Hi
Hi Eik,
greetings to Hamburg! :-) Thanks for the fast and helpful answer
Eik Vettorazzi-2 wrote:
#compare
str(red[,2])
str(red[2,])
I understand that the first is a real vector of nums in R and the second
is
a ?? matrix/list/data.frame ?? of single ? entries? Can I
. wrote:
Hi, continuing the improvements...
I've prepared a new code:
ddae - function(individuals, frac, sad, samp=pois, trunc=0, ...) {
dots - list(...)
Compound - function(individuals, frac, n.species, sad, samp, dots) {
print(c(Size:, length(individuals), Compound
Hi, I am wondering if anyone can suggest how to test the equality of 2
proportions. The caveat here is that the 2 proportions were calculated from the
same number of samples using 2 different tests. So essentially we are comparing
2 accuracy rates from same, say 100, samples. I think this is
Does file.exists answer to your question?
file.exists(.RData)
If you are not sure of the exact name of the file but know it contains
.RData, you can try:
List.files(directory,.RData)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Dear Colleagues:
Through your help an R related installation issue was resolved, but I now
have the following usage issue.
On any get.var.ncdf usage I am seeing:
Error: in routine alloca() there is a stack overflow: thread 0, max
535822282KB, used 0KB, request 24B
The same error is
Hello
I have a barplot where I get a grey gradient colors for each bar, I want to
append a legend with the names of each bar and next to each name I need a box
with the corresponding grey color.
I managed to get the legend without boxes, but never got the wanted grey
gradient in the legend:
Im doing test on my sample using 1-sample Kolmogorov-smirnov test(poisson
distribution) to specify my sample obtain avalanche effect. before i can
compare my D value with the critical value(Dn, alpha), i have to specify
which level of significance to use (eg. alpha = 0.01, 0.05, etc) and i don't
Hello, I am having some issues with the order write.table.
The fact is that I need to use , as the decimal character and not
. as default, and when I use:
write.table(Sales,file=Sales.xls,quote=FALSE, sep = \t, dec = ,,
row.names = FALSE, col.names = TRUE)
It does it perfectly, but then, in a
Dear Arne,
Thank you for fixing the package.
However I am still struggling to obtain model estmates.
The same code:
UpC - censReg(Power ~ Windspeed, left = -Inf, right =
2000,data=PData_In,method=BHHH,nGHQ = 4,start=c(-691.18,186.79,3.9,3.9))
Error in maxNRCompute(fn = logLikAttr, fnOrig =
Esteemed UseRs and DevelopeRs,
Apologies if this question belongs else where, but it does concern R's
package installation/maintenance.
How does one start R within Emacs/ESS with root privileges?
I tried without success:
M-x sudo R
Why i'm motivated to do so:
It seems logical to me, as
On 09/07/2011 08:54 AM, Karl Brand wrote:
Esteemed UseRs and DevelopeRs,
Apologies if this question belongs else where, but it does concern R's
package installation/maintenance.
How does one start R within Emacs/ESS with root privileges?
I tried without success:
M-x sudo R
Why i'm
On Wed, Sep 7, 2011 at 4:28 PM, Suphajak Ngamlak
supha...@phatrasecurities.com wrote:
Dear all,
I am trying to do weighted regression using lm function in R. However, I have
a question why the results from
1) lm(formula = Y~aX, weight = w)
2) lm(formula = wY~waX)
are
What's wrong with that?
(The values you submit as scale in prior are not fixed variances, but
parameters of the prior distribtion - your problem may be that you
believe that they are meant to be variances fixed by you!?)
Yes I did, so I think it is not possible to fix the variance. Anyway,
Hi
I want to make a grid map using diversity indices. I have a soil grid map with
7 class in r.
how can I make a diversity map based on soil map in r? I found something for
calculation diversity in a table (.csv).
Thanks
[[alternative HTML version deleted]]
If your matrix is being converted to characters, then the conversion
(and the use of '.' for decimal) is happening before the the
write.table call. Do the cbinding without putting titles (characters)
in the first line and then once the table is build, use 'colnames' to
add the titles. It sounds
Hello,
I've some questions concerning log-transformations and plotting of the
regression lines. So far as I know is it a problem to log-transform values
smaller than 1 (0-1). In my statistics lecture I was told to do a log(x+1)
transformation in such cases. So I provide here a small example to
Dear Igors
On 7 September 2011 10:39, Igors igors.lahanc...@gmail.com wrote:
However I am still struggling to obtain model estmates.
The same code:
UpC - censReg(Power ~ Windspeed, left = -Inf, right =
2000,data=PData_In,method=BHHH,nGHQ = 4,start=c(-691.18,186.79,3.9,3.9))
Error in
On 09/07/2011 09:54 AM, azam jaafari wrote:
Hi
I want to make a grid map using diversity indices. I have a soil grid map
with 7 class in r.
how can I make a diversity map based on soil map in r? I found something for
calculation diversity in a table (.csv).
Thanks
From you description, you should not used a paired Student's t-test. One uses
a paired test when pairs of observations come from the same experimental unit
(and thus are correlated). You describe a study where each experimental unit
is tested once and where there are two independent groups of
Cheers Paul.
Its a very good point. Although i am curious how badly i can damage my R
install by running as root. I always ran R in windows with admin.
privileges without problems (touch wood). Probably best to never find
out by sticking with user privileges.
However, even for taking care
Hi,
I have the following data-frame:
x - data.frame(first = c('a','c','k','b'), second = c('b','k','a','j'), third
= c('f','a','h','b'))
first second third
1 a b f
2 c k a
3 k a h
4 b j b
Now I would like to see wether there are entries
On Sep 7, 2011, at 3:02 AM, goong001 wrote:
Im doing test on my sample using 1-sample Kolmogorov-smirnov
test(poisson
distribution) to specify my sample obtain avalanche effect. before i
can
compare my D value with the critical value(Dn, alpha), i have to
specify
which level of
On Sep 7, 2011, at 8:15 AM, Martin Batholdy wrote:
Hi,
I have the following data-frame:
x - data.frame(first = c('a','c','k','b'), second =
c('b','k','a','j'), third = c('f','a','h','b'))
first second third
1 a b f
2 c k a
3 k a h
4 b j
I think that you have not understood your lecturer. You can log transform
any positive number. You can not log transform a negative number. Adding a
constant to a negative number to make it constant before log transformation
is sometimes suggested by those who do not understand what they are
Please! ... ?prop.test
not t tests.
-- Bert
--
On Wed, Sep 7, 2011 at 4:21 AM, John Sorkin jsor...@grecc.umaryland.edu wrote:
From you description, you should not used a paired Student's t-test. One
uses a paired test when pairs of observations come from the same
experimental unit (and
Acutally,
?mcnemar.test
since it is paired data.
Best,
Wolfgang
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Bert Gunter
Sent: Wednesday, September 07, 2011 15:34
To: John Sorkin
Cc: r-help@r-project.org
Subject: Re:
Hello List,
I am trying to do the following:
1. Use a BandPassFilter h[n] on a series x[n]
2. Then Decimate the series by a factor D such that y[k]=x[k*D]
The decimation factor is considerable (10,000) so that filtering before the
decimation, using convolve(), seems foolish.
Can you think of an
I have the following data (see RawData using dput below)
How do I get it in the following 3 column format (CO2 measurements are the
elements of the original data frame). I'm sure the package reshape is where
I should look, but I haven't figured out how.
Thanks ahead of time
Month Year CO2
Wolfgang:
On Wed, Sep 7, 2011 at 7:28 AM, Viechtbauer Wolfgang (STAT)
wolfgang.viechtba...@maastrichtuniversity.nl wrote:
Acutally,
?mcnemar.test
since it is paired data.
Actually, it is unclear to me from the OP's message whether this is the case.
In one sentence the OP says that the
Oh ... I should have added that either option could be handled by
glm(), of course (provided that you're willing to accept the
approximate tests).
But this is getting OT.
-- Bert
On Wed, Sep 7, 2011 at 7:28 AM, Viechtbauer Wolfgang (STAT)
wolfgang.viechtba...@maastrichtuniversity.nl wrote:
Hi,
Using legend as an argument within the call to barplot will automatically
add the color key to the legend. I don't know how to add the colors using
legend separately. Try this revised code:
y -
Using R version 2.13.1
Windows 7 Ultimate 64-bit
At 2011-09-06 14:51:09,Hasan Diwan hasan.di...@gmail.com wrote:
On 6 September 2011 08:01, KnifeBoot knifeb...@163.com wrote:
Can't installe packag maDB or limma.
Which R version, and what platform are you using?
--
Sent from my mobile device
Dear Sir,
I am using R 2.13.1. Since yesterday when I start the Rgui the
message Save Workspace Image? shows up immediately on opening. Even
if I cancel, it does not get cancelled. If I select either no or yes,
it quits. I do not know all of a sudden what happened? Right from
yesterday I am
Hi,
I am new to R and here is what I am doing in it now. I am using machine
learning technique (svm) to do predictions. The data that I am using is one
that is bound to grow perpetually. what I want to know is, say, I fed in a
data set with 5000 data points to svm initially. The algorithm derives
Hi,
I am new to R and here is what I am doing in it now. I am using machine
learning technique (svm) to do predictive modeling. The data that I am using
is one that is bound to grow perpetually. what I want to know is, say, I fed
in a data set with 5000 data points to svm initially. The algorithm
It works now. It is much more appropiate to add the titles than to add
a column with strings, and, in fact, it were rownames, not column
names (thats why colnames was set to true). Anyway using rownames
instead of adding an aditional column and setting row.names to true in
the write.trable works
The code looks like this:
L1=list(a=1,b=2,c=3)
f1=as.factor(c)
L1[[f1]] returns 1
What happens if we give a factor as an index at a list?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hello,
I work on the Stochastic Frontier Analysis (SFA) and I am looking for a
function in a R package which done the multivariate interpolation.
My problem is: For all i=1, ..., n, I have values of (xi1, xi2, ...,
xi7)in IR^7 and f(xi1, xi2, ..., xi7)in IR and I have also values of (x'i1,
Dear Arne,
I have sent you an e-mail to your e-mail at gmail.com. There I have attached
my data set and the code for this particular problem. I hope to hear from
you soon.
Many thanks in advance!
Igors
--
View this message in context:
Dear all,
I estimate Fixed-effects model. Since it use within transformation, it
also reports within SSR in summary report.
The problem is that I need overall SSR. How can I quickly compute it?
Thanks in advance!
Best wishes,
Igors
--
View this message in context:
Ben - I'm sorry, you're right! The following website shows how to make a
volcano plot using ggplot2.
http://bioinformatics.knowledgeblog.org/2011/06/21/volcano-plots-of-microarray-data/
Enjoy!
Anna
--
View this message in context:
Bump, please help!
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-by-statement-for-spaghetti-plots-tp3788536p3796131.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Can anybody help me with this?
--
View this message in context:
http://r.789695.n4.nabble.com/Change-properties-of-line-summary-in-interaction-plot-tp3788614p3796133.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
I have a script that runs as a cron job every minute (on Ubuntu 10.10 and R
2.11.1), querying a database for new data. Most of the time it takes a few
seconds to run, but once in while it takes more than a minute and the next run
starts (on the same data) before the previous one has finished.
Hi,
I'm trying to reshape my data set from wide to tall format for multilevel
modeling. Unfortunately, the function I typically use (make.univ from the
multilevel package) does not appear to work with unbalanced data frames,
which is what I'm dealing with.
Below is an example of the columns of
look at the melt function in reshape, specifically ?melt.data.frame
require(reshape)
Raw.melt-melt(RawData,id.vars='Year',variable_name='Month')
there is an additional feature in the melt function for handling na values.
names(Raw.melt)[3]-'CO2'
head(Raw.melt)
Year MonthCO2
1 1958 J
Is this meant to be a follow up to something?
In short, the answer to your question is how factors are encoded as integers
in R. If you run something similar to this code (except that actually runs)
you'll note that as.numeric(f1) = 1.
Michael
On Wed, Sep 7, 2011 at 5:04 AM, Varsha Agrawal
Try this:
funa - function(n, y, a, rate, samp) {
lambda - a * n
dexp(n, rate) * do.call(paste(d, samp, sep=), y, lambda)
}
funb - function(y, a, rate, samp) {
integrate(f1, 0, Inf, y, a, rate)
}
funb(1, 0.1, 0.1, pois)
Jean
Hello guys,
I
Indeed, the original post leaves some room for interpretation. In any case, I
hope the OP has enough information now to figure out what approach is best for
his data.
Best,
Wolfgang
-Original Message-
From: Bert Gunter [mailto:gunter.ber...@gene.com]
Sent: Wednesday, September 07,
On Sep 7, 2011, at 6:04 AM, Varsha Agrawal wrote:
The code looks like this:
L1=list(a=1,b=2,c=3)
f1=as.factor(c)
L1[[f1]] returns 1
What happens if we give a factor as an index at a list?
In a virgin session of R there would be no 'c' for the second line to
work with. And since 'c' is a
Hi,
On Wed, Sep 7, 2011 at 5:25 AM, Divyam divyamural...@gmail.com wrote:
Hi,
I am new to R and here is what I am doing in it now. I am using machine
learning technique (svm) to do predictive modeling. The data that I am using
is one that is bound to grow perpetually. what I want to know is,
Varsha Agrawal wrote on 09/07/2011 05:18:10 AM:
The code looks like this:
L1=list(a=1,b=2,c=3)
f1=as.factor(c)
L1[[f1]] returns 1
What happens if we give a factor as an index at a list?
L1=list(a=1,b=2,c=3)
f1=as.factor(L1$c)
L1[[f1]]
When you use a factor (e.g., f1, corrected
On Sep 7, 2011, at 9:46 AM, dadrivr wrote:
Can anybody help me with this?
Your chances of a meaningful reply would increase considerably if you
first read the Posting Guide and also followed the other guidance seen
at the bottom of every rhelp posting.
--
View this message in
The terminology (melt, cast, recast) just isn't intuitive to me; but I
understand how to use melt now.
Thanks!
Justin Haynes wrote:
look at the melt function in reshape, specifically ?melt.data.frame
require(reshape)
Raw.melt-melt(RawData,id.vars='Year',variable_name='Month')
there
On Sep 7, 2011, at 9:45 AM, dadrivr wrote:
Bump, please help!
When I encounter one of these Nabble bump messages doing my
moderation duties I simply discard it. This is not a website. This is
a mailing list.
If you didn't get an answer it suggests that your example was not
Mikkel Grum mi2kelgrum at yahoo.com writes:
I have a script that runs as a cron job every minute (on Ubuntu 10.10 and R
2.11.1), querying a database for new
data. Most of the time it takes a few seconds to run, but once in while it
takes more than a minute and the next
run starts (on the
Under Vista and Windows 7, I install R in a local directory pgms so I
never have to worry about permissions.
Under Linux, I use su not sudo. Then I call R and install.packages.
Then I quit R and su and continue with what I want.
hope this helps.
Spencer
On 9/7/2011 5:11 AM, Karl Brand
. wrote:
Hello guys,
I would like to ask for help to understand what is going on in
func2. My plan is to generalize func1, so that are expected same
results in func2 as in func1. Executing func1 returns...
0.25 with absolute error 8.4e-05
But for func2 I get...
Error in dpois(1,
To follow up (because I received a few emails off-list), things work now.
I'm not sure what I did differently. In case it is helpful:
I reinstalled R tools, and edited by path so that
C:\Rtools\bin;C:\Rtools\perl\bin;C:\Rtools\MinGW\bin;C:\Program
files\R\R-2.13.1\bin;C:\Program Files\HTML Help
Hello,
I am wondering if it is possible, or what the correct way to code a three-way
crosstab in R using the survey package?
I have been using the following code to complete two way crosstabs, but have
not seen any three-way code.
Two-Way: svyby(~factor(a), ~factor(b), data, svymean)
Thanks!
Yes, numDeriv::hessian is very accurate. So, I normally take the output from
the optimizer, *if it is a local optimum*, and then apply numDeriv::hessian to
it. I, then, compute the standard errors from it.
However, it is important to know that you have obtained a local optimum. In
fact, you
However, it is not known whether the standard errors obtained from this
Hessian are asymptotically valid.
Let me rephrase this. I think that as a measure of dispersion, the standard
error obtained using the augmented Lagrangian algorithm is correct. However,
what is *not known* is the
I have the following results for an ANOVA comparing two nested models. I
wasn't sure how I am supposed to report this result in the area of
psychology. Specifically, am I supposed to report the DF's or just the F
ratio? I could manually calculate the degrees of freedoms, but there must be
a reason
Well that is really on the windows end, not the R end (which may not make it
any less surprising). I would not be surprised if there was an easier way
using .net or vbs which I get the sense is partly where some things are going
(e.g., with the so-called PowerShell), but I am not terribly
I am trying to create a classification tree using either tree or rpart
but when it comes to plotting the results the formatting I get is
different than what I see in all the tutorials. What I would like to
see is the XX/XX format but all I get is a weird decimal value. I was
also wondering how
Dear all,
This relatively routine analysis has left me frustrated and in a rut. I
have a dataset (data1), which I subset in order to remove rows where
HabitatDensity=Med. This dataset looks correct when I call it up,
however, when I create a table out of the new subset (data2), my table
Sorry, I thought the link would work for people because it is a public link
and it works for me when I run it in R. Anyways, here is an example set of
data that I am having trouble with:
/id -
c(230017,230017,230017,230018,230018,230018,230019,230019,230019,230020,230020,
Thank you. This is a nice trick, and it works fine for my needs.
It's surprising that there isn't a simple way to get to the drive name.
On Tue, Sep 6, 2011 at 10:41 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:
Hi,
This comes with absolutely no guarantees (and a good recommendation to
be
hi,
say I have consecutively numbered objects obj1, obj2, ... in my R workspace.
I want to acces one of them inside a function, with the number given as an
argument.
Where can I find help on how to do that? Somebody must have been trying to do
this before...
Some keywords to start a search
I wish to test the hypothesis of mu equal to or less than 5 against the
specific alternative mu equal to or greater than 7. I am unable to find how to
persuade R to do this with any function (e.g. t.test). Suggestions?
Michael Grant
[[alternative HTML version deleted]]
Hello, I am attempting to train a random forest model using the
randomForest package on 500,000 rows and 8 columns (7 predictors, 1
response). The data set is the first block of data from the UCI
Machine Learning Repo dataset Record Linkage Comparison Patterns
with the slight modification that I
Here's an example:
/id - c(17,17,17,18,18,18,19,19,19,20,20,20,21,21,21,22,22,22,23,23,23,24,
24,24,25,25,25,26,26,26)
age - rep(c(30,36,42),10)
outcome -
c(12,17,10,5,5,2,NA,NA,NA,8,6,5,11,13,10,15,11,15,13,NA,9,0,0,0,20,
14,16,1,2,2)
mydata -
Dear R-users,
I'm running a maximization problem in which I want to impose a condition on
the relationship between 2 parameters.
The condition is that w[4] = (1+eps)/(2*w[1]), or equivalently w[4]*w[1] =
(1+eps)/2 , where eps is some small positive constant.
I've been trying to formulate a
Scott Bearer sbearer at TNC.ORG writes:
Dear all,
This relatively routine analysis has left me frustrated and in a rut. I
have a dataset (data1), which I subset in order to remove rows where
HabitatDensity=Med. This dataset looks correct when I call it up,
however, when I create a
The get() function should get you where you need.
Michael Weylandt
On Wed, Sep 7, 2011 at 12:53 PM, Berry Boessenkool
berryboessenk...@hotmail.com wrote:
hi,
say I have consecutively numbered objects obj1, obj2, ... in my R
workspace.
I want to acces one of them inside a function, with
On Sep 7, 2011, at 3:46 PM, dadrivr wrote:
Here's an example:
/id -
c(17,17,17,18,18,18,19,19,19,20,20,20,21,21,21,22,22,22,23,23,23,24,
24,24,25,25,25,26,26,26)
age - rep(c(30,36,42),10)
outcome -
c(12,17,10,5,5,2,NA,NA,NA,8,6,5,11,13,10,15,11,15,13,NA,9,0,0,0,20,
On 07/09/2011 3:46 PM, Michael Grant wrote:
I wish to test the hypothesis of mu equal to or less than 5 against the
specific alternative mu equal to or greater than 7. I am unable to find how to
persuade R to do this with any function (e.g. t.test). Suggestions?
According to the standard
Hi:
Your code doesn't work, but if you want to do spaghetti plots, they're
rather easy to do in ggplot2 or lattice.
The first problem you have is that one group in your test data has all
NA responses, so by() or any other groupwise summarization function is
going to choke unless you give it a
fit1, the unrestricted model, includes 1 more regressor than fit2, the
restricted model. Testing the models against each other means that fit2 is
equal to fit1, assuming that the coefficient on the additional regressor
that is included in fit1 is restricted to zero.
Your F-test is thus whether
Hello,
I would like to write a function where substitute operates on the
parameter, but ...
Expression = function(o,l) substitute(o, l)
Expression({x=.(FOO)}, list(FOO=2))
o
How do i get substitute to work on the contents of o.
Regards
Saptarshi
I have used do.call(substitute, ...) to work around the
fact that substitute does not evaluate its first argument:
R z - quote(func(arg))
R do.call(substitute, list(z, list(func=quote(myFunction),
arg=as.name(myArgument
myFunction(myArgument)
S+'s substitute (following S version 4)
Hi:
Having just seen your follow-up question, here's how it could be done
in ggplot2 or lattice:
# ggplot2 version:
ggplot(w, aes(x = age, y = outcome)) + geom_line(aes(group = .id)) +
stat_summary(fun.y = 'mean', geom = 'line', color = 'blue', size = 2)
# lattice version:
xyplot(outcome ~
On Sep 7, 2011, at 5:20 PM, Saptarshi Guha wrote:
Hello,
I would like to write a function where substitute operates on the
parameter, but ...
Expression = function(o,l) substitute(o, l)
Expression({x=.(FOO)}, list(FOO=2))
o
How do i get substitute to work on the contents of o.
I
On Sep 7, 2011, at 20:57 , zugi young wrote:
I have the following results for an ANOVA comparing two nested models. I
wasn't sure how I am supposed to report this result in the area of
psychology. Specifically, am I supposed to report the DF's or just the F
ratio? I could manually calculate
Hi all, thanks very much for sharing your thoughts. and sorry for my describing
the problem not clearly, my fault.
My data is paired, that is 2 different diagnostic tests were performed on the
same individuals. Each individual will have a test results from each of the 2
tests. Then in the end,
Hi all,
Although I sent the mail to Piotr, the author of FSelector, it should be better
to ask here to let others know.
Yanwei
Begin forwarded message:
From: Yanwei Song yanwei.s...@gmail.com
Date: September 7, 2011 4:41:58 PM EDT
To: p.roman...@stud.elka.pw.edu.pl
Subject: FSelector and
Hi Jean,
Thanks for the reply..
Using your suggestion, I end up in in the source code (Fortran 77 i
believe). At first look, it seems a bit more tedious to implement than I
expected.
--
View this message in context:
Hi,
Premises: I have a database which contain the list of events and their time
stamps (This is a Unix time stamps)
What I want to do : I want know how much is the maximum occurrence of this
in any a time period of 7 days or does a event occur es more than N (say
5) times in a period of 7 days.
Thanks!
That's exactly it.
get gets it ;-)
Berry
From: michael.weyla...@gmail.com
Date: Wed, 7 Sep 2011 14:54:48 -0500
Subject: Re: [R] access objects
To: berryboessenk...@hotmail.com
CC: r-help@r-project.org
The get() function should get you where you need.
Michael Weylandt
On Wed, Sep
Hi,
I am using the package ncdf to create netCDF files and I want to mimic the the
header of an exiting netCDF file created outside of R. Below is what the
existing header looks like (part of it that is different):
netcdf ccd1984_05_08 {
dimensions:
lat = 1974 ;
lon = 1894 ;
David Winsemius wrote:
What mean summary line? I count 8 lines and that matches the number
if id's with complete data.
Yea, I don't know why there is no mean summary line showing up. I requested
it in the interaction.plot statement (fun = mean), and I don't get any
errors. Any ideas?
--
Hi all,
This seems like a basic problem, but no amount of playing with the code has
solved it. I have a time-series data set like that shown below (only longer)
and am seeking to plot the data with filled, circular points and error bars.
I would like the error bars to be behind the points
You could phrase this in most dialects of SQL, but you don't say what kind of
database you are using and this is not a SQL help list.
You could also look at the rollapply function in the zoo package.
---
Jeff Newmiller The
On Sep 7, 2011, at 4:49 PM, dadrivr wrote:
David Winsemius wrote:
What mean summary line? I count 8 lines and that matches the number
if id's with complete data.
Yea, I don't know why there is no mean summary line showing up. I
requested
it in the interaction.plot statement (fun =
Add geom_errorbar() before adding geom_point() ?
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries
I think (no promises) roll.apply() from the zoo package with sum can count
occurrences in whatever time period. Then use a logical test to identify
sufficiently active periods.
Hope this helps,
Michael Weylandt
On Sep 7, 2011, at 4:07 PM, Vineet Shukla shuklvin...@gmail.com wrote:
Hi,
On Wed, Sep 7, 2011 at 2:25 PM, Ross Maidment
r.i.maidm...@pgr.reading.ac.uk wrote:
Hi,
I am using the package ncdf to create netCDF files and I want to mimic the
the header of an exiting netCDF file created outside of R. Below is what the
existing header looks like (part of it that is
Hi Jeff,
My question is not related with SQL.
I have all the data available in a csv file.
I am looking for R algorithm where can specify a time period and then R
essentially uses this time period as a sliding window for all the event-time
stamps and gives
a significant distribution of
Hi, Jesper:
Where Outside R do you want to use it?
There are R interfaces to many other software packages, and many
that cannot easily link to R can link to Fortran. You could use
library(sos) to search for R packages to connect to whatever.
Hope this helps.
Doing a visual graphic and trying to make it pretty
Here's simple chart to play with:
library(rgl)
dotframe-data.frame(x=c(0,7,0,0,-7,0),y=c(0,0,7,0,0,-7),z=c(7,0,0,-7,0,0))
dotframe plot3d(dotframe$x,dotframe$y,dotframe$z, radius=3,
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