Dear R-users,
When I tried to import a text file (tab delimited) which has 2000+ rows with
the following command (With the importData in S, it works though),
x - read.table(textfile, sep= \t, skip=5, stringAsFactors=F)
I received the following warning message: Error in scan(file, what, nmax,
Hi R-experts,
I'm new to R in mle. I tried to do the following but just couldn't get it
right. Hope someone can point out the mistakes. thanks a lot.
t - c(1:90)
y -
c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,217,226,230,
If I am right informed, 'is.binary.tree' cannot test for root
polytomies.
Consider this example:
tree.hiv - read.tree(text=((rat,mouse,(human,chimp)), kangaroo);)
is.binary.tree(tree.hiv)# will yield 'FALSE'
For further questions you might be better advised to use the
Brian,
The easiest way is to create the entire timeseries and then set the
missing values to NA. The NA values will lead to the gaps you want.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en
Please do read the help page (as you were asked to do before posting).
See the 'quote' argument.
This is also covered in the 'R Data Import/Export Manual'.
On Thu, 4 Sep 2008, Weiyang Lim wrote:
Dear R-users,
When I tried to import a text file (tab delimited) which has 2000+ rows with
the
See MASS (the book) for examples.
BTW, R _is_ case sensitive.
On Wed, 3 Sep 2008, Warren Schlechte wrote:
I am using the tree package. One option is the cv.tree, which is
supposed to run cross-validations on your tree models. Is there
somewhere I can find some documentation on this
2008/9/4 Ted Byers [EMAIL PROTECTED]:
Erin,
I trust you know what you risk when you assume. ;-)
There IS a license, but it basically lets you copy or distribute it, or, in
your case, install on as many machines as you wish. It is the GNU GENERAL
PUBLIC LICENSE.
Like most open source
Not sure if you can do it from within R, but if not surely you can just
go to the website
www.r-project.org
Download and install the latest version and remove older versions if you
wish.
Robin Williams
Met Office summer intern - Health Forecasting
[EMAIL PROTECTED]
-Original
On Wed, 3 Sep 2008, Marc Schwartz wrote:
on 09/03/2008 04:56 PM Michael Friendly wrote:
I'm trying to develop some graphic methods for glm objects, but they
only apply for models
where all predictors are discrete factors. How can I test for this in a
Is an ordered factor a 'discrete
Hi,
Is there a function in R to calculate the coefficient of skewness of
some data? I had expected there to be one, but can find no information
about it.
Thanks for any pointers.
Robin Williams
Met Office summer intern - Health Forecasting
[EMAIL PROTECTED]
[[alternative HTML
Thanks, Prof Ripley.
I did read about the quote argument in the 'R Data Import/Export Manual'. But
unfortunately, I did not really understand what it means and did not adjust
that argument. But now after adjusting for it, it works for me.
Thanks.
Best Regards,
wy
-Original Message-
Dear Robin,
RSiteSearch(skewness, restrict = functions) gave me 222 hits. There
are several functions that calculate skewness.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research
I see nothing here to do with the 'bootstrap', which is sampling with
replacement.
Do you know what you mean exactly by 'randomly sample'? In general the
way to so this is to sample randomly (uniformly, whatever) and reject
samples that do not meet your restriction. For some restrictions
It exists in many packages: e1071 was one of the earliest.
Other packages include
GLDEX, HyperbolicDist, QuantPsyc, TSA, agricolae, fUtilities, modeest,
moments, npde, s20x
It is also a common exercise in basic R programming courses.
On Thu, 4 Sep 2008, Williams, Robin wrote:
Hi,
Is there
Hi,
Sorry if this may be a simple question. Is there any function to calculate
returns (percentage or non-percentage) directly like the function getReturns()
in S Finmetrics?
Thanks.
Best Regards,
wy
**
The information provided in this
Thanks for that, looks like pretty much what I'm after.
Kind Regards
Chibisi
On Wed, Sep 3, 2008 at 9:17 PM, Iain Gallagher
[EMAIL PROTECTED] wrote:
Hi Chibisi
I'm sort of jumping into this thread but from your description of charts /
plotting below I thought you might like to take a
B == Birgitle [EMAIL PROTECTED]
on Tue, 2 Sep 2008 03:02:31 -0700 (PDT) writes:
B I try to perform a clustering using an existing dissimilarity matrix
that I
B calculated using distance (analogue)
B I tried two different things. One of them worked and one not and I don`t
B
From ?optim
fn: A function to be minimized (or maximized), with first
argument the vector of parameters over which minimization is
to take place. It should return a scalar result.
I think you intended to optimize over c(a,b,p, lambda), so you need to
specify them as
On Wed, 3 Sep 2008, Ranney, Steven wrote:
and I'm trying to fit a simple Von Bertalanffy growth curve with program:
VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4,
start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE)
...
Everything works as it should right up until the confint(VonB)
Hi,
I agree with you that Excel is not the best tool for fittings, that's why I
try to handle R.
But I need to use this specific model (LgmAltFormula) and not a polynomial
expression with the different parameters even if your method produced
correct fitting.
The parameters a and b are the
Dear list members,
I am trying, within a lapply command, to print the name of the objects
in list or data frame. This is so that I can use odfWeave to print out a
report with a section for each object, including the object names.
I tried e.g.
a=b=c=1:5
lis=data.frame(a,b,c)
lapply(
lis,
On Thu, 4 Sep 2008, Steve Powell wrote:
Dear list members,
I am trying, within a lapply command, to print the name of the objects
in list or data frame. This is so that I can use odfWeave to print out a
report with a section for each object, including the object names.
I tried e.g.
a=b=c=1:5
Well, the model you try to fit is almost independent of b
try
plot(Qe,fff1(364,0.0126))
points(Qe,fff1(364,1), col=2)
So you can quite freely choose b without any substantial improvement to
fit. I am not an expert in nonlinear fitting but seems to me that your
data does not follow your
Dear all,
I have a tab-delimited text (.txt) file which I'm trying to read into R. This
file is of column format - there are in fact 3 columns and 259201 rows
(including the column headers). I've been using the following commands, but
receive an error each time which prevents the data from
On Thu, 4 Sep 2008, Steve Murray wrote:
Dear all,
I have a tab-delimited text (.txt) file which I'm trying to read into R.
This file is of column format - there are in fact 3 columns and 259201
rows (including the column headers). I've been using the following
commands, but receive an
Hello Professor Ripely,
Sorry for not being clear. I posted after a long day of struggling. Also
my toy distance matrix should have been symmetrical.
Simply put I have spatially autocorrelated data collected from many points.
I would like to do a linear regression on these data. To deal with
Hello,
I have a survivor curve that shows account cancellations during the past 3.5
months. Â Fortunately for our business, but unfortunately for my analysis, the
survivor curve doesn't yet pass through 50%. Â Is there a safe way to extend
the survivor curve and estimate at what time we'll hit
Thanks Prof. Ripley! I knew it would be something simple - I'd missed the \t
from the read.table command! I won't be doing that again...!!
Thanks again,
Steve
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
Dear all,
Sorry if this is too obvious.
I am trying to fit my multiple regression model using lm()
Before starting model simplification using step() I checked whether the
model presented heteroscedasticity with ncv.test() from the CAR package.
It presents it.
I want to correct for it, I used
I have a survivor curve that shows account cancellations during the
past 3.5 months. Â Fortunately for our business, but unfortunately
for my analysis, the survivor curve doesn't yet pass through 50%.
 Is there a safe way to extend the survivor curve and estimate at
what time we'll hit
Hi Benoit,
another way of making Petr's point is by looking at the profile log likelihood
function
for b; that is, only estimating the a parameter for a grid of b values:
## Defining mean function for fixed b
lgma - function(b){
function(C0, m, V, a){ (V + b * m * a + C0 * V * b - ((C0 *
Thanks Prof Ripley! How obvious in retrospect!
Prof Brian Ripley wrote:
On Thu, 4 Sep 2008, Steve Powell wrote:
Dear list members,
I am trying, within a lapply command, to print the name of the objects
in list or data frame. This is so that I can use odfWeave to print out a
report with a
Dear Users, I already posted this question: it either went unnoticed,
or it is to basic (if this is so, please sent me a hint).
I would like to know if there is a way to add a common
legend to an arrangement of plots. In the example below, I get four
plots in my device. each one has a density for
You might consider a probit analysis using ln(Time) as the dose.
At 09:24 AM 9/4/2008, [EMAIL PROTECTED] wrote:
I have a survivor curve that shows account cancellations during the
past 3.5 months. Â Fortunately for our business, but unfortunately
for my analysis, the survivor curve doesn't
On 9/4/2008 9:35 AM, Nelson Villoria wrote:
Dear Users, I already posted this question: it either went unnoticed,
or it is to basic (if this is so, please sent me a hint).
I would like to know if there is a way to add a common
legend to an arrangement of plots. In the example below, I get four
On Thu, 4 Sep 2008, [EMAIL PROTECTED] wrote:
I have a survivor curve that shows account cancellations during the
past 3.5 months. Â Fortunately for our business, but unfortunately
for my analysis, the survivor curve doesn't yet pass through 50%.
 Is there a safe way to extend the survivor
Duncan Murdoch wrote:
On 03/09/2008 4:34 PM, Kevin E. Thorpe wrote:
Hello.
I'm having trouble installing rgl. I have a theory as to the problem.
First, the error message and session info.
install.packages(rgl)
trying URL 'http://probability.ca/cran/src/contrib/rgl_0.81.tar.gz'
Content
no I mean maybe use a higher dimension solution- the other thing you
may wish to check out is package vegan which has an mds function and
very good viginettes, and is my favorite package for doing ordination
analysis (metaMDS uses isoMDS).
hope this helps
On Wed, Sep 3, 2008 at 9:03 PM, 陈武
Dear Roman,
You can use the coefficient-covariance matrix returned by hccm() for
calculating corrected standard errors for the coefficients. Alternatively,
if you know the pattern of heteroscedasticity [as you probably do if you
used ncv.test()], you could try to correct for it by a
Hi all,
I am drawing a ternary graph. Everything is fine with both ternaryplot
(package vcd) and triangle.plot (package ade4), but I want to present
scales in neither percents nor from 0 to 1 (this is actually the only
option I found in both functions). I want the scales to be in a
natural scale
On Thu, 4 Sep 2008, Carrasco-Torrecilla, Roman R wrote:
Dear all,
Sorry if this is too obvious.
I am trying to fit my multiple regression model using lm()
Before starting model simplification using step() I checked whether the
model presented heteroscedasticity with ncv.test() from the CAR
rXLSolutions Corporation (www.xlsolutions-corp.com) is proud to announce
our*** R/Splus Fundamentals and Programming Techniques***course at 3 USA
locations for September 2008.
(1) R/Splus Fundamentals and Programming Techniques
http://www.xlsolutions-corp.com/Rfund.htm
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# 1. Supplying the derivatives results in convergence:
lgmg - function(a, b, C0, m, V) {
+ e - expression((V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 *
+ C0 * b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m *
+ a * b + (b * m * a)^2))/(2 * b * m))
+ val - eval(e)
+ attr(val,
Hi there,
When I do bootstrap on a maximum likelihood estimation, I try the
following code, however, I get error:
Error in minuslogl(alpha = 0, beta = 0) : object x not found
It seems that mle() only get data from workspace, other than the
boot.fun().
My question is how to pass the data to
On 9/4/2008 10:54 AM, Jinsong Zhao wrote:
Hi there,
When I do bootstrap on a maximum likelihood estimation, I try the
following code, however, I get error:
Error in minuslogl(alpha = 0, beta = 0) : object x not found
It seems that mle() only get data from workspace, other than the
boot.fun().
Hi,
Is there any facility in R to perform a stepwise process on a model,
which will remove any highly-correlated explanatory variables? I am told
there is in SPSS. I have a large number of variables (some correlated),
which I would like to just chuck in to a model and perform stepwise and
see what
Hi all,
I used the function jarque.test (in the moments package) on my data set and
I obtained something like this:
Jarque-Bera Normality Test
data: x
JB = 4.8381, p-value = 0.089
alternative hypothesis: greater
or
Jarque-Bera Normality Test
data: x
JB = 2.6018, p-value =
Hello everyone,
I have a problem using a for-loop to go through a matrix. I want to remove all
rows with a sum of 0.
What I do is basically:
for(i in 1:length(data))
{
if(sum(add(data[i,]) == 0)
{
data - data[-i,]
}
}
I get a error message: Error in `[.data.frame`(data, i) :
Hi there,
I have a dataset stored in an object which has very huge volume of rows.I
want to reuse it for comparing with other datasets.I dont want it to reload
every time i run the script.Is there a way of saving a particular loaded
object in the workspace and reusing it.
Kindly help me.
?save
I assume you can connect to it from within the script, presumably by
supplying the path of the object to the appropriate argument in your
script. I'm no expert though.
HTH,
Robin Williams
Met Office summer intern - Health Forecasting
[EMAIL PROTECTED]
-Original Message-
Robin Williams wrote
Is there any facility in R to perform a stepwise process on a model,
which will remove any highly-correlated explanatory variables? I am told
there is in SPSS. I have a large number of variables (some correlated),
which I would like to just chuck in to a model and perform
On 04-Sep-08 15:33:06, Markus Mühlbacher wrote:
Hello everyone,
I have a problem using a for-loop to go through a matrix. I want to
remove all rows with a sum of 0.
What I do is basically:
for(i in 1:length(data))
{
if(sum(add(data[i,]) == 0)
{
data - data[-i,]
}
}
I
I have a table statement that returns the following:
[10.839,10.841] (10.841,10.843] (10.843,10.846] (10.846,10.848]
(10.848,10.85]
0 0 0 0
1
(10.85,10.852] (10.852,10.854] (10.854,10.857] (10.857,10.859]
(10.859,10.861]
There is also the tri function in the cwhtool package, triax.plot in the
plotrix package, and triplot in the TeachingDemos package (I think it is
between this plot and progress bars as to what functionality is reproduced in
the most packages). One of the others may do what you want, or be
Try this:
v - as.numeric(gsub(]|\\), , unlist(lapply(strsplit(names(tb), ,),
[, 2
Where tb is your table
On Thu, Sep 4, 2008 at 10:20 AM, Trubisz, Joseph
[EMAIL PROTECTED] wrote:
I have a table statement that returns the following:
[10.839,10.841] (10.841,10.843] (10.843,10.846]
Here is one way:
y - levels(cut(x,5))
y
[1] (0.0124,0.209] (0.209,0.405] (0.405,0.601] (0.601,0.797]
(0.797,0.993]
sapply(y, function(x) sub(.*,(.*)], \\1, x))
(0.0124,0.209] (0.209,0.405] (0.405,0.601] (0.601,0.797] (0.797,0.993]
0.2090.4050.6010.797
Thank you, Greg. After all I noticed that in ternaryplot it can be
easily made by setting a 'scale' parameter.
Thanks,
Marcin
On Thu, Sep 4, 2008 at 6:00 PM, Greg Snow [EMAIL PROTECTED] wrote:
There is also the tri function in the cwhtool package, triax.plot in the
plotrix package, and
Dear all,
I have a dataset of four columns, and I wish to plot (as a scatter graph) the
values of the third column where the values are greater than zero, and the
fourth column.
I tried doing this via the plot command itself, but got into a bit of a mess
(resulting in errors!). My
See http://gsubfn.googlecode.com for more on the gsubfn package.
Below we use strapply to pick out every string of digits and decimal points
converting each to numeric returning a list of pairs and rbinding that list
into a matrix.
x - c([10.839,10.841], (10.841,10.843], (10.843,10.846],
Try this:
plot(January[January[,3] 0, 3:4])
On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray [EMAIL PROTECTED] wrote:
Dear all,
I have a dataset of four columns, and I wish to plot (as a scatter graph)
the values of the third column where the values are greater than zero, and
the fourth
Alternatively you can use saveObject() and loadObject() of R.utils -
that will not hardwire the name of the loaded object avoiding name
conflicts, e.g.
library(R.utils);
foo - 1:10;
saveObject(foo, file=foo.RData);
bar - loadObject(foo.RData);
/HB
On Thu, Sep 4, 2008 at 8:44 AM, Williams, Robin
Ah that's great, thank you very much.
As a follow-on, in the same format, how would I plot where column 3 is greater
than 0 *but also less than 2*?
Once again, any help is much appreciated.
Thanks,
Steve
Date: Thu, 4 Sep 2008 13:39:12 -0300
From: [EMAIL
Use '':
plot(January[January[,3] 0 January[,3] 2, 3:4])
On Thu, Sep 4, 2008 at 1:47 PM, Steve Murray [EMAIL PROTECTED] wrote:
Ah that's great, thank you very much.
As a follow-on, in the same format, how would I plot where column 3 is
greater than 0 *but also less than 2*?
Once again,
I have packaged R base 2.7.2 for Mandriva 2008.1 x86_64 who should I send
it to so that it can be made available to everybody?
It is my first attempt and it works well on my computer, but it will need some
testing.
Best,
--
Corrado Topi
Global Climate Change Biodiversity Indicators
Hello friends!!!
I have a list of values called y.
The list is y=c(221.0, 212.0, 206.0, 202.7, 198.4, 195.1, 192.2, 189.7,
187.6, 185.8);
y is f(x) and x=c(10,20,30,40,50,60,70,80,90,100).
I only have x and y. I don´t know f(x). I would like interpolate f(x) to
obtain other values as f(15),
I think that you can use the splinefun function:
f - splinefun(x, y)
f(15)
On Thu, Sep 4, 2008 at 1:52 PM, ermimi [EMAIL PROTECTED] wrote:
Hello friends!!!
I have a list of values called y.
The list is y=c(221.0, 212.0, 206.0, 202.7, 198.4, 195.1, 192.2, 189.7,
187.6, 185.8);
y is f(x)
Thank you very much, You have helped me to resolve the problem.
Thank you!!
A greetings, Luismi
Henrique Dallazuanna wrote:
I think that you can use the splinefun function:
f - splinefun(x, y)
f(15)
On Thu, Sep 4, 2008 at 1:52 PM, ermimi [EMAIL PROTECTED] wrote:
Hello
Hello,
Is it possible to create confidence intervals for hazard rates? Â I'm creating
two muhaz objects:
haz1 - muhaz(NumDaysCustomer[cRV==true],status[cRV==true])
haz2 - muhaz(NumDaysCustomer[cRV==false],status[cRV==false])
and plotting them. Â There are many, many more observations in
giov biowoman at libero.it writes:
Hi all,
I used the function jarque.test (in the moments package) on my data set and
I obtained something like this:
Jarque-Bera Normality Test
data: x
JB = 4.8381, p-value = 0.089
alternative hypothesis: greater
or
Jarque-Bera
R-helpers:
I just updated from R 7.0 to R 7.2.2 today. I am using MAC OS X
version 10.5.4 on a Macbook to run R.
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-apple-darwin8.11.1
locale:
en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base
Peter Flom peterf at brainscope.com writes:
Robin Williams wrote
Is there any facility in R to perform a stepwise process on a model,
which will remove any highly-correlated explanatory variables? I am told
there is in SPSS. I have a large number of variables (some correlated),
which I
I do have a txt file where each row is a record and the first element of
each record is an id for an individual. I am looking into combing the
records into one row if the id is the same and save as a txt file. Notice
that the rows may not have the same length in the result file that I want to
I am trying to lookup a value in 1 of 10 loaded two column-data sets (Bins)
by displaying the value of the second column based on the value of the
first. For instance in
Bin1_Acres Bin1_parprobBin1_TAZ
[1,] 0.0004420.978 356
[2,] 0.0004530.954
Also consider the redun function in the Hmisc package, which does not
use the response variable but uses flexible nonlinear additive models to
predict each predictor variable from all the others, using a stepwise
procedure in a formal redundancy analysis.
Frank
Ben Bolker wrote:
Peter Flom
Your IF statement (if (x = 3:6)) does not work as you probably expect
it since it does not test for the range. You probably want to use
cut to convert a value to a bucket in a range:
x - 3.5
cut(x, breaks=seq(-3,5,.2), labels=FALSE)
[1] 33
seq(-3,5,.2)
[1] -3.0 -2.8 -2.6 -2.4 -2.2 -2.0 -1.8
Read your data in, parse off the ID, split the records (using 'split')
and then combine the like values. Since you did not read the posting
guide, it is hard to give explicit help.
On Thu, Sep 4, 2008 at 1:42 PM, kayj [EMAIL PROTECTED] wrote:
I do have a txt file where each row is a record
Hello all,
I've been working on a new R Graphics device that targets Adobe's
Shockwave Flash format (SWF for short). It uses http://www.libming.org/
on the backend. Here are some example outputs so far:
http://160.129.129.41/~hornerj/plots/
Once you click on the above, choose a directory
Hello,
I'm new to R (using it since about two weeks),
but absolutely a fan of it from the beginning on. :-)
Best tool for working with data I found. :-)
I tried using the fft() and other funcitons for
analysing time series.
What I would be glad to have, would be a
convenient way to display
Try another mirror: this is not the first time we've seen problems at
www.ibiblio.org.
On Thu, 4 Sep 2008, Brant Inman wrote:
R-helpers:
I just updated from R 7.0 to R 7.2.2 today. I am using MAC OS X
version 10.5.4 on a Macbook to run R.
sessionInfo()
R
Dirk, Gabor,
thanks for your advice, I have now tried to study these
vignettes, but I must say, as an aging Gynecologist,
I am facing an enormous learning curve :-)-O.
This works nicely:
rawData - data.frame(x = c(2008-03-01, 2008-03-21, 2008-03-23,
2008-04-08, 2008-04-20, 2008-05-10,
Oliver Bandel wrote:
Hello,
I'm new to R (using it since about two weeks),
but absolutely a fan of it from the beginning on. :-)
Best tool for working with data I found. :-)
I tried using the fft() and other funcitons for
analysing time series.
What I would be glad to have, would be a
Thanks again Henrique - that's really useful to know!
_
Discover Bird's Eye View now with Multimap from Live Search
__
R-help@r-project.org mailing list
Hi All-
I have a data set (spdco2)
spdco2
[,1] [,2] [,3]
[1,]1 5.4 382.4212
[2,]2 5.1 383.0315
[3,]3 4.8 383.9520
[4,]4 4.7 384.4376
[5,]5 4.7 384.5929
[6,]6 4.4 384.8864
[7,]7 4.1 385.2156
[8,]8 3.8 385.2919
[9,]9 3.7 385.5925
[10,] 10
Thank you for your suggestion. I did as you suggested and switched
from USA (NC) to USA (IA).
Again, the packages seemed to download OK but this time a single error
message appeared:
--
...snip...
==
downloaded 168 Kb
trying URL
Dear Sherri,
Perhaps:
spdco2=matrix(scan(),ncol=3,byrow=TRUE)
1 5.4 382.4212
2 5.1 383.0315
3 4.8 383.9520
4 4.7 384.4376
5 4.7 384.5929
6 4.4 384.8864
7 4.1 385.2156
8 3.8 385.2919
9 3.7 385.5925
10 3.9 385.6801
spdco2[,2]-ifelse(spdco2[,2]=4.7,spdco2[,2],NA)
spdco2
Sherri Heck wrote:
Hi All-
I have a data set (spdco2)
spdco2
[,1] [,2] [,3]
[1,]1 5.4 382.4212
[2,]2 5.1 383.0315
[3,]3 4.8 383.9520
[4,]4 4.7 384.4376
[5,]5 4.7 384.5929
[6,]6 4.4 384.8864
[7,]7 4.1 385.2156
[8,]8 3.8 385.2919
[9,]
Try this:
is.na(spdco2[,2]) - which(spdco2[,2] 4.7)
spdco2
On Thu, Sep 4, 2008 at 4:21 PM, Sherri Heck [EMAIL PROTECTED] wrote:
Hi All-
I have a data set (spdco2)
spdco2
[,1] [,2] [,3]
[1,]1 5.4 382.4212
[2,]2 5.1 383.0315
[3,]3 4.8 383.9520
[4,]4 4.7
Fantastic! Each suggestion worked beautifully. Thanks so much.
Henrique Dallazuanna wrote:
Try this:
is.na http://is.na(spdco2[,2]) - which(spdco2[,2] 4.7)
spdco2
On Thu, Sep 4, 2008 at 4:21 PM, Sherri Heck [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] wrote:
Hi All-
I have a
Is this what you want?
library(ggplot2)
rawData - data.frame(Date = c(2008-03-01, 2008-03-21, 2008-03-23,
2008-04-08, 2008-04-20, 2008-05-10, 2008-06-20), y = c(4, 6,8, 5, 7, 2
,1))
rawData$Date - as.Date(rawData$Date)
Hi everyone,
I have a matrix containing color values of an image same as following,
[1,] #44 #44 #434343 #404040 #3D3D3D #3D3D3D #3E3E3E
[2,] #414141 #414141 #414141 #404040 #3F3F3F #3F3F3F #3F3F3F
[3,] #3E3E3E #3E3E3E #3F3F3F #404040 #404040 #404040 #404040
[4,] #3E3E3E #3D3D3D #3E3E3E
I think being able to say lable your points with a third column of
data would be cool (which I am sure is already do-able), and also to
be able to hover over a point and find out the value. On a huge
wishlist that I don't have a clue how to implement- a zoom window
function close to kaliedagraphs
Zitat von Duncan Murdoch [EMAIL PROTECTED]:
Oliver Bandel wrote:
Hello,
I'm new to R (using it since about two weeks),
but absolutely a fan of it from the beginning on. :-)
Best tool for working with data I found. :-)
I tried using the fft() and other funcitons for
analysing
I am starting with a matrix in which rows are vegetation plots and
columns are various characteristics including ID# and elevation. I
removed elevation and ID columns to avoid having those characteristics
influence the distances between points which I calculated using the
dist command.
Hello,
This maybe a newbie question. I have a dataframe that looks like the sample
at the bottom of the email. I have monthly precipitation data from several
sites over several years. For each site, I need to extract years that have
a complete series of 12 monthly precipitation values, while
On 04/09/2008 4:44 PM, Oliver Bandel wrote:
Zitat von Duncan Murdoch [EMAIL PROTECTED]:
Oliver Bandel wrote:
Hello,
I'm new to R (using it since about two weeks),
but absolutely a fan of it from the beginning on. :-)
Best tool for working with data I found. :-)
I tried using the fft() and
I have a need to build a time series and there are a couple of aspects about
the time series object that are confusing me. First it seems that ts.union is
not doing what I would expect. For example:
x0 - rep(0,10)
x1 - rep(1,10)
xt0 - ts(x0, frequency=10)
xt1 - ts(x1, frequency=10)
st2 -
Andrew Barr wabarr at gmail.com writes:
This maybe a newbie question. I have a dataframe
that looks like the sample
at the bottom of the email. I have monthly
precipitation data from several
sites over several years. For each site,
I need to extract years that have
a complete series of
Here is one approach:
tmp - scan(what='')
1: #44 #44 #434343 #404040 #3D3D3D #3D3D3D #3E3E3E
8: #414141 #414141 #414141 #404040 #3F3F3F #3F3F3F #3F3F3F
15: #3E3E3E #3E3E3E #3F3F3F #404040 #404040 #404040 #404040
22: #3E3E3E #3D3D3D #3E3E3E #404040 #414141 #414141 #404040
29: #3E3E3E
Zitat von Duncan Murdoch [EMAIL PROTECTED]:
On 04/09/2008 4:44 PM, Oliver Bandel wrote:
Zitat von Duncan Murdoch [EMAIL PROTECTED]:
Oliver Bandel wrote:
Hello,
[...]
plot3d doesn't support that directly, but you could plot with
type='n',
then use segments3d to add the lines.
On Thu, 4 Sep 2008, [EMAIL PROTECTED] wrote:
I have a need to build a time series and there are a couple of aspects about
the time series object that are confusing me. First it seems that ts.union is
not doing what I would expect. For example:
x0 - rep(0,10)
x1 - rep(1,10)
xt0 - ts(x0,
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