Try
tempFun - function(x) sum(!is.na(x))
nonZeros - aggregate(pollution[pol],format(pollution[date],%Y-%j), FUN
= tempFun)
--- On Wed, 12/8/09, Tim Chatterton tim.chatter...@uwe.ac.uk wrote:
From: Tim Chatterton tim.chatter...@uwe.ac.uk
Subject: [R] Counting the number of non-NA values per
Alternatively download the xlsReadWrite package from
http://treetron.googlepages.com/
install it an proceed as in older version of R.
--- On Tue, 11/8/09, Inchallah Yarab inchallahya...@yahoo.fr wrote:
From: Inchallah Yarab inchallahya...@yahoo.fr
Subject: [R] Re : How to Import Excel file
On Tue, 11-Aug-2009 at 02:43PM -0500, Peng Yu wrote:
| Hi,
|
| I frequently need to open multiple help pages in R, which requires the
| start of multiple R sessions. I am wondering if there is a way to
| invoke the help page from the command line just like 'man'.
If you use ESS (with Emacs) you
Hi Jill,
Completely not elegant, but may be usefull.
Of course other colleagues will solve this with 1 line command :-)
cheers
milton
df-read.table(stdin(), head=T, sep=,)
V1,V2,V3,V4
DPA1*,DPA1*,DPB1*,DPB1*
0103,0104,0401,0601
0103,0103,0301,0402
df.new-as.matrix(df)
for (i in 2:dim(df)[1])
On Wed, 12-Aug-2009 at 12:38PM +0800, Leon Yee wrote:
Hi, all
I have a question of adjusting the output of a data frame with many
columns. By default, print() will print out several columns according to
the window size, and then it scrolls down and print out left columns.
How can
Hi Hitesh,
while the rigth response not arrive, you can play with:
id-1:10
x-runif(10)
y-runif(10)
plot(x,y)
identify(x, y , labels = id)
Try give a look on spatstat help. I am not quite sure, but may be there you
can find some solution (sorry if it is not true)
cheers
milton
On Wed, Aug 12,
Hello,
Consider MCMC sampling with metropolis / metropolis hastings proposals
and a density function with a given valid parameter space. How are MCMC
proposals performed if the parameter could be located at the very
extreme of the parameter space, or even 'beyond that' ? Example to
express
Patrick Connolly wrote:
On Wed, 12-Aug-2009 at 12:38PM +0800, Leon Yee wrote:
Hi, all
I have a question of adjusting the output of a data frame with many
columns. By default, print() will print out several columns according to
the window size, and then it scrolls down and print out
On Tue, 11-Aug-2009 at 06:48PM -0700, Jill Hollenbach wrote:
|
| Hi,
| I am trying to edit a data frame such that the string in the first line is
| appended onto the beginning of each element in the subsequent rows. The data
| looks like this:
|
| df
| V1 V2 V3 V4
| 1 DPA1*
Hadley's answer is probably better,
but 'The R Inferno' page 15 is an
answer to the literal question.
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)
hadley wickham wrote:
You might also want to
see
?dput
?save
e.g.
setClass(track, representation(x=numeric, y=numeric))
x - new(track, x=1:4, y=5:8)
# save as binary
fn - tempfile()
save(x, ascii=FALSE, file=fn)
rm(x)
load(fn)
x
# save as ASCII
save(x, ascii=TRUE, file=fn)
# ASCII text representation from which to regenerate the data
Jiucang Hao wrote:
Hi All,
I did a google search and could not find the answer. Thus I post this
message. I found runmax only work for positive numbers.
x = rep(-1,10)
runmax(x,3)
x = rep(0, 10)
runmax(x,3)
for 32-bit R, i got very small numbers: 2.121996e-314
for 64-bit R, i got NaN.
Hi,
I have a csv file with different datatypes:
2009-01-01, character1, 10, 20.1
2009-01-02, character2, 11, 21.1
(I have attached the file to this post)
I read this file with read.zoo as I want a zoo/xts timeseries:
t = read.zoo(./data.txt, sep=,, dec = ., header=FALSE)
If I look at the zoo
Thomas Mang wrote:
Hello,
Consider MCMC sampling with metropolis / metropolis hastings proposals
and a density function with a given valid parameter space. How are MCMC
proposals performed if the parameter could be located at the very
extreme of the parameter space, or even 'beyond that' ?
Dear R-users,
was anybody able to install the mlica package succesfully? Or other ICA
(Independent Component Analysis) package? But not the e1071.
I have found mlica in the archive
http://lib.stat.cmu.edu/R/CRAN/src/contrib/Archive/mlica/, repacked it to
ZIP, installed and got an updating HTML
On Wed, 12 Aug 2009, Mark Breman wrote:
Hi,
I have a csv file with different datatypes:
2009-01-01, character1, 10, 20.1
2009-01-02, character2, 11, 21.1
(I have attached the file to this post)
I read this file with read.zoo as I want a zoo/xts timeseries:
t = read.zoo(./data.txt, sep=,,
I just grabbed the new EL5 binary's for my x64 VPS from CRAN. However, where
updates usually go smoothly, I now get these errors:
-bash-3.2# rpm -i R-2.9.1-1.el5.x86_64.rpm
error: Failed dependencies:
R-devel = 2.9.1-1.el5 is needed by R-2.9.1-1.el5.x86_64
-bash-3.2# rpm -i
Dear List,
Does anyone know how to simulate data from a GLM object correponding
to values of the independent (x) variable that do not occur in the
original dataset?
I have tried using simulate(), but it generates a new value of the
dependent variable corresponding to each of the original
Hello people,
A while back I wanted to plot boxplots with interactions that will have a
dot for the mean of the sample + bars for the SE.
After searching for some code, I found something that did it for one level,
but couldn't find something that will allow for interactions the way the
original
try this , save your data in C:/
and write this
(data - read.csv2(c:/jill.csv, sep=,))
(C-data[1,])
A - numeric(4)
B - numeric(4)
for (i in 1 :4){
A[i] - paste(data[1,i],data[2,i])
B[i] - paste(data[1,i],data[3,i])
}
A
B
(data1 - rbind(as.character(A),as.character(B)))
(data2 - rbind(C,data1))
On Wed, 2009-08-12 at 02:28 -0700, Jeroen Ooms wrote:
I just grabbed the new EL5 binary's for my x64 VPS from CRAN. However, where
updates usually go smoothly, I now get these errors:
-bash-3.2# rpm -i R-2.9.1-1.el5.x86_64.rpm
error: Failed dependencies:
R-devel = 2.9.1-1.el5 is
On Wed, 2009-08-12 at 09:52 +, Inchallah Yarab wrote:
try this , save your data in C:/
and write this
(data - read.csv2(c:/jill.csv, sep=,))
Why use read.csv2 yet change the separator to be the decimal point
indicator in that function. You are really asking for trouble reading
data that
On Wed, 2009-08-12 at 10:55 +0100, Gavin Simpson wrote:
snip /
What could be the problem?
Do you have perl(File::Copy::Recursive) or pcre-devel installed? If not,
yum install perl(File::Copy::Recursive) pcre-devel
as route in a terminal will install those packages and their
^^
why you don't use read.csv2 (you save your file.csv) and you write
read.csv2(path file,sep=,)
hope this helps
De : Mark Breman breman.m...@gmail.com
à : r-h...@stat.math.ethz.ch
Envoyé le : Mercredi, 12 Août 2009, 10h46mn 43s
Objet : [R] Zoo and numeric data
Hello,
I'd like to normalize a data frame which are 45000 obs. of 212 variables.I
want to normalize the data frame by each row. But I failed.Here is my R code,
can you tell me where am I wrong? bb-function(normalize)
(dat[i,]-min(dat[i,]))/(max(dat[i,])-min(dat[i,])) dat1-apply(dat,1,bb)
This can be done much easier and transparent with ggplot2
library(ggplot2)
ggplot(mtcars, aes(x = factor(round(wt)), y = mpg, colour = factor(am)))
+ geom_boxplot() + geom_point(stat = summary, fun.y = mean, position
= position_dodge(width = 0.75))
HTH,
Thierry
Hi Gina.
I am not sure your normalization process does what you want, but just to get
the code working, here is how it can be done:
# example data
my.data - data.frame(rnorm(100,3,4), rnorm(100,3,4))
bb-function(dat)
{
(dat-min(dat))/(max(dat)-min(dat))
}
dat1-t(apply(my.data,1,bb))
Best,
Thanks Thierry,
I didn't know about that, it's always nice to see different ways of doing
things.
Question: How would you add the bars ?
Hope to see more interesting replies,
Tal
On Wed, Aug 12, 2009 at 1:29 PM, ONKELINX, Thierry thierry.onkel...@inbo.be
wrote:
This can be done much
First of all, sorry for not giving all information.
Secondly, thanks a lot. This is a real help!! I did not know, that you can
use names...
This is really simple and works great!!!
If anyone is close enough to the people writing the help in R, please tell
them that they should write a tutorial
Does anyone know of an R-function or method to compare two C-statistics
(Harrells's C - rcorr.cens) obtained from 2 different models in
partially paired datasets (i.e. some similar and some different cases),
with one continuous independent variable in each separate model? (in a
survival
Dear Sirs,
I would like to ask you, what function can I use for matrices addition?
I couldn't find any information about it in the manual or in the
internet.
(A+B suits, when the number of matrixes is small, function sum() doesn't
suit for matrices addition, because it sums all variables in the
Thanks milton,
But that does not work in my case.. Any other suggestion be appreciated.
Thanks and regards,
Hitesh Singla
Hi Hitesh,
while the rigth response not arrive, you can play with:
id-1:10
x-runif(10)
y-runif(10)
plot(x,y)
identify(x, y , labels = id)
Try give a look on
Would the predict routine (using 'newdata') do what you need?
Cliff Long
Hollister Incorporated
On Wed, Aug 12, 2009 at 4:33 AM, Jacob Nabe-Nielsenjacobn...@me.com wrote:
Dear List,
Does anyone know how to simulate data from a GLM object correponding
to values of the independent (x)
I have been struggling trying to write some code to produce all
combinations subject to some restrictions. I thought someone might have
some bright ideas.
I have 11 values which fall into 5 groups. I want all combinations of
2,3, and 4 values where each value must be from a different group.
Hello All,
Could any one of you kindly let me know , if there is an R package which has
Cointegrated Regression Durbin Watson test ?
lmtest package has a function dwtest but this is not for cointegrated
systems, which uses different critical values
Regards
Radha
[[alternative HTML
you could try something like the following:
groups - list(gp1 = 1:3, gp2 = 4:5, gp3 = 6:7,
gp4 = 8:10, gp5 = 11)
combn(5, 2, function (x) expand.grid(groups[x]), simplify = FALSE)
combn(5, 3, function (x) expand.grid(groups[x]), simplify = FALSE)
combn(5, 4, function (x)
Try this:
data$x[data$x == 6] - 66
On Tue, Aug 11, 2009 at 3:05 PM, SNN s.nan...@yahoo.com wrote:
Hi All,
this could be a simple question but I am looking into modifying a data
frame
using a condition without the need to loop over that data, would that be
possible?
I have tried the
Hanneke Wijnhoven wrote:
Does anyone know of an R-function or method to compare two C-statistics
(Harrells's C - rcorr.cens) obtained from 2 different models in
partially paired datasets (i.e. some similar and some different cases),
with one continuous independent variable in each separate
Rusyte, Lina wrote:
Dear Sirs,
I would like to ask you, what function can I use for matrices addition?
I couldn't find any information about it in the manual or in the
internet.
(A+B suits, when the number of matrixes is small, function sum() doesn't
suit for matrices addition,
Dimitris Rizopoulos wrote:
you could try something like the following:
groups - list(gp1 = 1:3, gp2 = 4:5, gp3 = 6:7,
gp4 = 8:10, gp5 = 11)
combn(5, 2, function (x) expand.grid(groups[x]), simplify = FALSE)
combn(5, 3, function (x) expand.grid(groups[x]), simplify = FALSE)
Try this , save your data in C:/
and write this
(data - read.csv2(c:/jill.csv, sep=,))
(C-data[1,])
A - numeric(4)
B - numeric(4)
for (i in 1 :4){
A[i] - paste(data[1,i],data[2,i])
B[i] - paste(data[1,i],data[3,i])
}
A
B
(data1 - rbind(as.character(A),as.character(B)))
(data2 -
Thanks very much Deepayan and Duncan. Both of your suggestions are very
helpful and should do what I need.
Frank
Deepayan Sarkar wrote:
On Tue, Aug 11, 2009 at 6:30 AM, Frank E Harrell
Jrf.harr...@vanderbilt.edu wrote:
Dear Group:
I want to use lattice with a formula such as y ~ x | v to
See zoo-faq number 4:
vignette(zoo-faq)
On Wed, Aug 12, 2009 at 4:46 AM, Mark Bremanbreman.m...@gmail.com wrote:
Hi,
I have a csv file with different datatypes:
2009-01-01, character1, 10, 20.1
2009-01-02, character2, 11, 21.1
(I have attached the file to this post)
I read this file
Two solutions with errorbars
library(ggplot2)
#option 1
lowCI - function(x){mean(x) + qt(0.025, length(x) - 1) * sd(x) /
sqrt(length(x))}
highCI - function(x){mean(x) + qt(0.975, length(x) - 1) * sd(x) /
sqrt(length(x))}
ggplot(mtcars, aes(x = factor(round(wt)), y = mpg, colour = factor(am)))
+
It would appear that if you were extracting the values at the indices
in 'res' that you should get: 2,2,2
x - list(c(1,2,3,4), c(1,2,4,3), c(1,4,2,3))
res - c(2,2,3)
x
[[1]]
[1] 1 2 3 4
[[2]]
[1] 1 2 4 3
[[3]]
[1] 1 4 2 3
?mapply
mapply(function(a,b)a[b], x, res)
[1] 2 2 2
On Tue,
Hallo,
I'm dynamically generating buttons depending on the number of rows of
my dataframe. Every button is supposed to call a function which
generates a plot with the values of one of my dataframe rows.
My code looks like this:
base - tktoplevel()
plotten - function(mat, namen, titel) {
On Wed, 2009-08-12 at 10:03 +, Inchallah Yarab wrote:
why you don't use read.csv2 (you save your file.csv) and you write
read.csv2(path file,sep=,)
No you don't!!! Please understand what read.csv2 is for. It is for
locales where the , is used as the decimal point, e.g. 5,2323 ==
5.2323. As
Dear Nancy,
Try also:
x - c(4,5,6,6,8)
y - c(a,b,b,b,c)
mydata - data.frame(x,y)
mydata
mydata$x - with(mydata, ifelse(x == 6, 66, x))
mydata
See ?ifelse for more details. BTW, be careful: data is an R reserved name
:-)
HTH,
Jorge
On Tue, Aug 11, 2009 at 2:05 PM, SNN wrote:
Hi All,
All of the data frames in the list (list of 74) have the same column
names, and some have the exact same index values, while others have
out of sync, but in the same range. I can send a small example off
list. I will look at it again when I get into work this morning to
see if I can fix it.
Hi,
Can you please tell me what does the the function logged2 in R do im list or..?
As I have
?logged2
No documentation for 'logged2' in specified packages and libraries:
you could try '??logged2'
??logged2
No help files found with alias or concept or title matching âlogged2â
using fuzzy
Hi,
On Aug 12, 2009, at 9:26 AM, amor Gandhi wrote:
Hi,
Can you please tell me what does the the function logged2 in R do im
list or..? As I have
?logged2
No documentation for 'logged2' in specified packages and libraries:
you could try '??logged2'
??logged2
No help files found with
Hitesh Singla wrote:
Thanks milton,
But that does not work in my case.. Any other suggestion be appreciated.
You can't do what you want with standard graphics or
grid/lattice/ggplot2, as far as I know. But there are various other
graphics systems in contributed packages that can
(I think) I'd like to use the hmm.discnp package for a simple discrete,
two-state HMM, but my training data is irregularly shaped (i.e. the
observation chains are of varying length). Additionally, I do not see how
to label the state of the observations given to the hmm() function.
Ultimately, I'd
Untested, but try something like this:
for(i in 1:(reihen-1)) {
but - tkbutton(base, text = classi[i], command = local({
anzeige - data.matrix(dataframe[i,-c(spalten)])
namen - names(anzeige)
tit - paste(classi[i],
Thank you, the x-axis includes zero, but even, when I try to plot it not
including zero, the y-axis keeps
disappearing. One can't even add the axis afterwards!
Here is an example:
library(rmeta)
set.seed(123)
# simulated data:
theta - 0.5
sds - runif(50,0.8,8)
b - rnorm(50,0,0.3)
Dear All,
I'm trying to plot a histogram (with the relative frequencies as the Y axis),
But the scale of the y axis is given by
0e+00, 1e-04, 2e-04, 3e-04,.
Now, I have 2 questions
1- Does (1e-04=0.01831563)?
2- If this true,how can i change the given scale to (0.01,0.03,0.05,0.07,0.09)?
Hello,
I have a dataframe containing various time series (not time series objects
though!)with hourly time steps. I´d like to perform ccf for I need to know the
correlation factors for different lags.
Here is an example:
x-as.POSIXct(c(2008-12-25 16:00:00, 2008-12-25 17:00:00, 2008-12-25
I sometimes use the View() (note the capital V) function to view long/wide
data.frames.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Leon Yee
Sent: Wednesday, August 12, 2009 2:12 AM
To: Patrick Connolly
Cc:
Thank you Gerrit,
Your suggestion proved right after all:
This does put error bars (dy) on the points. You only need to figure out
the proper y axis scaling.
xyplot(y~x|f,data=xy,groups=dy,panel=function(x,y,groups,subscripts,...)
{panel.xyplot(x,y,...)
Hello All,
I am trying to write a function which would operate on columns of a
dataframe specified in parameters passed to that function.
f = function(dataf, col1 = column1, col2 = column2) {
dataf$col1 = dataf$col2 # just as an example
}
The above, of course, does not work as
Thanks, works just fine! Great!
--
Anne Skoeries
Am 12.08.2009 um 16:04 schrieb jim holtman:
Untested, but try something like this:
for(i in 1:(reihen-1)) {
but - tkbutton(base, text = classi[i], command = local({
anzeige - data.matrix(dataframe[i,-c(spalten)])
you could stick everything in a 1-liner, but that would make it less
readable:
myf - function(x){
tmp - as.character(x)
c(tmp[1], paste(tmp[1], tmp[-1], sep=))
}
df2 - as.data.frame(sapply(df, myf))
b
On Aug 12, 2009, at 3:39 AM, milton ruser wrote:
Hi Jill,
Completely not elegant,
I think ONE answer to what you actually want to do might be
f - function(dataf, col1 = column1, col2 = column2) {
dataf[[col1]] - dataf[[col2]] # just as an example
dataf
}
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Thank you Gavin very much for this explication!!!
inchallah yarab
De : Gavin Simpson gavin.simp...@ucl.ac.uk
Cc : r-help@r-project.org
Envoyé le : Mercredi, 12 Août 2009, 14h50mn 37s
Objet : Re: [R] Re : Zoo and numeric data
On Wed, 2009-08-12 at 10:03
Hi All,
I am trying to lable the y-axis on my scatterplot with the following:
Soil moisture content (m3m-3)
I am using the following coding for plotting the graph:
plot(soilmoisture~gradientlevel, xlab=Levels of droughting gradient,
ylab=Soil moisture content (m3m-3), bty=l, font.main=2,
That's on the to do list :(
Hadley
On Tue, Aug 11, 2009 at 10:00 PM, Ben Bolkerbol...@ufl.edu wrote:
Thanks. I can get it to work for facet_grid (which will do for my current
purposes) but am curious about whether there's a way to do the same
for facet_wrap (which doesn't have a labeller
Hi Andreas, thank you for replying so soon. You are right, your code works
perfectly for my example. Unfortunately, the real data contains inner NAs,
which I forgot to include. Sorry about that. Any other suggestions?
Thanks,
Katharina
Original-Nachricht
Datum: Wed, 12 Aug
Returning the changed value as in Erik's answer is probably the
most common and R-like solution but here are two others. The
assign/get approach is perhaps the closest to what you are asking
for.
# replacement function approach
# replacement function - rhs is formula whose response is assigned
Hi Ferdogan,
Sorry for the late response.
On Thu, Jul 23, 2009 at 8:29 AM, Ferdoganferdog...@hotmail.com wrote:
I have two products which are substitudes. I try to fix a system as below to
mydata.
Demand1 = A1 -B1*Price1 + C1*Price2
Demand2 = A2 +B2*Price1 - C2*Price2
I would expect C1
Dear Sirs,
Â
I would like to ask you, what function can I use for matrices addition? I
couldnât find any information about it in the manual or in the internet.
(A+B suits, when the number of matrixes is small, function sum() doesnât suit
for matrices addition, because it sums all variables
Without a short reproducible example your question is difficult to
understand, and difficult to give a useful answer.
See comments below.
-Don
At 7:13 AM -0700 8/12/09, maram salem wrote:
Dear All,
I'm trying to plot a histogram (with the relative frequencies as the
Y axis), But the scale of
Dear R-users
I want to generate data for a logistic regression for an epidemiological
simulation.
First, I created a disease-vector containing a 1 if a subject is a
cases (i.e. has the disease) and a 0 if a subject is a control. E.g.:
disease - as.factor( c(rep(1, n.cases), rep(0,
Let's start with something simple and relatively easy to understand,
since you're new to this.
First, here's an example of the core of the idea:
paste('a',1:4)
[1] a 1 a 2 a 3 a 4
Make it a little closer to your situation:
paste('a*',1:4, sep='')
[1] a*1 a*2 a*3 a*4
Sometimes it helps
Frank,
Thank you for your quick response!
I want to compare the discriminative capacity of different
anthropometric measures in predicting mortality, focussing on the thin
site of these measures.
Since these associations are not linear (U shaped for BMI and inversily
J-shaped for mid-upper
Your running into the pretty common factor vs character problem in R.
By default data.frame turns character vectors into factor (sort of
like ENUM in mysql) vectors. Since you only have 1 factor (empty
string '') in your starting dataframe, when you go to insert new data
R sees a new value and
I am calculating some values that I am inserting into a data frame. From
what I have read, creating the dataframe ahead of time is more efficient,
since rbind (so far the only solution I have found to appending to a data
frame) is not very fast.
What I am doing is the following:
# create data
Hi Sarah,
expression works well, but you have to use it as function.
plot(1,ylab=expression(Soil moisture content
~~bgroup((,m^3*m^{-3},
see
?plotmath
for further examples
hth.
Buckmaster, Sarah schrieb:
Hi All,
I am trying to lable the y-axis on my scatterplot with the following:
Hi you can use newvariable=as.numeric(variablename). This converts your
factors into numeric variables, but not always with the desired result. So
make sure that you check whether newvariable gives you what you want.
Otherwise recoding by hand is indicated.
Best,
Daniel
Noah Silverman-3
It's generally safer to use
as.numeric(as.character(variablename))
since it eliminates problems associated with factors.
- Phil Spector
Statistical Computing Facility
oops, there was a little problem with an automated text replacement in
my last post. m^3 should be read as m ^ 3 (without white space).
or, using curly brackets for safety, try
plot(1,ylab=expression(Soil moisture
content~~bgroup((,m^{3}*m^{-3},
hth.
Buckmaster, Sarah schrieb:
Thanks
I'm trying to annotate a density plot and I'm using bquote to paste the sigma
symbol next
to the numeric text of the standard deviation calculation that I am performing.
I have been able to successfully turn the sigma symbol and numeric output the
color blue,
but when I try to change the font of
Hi,
I have a ts object called data.gas that has three economic variables:
gasoline consumption (gas), price per gallon (price), and household
income (income) by month from January 1970 to December 2008. I want to
plot this gas and price variables using ggplot2, but ggplot2 will not
allow an
time(obj) gives the times of a ts object, obj.
On Wed, Aug 12, 2009 at 12:51 PM, Data Analytics
Corp.dataanalyt...@earthlink.net wrote:
Hi,
I have a ts object called data.gas that has three economic variables:
gasoline consumption (gas), price per gallon (price), and household income
Thanks so much everybody, this has been incredibly helpful--not only is my
immediate issue solved but I've learned a lot in the process. The lapply
solution is best for me, as I need flexibility to edit df's with varying
numbers of columns.
Now, one more question: after appending the string
Hi,
I have a text (.dat) file, in which each row contains several long numeric
strings. One of the strings is 38 digits long, for example:
03200801200801172008011720092904008901
When I read in the data file, this string shows up as 3.200801e+36. To get
rid of the scientific notation, I used
Use colClasses argument in read.table to set the class of column:
For a file with two columns, where the first is string and the other is
numeric:
read.table('your_file.dat', colClasses = c('character', 'numeric'))
On Wed, Aug 12, 2009 at 1:43 PM, Andrew C acarr...@gmail.com wrote:
Hi,
I
The illustration you show is for the so-called traditional or historical
counties of England, which may be available somewhere. There are
non-georeferenced PNG files on Wikipedia, which might be used, but as far as
I can see, only UK-based academics can register for access to the edina UK
borders
I have some data fit to a smooth.spline object as follows: (x=vector of data
for the predictor variable, y=vector of data for the response variable)
fit - smooth.spline(x,y)
Now, given a spline fit point y_new, I want to be able to find out what
value of x_new yielded this fit value. How to do
You need to read up about finite precision arithmetic and floating point
representation. In brief, note that 2^64 requires 20 decimal digits, and
some bits in double precision must be given up to sign, exponent, etc.
leaving 53 bits for precision = 16 decimal digits. This is exactly the
number of
From ?plotmath, it looks like when using expressions you set the font
inside the expression (e.g. bold(x)). It looks you tried this already
but I wonder if there was something tiny out of place since the
following works for me:
Hi,
Thanks for the tip,
Neither method works as my data is truly nominal
-
foo - c(blue, red, green)
as.numeric(foo)
[1] NA NA NA
Warning message:
NAs introduced by coercion
Rapid Miner has a function that will
another approach may be:
foo - c(blue, red, green)
s.foo - sample(foo, 20, TRUE)
s.foo
match(s.foo, foo)
I hope it helps.
Best,
Dimitris
Noah Silverman wrote:
Hi,
Thanks for the tip,
Neither method works as my data is truly nominal
-
foo - c(blue,
Hi everybody - this is an oddball question.
I wonder if anybody has programmed any games in R, such as Sudoku,
Tic-Tac-Toe and the like. Or even a flight simulator...
R mateys! Let's make some t-tests!
Regards, David
__
R-help@r-project.org
I think ncdf is broken now, and so is Rnetcdf. I can't build from source,
either. If I find a solution, I'll post it here. FYI, I have an
intel-based Mac running the latest OS and with R 2.9.1
--
View this message in context:
Scott,
Your suggestion works great for changing the numbers to bold font, but is it
possible to change the sigma symbol and the equals sign to bold font as well?
I've poked around ?plotmath and am I right in saying that there is a different
method for controlling symbol fonts?
Thank you for
Hello.
shapefiles in geographic coordinates for Epi Info
http://www.cdc.gov/epiinfo/europe.htm second-level at
1998, free available but not public domain
Regards.
--- Roger Bivand roger.biv...@nhh.no wrote:
The illustration you show is for the so-called
traditional or historical
I am trying to install and use the biOps package on my x86_64 GNU/Linux
ssytem.
I have installed the fftw-3.2.2.tar.gz
and
biOps_0.2.1.tar.gz
but when I start R and try to load the library,
I get this message:
##
library(biOps)
Error in dyn.load(file,
Hello,
I'm working with a package that uses download.file in functions to
extract information from remote databases. My current environment is
Windows XP Pro SP3, R 2.7. A full extraction can be a great deal of
data, so the download is accomplished in generally manageable packets,
such that a
Hi,
On Aug 12, 2009, at 2:23 PM, Dan Kelley wrote:
I think ncdf is broken now, and so is Rnetcdf. I can't build from
source,
either. If I find a solution, I'll post it here. FYI, I have an
intel-based Mac running the latest OS and with R 2.9.1
While I haven't compiled it myself, it
Hi,
There's a couple of games listed on crantastic: http://crantastic.org/tags/games
-Bjorn
2009/8/12 David Croll david.cr...@gmx.ch:
Hi everybody - this is an oddball question.
I wonder if anybody has programmed any games in R, such as Sudoku,
Tic-Tac-Toe and the like. Or even a flight
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