Canada
> -Original Message-
> From: stude
Hello,
I have a question on how to tabulate the time series data. I use
RStudio, but if can be done in any other R editor, it should work in
RStudio as well.
> a1<-11:22
> a1ts<-ts(a1, frequency=4, start=c(1978,1))
> a1ts Qtr1 Qtr2 Qtr3 Qtr4
1978 11 12 13 14
1979 15 16 17
On Thu, Feb 16, 2012 at 7:31 PM, Henry wrote:
> Newbie question - mechanical engineer trying to learn R
> I've had success with plotting time series data and even made a heat map
> using R Graphs Cookbook by Mittal.
> I have a new problem - I need to align a number of time series data
> "columns"
Thank you very much for your reply. The command "window" does help.
> a1<-11:22> a1ts<-ts(a1, frequency=4, start=c(1978,1))> a1tsw<-window(a1ts,
> c(1978,3), c(1979,2) )> a1tsw Qtr1 Qtr2 Qtr3 Qtr4
1978 13 14
1979 15 16 > a1tsw[2:3][1] 14 15
>
2012/2/16 R. Michae
Sir,
On 16 February 2012 16:55, Henry wrote:
> Please see my question from a few minutes ago - I wanted to improve the
> title.
xts über alles -- H
--
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Please see my question from a few minutes ago - I wanted to improve the
title.
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Newbie question - mechanical engineer trying to learn R
I've had success with plotting time series data and even made a heat map
using R Graphs Cookbook by Mittal.
I have a new problem - I need to align a number of time series data
"columns" to the desired regular exact time stamp vector. The targe
ggplot is looking for thisData as a column of coffs. the most
'ggplotesque' way of doing this would be:
# melt your data to a "long" format:
coffs.melt <- melt(coffs, id.vars = 'levels')
# plot using colour aes parameter:
ggplot(coffs.melt, aes(x=levels, y=value, colour=variable)) + geom_line()
It's been a long time since this topic was posted, but I recently had
occasion to use stats::reshape again. This time, I looked closer at the
code for the function so I could understand what was going on.
I now realize that if the argument "varying" is a vector of names (as I
had) rather than a l
Folks,
I want to automate some graphing using ggplot.
Here is my code
graphChargeOffs2<-function(coffs) {
ggplot(coffs, aes(levels))
dataNames<-names(coffs)[!names(coffs) == "levels"]
for(i in dataNames) {
thisData<-coffs[[i]]
last_plot() + geom_line(aes(y = thisData, colour = i))
On Feb 16, 2012, at 2:42 PM, Jordan Patricia Sinclair wrote:
Hello all.
I need to graph multiple lines of different lengths on the same graph.
?segments
When I try to add lines I get an error due to different lengths.
The only thing I could find when reading up on it was that 'if the
Hi Taby,
i could not work out how to add to a table, so what i did was to kill
the old one and write a new one --
here is a snippet of my code:
# delete the old table
sqlDrop(con, sqtable="__mytable__") # note this will hang if the table
cannot be found
# insert the new table
sqlSave(con
On Feb 16, 2012, at 11:43 AM, protoplast wrote:
hello mailing list!
i still consider myself an R beginner, so please bear with me if my
questions seems strange.
i'm in the field of biology, and have done consecutive hydraulic
conductivity measurements in three parallels ("Sample"), resulting
# All days in years 2006 to 2009 by month in 48 (12x4) files.
days <- seq(as.Date("2006/1/1"), as.Date("2009/12/31"),by="day") # one
long vector
out <- paste(rep(format(days,'%d%m%y'),each=2),c('aaa','bbb'),sep='_')
# reformat to style
month <- factor(rep(format(days,'%B%y'),each=2)) # group by
On Feb 16, 2012, at 6:50 AM, David Zastrau wrote:
Hello everyone,
i’ve got a problem with my diagram’s legend. I know i should be able
to figure it out by reading the ‘plot’ and ‘legend’ reference.
However nothing works so it would be kind if anyone could point me
to the necessary parame
Also look at the zapsmall function. A useful but often overlooked tool.
On Thu, Feb 16, 2012 at 2:54 AM, Petr Savicky wrote:
> On Thu, Feb 16, 2012 at 10:17:09AM +0100, Gian Maria Niccolò Benucci wrote:
>> Dear List,
>>
>> I will appreciate any advice regarding how to convert the following numbe
I didn't think through mine all the way -- you do need the cbind()
call to do the indexing like I was thinking -- so mine when corrected
just turns into David's.
Michael
On Thu, Feb 16, 2012 at 5:26 AM, nymphita
wrote:
> Hi Michael,
> Your answer was very interesting, thank you!
> However, I tri
Ah, yes. If you can't find the answer to your question, ask a different
question!
sqldf does, indeed, do what I want. Thank you
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-O
On Thu, Feb 16, 2012 at 05:32:15PM +0100, Nino Pierantonio wrote:
> Dear All,
>
> I am pretty new to R and thus my question may sound silly.
>
> Is there a way to automatically generate a series of separate vectors
> (so not arranged in a matrix), without typing and changing every time
> the va
On Feb 16, 2012, at 3:53 AM, Gundala Viswanath wrote:
Dear expert,
Given such data:
#Cutpoint SN (1-PPV)
5 0.560.01
7 0.780.19
9 0.910.58
How can I plot ROC curve with R that produce similar result like the
attached file?
I know ROCR package but it do
To whom it may concern:
Please take my off the mailing list for R related emails. Thanks.
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PLEASE do read the posting guide htt
On 16-02-2012, at 12:31, Hans W Borchers wrote:
> ACM Software Copyright and License Agreement
>
> I have often seen the use of routines from the ACM Collected Algorithms, i.e.
> (CALGO, or Trans. On Math. Software, TOMS), in Open Source
> programs, maybe also in some R packages --- and sometim
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
On Thu, Feb 16, 2012 at 12:42 PM, Jordan Patricia Sinclair
wrote:
> Hello all.
> I need to graph multiple lines of different lengths on the same graph.
VictorDelgado wrote
>
> gsa <- function(m, n, preference.row, preference.col, first)
> {
> # m: number of rows (men)
> # n: number of columns (women)
> # first 1 for row (men); and 2 for column (women)
> #
> # Two Auxiliary functions:
> # 1:
> min.n <- function(x,n,value=TRUE){
> s <- sort(x, ind
Hello all.
I need to graph multiple lines of different lengths on the same graph. When I
try to add lines I get an error due to different lengths. The only thing I
could find when reading up on it was that 'if the data are inputted separately
they must be of the same length'.
Could someone ple
I think this is what you are getting at. Hope this helps.
#index data to determine what low and high levels of tvHrs are (I said
anything over 15 hours is considered 'high')
index=which(tvHrs > 15)
#Plot the first plot, which is the 'high', only using values from crimeDvp
that are in 'index' (c
The default units of polygonGrob are "npc", I think you want "native" instead.
Try the following,
library(grid)
d = data.frame(x=rnorm(100, 10), y=rnorm(100, -100))
v = dataViewport(xData=d$x, yData=d$y)
grid.points(d$x,d$y, default.units="native", vp=v)
HTH,
b.
On 17 February 2012 02:47,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jun Shen
> Sent: Thursday, February 16, 2012 10:16 AM
> To: R-help
> Subject: [R] Wilcoxon test p value with one decimal place
>
> Dear list,
>
> Let's say I have data
>
> a=c
On 16-Feb-2012 Bert Gunter wrote:
> On Thu, Feb 16, 2012 at 10:15 AM, Jun Shen wrote:
>> Dear list,
>>
>> Let's say I have data
>>
>> _a=c(37.961,38.214,57.68)
>> _b=c(77.56,61.875,67.683)
>>
>>
>>
>> the wilcoxon test only gives me a p value with one decimal place. Is this
>> normal?
>
> No, it
On Thu, Feb 16, 2012 at 10:15 AM, Jun Shen wrote:
> Dear list,
>
> Let's say I have data
>
> a=c(37.961,38.214,57.68)
> b=c(77.56,61.875,67.683)
>
> wilcox.test(a,b)
>
> the wilcoxon test only gives me a p value with one decimal place. Is this
> normal?
No, it's discrete :-)
(Actually, that's
Hey,
You could also use (after initializing x):
x[lower.tri(x)] <- data$k
x <- t(x)
Cheers,
Tsjerk
On Feb 16, 2012 6:59 PM, "Rui Barradas" wrote:
Hello,
I'm glad it helped.
The difference in the ordering is due to the fact that R defaults to
column-first ordering.
David's solution uses row
Dear list,
Let's say I have data
a=c(37.961,38.214,57.68)
b=c(77.56,61.875,67.683)
wilcox.test(a,b)
the wilcoxon test only gives me a p value with one decimal place. Is this
normal? Thanks.
Jun
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R-he
On Thu, Feb 16, 2012 at 10:12 AM, Nutter, Benjamin wrote:
> I'm thoroughly stumped. I've been playing with RODBC and wanted to see if I
> could retrieve data from text files using this package as well (for the most
> part, this is an intellectual exercise, but occasionally I do get data files
Read the Details section in ?viewport carefully. You are treating
xscale/yscale as if they are xlim/ylim in base graphics. They are not.
It may take some trial and error on your part to figure out how
exactly this works, in general you are setting the size and location
of each polygon relative to t
Hello,
> I read a .csv file into R
>
> with the following command:
>
> A<-read.csv2(file="Mappe3.csv")
>
> It worked fine, except that I would like to get rid of the points between
> the words and get spaces instead like I have got in the .csv file.
Try
gsub('\\.', ' ', A)
(And see ?regexp
Hello,
I'm glad it helped.
The difference in the ordering is due to the fact that R defaults to
column-first ordering.
David's solution uses row-first (which is what you wanted).
Rui Barradas
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Dear R users
I'd like to hear from someone if there is a function to do a repeated k-fold
cross-validation for a lm object and get the predicted values for every
observation. The situation is as follows:
I had a data set composed by 174 observations from which I sampled randomly
a subset composed
hello mailing list!
i still consider myself an R beginner, so please bear with me if my
questions seems strange.
i'm in the field of biology, and have done consecutive hydraulic
conductivity measurements in three parallels ("Sample"), resulting in three
sets of conductivity values ("PLC" for perce
Dear All,
I am pretty new to R and thus my question may sound silly.
Is there a way to automatically generate a series of separate vectors
(so not arranged in a matrix), without typing and changing every time
the values, and store them as separate *xlsx file, where the "*" is
replaced by the
If you don't need to refer to colnames you should be fine.
If you do need to subset by colnames, you can still do it with `[`,
but you'll have trouble if you want to use the $ trick, unless you use
back-ticks. Don't worry about these things if they aren't troubling
you now though.
Michael
On Thu
David, thanks for the tips, though, I tried these arguments but it didn'
work.
Michael, thanks for your advice which worked, so far. I am using my object
to create a barplot afterwards, which automatically puts the colnames
besides the bars. Maybe this is still a good solution for my purpose as I
If you want a row for each unique item:
> example <- data.frame(id=rep(
+ ( abs(round(rnorm(50,mean=500,sd=250),digits=0)))
+ ,3), group=rep(1:15,10))
> example <-example[with(example,order(id,group)),]
>
> uniqueIDs <- do.call(rbind
+ , lapply(split(example, example$group), function(.grp){
+
Is this close to what you want:
> example <- data.frame(id=rep(
+ ( abs(round(rnorm(50,mean=500,sd=250),digits=0)))
+ ,3), group=rep(1:15,10))
> example <-example[with(example,order(id,group)),]
>
> uniqueIDs <- do.call(rbind
+ , lapply(split(example, example$group), function(.grp){
+
check.names=FALSE
but it's discouraged because it will make it hard to refer to column
names inside R.
Michael
On Thu, Feb 16, 2012 at 11:18 AM, Marion Wenty wrote:
> Dear r-helpers,
>
> I read a .csv file into R
>
> with the following command:
>
> A<-read.csv2(file="Mappe3.csv")
>
> It worked
Hello everybody!
I have again another newbie-question. I was trying to plot three curves
within one single
plot: Crime development (relative frequencies) according to the hours of tv
consume per week (high/low/all together).
Here are the data:
par(mfrow=c(1,1))
# Data input
tvHrs<-c(21,22,23,24,2
Dear r-helpers,
I read a .csv file into R
with the following command:
A<-read.csv2(file="Mappe3.csv")
It worked fine, except that I would like to get rid of the points between
the words and get spaces instead like I have got in the .csv file.
At the moment it looks like the following:
habe.m
Dear R experts,
I am having difficulty using loops productively and would like to please
ask for advice. I have a dataframe of ids and groups. I would like to
break down the dataframe into groups, find the unique sets of ids, then
reassemble. My thought was to use a loop, but I have been unable
hi,
I have a question regarding cp values in rpart(). When I use plotcp() I get
a figure with cp values on the x-axsis, but then I use printcp() the cp
values in that list are different from the values in the figure by
plotcp(). Does someone know why?
Silje
[[alternative HTML version del
Dear R-users,
How can I extract the weighting values (w-values) for each feature from a
rvm-object of the kernlab package?
I can access the relevance vectors but how can I get the weighting values of
each feature?
example code:
# create data
x1 <- rnorm(401,sd=1.10)
x2 <- rnorm(401,sd=1.10)
x3
I figured it out:
you have to pass the mar-argument to the plot function and than you may
position the legend via inset.
Have a nice day!
From: David Zastrau
Sent: Thursday, February 16, 2012 12:50 PM
To: r-help@R-project.org
Subject: Legend vanishes when placed outside the graph
Hello ever
Hello everyone,
iâve got a problem with my diagramâs legend. I know i should be able to
figure it out by reading the âplotâ and âlegendâ reference. However
nothing works so it would be kind if anyone could point me to the necessary
parameters.
The problem is that iâm moving the l
Dear all,
I am wanting to plot points that have a defined absolute size relative
to the scale of the axis.
For example, the following gives me a plot with sizes relative to the
defined "size" variable but are not of the specified absolute size
(relative to the axis)
coords <- as.data.frame(matrix
Hi again,
I just realized that in this solution there is something funny on the
position the values in the matrix, they don't really correspond to the
position indicated in the subscripts... However, David Winsemius has given a
valid solution.
Thank you for all your ideas!
--
View this message i
Problem solved.
Many many thanks for your ideas!!
(this site is very stimulant) :)
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_
Hi Rui,
Thank you very much for your idea. It works!!!
I converted my dataframe into a vector (I first removed the header and the
first and second column) and then tried your solution:
> data <- as.vector(as.matrix(read.table(file="data.txt", head=F,
> sep="\t")[-c(1,2)]))
> data
[1] 5.2 9.1 8.
Hi Michael,
Your answer was very interesting, thank you!
However, I tried it and the result was:
> df <- read.table(file="df.txt", head=T, sep="\t")
> df
i j k
1 1 2 5.2
2 1 3 9.1
3 1 4 8.0
4 1 5 2.3
5 1 6 8.4
6 2 3 6.6
7 2 4 7.4
8 2 5 7.1
9 2 6 5.5
10 3 4 4.1
11 3 5 3.9
12 3 6 9.2
1
Hi David,
What an good solution. It works perfectly and it's really simple.
(I only removed the "1+" in ncol=1+max(j), it already has 6 columns)
My result has been:
> df <- read.table(file="df.txt", head=T, sep="\t")
> df
i j k
1 1 2 5.2
2 1 3 9.1
3 1 4 8.0
4 1 5 2.3
5 1 6 8.4
6 2 3 6
Am just feeling my way into the grid library, and cannot figure out how to
define the plot limits. 3/5 of the example polygons below plot in the
default 0-1 range viewport. But when I try to redefine the viewport the
polygons plot in the same places. I also get the same result without
employing
Hi Petr,
You advice options(scipen=20) gave me the expected result and fix the
problem.
Thanks a lot!
Gian
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PLEASE do read th
I'm thoroughly stumped. I've been playing with RODBC and wanted to see if I
could retrieve data from text files using this package as well (for the most
part, this is an intellectual exercise, but occasionally I do get data files
large enough in CSV format RODBC could be helpful) .
I set up a
On Thu, 16 Feb 2012, buehlerman wrote:
Achim Zeileis-4 wrote
The reason for the various approaches is that efp() was always confined to
the linear model and gefp() then extended it to arbitrary estimating
function-based models. And for the linear model this provides the option
of treating the
On Feb 16, 2012, at 12:31 , Hans W Borchers wrote:
> ACM Software Copyright and License Agreement
>
> I have often seen the use of routines from the ACM Collected Algorithms, i.e.
> (CALGO, or Trans. On Math. Software, TOMS), in Open Source
> programs, maybe also in some R packages --- and some
Achim Zeileis-4 wrote
>
>
> The reason for the various approaches is that efp() was always confined to
> the linear model and gefp() then extended it to arbitrary estimating
> function-based models. And for the linear model this provides the option
> of treating the variance of a nuisance par
Dear Steve,
I'm very busy this morning, so will reply briefly: the error message is
self-explanatory (though the error message itself has an error in it, which
prevented the numbers from being printed as intended): you have to define
more colours -- one for each level of VegtypeID; if you want to
On 16-02-2012, at 12:31, Hans W Borchers wrote:
> ACM Software Copyright and License Agreement
>
> I have often seen the use of routines from the ACM Collected Algorithms, i.e.
> (CALGO, or Trans. On Math. Software, TOMS), in Open Source
> programs, maybe also in some R packages
I use a modifi
After one month of struggle, I lost confidence on Spotfire, it's totally
not user-friendly. Yes it is very good for certain data presentation, but
you still need do heavy lifting on data processing by other software. For
me, R is the perfect tool and has best user supporting group. I send this
emai
No I'm talking about changing the order (the first part of your
question). What you see as dates on the axis aren't actually "Date"s
(in the sense of the R class of that name) but rather factors. To R
they are just labelled categories so it just uses an arbitrary order
to plot them: if you chan
John,
I would like ellipsoids to represent the distributions relative to each of
the vegetation types.
I tried your approach thinking that it would work as well, but I'm getting
an error
scatter3d(MADep2004 ~ Dhydro2004 + MedWet2004 | VegtypeID, ellipsoid=TRUE.
data= CSSS)
Error in sprintf ("Nu
?window may help.
Michael
On Thu, Feb 16, 2012 at 3:08 AM, jpm miao wrote:
> Hello,
>
> Let us convert a vector to a time series object starting in 1978Q1:
> FRW<-ts(FRW0, frequency=4, start=c(1978,1))
> FRW[3:6] represents the data from 1978Q3 to 1979Q2. Could we access the
> data by the
Dear Steve,
It's not obvious to me what you want to do. If you want ellipsoids for
MeanAnnualDepth, Hydroperiod, and MedianWet, which sound as if they are
numeric variables, separately for each VegtypeID, which is presumably a
factor with several sets of measurements for each value, then
MeanAnnua
Hello,
I'm working with a series (30+) of hydrologic metrics and 10 vegetation
communities and I need to determine which of the metrics provide the best
separability for each of the vegetation communities. The hydrologic
metrics are highly correlated, therefore the need to reduce the number of
2012/2/15 gab
>
> Errore in scale(newdata[, object$scaled, drop = FALSE], center =
> object$x.scale$"scaled:center", :
> (subscript) indice logicol troppo lungo
>
I'm pretty sure the problem is with your data frame. Maybe if you share the
result of
dput(training[1:10, ])
# (make sure to include
On 12-02-15 11:58 PM, Ben quant wrote:
Thank you Duncan. Interesting. I find it strange that you can't get a list
of the environments. But I'll deal with it...
I'd advise thinking really carefully about this, because I think it
indicates you've got a mental model of R internals that isn't a go
Hi,
On Wednesday, February 15, 2012 08:02:44 AM Nerak wrote:
> Hi all,
>
> I was wondering if there is a function kind of similar that splits a
> dataframe, applies a function to each row and returns in a data frame. I
> know ddply but this one isn’t useful in this situation.
Why not? Sounds li
Hello Rolf,
thank you for your advice!
Though, I find there is no need to get personal about my stage of
devolopment in R.
Cheers,
Marion
2012/2/15 Rolf Turner
> On 16/02/12 03:59, Marion Wenty wrote:
>
>
>
> Thank you for your answers!
>>
>> Jeff, thanks very much for the tip with the<<-
Hi vd,
The answer to the first part of your question is reorder. To continue the
learnr example:
df.m <- transform(df.m, Period = reorder(Period, -1*value))
ggplot(df.m, aes(x = Period, y = value/1e+06, fill = Region)) +
geom_bar(stat = "identity", position = "stack")
For the second questi
Hi Mj,
did you get a solution for this? I am having the same error
Error in odbcUpdate(channel, query, mydata, coldata[m, ], test = test, :
missing columns in 'data'
I would appreciate if you can share the solution with me.
Kind regards,
Taby
From: ma
On 02/16/2012 08:08 AM, jpm miao wrote:
> FRW<-ts(FRW0, frequency=4, start=c(1978,1))
Please make your example reproducible, we do not have FRW0.
cheers,
Paul
--
Paul Hiemstra, Ph.D.
Global Climate Division
Royal Netherlands Meteorological Institute (KNMI)
Wilhelminalaan 10 | 3732 GK | De Bilt |
Hello people,
I have got a question concerning reading spss files into R:
I used the package foreign and the following command:
read.spss("C:/Eigene Dateien/myspssfile.sav",to.data.frame=T)
I have read two different files into R (several times) and always got the
following two warning messages:
ACM Software Copyright and License Agreement
I have often seen the use of routines from the ACM Collected Algorithms, i.e.
(CALGO, or Trans. On Math. Software, TOMS), in Open Source
programs, maybe also in some R packages --- and sometimes these programs are
distributed under the GPL license, som
On 02/16/2012 05:22 AM, Colin Wahl wrote:
Thank you,
Its looking like your package will work for me. I have two questions.
First, how do I rotate the plot 90 degrees so the group labels are on
the x axis and the response value on the y axis?
Second, I'm having trouble with the group labels. I n
> Hi
>
> We've found that when using parse_args(..., positional_arguments=FALSE), it
> is permissible to invoke our script with either "--myfoo=bar" or "--myfoo
> bar"; that is, whether or not the equals sign is present makes no
> difference, and in fact both usage forms are demonstrated in the opt
On Thu, Feb 16, 2012 at 10:17:09AM +0100, Gian Maria Niccolò Benucci wrote:
> Dear List,
>
> I will appreciate any advice regarding how to convert the following numbers
> [I got in return by taxondive()] in numeric integers without the e.g.
> 6.4836e+01
> abbreviations.
> Thank you very much in ad
On 16.02.2012 01:42, Duncan Murdoch wrote:
On 12-02-15 5:26 PM, wpcmba wrote:
Issue: Connecting to a computer in a network.
I would like to connect to a file on a computer on my internal network
that
has a different password. I have tried download.file and read.csv to no
avail.
When use this
On 16-02-2012, at 09:01, Petr Savicky wrote:
>
> Hi.
>
> There were several solutions in this thread. Their speed differs
> quite significantly. Here is a comparison.
>
> patrn <- 1:4
> exmpl <- sample(1:4, 1, replace=TRUE)
>
> occur1 <- function(patrn, exmpl)
> {
>m <- length(pat
On 02/16/2012 01:35 AM, agent dunham wrote:
Dear community,
I've plotted data and coloured depending on the factor variable v3.
In the legend, I'd like to assign properly the same colors than in the
factor (the factor has 5 levels).
I've been trying this but it doesn't work.
plot(var1, var2,
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