Dear R users,
I'm using graph library to create a mesh-like network topology and
implement a load balance routing algorithm. The current implementation
uses graph, RBGL, and Rgraphviz libraries. I have a few attributes on
every edge to represent the network loading and capacity, and I
frequently u
On Fri, Feb 04, 2011 at 04:35:00AM -0800, dpender wrote:
>
> Hi,
>
> I am using the uniroot function in order to carry out a bivariate Monte
> Carlo simulation using the logistics model.
>
> I have defined the function as:
>
> BV.FV <- function(x,y,a,A)
> (((x^(-a^-1)+y^(-a^-1))^(a-1))*(y^(a-1/
You should be able to use 'ifelse'
Os.chr4.gene.new$color <-
ifelse(Os.chr4.gene.new$if_TE_related == "TE_related", "black", "orange")
On Fri, Feb 4, 2011 at 7:09 PM, Tae-Jin Lee wrote:
> Hello,
>
> I'm trying to add a column to the following data frame. The new column
> will contain "blac
Sorry. I made a typo. It is the 5th column (not 4th). Thank you.
Tae-Jin
Begin forwarded message:
> From: Tae-Jin Lee
> Date: February 4, 2011 7:09:38 PM EST
> To: r-help@R-project.org
> Subject: Help!!! from R beginner
>
> Hello,
>
> I'm trying to add a column to the following data frame. The
Danny,
sounds like you already have a certain idea how a 'nugget' distribution could
look like. Maybe you also could intentionally produce some experimental
data having such distributions, harvest the related patterns from the
microarray and then apply a method as it was described in
http://ww
Hello,
I'm trying to add a column to the following data frame. The new column
will contain "black" when the 5th column(if_TE_related) is
"TE_related", or "orange" when the 4th column is " " (space).
"chromo""MSU_locus" "end5" "end3" "if_TE_related"
"chr04" "LOC_Os04g01006"
Dear list:
I have tried MASS's mca function and SAS's PROC corresp on the
farms data (included in MASS, also used as mca's example), the
results are different:
R: mca(farms)$rs:
1 2
1 0.059296637 0.0455871427
2 0.043077902 -0.0354728795
3 0.059834286
Ista,
Thank you again.
I had figured that out... and was crafting another message when you replied.
The NAs do come though on the variable that is being aggregated,
However, they do not come through on the categorical variable(s).
The aggregate function must be converting the data frame variabl
Just to be clear:
This works:
> set.seed(100)
> dat=data.frame(
+ x1=sample(c(NA,'m','f'), 100, replace=TRUE),
+ x2=sample(c(NA, 1:10), 100, replace=TRUE),
+ x3=sample(c(NA,letters[1:5]), 100, replace=TRUE),
+ x4=sample(c(NA,T,F), 100, replace=TRUE),
+ y=sam
Hi!
Does anyone have a numeric example for calculating BLUE and BLUP, please?
Thanks,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
oops. For clarity, that should have been
sum(ddply(dat, .(x1,x2,x3,x4), function(x){data.frame(y.sum=sum(x$y,
na.rm=TRUE))})$y.sum)
-Ista
On Fri, Feb 4, 2011 at 7:52 PM, Ista Zahn wrote:
> Hi again,
>
> On Fri, Feb 4, 2011 at 7:18 PM, Gene Leynes wrote:
>> Ista,
>>
>> Thank you again.
>>
>> I
Hi again,
On Fri, Feb 4, 2011 at 7:18 PM, Gene Leynes wrote:
> Ista,
>
> Thank you again.
>
> I had figured that out... and was crafting another message when you replied.
>
> The NAs do come though on the variable that is being aggregated,
> However, they do not come through on the categorical va
No, your approach is not correct. For one you have not taken the covariance
between B and C into account, another thing is that the ratio of 2 normal
random variables is not necessarily normal, in some cases it can even follow a
Cauchy distribution. Also note that with only 1 degree of freedom
Have you looked into bioconductor? There is a separate mailing list and many
packages designed for genetic analysis within the bioconductor project.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
>
Hi,
On Fri, Feb 4, 2011 at 6:33 PM, Gene Leynes wrote:
> Thank you both for the thoughtful (and funny) replies.
>
> I agree with both of you that sum is the one picking up aggregate. Although
> I didn't mention it, I did realize that in the first place.
> Also, thank you Phil for pointing out th
On Feb 4, 2011, at 4:24 PM, Mcmahon, Kwyatt wrote:
Hello compadRes,
I'm developing a script that selects "cells" over a certain
metabolic rate to kill them. A rate between 9 and 12 means that the
cells are candidates for death.
I'll show you what I mean:
# a would be a vector of cell m
Hi Mario,
>From the help file for barplot:
"Specifying a single value will have no visible effect unless 'xlim'
is specified."
OK, so:
barplot(d,col=barcol,ylim=c(min(d-s*1.25),max(d+s*1.25)), xlim=c(0,1),
width=.1, space = 2)
Best,
Ista
On Fri, Feb 4, 2011 at 4:15 PM, Mario Beolco wrote:
> De
Thank you both for the thoughtful (and funny) replies.
I agree with both of you that sum is the one picking up aggregate. Although
I didn't mention it, I did realize that in the first place.
Also, thank you Phil for pointing out that aggregate only accepts a formula
value in more recent versions!
A bioligist colleague sent me the following data.
x Y
3 1
7 5
148
240
(Yes, only four data points.) I don't know much about the
application, but apparently there are good empirical
reasons to use a quadratic model.
The goal is to find the X value which maximizes the
respons
Dear R-users,
apologies for the total beginner's question, but I have been trying to
solve this problem for ages and I seem to be getting nowhere. I also
have tried to search through the archives of the R mailing list, but I
am still left with my problem. How do I change the width of the bars
for
Hello compadRes,
I'm developing a script that selects "cells" over a certain metabolic rate to
kill them. A rate between 9 and 12 means that the cells are candidates for
death.
I'll show you what I mean:
# a would be a vector of cell metabolic rates.
a<-c(8, 7, 9, 6, 10, 11, 4, 5, 6)
#now id
I created a file called .Renviron and set R_LIBS_USER to the same path I had
set R_LIBS to. I put this file in
C:/Users/myusername/Documents/mysubdirectory. I also commented out the line
I had put in the Renviron.site file. Now I get the same error as I
originally got. When I put .Renvrion in C:/U
Greg, Dennis - thanks for your input, I really appreciate the feedback, as it
is not easy to source.
In terms of the data; I've described it as 20 columns, which is the smallest
dataset, but this can run to 320 columns, so in some cases there is likely
to be enough power to detect non-normality.
Sorry, I didn't see Phil's reply, which is better than mine anyway.
-Ista
On Fri, Feb 4, 2011 at 5:16 PM, Ista Zahn wrote:
> Hi,
>
> Please see ?na.action
>
> (just kidding!)
>
> So it seems to me the problem is that you are passing na.rm to the sum
> function. So there is no missing data for th
Hi,
Please see ?na.action
(just kidding!)
So it seems to me the problem is that you are passing na.rm to the sum
function. So there is no missing data for the na.action argument to
operate on!
Compare
sum(aggregate(y~x1+x2+x3+x4, data=dat, sum, na.action=na.fail)$y)
sum(aggregate(y~x1+x2+x3+x4
Gene -
Let me try to address your concerns one at a time:
Since the formula interface to aggregate was introduced
pretty recently (I think R-2.11.1, but I might be wrong)
so when you try to use it in an R-2.10.1 it won't work.
Now let's take a close look at the help page for aggregate.
The
Hi:
The problem you have, IMO, is the multiplicity of tests and the p-value
adjustments that need to be made for them. Here's a little simulation using
normal and exponential distributions of the size you're contemplating.
# Normal data
d <- matrix(rnorm(40), nrow = 2)
# Function to compu
It is fine, you just overthought the solution and used both the applys and for
loops (see another thread today where I made the same mistake of overthinking
and combining 2 different methods). I was just pointing out the errors so you
could improve for next time.
But here is some things to thi
data2elrm<-cbind(mydata,n=rep(1,dim(mydata)[1]))
>
More logic would be:
data2elrm2<-cbind(mydata,n=rep(1,nrow(mydata)))
Sorry for obfuscation.
--
Mi³ego dnia
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://
2011/2/4 Den
> To use elrm() I have to aggregate my data,which is really time consuming
> when I look for the way out through many variables.
You don't have to do that. *One exception is that the binomial response
should be specified as success/trials, where success gives the number of
successes
Search the help archive and you'll find dozens of suggestions about beginner
manuals
You can search the archive at nabble.com or markmail.com
http://r-project.markmail.org/
http://r.789695.n4.nabble.com/R-help-f789696.html
(I don't know why the nabble URL is so complicated)
For seeing R examples
Can someone please tell me what is up with na.action in aggregate?
My (somewhat) reproducible example:
(I say somewhat because some lines wouldn't run in a separate session, more
below)
set.seed(100)
dat=data.frame(
x1=sample(c(NA,'m','f'), 100, replace=TRUE),
x2=sample(c(NA, 1:10
Darn, Good catch, I fell victim to overthinking the problem.
I think I was more thinking of:
'[0-9]+(?=/)'
Which uses the whole match (then I switched thinking and captured the number,
but did not simplify the other part). Yours is the best.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data C
On 04.02.2011 01:42, Longe wrote:
On 02/03/2011 01:57 PM, Peter Alspach wrote:
Tena koe
?par
and check the las argument.
HTH
Peter Alspach
Thank you, Peter, and also David and William. The las argument indeed
helps in setting the tick labels horizontally. Beautiful! But how do I
rot
On Fri, Feb 04, 2011 at 02:03:22PM -0500, sudhir cr wrote:
> Hello,
>
> I have a R code for doing convolution of two functions:
>
> convolveSlow <- function(x, y) {
> nx <- length(x); ny <- length(y)
> xy <- numeric(nx + ny - 1)
> for(i in seq(length = nx)) {
> xi <- x[[i]]
>
Thanks Peter.
I understand your point, and that there is potentially a high false
discovery rate - but I'd expect the interesting data points (genes on a
microarray) to be within that list too. The next step would be to filter
based on some greater understanding of the biology...
Alternative ap
Yes, that was dumb - I got that...
--
View this message in context:
http://r.789695.n4.nabble.com/Finding-non-normal-distributions-per-row-of-data-frame-tp3259439p3260843.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-projec
On 2011-02-04 11:00, DB1984 wrote:
Hi Greg,
In addition to the reply above, to address your questions - I fully
appreciate that my understanding of the code is basic - this is my first
attempt at putting this together...
My starting point is a data frame with numeric and text columns, but I ca
On Fri, Feb 4, 2011 at 1:27 PM, Greg Snow wrote:
> Try this:
>
>> x <- c("349/077,349/074,349/100,349/117",
> + "340/384.2,340/513,367/139,455/128,D13/168",
> + "600/437,128/903,128/904")
>>
>> library(gsubfn)
>> out <- strapply(x, '([0-9]+)(?=/)')
>> out
> [[1]]
> [1] "349" "349
Hello, All,
GWAF 1.2
R.Version() is below.
system(lme.batch.imputed(
phenfile = 'phenfile.csv',
genfile = 'CARe_imputed_release.0.fhsR.gz',
pedfile='pedfile.csv',
phen='phen1',
covar=c('covar1','covar2'),
kinmat='imputed_fhs.kinship.RData',
outfile='imputed.FHS.IBC.GWAF.LME.output.0.txt'
))
Give
On 4 February 2011 at 14:03, sudhir cr wrote:
| Hello,
|
| I have a R code for doing convolution of two functions:
|
| convolveSlow <- function(x, y) {
| nx <- length(x); ny <- length(y)
| xy <- numeric(nx + ny - 1)
| for(i in seq(length = nx)) {
| xi <- x[[i]]
| for(
Hello,
I have a R code for doing convolution of two functions:
convolveSlow <- function(x, y) {
nx <- length(x); ny <- length(y)
xy <- numeric(nx + ny - 1)
for(i in seq(length = nx)) {
xi <- x[[i]]
for(j in seq(length = ny)) {
ij <- i+j-1
xy[[ij
Hi Greg,
In addition to the reply above, to address your questions - I fully
appreciate that my understanding of the code is basic - this is my first
attempt at putting this together...
My starting point is a data frame with numeric and text columns, but I can
cut columns to make a fully numeric
On Feb 4, 2011, at 1:41 PM, DB1984 wrote:
Thanks David - but '1' (if I understood correctly) returns the same
value for
each row, which I took to be an error.
And exactly what were you expecting with that data?
nt
V1V2V3V4V5V6
1 24.71 23.56 24.71 23.56 24.71 23.
Try this:
strsplit(x, "/\\d+\\.\\d+,|/\\d+,|/\\d+")
On Fri, Feb 4, 2011 at 1:37 PM, Dick Harray wrote:
> Hi there,
>
> I have a problem about lapply, strsplit, and accessing list elements,
> which I don't understand or cannot solve:
>
> I have e.g. a character vector with three elements:
>
> x
Thanks David - but '1' (if I understood correctly) returns the same value for
each row, which I took to be an error.
nt
V1V2V3V4V5V6
1 24.71 23.56 24.71 23.56 24.71 23.56
2 25.64 25.06 25.64 25.06 25.64 25.06
3 21.29 20.87 21.29 20.87 21.29 20.87
4 25.92 26.92 25.92 26.92
This is the qqmath example from the lattice package. I added the scales to the
example. I would like to switch the axis and not sure how? Meaning I would like
the "height" on the x-axis and the probability on the y-axis. Will you show me
the correct syntax for this switch thanks.
qqmath(~ hei
Of course. Thanks.
On 04/02/2011 18:32, Greg Snow wrote:
Try:
do.call(test,point)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and pro
Try:
> do.call(test,point)
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Florian Burkart
> Sent: Friday, F
Try this:
> x <- c("349/077,349/074,349/100,349/117",
+ "340/384.2,340/513,367/139,455/128,D13/168",
+ "600/437,128/903,128/904")
>
> library(gsubfn)
> out <- strapply(x, '([0-9]+)(?=/)')
> out
[[1]]
[1] "349" "349" "349" "349"
[[2]]
[1] "340" "340" "367" "455" "13"
[[3]]
[1]
Chris -
You can solve your problem by removing the print
statements and replacing them with
answer = list(org.plot[org.plot$sub %in% ex.plot,],
new.pl[new.pl$yar %in% "1991",],
new.pl[new.pl$yar %in% "1993",])
prefix = sub('-','.',ex.plot)
names(answer) = paste(pr
Try this:
do.call(test, point)
On Fri, Feb 4, 2011 at 12:35 PM, Florian Burkart wrote:
> Hey,
>
> this may be a simple question, but I am struggling to apply a list of
> parameters to a function.
>
>
> Say I have the following function:
>
> test<-function(a=1,b=2,c=3,d=4){a+b+c+d}
>
> And the f
Dear all,
I collected my data from the different agricultural fields every week over a
period of a month.
how can I test for spatial autocorrelation in R with data that are temporally
pseudoreplicated?
I used lme with correlation=corCompSymm(form=~Date) to model temporal
pseudoreplication.
Reg
I actually have several questions revolving around the generation of
wireframe plots.
The most pressing is that which is described by the subject line.
I am trying to produce a figure with x, y, and z labels, but only tick marks
on the x, and y axes. I have supplied sample code below subst
(Apologies to the cc-list: I'm resending from a different address because
I didn't realize it was going to r-help.)
I'm also not an expert on this topic. I just wanted to list a couple of
ways that non-PD matrices might arise. I'll just add now a couple of
pointers:
First, I believe the te
1. Martin Maechler's comments should be taken as replacements
for anything I wrote where appropriate. Any apparent conflict is a
result of his superior knowledge.
2. 'eigen' returns the eigenvalue decomposition assuming the
matrix is symmetric, ignoring anything in m[upper.tri(
On Feb 4, 2011, at 10:48 AM, DB1984 wrote:
Thanks for the feedback Patrizio - but your function is performing the
shapiro.test on columns instead of rows...
I tried:
nt<-data.frame(#a dataframe with 6 columns and 9 rows)
nr <- nrow(nt)
test <- apply(nt, nt[1:nr,], shapiro.test)
Error in d
1. Martin Maechler's comments should be taken as replacements
for anything I wrote where appropriate. Any apparent conflict is a
result of his superior knowledge.
2. 'eigen' returns the eigenvalue decomposition assuming the
matrix is symmetric, ignoring anything in m[upper.tri(
Hi Marianne,
The quick-and-dirty solution is to add one character and make ns global:
ns <<- nrow(x)
Poor practice, but OK temporarily if you're just debugging.
This is an issue of "scope". You are assuming dynamic scope, whereas R uses
static scope. 'ns' was not defined when you said paste("n=
The options showWarnCalls and showErrorCalls may also help --
they can be use do enable automatic printing of a call stack
summary. From ?options:
‘showWarnCalls’, ‘showErrorCalls’: a logical. Should warning and
error messages show a summary of the call stack? By default
I'm also not an expert on this topic. I just wanted to list a couple of
ways that non-PD matrices might arise. I'll just add now a couple of
pointers:
First, I believe the term "semipositive definite" is considered ambiguous
because in some literature it means that the matrix the smallest
e
Hate to say that, but it looks like Stata is way above R, considering
exact logistic regression.
To use elrm() I have to aggregate my data,which is really time consuming
when I look for the way out through many variables. But the worst thing
is that I am not not sure if I can trust to p-values in
attempting to do multivariate modelling in R with known future
conditions (in this case variable 'b') using MSBVAR and hc.forecast.
The sample code (a paired down representation) does not give anywhere
near the expected results - I am assuming that a forecast 8 steps out
would approximate 'a' as th
Thanks for the feedback Patrizio - but your function is performing the
shapiro.test on columns instead of rows...
I tried:
nt<-data.frame(#a dataframe with 6 columns and 9 rows)
nr <- nrow(nt)
test <- apply(nt, nt[1:nr,], shapiro.test)
Error in ds[-MARGIN] : invalid subscript type 'list'
fr
'Aggregate multiple functions into a single function. Combine multiple
functions to a single function returning a named vector of outputs'
This is a short description of each() function from plyr package
Here is an example from help
each(min, max)(1:10)
Hope this helps
Regards
Denis
У Пят
Hello,
I'm still trying to modify my script in order to use RODBC instead of RMySQL
(no more ready-to-use package for windows).
My new problem is the following one (not seen with RMySQL):
I'd like to copy a data.frame to a mysql table. One column is a numeric one
with big integer like :
2005000
Hey,
this may be a simple question, but I am struggling to apply a list of
parameters to a function.
Say I have the following function:
test<-function(a=1,b=2,c=3,d=4){a+b+c+d}
And the following list:
point<-list(a=3,d=2)
Is there a way I can evaluate function test at point?
(Apart fro
Dear Łukasz
Thank you very much for information
Dear R people could you please help please with following questions
Sorry for my silly questions, because I am not a mathematician.
1. Is elrm() works similar as exact regression in SAS or Stata? After
double check in Stata and R some results from m
Hi,
I'm trying to create a function to return three dataframes for later use in
a graphic so I want the return from the function to give me dataframes and
with unique names relating to the variable they are based on.
For example.
sub<-c("6-1a","6-1a","6-1a","9-2b","9-2b","9-2b","7c","7c","7c"
You can do this with regular expressions, since you want to extract specific
values from the string I would suggest learning about the gsubfn package, it is
a bit easier with gsubfn than with the other matching tools.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healt
So you want to combine multiple columns back into a single column with the
strings pasted together? If that is correct then look at the paste and sprintf
functions (use one or the other, not both).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail
Thank you. This is what I want.
I need to study about attr :).
Soyeon
On Fri, Feb 4, 2011 at 11:07 AM, David Winsemius wrote:
>
> On Feb 4, 2011, at 11:56 AM, Soyeon Kim wrote:
>
>> Dear All,
>>
>> I used glm and then used step function for stepwise regression.
>> Now, I want to store the variab
On Feb 4, 2011, at 11:56 AM, Soyeon Kim wrote:
Dear All,
I used glm and then used step function for stepwise regression.
Now, I want to store the variables used in the stepwise regression.
I am not sure that what you ask here is actually what you eventually
ask and will limit my answer to
Dear All,
I used glm and then used step function for stepwise regression.
Now, I want to store the variables used in the stepwise regression.
The code is the following.
m_logistic <- glm(y ~ . + M1:T + M2:T + M3:T+ M4:T +M5:T,
family=binomial("logit"), data = data)
step_glm <- step(m_logistic
I get a different set of errors than you do (what version of R are you using?).
Patrizio showed one way to do what you want. But, what is it that you are
really trying to accomplish? What do you think the result of 20,000 normality
tests (each of which may not be answering the real question on
Dear R users?
I want to know how to use strata in survival analysis, I want to combine
multiple dataset:
Coxph(Surv(merged.time.v, merged.event.v)~merged.score+
strata(dataset.v))
So merged.time.v and merged.event.v are from multiple datasets.
How should I define dataset.v here?
Ma
Did you look at the help? ?formula has a whole section about why
formulas have environments and what they do.
On Fri, 4 Feb 2011, Aviad Klein wrote:
# Hi all,
# I've made a function to make a formula out of a data.frame without columns
which contain a constant value.
# The function "which.c
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Duncan Murdoch
> Sent: Friday, February 04, 2011 6:03 AM
> To: Ernest Adrogué
> Cc: r-h...@stat.math.ethz.ch
> Subject: Re: [R] get caller's name
>
> On 03/02/2011 10:27 AM, Erne
dpender wrote:
>
> Hi,
>
> I am using the uniroot function in order to carry out a bivariate Monte
> Carlo simulation using the logistics model.
>
> I have defined the function as:
>
> BV.FV <- function(x,y,a,A)
> (((x^(-a^-1)+y^(-a^-1))^(a-1))*(y^(a-1/a))*(exp(-((1^(-a^-1)+y^(-a^-1))^a)+y^-1
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Dick Harray
> Sent: Friday, February 04, 2011 7:37 AM
> To: r-help@r-project.org
> Subject: [R] lapply, strsplit, and list elements
>
> Hi there,
>
> I have a problem about lap
# Hi all,
# I've made a function to make a formula out of a data.frame without columns
which contain a constant value.
# The function "which.constant" returns the indices of colums with constant
values:
which.constant <- function(data.frame) { # counts the number of columns in a
data.frame which
On Feb 4, 2011, at 8:26 AM, Denis Kazakiewicz wrote:
Dear R people
Could you please help
I have similar but opposite question
How to reshape data from DF.new to DF from example, Mark kindly
provided?
Well, I don't think you want a random order, right? If what you are
asking is for a singl
Hi there,
I have a problem about lapply, strsplit, and accessing list elements,
which I don't understand or cannot solve:
I have e.g. a character vector with three elements:
x = c("349/077,349/074,349/100,349/117",
"340/384.2,340/513,367/139,455/128,D13/168",
"600/437,128/903,1
--- begin included message
I'm trying to do some analysis using survreg function. I need to
implement
there my own distribution with density:
lambda*exp(-lambda*y), where y = a1/(1+exp(-a2*x)).
a1, a2 are unknown parameters and x >0.
--- end inclusion --
The survreg
I would probably use a Dynamic Linear Model, combining seasonal
component and regression. See package dlm and its vignette for
examples.
HTH,
Giovanni
On Tue, 2011-02-01 at 11:01 +, Paolo Rossi wrote:
> Hi list,
>
>
> I would like to estimate and forecast the seasonal component of a series
Hi
I have a list of nodes and edges.
I want to draw a graph from two different direction left and right. I
am drawing the required graph which is growing from left. Similar will be
the case that will grow from right. the one that grows from right may have
few nodes present in left subtree.
Hi,
Got my issues sorted.
Error message solved:
I heard from the guy who developed MuMIn and his suggestion worked.
"As for the error you get, it seems you are running an old version of MuMIn.
Please update the package first."
I did (I was only 1 version behind in both R and in MuMIn) and erro
Thanks Ista,
Very instructive and works like a charm!
Cheers!
Olga
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, mi
Hi Olga,
Here is how I would approach this problem. I hope it is instructive.
library(vegan)
data(BCI)
mod <- radfit(BCI[1,])
class(mod) #find out what mod is
methods(class="radfit") #find out what methods are available for
objects of this class. Nothing looks promising, so define a summary
metho
as.POSIXlt(book$DATE, origin="1582/10/14") works.
The Gregorian calendar was the kicker, thanks to ggrothendieck at gmail.com
Ross
--
View this message in context:
http://r.789695.n4.nabble.com/Importing-dates-from-SPSS-file-tp3260293p3260329.html
Sent from the R help mailing list archive at
On Fri, Feb 4, 2011 at 9:47 AM, dunner wrote:
>
> Hello all, kind regards,
>
> I have imported a data.frame from SPSS using "foreign":read.spss but
> unfortunately it is reading dates in a way neither R nor myself can
> understand.
>
>> book$DATE
> [1] 13502246400 13443321600 13477795200 13472956
Responding to T. Therneau:
> sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252
LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C LC_TIME=English_United States.125
What do the dates look like in the original file?
On Fri, Feb 4, 2011 at 9:47 AM, dunner wrote:
>
> Hello all, kind regards,
>
> I have imported a data.frame from SPSS using "foreign":read.spss but
> unfortunately it is reading dates in a way neither R nor myself can
> understand.
>
>> book$DATE
Hello all, kind regards,
I have imported a data.frame from SPSS using "foreign":read.spss but
unfortunately it is reading dates in a way neither R nor myself can
understand.
> book$DATE
[1] 13502246400 13443321600 13477795200 13472956800 13501728000 13445395200
13501382400 13502851200 134441856
On Wed, Feb 2, 2011 at 7:59 AM, Karl Ove Hufthammer wrote:
> Dear list members,
>
> I recall seeing a convenience function for applying multiple functions to
> one object (i.e., almost the opposite of 'mapply’) somewhere.
> Example: If the function was named ’fun’ the output of
>
> fun(3.14, mode
Dear Jim,
One touch!! Thank you very much
Best regards
Ogbos
On 4 February 2011 13:47, Jim Lemon wrote:
> On 02/04/2011 09:14 PM, ogbos okike wrote:
>
>> Dear all,
>> Using the code I got from the link (
>> http://www.phaget4.org/R/image_matrix.html), I obtained a nice plot that
>> suits
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
I hava a simulation model, of which I want to do a sensitivity analysis.
I have identified a number of input variables and my response variable.
What I hava done so far:
1) I created a Latin Hypercube with the lhs package (10.000 simulations)
2) s
Dear R people
Could you please help
I have similar but opposite question
How to reshape data from DF.new to DF from example, Mark kindly
provided?
Thank you
Denis
On Пят, 2011-02-04 at 07:09 -0600, Marc Schwartz wrote:
> On Feb 4, 2011, at 6:32 AM, D. Alain wrote:
>
> > Dear R-List,
> >
> >
Dear all,
Using:
library(vegan)
data(BCI)
mod <- radfit(BCI[1,])
mod
RAD models, family poisson
No. of species 93, total abundance 448
par1 par2 par3Deviance AIC BIC
Null 39.5261 315.4362 315.4362
Preemption 0.042797
On 03/02/2011 10:27 AM, Ernest Adrogué wrote:
Hi,
Suppose a function that checks an object:
stop.if.dims<- function(x) {
if (! is.null(dim(x))) {
stop("cannot handle dimensional data")
}
}
This would be used by other functions that can only work with
dimensionless objects. The proble
1 - 100 of 136 matches
Mail list logo
