On Oct 19, 2012, at 06:05 , Thomas Lumley wrote:
> On Fri, Oct 19, 2012 at 12:21 PM, Sheng Liu wrote:
>> Thanks a lot. It's very helpful.
>> I've read through the c code. Just FYI and for my completion of the
>> question, I post some of my thought on it:
>> To me it looks like the algorithm is a
I am aware of the most basic stuff, especially vectorization and
subscripting.
I only gave a simple example with length 4. I need to do that for vectors of
length 190.
Are there any in-built commands?
Or should I write the loops myself?
Thank you.
--
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I need to estimate the parameters for negative binomial distribution (pdf)
using maximun likelihood, I also need to estimate the parameter for the
Poisson by ML, which can be done by hand, but later I need to conduct a
likelihood ratio test between these two distributions and I don't know how
to st
Have you read the Introduction to R tutorial? You seem unaware of even
the most basic stuff, especially vectorization and subscripting.
Also, this smells like homework, and we don't do homework here.
-- Bert
On Thu, Oct 18, 2012 at 12:33 PM, djbanana wrote:
> I would like to code the following
I'm new to R and igraph and I was wondering if anybody can help me with the
following.
I want to find the edge weight between two vertices in a graph. My graph
structure is defined by the normal ego (node1), alter (node2) and the weight
of the edge between them.
I know that I can get the weight f
Dear R users,
I am trying to find the mle that involves integration.
I am using the following code and get an error when I use the nlm function
d<-matrix(c(1,1,0,0,0,0,0,0,2,1,0,0,1,1,0,1,2,2,1,0),nrow=10,ncol=2)
h<-matrix(runif(20,0,1),10)
integ<-matrix(c(0),nrow=10, ncol=2)
ll<-function(p){
I am a new user of R and am crunching through the system. I have reached an
impasse with mapping; I want to make a bubble map and lay it over a grid that
is composed of a standard x,y axis. Within this, are 16 (4x4) gridded blocks,
numbered 1-16. And, within these individual hectares(1-16) i
I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) +
a3(b1+b2+b4) + a4(b1+b2+b3)
or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i
I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i
* b_i
would appreciate some help.
Thank you.
--
View this
I am using DIANA/DAISY/GOWER. Some of my categorical data include NULLS.
When assessing disimilarity, these NULLS are considered similar. I do not
want these NULLS to contribute towards similarity. Instead is there a way
for these NULLS to each be considered different so as to contribute to
disimi
Hi
I have created a loop to obtain data from several webpages
but the loop keeps crashing with the error
"Error in function (type, msg, asError = TRUE) :
Operation timed out after 5000 milliseconds with 9196 bytes received"
Page = getURLContent(page[i], followlocation=TRUE, curl = curl,.o
On Fri, Oct 19, 2012 at 12:21 PM, Sheng Liu wrote:
> Thanks a lot. It's very helpful.
> I've read through the c code. Just FYI and for my completion of the
> question, I post some of my thought on it:
> To me it looks like the algorithm is actually inquiring through an
> approximation table (the a
HI,
You can also try this.
Should remove the problematic lines, but it may not match up to Rui's method in
terms of speed.
input<-readLines(textConnection("TABLE NO. 1
COL1 COL2 COL3 COL4 COL5 COL6
COL7 COL8 COL9 COL10 C
Hi Dr. Lumley,
Further thoughts: To get the histogram of age with proportions (relative
frequencies) on y-axis, I probably need to rescale the weight for each subgroup
separately so that the rescaled weight would sum to 1 for the respective
subgroup. Am I correct?
Thanks,
Pradip Muhuri
r-sig-geo is a better place to ask this question.
There doesn't appear to be anything wrong with how you are using
spTransform().
"x" should be longitude
"y" should be latitude
More precisely, the first column of your matrix
matrix(c(x,y), ncol=2)
should be longitude, the second column la
r-sig-geo would be a better place to ask this question.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 10/18/12 6:52 AM, "Filoche" wrote:
>Dear R users,
>
>I'm currently trying to re-project a geotiff in another coord
Thanks a lot. It's very helpful.
I've read through the c code. Just FYI and for my completion of the
question, I post some of my thought on it:
To me it looks like the algorithm is actually inquiring through an
approximation table (the approximations, at least for pnom, is "derived
from those in "R
i believe ff has a dataframe class. as for your object data im less clear.
how big is it
On Oct 18, 2012 12:45 PM, "Alexander Shenkin" wrote:
> Hi Folks,
>
> I've been bumping my head against the 4GB limit for 32-bit R. I can't
> go to 64-bit R due to package compatibility issues (ROBDC - possib
Hi Mihnea,
I'm afraid I don't have much time to look at this right now -- I'm
sort of swamped for the next 5 days -- so I'm forwarding back to the
list so more eyeballs pass over it. In general, you should always cc
the list so the archives remain complete and so that you get a quicker
response.
Hi Li,
I'm afraid I don't have much time to look at this right now -- I'm
sort of swamped for the next 5 days -- so I'm forwarding back to the
list so more eyeballs pass over it. In general, you should always cc
the list so the archives remain complete and so that you get a quicker
response.
On T
Hello Dr. Lumley,
Thank you for your advice/suggestions.
I have rescaled the weight (i.e., "original weight" divided by "total weighted
count" averaged across 8 surveys - NHIS). As can be seen below (R console), the
new weight sums to 1.
I have used the freq=TRUE argument in the svyhist () fun
Here's an example. The problem is that hexbin uses grid graphics, not
the base graphics that plot.svysmooth() uses, so you need to do the
plotting yourself with grid. The necessary viewport is returned by
svyplot()
hexstuff<-svyplot(api00~api99, design=dstrat, style="hex", xlab="1999
API",ylab="
System Info:
R 2.14.2
Windows 7 Pro x64 SP1
8GB RAM
On 10/18/2012 3:42 PM, Alexander Shenkin wrote:
> Hi Folks,
>
> I've been bumping my head against the 4GB limit for 32-bit R. I can't
> go to 64-bit R due to package compatibility issues (ROBDC - possible but
> painful, xlsReadWrite - not possi
Hi Folks,
I've been bumping my head against the 4GB limit for 32-bit R. I can't
go to 64-bit R due to package compatibility issues (ROBDC - possible but
painful, xlsReadWrite - not possible, and others). I have a number of
big dataframes whose columns all sorts of data types - factor,
character,
Hi
On 19/10/2012 6:49 a.m., Durant, James T. (ATSDR/DCHI/SSB) wrote:
Hi all-
So sorry to bother you all with something pretty basic.
I am trying to add the lines method output from svysmooth to a
svyplot with style="grayhex". However, the line either appears in
the wrong place or if I am runn
So I do not find example what I expect.
I plan to estimate the multi-factor model for Kalman Filter Mean Reverting,
Random Walk and Random Coefficient.
For example:
R(it)= Alpha(it)+ Beta(it)R(mt)+Gamma(it)(R(mt)^2)+delta(it)(R(mt)^3)+ V(it)
KF Random walk
Alpha(it)= Alpha(it-1)+W(i1t)
Beta
Hi Peter,
Thank you for your time and input on this matter. I will take your advice
and reconstruct the blocks using a different maximum cutoff and plot the
figures next to one another. I also wanted to thank you for creating the
WGCNA package and including the section on exporting to other visual
I know these package but I plan to analyse financial multi factorial data
set, and also estimate diffuse initial values for these
models.
I generated my own code, but I had problem with optim() package problem. I
need some constraints and I do not apply it
in my code.
Do you have any suggesti
I have generated plot using ggplot, and i would like to add mean value to the
regression line as a marker. how to do it?
p <- ggplot(data, aes(x = x, y = y,
color = a, shape = factor(sex),
linetype = factor(sex)))
p0 <- p +
scale_color_man
Hello,
I am trying to set up a loop that can run the survdiff function with the
ultimate goal to generate a csv file with the p-values reported. However,
whenever I try a loop I get an error such as "invalid type (list) for
variable 'survival_data_variables[i]".
This is a subset of my data:
str
Hello,
Time down by a factor of 4. It still takes some minutes, 2 mins for a
file of 380Mb/3.6M lines. So maybe system commands (maybe awk?) can do
the job better.
fun <- function(infile, outfile, lines = 1L){
remove <- function(x){
i1 <- grep("TABLE", x)
i2 <- grep("C
Good suggestion - I tried it - it does not work (line appears too high with the
data I am working with).
VR
James
James T. Durant, MSPH CIH
Environmental Health Scientist
US Agency for Toxic Substances and Disease Registry
Atlanta, GA 30341
770-488-0668
From: Anthony Damico [mailto:ajdam...@
try adding legend = 0 to your svyplot()
On Thu, Oct 18, 2012 at 1:49 PM, Durant, James T. (ATSDR/DCHI/SSB) <
h...@cdc.gov> wrote:
> Hi all-
>
> So sorry to bother you all with something pretty basic.
>
> I am trying to add the lines method output from svysmooth to a svyplot
> with style="grayhex"
Hi,
I agree with Sarah that read.table() will much easier in this case.
If you want scan(), then you can try this:
dat1<-scan(what=list("numeric","character","character","numeric"),text="
id name sex age
111 HELEN f 22
112 DONNA f 22
113 ERIC m 21
114 LINDA
Hi,
You can also try this:
a <- "1,2"
as.numeric(c(gsub("(.*)\\,(.*)","\\1",a), gsub("(.*)\\,(.*)","\\2",a)))
#[1] 1 2
- Original Message -
From: BenM
To: r-help@r-project.org
Cc:
Sent: Thursday, October 18, 2012 10:17 AM
Subject: Re: [R] converting a string to an integer vector
T
Hi all-
So sorry to bother you all with something pretty basic.
I am trying to add the lines method output from svysmooth to a svyplot with
style="grayhex". However, the line either appears in the wrong place or if I
am running in R Studio it causes the system to crash.
I know this is somethi
Hi Rainer,
Thanks for notifying me. You are right.
Sorry, I was working with library(reshape) instead of reshape2. So, I guess
the cast() will not work if we load only reshape2.
A.K.
- Original Message -
From: Rainer Schuermann
To: arun
Cc:
Sent: Thursday, October 18, 2012 1:33
Another option would be to read the data using read.table or similar
to get the data into a data frame then use the xtabs function,
something like:
result <- xtabs( count ~ docID + wordID, data=mydf)
On Thu, Oct 18, 2012 at 6:44 AM, Rui Esteves wrote:
> Hi,
>
> I downloaded a dataset from UCI
You have a list of models, the coef and confint functions only work on
a single model, so you need to use lapply or sapply to get the
information from each model. Possibly something like (untested):
tableOfOddsRatios <- sapply( models, function(x) exp(c( coef(x)[2],
confint(x)[2,]) )
I included
Hi Derek,
the simple answer is that the block-specific dendrograms cannot be
meaningfully combined into a single dendrogram. You have to plot them
separately. You can create a multi-panel figure that shows all block
dendrograms in one big figure, although with 10 blocks I would not
necessarily att
Possible? Yes. (see fortune("Yoda"))
Automated using the legend function? No
Automated using another function? possbly somewhere in the 4,000+
packages on CRAN, but I don't know which.
It is doable with the basic tools. You could either find a part of
your graph with open area to put the legend i
On 18/10/2012 12:56 PM, Paul Johnson wrote:
What is the correct format for the shebang line and what options are
allowed or necessary along with this?
I find plenty of blogs and opinions, but few authoritative answers.
The authoritative source for the options is the R help page ?Rscript.
The
What is the correct format for the shebang line and what options are
allowed or necessary along with this?
I find plenty of blogs and opinions, but few authoritative answers.
I want an R script to run and update packages periodically, with a
cron job that launches it. What is necessary to put in
On 18-10-2012, at 14:16, Martin Hehn wrote:
> Dear all,
>
> I am using the nlsLM function to fit a Lorentzian function to my experimental
> data.
> The LM algorithm should allow to specify limits, but the upper limit appears
> not to work as expected in my code.
> The parameter 'w', which is p
Hi,
You can also try this:
dat1<-read.table(text="
1 1 3
1 2 54
1 3 11
1 4 17
2 1 5
2 4 78
2 5 20
",sep="",header=FALSE)
library(reshape2)
dat2<-cast(dat1,V1~V2)
dat2<-dat2[,-1]
dat2[is.na(dat2)]<-0
dat3<-as.matrix(dat2)
dat3
# [,1] [,2] [,3] [,4] [,5]
#[1,] 3 54 11 17 0
#[2,] 5
Package dlm does it, as well as other contributed packages (KFAS, sspir,
dse,...)
Best,
Giovanni
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] on behalf of
nserdar [snes1...@hotmail.com]
Sent: Thursday, October 18, 2012 8:40 AM
To: r-
On Thu, Oct 18, 2012 at 4:20 PM, Sam Steingold wrote:
>> * R. Michael Weylandt [2012-10-18 16:01:37
>> +0100]:
>>
>> On Thursday, October 18, 2012, Sam Steingold wrote:
>>
>>> > * Bert Gunter [2012-10-17 23:21:44 -0700]:
>>> >
>>> > However, Is level "5" in 'a' the same as level "5" in 'b' ?
>>
And there's a wealth of textbooks on time series based in R. Two to start with:
Paul S.P. Cowpertwait and Andrew V. Metcalfe (2009). Introductory Time Series
with R. Use R! Series. Springer.
Robert H. Shumway and David S. Stoffer (2006). Time Series Analysis and Its
Applications With R Examples
Replace the y[1] with x[2] in the apply function - there is no 'y' defined in
the function passed to apply. In the apply function, x is an 2-element vector
representing a single row of the data (since the 1 in the second argument of
apply says that it is applied row-wise). So the second element,
Hi,
Are you looking for something like this?
#dat1
aggregate(dat1[,2:4],by=list(dat1[,1]),FUN=mean)
# Group.1 val1 val2. val3
#1 AKT 2.2 5.00 6
#2 CKT 4.5 5.75 7
A.K.
- Original Message -
From: Nico Met
To: R help
Cc:
Sent: Thursday, October 18, 2012 11:59 AM
Sub
> * William Dunlap [2012-10-18 15:33:38 +]:
>
> c() has an unfortunate history.
:-)
ISTR reading in the R manual ~15(?) years ago that the language was in a
flux and one could not expect code written for the current release to
work in the next release. I was considering R as the graphing bac
? .onAttach
? .onLoad
at slightly different points in the load process.
Cheers,
Michael
On Wed, Oct 17, 2012 at 10:12 PM, Marc Girondot wrote:
> Dear List Member
>
> Is there a mechanism to automatically run a piece of R code when a package
> is loaded using library("xxx") ? Preferentially that
Dear all,
I want to calculate mean values for multiple rows:
structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L), .Label = c("AKT", "CKT"), class = "factor"), val1 = c(2,
3, 2, 2, 2, 5, 3, 8, 2), val2. = c(4, 5, 4, 8, 4, 8, 4, 7, 4),
val3 = c(5, 6, 5, 9, 5, 9, 5, 9, 5)), .Na
There are a bunch of things wrong with your code, from not
understanding subsetting to only returning a single value rather than
a vector of values. I highly recommend that you read the Introduction
to R document that came with your R installation and is readily
available online.
In the meantime:
Sure:
http://cran.r-project.org/web/views/TimeSeries.html
Next time please try google first.
Best,
Ista
On Thu, Oct 18, 2012 at 11:29 AM, vas wrote:
> Hello,
>
> I am totally new in the field of time series analysis and forecasting and R.
>
> I read that R is a powerful tool for time series. C
Why do you need to use scan()? read.table() would be much easier.
Sarah
On Thu, Oct 18, 2012 at 9:14 AM, killerkarthick wrote:
> Hi
> I have one text file which containing 4 variables with 10 observations.
> I would like to import with scan() function. Please give some
> suggestion.
c() has an unfortunate history. Originally, c(x) stripped the attributes,
except names but including dim, dimnames, and class, from x.
Also, c(x,y) stripped the attributes from both x and y and concatenated
them. Also, c(nameA=1,nameB=2) constructed a vector with a names attribute.
Then c() bec
Hello!
I am trying to model data on species abundance (count data) with a poisson
error distribution. I have a fixed and a random variables and thus needs a
mixed model. I strongly doubt that my model is overdispersed but I don't
know how to get the overdispersion parameter in a mixed model. Maybe
Hello,
Try the following, readaing your file into 'x', using readLines.
tc <- textConnection("
TABLE NO. 1
COL1COL2COL3COL4COL5 COL6
COL7COL8COL9COL10 COL11 COL12
1.0010E+05 0.E+00 1.E+00 1.E+03 -1.E+0
I analyzed the kalman filter based approaches like mean reverting, random
coefficient and random walk.
At this point Automatic package is inadequate and need some constraints. I
also found Kalman Filter code
in Shumway$Stoffer book, but it did not provide the correct optimization.
Can you su
Hi Arun,
Thanks, it works now.
Hi Rui,
Your code also works.
Thanks,
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Dear all,
I am trying to project my LongLat-maps to a plane.
The ultimate purpose is to do a search of points in vicinity of other points
using overlay-commands (sp) with radius in km.
I am applying spTransform (package rgdal) and it gives my some curious results.
An example.
Let's take a poin
Hello,
I am totally new in the field of time series analysis and forecasting and R.
I read that R is a powerful tool for time series. Could anyone give me
navigation what models of time series are availiable in R etc?
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> * Jeff Newmiller [2012-10-18 07:53:24 -0700]:
>
> If you HAVE defined your factors using explicit levels definitions, you
> should have no trouble combining them.
http://article.gmane.org/gmane.comp.lang.r.general:277719
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11
Hello fellow R users,
I am currently learning to use R, so please forgive me if there is an
obvious explanation for the following problem. My goal is to perform WGCNA
on a dataset of 19776 genes, so I opted to follow the block-wise network
construction (Section 2c) in the WGCNA R Tutorial by Peter
Dear R users,
I'm currently trying to re-project a geotiff in another coordinate system.
For instance, I have a tif image in UTM 19 zone which I would like to
reproject into UTM 18. I was wondering if it was possible in R.
Furthermore, I looked into 'rgdal' package, but I can't really find out if
Hi
I have one text file which containing 4 variables with 10 observations.
I would like to import with scan() function. Please give some
suggestion
Thanks...
Mydata set is.
id namesex age
111 HELEN f 22
112 DONNA f 22
113 ERICm
Dear all,
Thanks for the many replies!
Indeed, the function does not represent my explanation. It should have been:
if (x[i] > 170 && x[i] < 1250 && y[j] > 150 && y[j] < 480)
Sorry, my mistake.
My data looks like this:
head(input[,3:4])
x y
1 701 209
2 685 209
3 566 248
4 562 234
5
Depending on what exactly you are trying to accomplish:
> as.numeric(unlist(strsplit(a, ",")))
[1] 1 2
> read.csv(textConnection(a), header=FALSE)
V1 V2
1 1 2
Sarah
On Thu, Oct 18, 2012 at 9:08 AM, BenM wrote:
> Hi All,
> Thanks in advance for your help. I'm trying to convert a strin
> * R. Michael Weylandt [2012-10-18 16:01:37
> +0100]:
>
> On Thursday, October 18, 2012, Sam Steingold wrote:
>
>> > * Bert Gunter [2012-10-17 23:21:44 -0700]:
>> >
>> > However, Is level "5" in 'a' the same as level "5" in 'b' ?
>>
>> yes, of course.
>> would anyone want to _different_ factors
Hi,
I'm trying to obtain a table of coefficients and confidence intervals from a
logistic regression analysis in R. My code is as follows:
# read in csv file
datafile<-read.csv("file.csv", row.names=1)
# read in the variable list
varlist<-names(datafile)[66:180]
models<-lapply(varlist, functio
Hi All,
Thanks in advance for your help. I'm trying to convert a string to an
integer vector. For instance, I will start with
a <- "1,2"
The result I want to end up with will be the equivalent of
c(1,2)
What's the best way to make the conversion? I've tried using as.integer(a),
but R s
Thank you very much. That appears to be what I wanted.
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__
R-help@
Dear R community,
is there any efficient way to use aaply on different datacubes? I have 3
dimesniolan datacubes/arrays with dimensions lon x lat x time. Now I
would like to do caclulations on each individual time series (e.g. all
vectors along the third dimension) using a time series (or mor
hi Jorge,
> * Jorge I Velez [2012-10-18 16:43:58 +1100]:
>
>> a <- factor(5:1,levels=1:9)
>> b <- factor(9:1,levels=1:9)
>> lev <- sort(unique(f <- c(a, b)))
>> f <- factor(f, levels = lev)
>> str(f)
> Factor w/ 9 levels "1","2","3","4",..: 5 4 3 2 1 9 8 7 6 5 ...
is sort(unique()) really neces
On Thursday, October 18, 2012, Sam Steingold wrote:
> > * Bert Gunter [2012-10-17 23:21:44 -0700]:
> >
> > However, Is level "5" in 'a' the same as level "5" in 'b' ?
>
> yes, of course.
> would anyone want to _different_ factors with identical string
> representations?!
>
>
Off the cuff, studyin
mood <- factor(c("blue", "sunny"))
skycolor <- factor(c("azure","blue","teal")
If factors are not defined with levels specifications, automatic merging should
never be allowed. The fact that read.table automatically generates factors
using default levels is why I nearly always import using as.is
> * Bert Gunter [2012-10-17 23:21:44 -0700]:
>
> However, Is level "5" in 'a' the same as level "5" in 'b' ?
yes, of course.
would anyone want to _different_ factors with identical string representations?!
--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http:
Jason
Are you suggesting grep in R or grep in the system? If the latter, this won't
work because I need to implement this same procedure in Windows (sorry about
not mentioning this), in which grep does not exist. If in R, the syntax is not
obvious -- could you provide an example?
Dennis
Den
On 10/18/2012 09:57 AM, Fisher Dennis wrote:
R 2.15.1
OS X
Colleagues,
I am reading a 1 GB file into R using read.table. The file consists of 100
tables, each of which is headed by two lines of characters.
The first of these lines is:
TABLE NO. 1
The second is a list of column header
Still so perfect Rui! A bit much more complicated as what I thought,
nevertheless it's what I want!
Thank you Rui!
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Hello,
Try this, It'll maybe help you:
a <- "1,2"
b <- strsplit(a,",") #split your data according to ","
b <- unlist(b) # it creates a list, so we unlist the result to obtain a
vector like c(1,2)
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R 2.15.1
OS X
Colleagues,
I am reading a 1 GB file into R using read.table. The file consists of 100
tables, each of which is headed by two lines of characters.
The first of these lines is:
TABLE NO. 1
The second is a list of column headers.
For example:
TABLE NO. 1
COL1CO
Hello,
The function below fails if any of the followind conditions is met.
1. The first value in the vector is NA
2. There are less than n = 5 (or 500) values before/after the
first/last, respectively, NA.
fun <- function(x, n){
na <- is.na(x)
rna <- rle(na)
sna <- cumsum(rna$len
Hi everybody,
I have a little problem about filling some gaps of NAs in my data.
These gaps are between nearly constant data (temperature under snow). Here's
a fake example to illustrate how it looks like approximately:
DF <-
data.frame(data=c(-0.51,-0.51,-0.48,-0.6,-0.54,-0.38,-0.6,-0.42,NA,NA,
Hello,
It's much easier than you think, the first two columns of the input
matrix are the row and column numbers into the output matrix, therefore
those columns form an index matrix. Just see:
x <- scan(text="
1 1 3
1 2 54
1 3 11
1 4 17
2 1 5
2 4 78
2 5 20
")
mat <- matrix(x, ncol = 3, byro
Better would be to use interval censored data. Create your data set so that
you have
(time1, time2) pairs, each of which describes the interval of time over which
the tag was
lost. So an animal first captured at time 10 sans tag would be (0,10); with
tag at 5 and
without at 20 would be (5,2
- Why are you making a matrix for lookzone if it is ever only one number?
- changing a variable used for iteration is bad practice (input in this case)
though you aren't changing the rows i guess.
- There are too many x and too few y in those ifs.
Plus this is shorter:
test<-apply(input,
Hi,
I downloaded a dataset from UCI repositories named Bag of Words:
http://archive.ics.uci.edu/ml/machine-learning-databases/bag-of-words/readme.txt
The dataset is in a text file with the following structure:
---
docID1 wordID1 count
docID1 wordID2 count
docID1 wordID3 count
docID1 wordID4 cou
If you use ifelse() you don't need the loops at all since ifelse() is
vectorized.
But I don't have any idea why you need two loops: do you really need to
compare all pairs of observations?
Also, the logic in your if statements is not the same as the logic in your
problem description.
Sarah
On T
Hi,
May be this also works:
a <- factor(c(1,3,5))
b <- factor(c(5,7))
f1<-as.numeric(c(as.character(a),as.character(b)))
lev<-as.numeric(c(levels(a),setdiff(levels(b),levels(a
f2<-factor(f1,levels=lev)
f2
#[1] 1 3 5 5 7
#Levels: 1 3 5 7
a1<-factor(5:1,levels=1:9)
b1<-factor(9:1,levels=1:9)
Dear all,
I am using the nlsLM function to fit a Lorentzian function to my experimental
data.
The LM algorithm should allow to specify limits, but the upper limit appears
not to work as expected in my code.
The parameter 'w', which is peak width at half maximuim always hits the upper
limit if t
Without commenting on the method itself, shouldn't you be using both x and y
twice within each loop instead of x three times and y only once?
Please provide a snippet of your data by including the output of
dput(head(data,30)) for example.
> -Original Message-
> From: r-help-boun...@r-p
Hello,
I am trying to use this fix for the convergence problem in polr, but I don't
seem to get the change of code right. I redefine the function polr by the
lines
tjb wrote
> if(missing(start)) {
> # try something that should always work -tjb
> u <- as.integer(table(y))
> u
Hi List,
I want to show three matrices in a 3D plot, one horizontal, and the other
two as vertical slices.
Ideally this would use the persp or drape.plot functions.
I have found a means of assigning one matrix to the lower xy plane (like a
carpet), thus:
mat_2D <- matrix(runif(100),nrow=10,ncol=1
I generated maps with the function symbols (graphics). These are basic
maps generated with :
symbols(x,y,circles=myvariable)
where x et y are spatial coordinates corresponding to replicates of
"myvariable".
I would associate legend to this kind of maps, is it possible?
Regards,
Marion.
--
Dear all!
I'm quite new to R and at this moment a little bit lost in trying to make a
function work with a for loop
that has two variables (i and j) and multiple if statements..
I hope that someone of you could help me. I would very appreciate it!
What do I want to do? I am analysing the data o
Thanks for the reply!!
Inga
--
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Hello,
I am trying to run the demoscript for the package "siar". However, using
the example script and example data supplied I cannot get 1) the model to
run unless I specify iterations =500 and burnin = 500 and 2) if the model
does work without crashing R then the function "siarhistograms" does
As Rui said, this is a pretty vague question... Nonetheless, if you are using
lm() for example you should use the newdata argument in predict(). Have a look
at ?predict.lm for example.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On B
Hi,
I agree with Ista that you need the second option. So long as you don't want to
look at interactions between condition and item it shouldn't be a problem that
you don't have the same items in each condition. However, it may be difficult
to find a main effect of condition that couldn't also
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