Hi,
May be this helps:
tb[1,]%*%(((val-rep(meansb79[1,],5))^2)/6)
# [,1]
#[1,] 1.47619
tryvarb-c(1,2,3,4,4,4,4)
var(tryvarb)
#[1] 1.47619
tb[2,]%*%(((val-rep(meansb79[2,],5))^2)/6)
# [,1]
#[1,] 1.904762
sapply(seq_len(nrow(tb)),function(i)
Hi R forum,
Each row of my data (below) show a new contact event between animals.
In order to ultimately look at the patterns of intervals between
contacts, I need to calculate a contact end time. The contact starts
at the date and time shown in V4 and V5, and lasts for the duration
shown IN
Hello, Cat,
see inline below.
Hth -- Gerrit
On Fri, 12 Apr 2013, Cat Cowie wrote:
Hi R forum,
Each row of my data (below) show a new contact event between animals.
In order to ultimately look at the patterns of intervals between
contacts, I need to calculate a contact end time. The
Thank you all for your useful suggestions.
Jean, I had already tried to use the maxiter command in that way, but it simply
told me that the model had not converged.
Based on Profs Ripley and Nash's comments, I have opted to use an alternative
approach, and am creating a distribution of
Hi Gerrit,
Thanks for your quick reply - this, in combination with reversing the
date as it is shown in my data, worked perfectly.
startt- strptime((paste(data2$V4, data2$V5)), %Y-%m-%d %H:%M:%S)
Apologies for the silly question it seems, but you've saved me a lot
of time. Always learning
Hi,
The following should work:
x - paste(paste(data2$V4, data2$V5), data2$V6, sep='.')
startt - strptime(x, %Y-%m-%d %H:%M:%OS)
Regards,
Pascal
2013/4/12 Cat Cowie cat.e.co...@gmail.com
Hi R forum,
Each row of my data (below) show a new contact event between animals.
In order to
Micah Bennett micahgbennett at yahoo.com writes:
Hi,
I am working with a species-by-trait .csv file (columns=traits,
rows=species) and get the following
warning message when trying to plot results of both metaMDS and pcoa:
Warning message:
In ordiplot(x, choices = choices, type =
avoid using F, spell it out
Use an appropriate TZ... e.g. use Sys.setenv( TZ=Etc/GMT+5 ) before
converting if your time is local standard time year round
Use difftime to add to time values for consistent results
endt - startt + as.difftime( data2$V6, units=secs )
Use the str function to
Dear Leonore,
On Wed, Apr 10, 2013 at 5:51 PM, Leonore Wigger leonore.wig...@unil.ch wrote:
Question: I have specified a blacklist. I would have expected this to
completely disallow the arcs on the blacklist. But the result shows that
some of the blacklisted arcs have a strength 0 (rows
Hello
I am studying ANOSIM (analysis of similarities) using a package(vegan)
provided in r statistics.
I am wondering how to do pairwise ANOSIM test among three groups.
Can anybody please explain this question.
Thanks.
[[alternative HTML version deleted]]
http://aragon-vida.com/11a-hdnsdsjreunx-1.php
well, good night and thank you very much
932d4a1dff1375c06208c4cf2a7675e527431024e7d241382978df47d0f8c43f
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Dear Members,
I upgraded R from 2.15.3 to 3.0.0 (binary) on my WinXP and now I can't
load my workspaces. The error message is:
Error: requested primitive type is not consistent with cached value
During startup - Warning message:
unable to restore saved data in .RData
Any idea to fix that?
Hm. Sometimes workspace cannot be loaded when some packages which vere used for
creationd objects in workspaces are missing, but in that case the error message
is different.
So you either can delete .RData but it obviously **delete** all objects or you
need to find which package is missing
Thanks a lot, Marco. This works. And your remark regarding less reliable
arc strengths with cpdag=FALSE was very helpful.
Leonore
Le 12/04/2013 11:08, Marco Scutari a écrit :
Dear Leonore,
On Wed, Apr 10, 2013 at 5:51 PM, Leonore Wigger leonore.wig...@unil.ch wrote:
Question: I have
Berend Hasselman bhh at xs4all.nl writes:
Yes. curve expects the function you give it to return a vector if the input
argument is a vector.
This is clearly documented for the argument expr of curve.
Thanks a lot, Berend!
In fact, I didn't read carefully the documentation of curve. Anyway,
Greetings:
I am trying to query an MSSQL database which contains time series data
using RODBC.
Using Server Management Studio, I can retrieve the data with this query:
Select * from tb_ace_mag_1m where time_tag='2012-01-08 00:00:00' AND
time_tag'2012-01-08 03:00:00'
However, when I try to
Hi,
I am trying to predict the habitat suitability of lobster with GAMs. I need to
produce a map in ArcGis of the predicted densities. I did some search and the
function predict.gam seems to do the job.
Is this the right way to do it? when you apply the function you get the
predicted values
Le jeudi 11 avril 2013 à 13:40 -0700, Paul Miller a écrit :
Hello All,
Learning to use the odfWeave package. I really like the package. It has good
documentation, makes some very nice looking tables, and seems to have lots of
options for customizing output.
There are a few things I'd
Hello,
I'm trying to use R with an distributed File system (HDFS) and when I launch
a MapReduce job with DSL package, I've got this message :
*Streaming Command Failed!
Error in !unlist(lapply(chunks, is.null)) : invalid argument type*
I am using the Cloudera CDH4 Virtual Machine, R version is
Folks,
I have some computations in a function that create some large matrices.
I have been in the habit in these circumstances to call null out a matrix
once used and call gc().
Some pseudo code:
theFunction-function(x, y, z, 1) {
myMatrix - black.box.function(x, y, z, 1)
On Fri, Apr 12, 2013 at 10:37 AM, Lauria, Valentina
valentina.lau...@nuigalway.ie wrote:
Hi,
I am trying to predict the habitat suitability of lobster with GAMs. I need
to produce a map in ArcGis of the predicted densities. I did some search and
the function predict.gam seems to do the job.
Hi,
table.x- as.table(as.matrix(read.table(text=
A B C D
2 4 6 5
,sep=,header=TRUE)))
table.y- as.table(as.matrix(read.table(text=
C D
5 1
,sep=,header=TRUE)))
library(plyr)
library(reshape2)
vec1-rowSums(join(melt(table.x)[,-1],
Hello all,
I have a question: I am using the interquantile method to spot outliers
it gives me values of say 234 -120 or for the higher lower benchmarks.
I don't have any issues w/ the higher end. However I don't have any
negative values. My lowest possible value is 0. Should I consider 0
Hi Arun,
Thank you so much for your answer. It surely does help. Having to know
different approaches for the same problem has given me more insights on R
language. I appreciate your time and effort.
Best,
Janesh Devkota
On Thu, Apr 11, 2013 at 10:00 PM, arun smartpink...@yahoo.com wrote:
Berend Hasselman bhh at xs4all.nl writes:
Your function miBeta returns a scalar when the argument mu is a vector.
Use Vectorize to vectorize it. Like this
VmiBeta - Vectorize(miBeta,vectorize.args=c(mu))
VmiBeta(c(420,440))
and draw the curve with this
Dear all,
I would like to find out if anyone is aware of the existence of
implemented functions, or codes, for running multivariate AICc as it was
developed by* *Fujikoshi Satoh (1997)* *. I've been searching for it,
but goggle, rseek, and all other searching mechanism returns nothing.
Dear Barry,
Thank you very much for your reply. I am trying to create a map in GIS of the
habitat prediction of lobster using environmental predictors. I am attaching
here a brief example of my script and data.
I am OK for the GAM fitting and evaluation (residuals plot etc) but I am having
Thanks so much Arun. If you have time, what reference do you use to learn
seq_len(nrowâ¦etc and other aspects of your code?
Again. Thanks very much. Jan Beckstrand
From: arun kirshna [via R] [mailto:ml-node+s789695n4664037...@n4.nabble.com]
Sent: Friday, April 12, 2013 2:01 AM
To:
See below
On Fri, 12 Apr 2013, Julio Sergio wrote:
Berend Hasselman bhh at xs4all.nl writes:
Your function miBeta returns a scalar when the argument mu is a vector.
Use Vectorize to vectorize it. Like this
VmiBeta - Vectorize(miBeta,vectorize.args=c(mu))
VmiBeta(c(420,440))
and draw
On Apr 12, 2013, at 7:58 AM, Julio Sergio wrote:
Berend Hasselman bhh at xs4all.nl writes:
Your function miBeta returns a scalar when the argument mu is a
vector.
Use Vectorize to vectorize it. Like this
VmiBeta - Vectorize(miBeta,vectorize.args=c(mu))
VmiBeta(c(420,440))
and draw
Hi,
This is not an R question and is therefore not appropriate for this
list. You should post to a statistics forum such as
http://stats.stackexchange.com/. But the answer is NO!
Best,
Ista
On Fri, Apr 12, 2013 at 9:49 AM, ramoss ramine.mossad...@finra.org wrote:
Hello all,
I have a
-Original Message-
I have a question: I am using the interquantile method to
spot outliers it gives me values of say 234 -120 or for
the higher lower benchmarks.
I don't have any issues w/ the higher end. However I don't
have any negative values. My lowest possible value
Hi Janis,
For example
The goal here is to get results per row
BTW, your original posting had tb[,1]
tb[,1]%*%(((val-rep(meansb79[1,],5))^2)/6)
#Error in tb[, 1] %*% (((val - rep(meansb79[1, ], 5))^2)/6) :
# non-conformable arguments
I guess it should be:
Jeff Newmiller jdnewmil at dcn.davis.ca.us writes:
system.time( miBeta( seq( 370, 430, length.out=1e5 ) ) )
user system elapsed
1.300 0.024 1.476
system.time( miBetav( seq( 370, 430, length.out=1e5 ) ) )
user system elapsed
This is very interesting, Jeff. Of course, I
But is it supposed to be t^{-3/2} or t^{-0.5}?? The formula has the former and
the code the latter, and the integral is clearly divergent with the former.
-pd
On Apr 12, 2013, at 04:51 , Thomas Lumley wrote:
I don't get an error message (after I correct the missing line break after
the
On 12/04/2013 7:14 AM, Tamas Barjak wrote:
Dear Members,
I upgraded R from 2.15.3 to 3.0.0 (binary) on my WinXP and now I can't
load my workspaces. The error message is:
Error: requested primitive type is not consistent with cached value
During startup - Warning message:
unable to restore saved
Hi Peter,
It is supposed to be t^{3/2} instead of t^{0.5}. The code had a typo. Yes, I
figured out that when x =0, the solution is divergent because it is like
dividing something by zero.
Thanks.
Best,
Janesh
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent:
Dear list,
I want the 1st, 2nd, 5th, and 6th columns of mtcars. After copying them,
the columns become numeric class rather than data frame.
But, when I copy rows, they data frame retains its class. Why is this? I
don't see why copying rows vs columns is so different.
class(mtcars)
[1]
It's another reason not to use $ for extraction.
By default, R reduces dimensionality when subsetting, so mtcars$mpg
actually returns a numeric vector. With $, there's no way to override
the default behavior.
data(mtcars)
a - mtcars$mpg
class(a)
[1] numeric
dim(a)
NULL
a - mtcars[, mpg,
class(mtcars[1:4,]) # select some rows
[1] data.frame
class(mtcars[,1:4]) # select some columns
[1] data.frame
class(mtcars[,3]) # select one column
[1] numeric
class(mtcars[, 3, drop=FALSE]) # select one column
[1] data.frame
I cannot give a definitive reason why it is done this way, but you
YES, it's work! Fine! Thank you! I wait for the R-patch.
Regards,
Tamas
2013/4/12 Duncan Murdoch murdoch.dun...@gmail.com
On 12/04/2013 7:14 AM, Tamas Barjak wrote:
Dear Members,
I upgraded R from 2.15.3 to 3.0.0 (binary) on my WinXP and now I can't
load my workspaces. The error message
On 12-04-2013, at 21:32, C W tmrs...@gmail.com wrote:
Dear list,
I want the 1st, 2nd, 5th, and 6th columns of mtcars. After copying them,
the columns become numeric class rather than data frame.
But, when I copy rows, they data frame retains its class. Why is this? I
don't see why
Hi,
I am not sure I understand your question correctly.
dat1- read.table(text=
id responsed_at number_of_connection
scores
1 12-01-2010 1
2
1
What if my data is much larger, and I don't know what column number but
know its name? Do I have to grep its by name?
How about subset()? Is that what people commonly use?
Mike
On Fri, Apr 12, 2013 at 3:44 PM, Berend Hasselman b...@xs4all.nl wrote:
On 12-04-2013, at 21:32, C W
On 12-04-2013, at 21:56, C W tmrs...@gmail.com wrote:
What if my data is much larger, and I don't know what column number but know
its name? Do I have to grep its by name?
How about subset()? Is that what people commonly use?
Continuing with mtcars and the desired columns
b -
Hi,
From your example data,
dat1- read.table(text=
id1 id2 value
a b 10
c d 11
b a 10
c e 12
,sep=,header=TRUE,stringsAsFactors=FALSE)
#it is easier to get the output you wanted
dat1[!duplicated(dat1$value),]
# id1 id2 value
#1 a b 10
I was thinking,
b - subset(mtcars, select=c(mpg, cyl, drat, wt))
Yours is even simpler. :)
Thanks
Mike
On Fri, Apr 12, 2013 at 4:06 PM, Berend Hasselman b...@xs4all.nl wrote:
On 12-04-2013, at 21:56, C W tmrs...@gmail.com wrote:
What if my data is much larger, and I don't know what column
Hi,
May be this helps:
Not sure how you wanted to select those two letters.
dat1- read.table(text=
Seq,Output
A B B C D A C,Yes
B C A C B D A C,Yes
C D A A C D,No
,sep=,,header=TRUE,stringsAsFactors=FALSE)
library(stringr)
lapply(str_split(str_trim(dat1$Seq),
Hi R experts,
For example I have a dataset looks like this:
Number TimeStamp Value
1 1/1/2013 0:00 1
2 1/1/2013 0:01 2
3 1/1/2013 0:03 3
How can I split the TimeStamp Column into two and return a new table like
this:
Number Date Time Value
1
HI,
In case if you wanted to take BC and CB as the same.
dat1- read.table(text=
Seq,Output
A B B C D A C,Yes
B C A C B D A C,Yes
C D A A C D,No
,sep=,,header=TRUE,stringsAsFactors=FALSE)
lapply(str_split(str_trim(dat1$Seq), )[dat1$Output==Yes],function(x)
{x1-t(combn(x,2));
Hi,
dat1- read.table(text=
Number,TimeStamp,Value
1,1/1/2013 0:00,1
2,1/1/2013 0:01,2
3,1/1/2013 0:03,3
,sep=,,header=TRUE,stringsAsFactors=FALSE)
dat2-data.frame(Number=dat1[,1],do.call(rbind,strsplit(dat1[,2], )),
Value=dat1[,3])
names(dat2)[2:3]- c(Date,Time)
dat2
# Number Date Time
- Forwarded Message -
From: arun smartpink...@yahoo.com
To: Ye Lin ye...@lbl.gov
Cc:
Sent: Friday, April 12, 2013 6:25 PM
Subject: Re: [R] split date and time
Hi Ye,
Is this okay?
dat2-cbind(dat1[,-2],do.call(rbind,strsplit(dat1[,2],
)),stringsAsFactors=FALSE)
dat2
# Number
Thanks!
On Fri, Apr 12, 2013 at 3:30 PM, arun smartpink...@yahoo.com wrote:
- Forwarded Message -
From: arun smartpink...@yahoo.com
To: Ye Lin ye...@lbl.gov
Cc:
Sent: Friday, April 12, 2013 6:25 PM
Subject: Re: [R] split date and time
Hi Ye,
Is this okay?
Hi all,
I'm trying to use kNNimpute in the imputation package to fill in missing
precipitation data for a data frame I have.
Example is:
okee:
Date rainfall
1997-05-010
1997-05-020
1997-05-03NA
1997-05-04
Hi, I'm new to R. I have some daily soil moisture data for the year 1979 in
netcdf format such as these
sm19790101.1.nc
sm19790102.1.nc
.
.
.
sm19791231.1.nc
I need to average a variable called sm to monthly resolution. I've done these
days = formatC(1:31, width=2, flag=0)
ncfiles =
Hi everyone,
I did a CCA with R in the vegan package and got all the outputs I need/want
to report except for one, that is the species environment correlation
values. I only know of them because of one of the sources I read up on about
CCAs reported it and I have the suspicion that it´s a value
Hello
I have a dataset with 151 dependent variables and estimated in 6 different
conditions, and each condition is repeated 3 times. It si proteomic and
therefore expensive, so few repetitions.
I want to conduct a stepwise discriminant analysis to identify the variables
that really matter. To do
Hello everyone,
I am trying to fit the following model
All X. variables are continuous, while the conditions are categoricals.
model - lm(X2
Thanks very much for your rapid help Arun.
Vince
On Apr 12, 2013, at 4:10 PM, arun kirshna [via R] wrote:
Hi,
From your example data,
dat1- read.table(text=
id1 id2 value
a b 10
c d11
b a 10
c e 12
,sep=,header=TRUE,stringsAsFactors=FALSE)
I'm using a multithreaded BLAS library with R and I see the expected
speed improvements with matrix multiplication, svd, etc. However,
glmnet continues to use only a single CPU. Since this package is
compiled from Fortran, is this the expected behavior or is there a way
to compile the glmnet
Hello all,
I have a problem with the way attributes are dealt with in the
function xmlToList(), and I haven't been able to figure it out for
days now.
Say I have a document (produced by nmap) like this:
mydoc - 'host starttime=1365204834 endtime=1365205860status
state=up reason=echo-reply
Trying to install rjsonio and I've run into a couple of issues.
1. When installing from R, it's starts to download, then I get
* installing *source* package 'RJSONIO' ...
ERROR: configuration failed for package 'RJSONIO'
* removing 'C:/Program Files/R/R-3.0.0/library/RJSONIO'
2. I've tried
Hi,
In the example you provided, it looks like the dates in Date2 happens first.
So, I changed it a bit.
DataA- read.table(text=
ID,Status,Date1,Date2
1,A,3-Feb-01,15-May-01
1,B,15-May-01,16-May-01
1,A,16-May-01,3-Sep-01
On Sat, Apr 13, 2013 at 6:31 AM, peter dalgaard pda...@gmail.com wrote:
But is it supposed to be t^{-3/2} or t^{-0.5}?? The formula has the former
and the code the latter, and the integral is clearly divergent with the
former.
I didn't look at that -- I just assumed that the code would
HI,
In cases like below:
DataA- read.table(text=
ID,Status,Date1,Date2
1,A,3-Feb-01,15-May-01
1,B,15-May-01,16-May-01
1,A,16-May-01,3-Sep-01
1,B,3-Sep-01,13-Sep-01
1,C,13-Sep-01,26-Feb-04
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