On 03/17/2011 07:46 AM, scarlet wrote:
I am new to the R language. I am trying to plot multiple figures on one page
through a loop, but the code just produce one graph on one page. Can someone
show some light on what's wrong?
Here is my code:
library("quantreg")
tcdata<-read.table("mydata.txt",
Some comments:
1. [^\s] matches everything up to a literal 's', unless perl=TRUE.
2. The (.*) is greedy, so you'll need (.*?)"\s"(.*?)"\s"(.*?)"$ or
similar at the end of the expression
With those changes (and removing a space inserted by the newsgroup
posting) the expression works for me.
Dear All,
I am trying to create a empty structure that I want to fill gradually
through the code.
I want to use something like rbind to create the basic structure first.
I am looking for a possibility to do an rbind where the columns names
dont match fully (but the missing columns can be defaulte
On 03/16/2011 08:20 PM, fre wrote:
I have the following problem:
I have some string with numbers like k. I want to have a table like the
function table() gives. However I am not able to call the first row, the 1,
2, 3, 5, 6 or 9. I tried to do that by a data.frame, but that doesn't seem
to work
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Subject: Re: [R] table() reading problem
Message-Id: <4d81c14a.2000...@bitwrit.com.au>
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fr
Ok thank you!
On Mar 17, 12:12 pm, wrote:
> subset(data,grepl("[1-5]", section) & !grepl("0", section))
>
> BTW
>
> grepl("[1:5]", section)
>
> does work. It checks for the characters 1, :, or 5.
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-proje
Thank you Thierry for your kind answer!
If you don't mind I would like to ask a follow-up question. In your
suggestions I get P-values for "Species". However, I am really not
interested in that factor per se. Would it make sense to use this
model instead if I am only interested in "Genotype"?
> m
It was a general question to find out if the caret framework supported
passing feature weights to any predictive model. But I guess, it may only
make sense in the context of the model. The ones I am interested in would be
svm and knn.
Thanks,
Kendric
On Wed, Mar 16, 2011 at 6:20 PM, Max Kuhn wro
I am working on a similar problem. I have to add two columns: one containing
the US state to which the origin belongs and another one to add the state in
to which destination belongs. All I have is the latitude and the longitude
of the origin and destination. Are there any packages in R that can do
Thanks !
You all made my day !
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Hi Alexx,
I don't see any problem in comparing models based on different distributions
for the same data using the AIC, as long as they have a different number of
parameters and all the constants are included.
For example, you can compare distribution mixture models with different number
of com
Hello,
it seems that rJava tries to detect the path to the Java Virtual Maschine
from a registry key which is not installed. I gues that the HLM means
HKEY_LOCAL_MASCHINE where you find normally the path to your installed Java.
On my Windows System i have for example the path
HKEY_LOCAL_MASCHINE\S
On 03/17/2011 02:47 AM, derek wrote:
I know I can add line to graph with abline(), but I would like to print
R-squared, F-test value, Residuals and other statistics from lm() to a
graph. I don't know how to access the values from summary(), so that I can
use them in a following code or print the
The first line of this reply is a definite candidate for the fortunes package!
best
i
--- On Thu, 17/3/11, bill.venab...@csiro.au wrote:
> From: bill.venab...@csiro.au
> Subject: Re: [R] Why doesn't this work ?
> To: ericst...@aol.com, r-help@r-project.org
> Date: Thursday, 17 March, 2011, 3:
Hello everybody,
I have a data frame in R which is similar to the follows. Actually my real 'df'
dataframe is much bigger than this one here but I really do not want to confuse
anybody so that is why I try to simplify things as much as possible.
So here's the data frame.
id <-c(1,1,1,1,1,1,1,1
And, just to make it really clear (I hope!):
Your original expression
z <-ifelse(t==1 || 2 || 3, 1,0)
looks like a transcription into "R" of the words
"If t equals 1 or 2 or 3 then z is 1 else z is 0"
However, your "t==1 || 2 || 3" has to be parsed in the correct
order according to operator
On 15/03/2011, Francisco Gochez wrote:
> Hi,
>
> What you are after is:
>
> datasubset <- dataset[ dataset[,3] == "text3", ]
Thank you. For the set
text1,23,text2,45
text1,23,text3,78
text1,23,text3,56
text1,23,text2,45
Is it possible to write a function that selects rows containing
'text3' and
Dear list,
I have to problems that are connected:
PROBLEM 1
I wonder if it is somehow possible to patch the function
'unlist(use.names=TRUE)' such that you can specify an arbitrary name
delimiter, e.g. "/" or "_". As I often name my variables "var.x.y", the
default delimiter makes it hard to dist
Hi!
Sorry, I made an error in the previous e-mail.
So try this:
by(df[,-1],df$id,function(x) apply(x,2,tabulate))
This gives you a list. You can rearrange it into a data frame or a 3d
array if you wish.
Regards,
Denes
> Hello everybody,
>
> I have a data frame in R which is similar to the
Hi again,
I'd like to ask you a question again.
I have a matrix like this:
a <-matrix(c(1,2,3,4,5,6,7,8,9,10,11,12))
a
[,1]
[1,]1
[2,]2
[3,]3
[4,]4
[5,]5
[6,]6
[7,]7
[8,]8
[9,]9
[10,] 10
[11,] 11
[12,] 12
Is there a proper way to change t
Is this what you want:
> x
V1 V2V3 V4
1 text1 23 text2 45
2 text1 23 text3 78
3 text1 23 text3 56
4 text1 23 text2 45
> str(x)
'data.frame': 4 obs. of 4 variables:
$ V1: Factor w/ 1 level "text1": 1 1 1 1
$ V2: int 23 23 23 23
$ V3: Factor w/ 2 levels "text2","text3": 1 2 2 1
$ V4
Using the summarise function in package plyr is one way; taking df to be
your data frame with variable names V1-V4,
library(plyr)
summarise(subset(df, V3 == 'text3'), sum = sum(V4))
sum
1 134
Another is to use the data.table package:
library(data.table)
dt <- data.table(df)
dt[V3 == 'text3', s
Hi list,
I am not a frequent user of R. Recently I used R in principal
component analysis and got the result as a class, which has information like
standard deviation and principal components from 1 to 10. How is it
possible to extract the column corresponding to first principal compon
I would like to assign an value to an element of a list contained in an
environment. The list will contain vectors and matrices. Here's a simple
example:
# create toy environment
testEnv = new.env(parent = emptyenv())
# create list that will be in the environment, then assign() it
x = list(a=1
t(matrix(a,3,4))
for more complex arrays, see ?aperm
> Hi again,
>
> I'd like to ask you a question again.
>
> I have a matrix like this:
> a <-matrix(c(1,2,3,4,5,6,7,8,9,10,11,12))
> a
>
> [,1]
> [1,]1
> [2,]2
> [3,]3
> [4,]4
> [5,]5
> [6,]6
> [7,]7
>
Hi Santosh,
May be you looking at something like this
merge(d,dNewTests,by="type",all=TRUE)
On Thu, Mar 17, 2011 at 1:03 PM, Santosh Srinivas <
santosh.srini...@gmail.com> wrote:
> Dear All,
>
> I am trying to create a empty structure that I want to fill gradually
> through the code.
> I want
On Thu, Mar 17, 2011 at 7:25 AM, Richard D. Morey wrote:
> I would like to assign an value to an element of a list contained in an
> environment. The list will contain vectors and matrices. Here's a simple
> example:
>
> # create toy environment
> testEnv = new.env(parent = emptyenv())
>
> # creat
On Thu, Mar 17, 2011 at 3:54 AM, wrote:
> It doesn't work (in R) because it is not written in R. It's written in some
> other language that looks a bit like R.
It parses in R, so I would say it was written in R.
To paraphrase Obi-wan, it's just not the R you are looking for.
__
matrix(a, ncol=3, nrow=4, byrow=TRUE)
or
> dim(a) <- c(3,4)
> a <- t(a)
> a
[,1] [,2] [,3]
[1,]123
[2,]456
[3,]789
[4,] 10 11 12
Depending on the context of the problem.
Sarah
On Thu, Mar 17, 2011 at 7:59 AM, Bodnar Laszlo EB_HU
wrote:
> Hi again
I have been trying to use lavaan (version 0.4-7) for a simple path model,
but the program seems to be computing far less degrees of freedom for my
model then it should have. I have 7 variables, which should give (7)(8)/2 =
28 covariances, and hence 28 DF. The model seems to only think I have 13
D
Arni Magnusson hafro.is> writes:
>
> I have been reading about autocorrelation in linear models over the last
> couple of days, and I have to say the more I read, the more confused I
> get. Beyond confusion lies enlightenment, so I'm tempted to ask R-Help for
> guidance.
>
> Most authors are
On Wed, Mar 16, 2011 at 12:52 AM, wrote:
> The Lattice auto.key argument has a bug in R.12.2.
>
> R version 2.12.2 (2011-02-25)
> Platform: i386-pc-mingw32/i386 (32-bit)
>
> other attached packages:
> [1] lattice_0.19-17
>
> loaded via a namespace (and not attached):
> [1] grid_2.12.2
>
> If
Dear R users,
we are currently trying to set up a R working group at Swansea University.
I would be very grateful to get some information and feedback from R
users that have done something similar, in particular regarding:
- help with giving a general overview of what R is capable of
- help wi
I've used barplot(), including the anmes.arg parameter, on data frames
successfully, but I'm even newer to using zoo than I am to R. :-}
I am working on a functon that accepts a data frame ("df") as its
primary argument, extracts information from it to create a zoo, then
generates a plot based on
library(mgcv)
b <- gam(arcsine.success ~ s(date.num,clutch.size,k=50))
vis.gam(b,theta=30)
will fit a thin plate spline and plot it. k is an upper limit on the
number of degrees of freedom for the TPS, but the actual degrees of
freedom are chosen automatically (use the 'method' argument of gam
> Date: Wed, 16 Mar 2011 13:50:37 -0700
> From: solomon.mess...@gmail.com
> To: r-help@r-project.org
> Subject: Re: [R] How to make sure R's g++ compiler uses certain C++ flags
> when making a package
>
> Looks like the problem may be that R is automatic
On Mar 17, 2011; 11:43am Baugh wrote:
>> Question: can I simply substitute a dummy var (e.g. populated by zeros)
>> for "ID" to run the model
>> without the random factor? when I try this R returns values that seem
>> reasonable, but I want to be sure
>> this is appropriate.
If you can fit the
Exuse me, I don't claim R^2 can't be negative. What I say if I get R^2
negative then the data are useless.
I know, that what Thomas said is true in general case. But in my special
case of data, using nonzero intercept is nonsense, and to get R^2 less than
0.985 is considered poor job (standard R^2>
LMM without Random effect:
I want to run an LMM both with and without the random factor (ID). And then
extract the log-lik values from the two models in order to generate a
p-value.
with random factor as:lmer(y~x+(1|ID),data)
Question: can I simply substitute a dummy var (e.g. populated by z
> 2) I don't want to fit data with linear model of zero intercept.
> 3) I dont know if I understand correctly. Im 100% sure the model for my data
> should have zero intercept.
> The only coordinate which Im 100% sure is correct. If I had measured quality
> Y of a same sample X0 number of times I wo
I rewrite my previous comand, CVlm works now. It was related to having the
same names in the df than in the formula included at CVlm .
Now I'd like to know if the seed is of any special importance, or I can type
just seed=29 like in the example.
Thanks in advance, u...@host.com
--
View this me
I frequently get a segmentation fault error when using the "plot" command.
It happens about half the time.
We are running an old version of R (R version 2.8.0 (2008-10-20) on Linux.
I did a quick search for this problem and didn't find anything. Apologies if
I missed it.
*** Process received
On Thu, Mar 17, 2011 at 2:42 PM, Joanne Demmler wrote:
> Dear R users,
Hi
>
> we are currently trying to set up a R working group at Swansea University.
> I would be very grateful to get some information and feedback from R users
> that have done something similar, in particular regarding:
We s
On Thu, Mar 17, 2011 at 9:38 AM, David Wolfskill wrote:
> I've used barplot(), including the anmes.arg parameter, on data frames
> successfully, but I'm even newer to using zoo than I am to R. :-}
>
> I am working on a functon that accepts a data frame ("df") as its
> primary argument, extracts i
Dear Mark,
You cannot compare lm() with lme() because the likelihoods are not the same.
Use gls() instead of lm()
library(nlme)
data("sleepstudy", package = "lme4")
fm <- lm(Reaction ~ Days, sleepstudy)
fm0 <- gls(Reaction ~ Days, sleepstudy)
logLik(fm)
logLik(fm0)
fm1 <- lme(Reaction ~ Days, r
On 2011-03-17 02:08, derek wrote:
Exuse me, I don't claim R^2 can't be negative. What I say if I get R^2
negative then the data are useless.
I know, that what Thomas said is true in general case. But in my special
case of data, using nonzero intercept is nonsense, and to get R^2 less than
0.985 i
On Thu, Mar 17, 2011 at 6:58 AM, cchace wrote:
>
> I frequently get a segmentation fault error when using the "plot" command.
> It happens about half the time.
>
> We are running an old version of R (R version 2.8.0 (2008-10-20) on Linux.
>
> I did a quick search for this problem and didn't find a
Hi Erin,
On Thu, Mar 17, 2011 at 1:27 AM, Erin Hodgess wrote:
> Dear R People:
>
> Hello again!
>
> I found something unusual in the behavior of the "endpoints" function
> from the xts package:
>
>> x1 <- ts(1:24,start=2008,freq=12)
>> dat <- seq(as.Date("2008/01/01"),length=24,by="months")
>> li
see inline.
On Thu, Mar 17, 2011 at 4:58 AM, Rubén Roa wrote:
> Hi Alexx,
>
> I don't see any problem in comparing models based on different distributions
> for the same data using the AIC, as long as they have a different number of
> parameters and all the constants are included.
> For example
Your model is saturated.
I think lavaan calculates the number of degrees of freedom this way:
DF = n*(n + 1)/2 - t - n.fix*(n.fix + 1)/2
n = number of variables
t = number of free parameters
n.fix = number of fixed exogenous variables
So, if you fix the exogenous variables, as in mimic = "Mplus
Dear Andrew,
The reported df in lavaan is 0 which is correct. It is because this
path model is saturated. "28" is not the df, it is the no. of pieces
of information. The no. of parameter estimates is also 28. Thus, the
df is 0.
However, you are correct that there are only 13, not 28, free
paramet
1. There is a CRAN mirror at Bristol (roughly 100km from
Swansea?). They must have (or have had) some very active R users there.
2. Sundar Dorai-Raj and I developed a local R Archive Network
and subversion repository internal to a company where we used to work.
I believe t
On Wed, Mar 16, 2011 at 6:58 PM, jctoll wrote:
> Hi,
>
> I'm struggling to figure out the way to change the name of a column
> from within a loop. The problem is I can't refer to the object by its
> actual variable name, since that will change each time through the
> loop. My xts object is A.
>
On Wed, Mar 16, 2011 at 3:49 PM, derek wrote:
> k=lm(y~x)
> summary(k)
> returns R^2=0.9994
>
> lm(y~x) is supposed to find coef. a anb b in y=a*x+b
>
> l=lm(y~x+0)
> summary(l)
> returns R^2=0.9998
> lm(y~x+0) is supposed to find coef. a in y=a*x+b while setting b=0
>
> The question is why do I g
Taby,
First, it is better to reply to the whole list (which I have included on
this reply); there is a better chance of someone helping you. Just
because I could help with one aspect does not mean I necessarily can (or
have the time to) help with more.
Further comments are inline below.
On
Hello Allan,
Thanks the response. Provides me hope. I appreciate [3], might even go with
that.
And for posterity, here's the code (assuming pastebin never expires)
[1] Test string : http://pastebin.com/FyAFzmTv
[2] Pattern (modified as per your suggestion) : http://pastebin.com/s7VT0r5K
pattern
Dear R Users, R Core
Team,
I have a two dimensional space where I measure a numerical value in two
situations at different points. I have measured the change and I would like to
test if there are areas in this 2D-space where there is a different amount of
change (no change, increase, decrease).
On Thu, 17 Mar 2011, Henrik Bengtsson wrote:
On Thu, Mar 17, 2011 at 6:58 AM, cchace wrote:
I frequently get a segmentation fault error when using the "plot" command.
It happens about half the time.
We are running an old version of R (R version 2.8.0 (2008-10-20) on Linux.
I did a quick sea
I rewrite my previous comand, CVlm works now. It was related to having the
same names in the df than in the formula included at CVlm .
Now I'd like to know:
Q1: if the seed is of any special importance, or I can type just seed=29
like in the example?
Q2: I typed:
CVlm(df = mydf, form.lm = fo
Thank you for your very comprehensible answer.
I a priori know that model y=a*x+0 is right and that I can't get x=constant
nor y=constant.
I'm comparing performance of data gathering in my data set to another data
sets in which performance gathering is characterized by R-squared . The data
in dat
Yes they are. I had edited the reply, but It didn't help.
Correction:
2)I meant zero slope, no zero intercept.
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On Mar 17, 2011; 04:29pm Thierry Onkelinx wrote:
>> You cannot compare lm() with lme() because the likelihoods are not the
>> same. Use gls() instead of lm()
Hi Thierry,
Of course, I stand subject to correction, but unless something dramatic has
changed, you can. gls() can be used if you need to
Thats exactly what I would like to do. Any idea on good text? I've consulted
severel texts, but no one defined R^2 as R^2 = 1 - Sum(R[i]^2) /
Sum((y[i])^2-y*)) still less why to use different formulas for similar model
or why should be R^2 closer to 1 when y=a*x+0 than in general model y=a*x+b.
fr
different
combinations with "" and () around samp, but I keep getting the error
"object 'samp.csv' not found".
samp <- "20110317"
read.csv(file=samp.csv,...)
#next R processes some code that works fine, and then should save the
figure:
pdf(file=samp.pdf,...)
de
On Mar 17, 2011; 04:29pm Thierry Onkelinx wrote:
>> You cannot compare lm() with lme() because the likelihoods are not the
>> same. Use gls() instead of lm()
And perhaps I should have added the following:
First para on page 155 of Pinheiro & Bates (2000) states, "The anova method
can be used to
Dear,
I'm trying to run diagnostics on MCMC analysis (fitting a log-linear
model to rates data). I'm getting an error message when trying
Gelman-Rubin shrink factor plot:
>gelman.plot(out)
Error in chol.default(W) :
the leading minor of order 2 is not positive definite
I take it that somewher
Hi,
I have the following type of data: 86 subjects in three independent groups
(high power vs low power vs control). Each subject solves 8 reasoning problems
of two kinds: conflict problems and noconflict problems. I measure accuracy in
solving the reasoning problems. To summarize: binary respo
me, and I would like to be able
> to type in the date (later perhaps automate this using list.files) and then
> read the csv and write the pdf automatically. I have tried different
> combinations with "" and () around samp, but I keep getting the error
> "object 'sa
I suggest that you post this on the R-sig-mixed-models list where you
are more likely to find those with bothe interest and expertise in
these matters.
-- Bert
On Thu, Mar 17, 2011 at 7:44 AM, Franssens, Samuel
wrote:
> Hi,
>
> I have the following type of data: 86 subjects in three independent
> paste(samp, ".pdf", sep="")
[1] "20110317.pdf"
> paste(samp, ".csv", sep="")
[1] "20110317.csv"
On Thursday, March 17, 2011 at 10:05 AM, pierz wrote:
I would like to use samp as a part of a filename that I can change. My source
&
Jim,
Thanks for looking into this. The c without paste works. If the rq model
overrides the mfrow, I think I will have to piece together individual plots
using other software.
scarlet
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It is all a matter of what you are comparing too, or what the null model is.
For most cases (standard regression) we compare a model with slope and
intercept to an intercept only model (looking at the effect of the slope), the
intercept only model fits a horizontal line through the mean of the
Hi all,
I would like to fit a gamm model of the form:
Y~X+X*f(z)
Where f is the smooth function and
With random effects on X and on the intercept.
So, I try to write it like this:
gam.lme<- gamm(Y~ s(z, by=X) +X, random=list(groups=pdDiag(~1+X)) )
but I get the error messag
Scarlet,
If the mfrow is being overridden, perhaps the rimage package might be
able to piece the individual plots...
--
Muhammad Rahiz
Researcher & DPhil Candidate (Climate Systems & Policy)
School of Geography & the Environment
University of Oxford
On Thu, 17 Mar 2011, scarlet wrote:
Jim,
Is X a numeric variable or a factor? If it's numeric try
gam.lme<- gamm(Y~ s(z, by=X), random=list(groups=pdDiag(~1+X)) )
... since otherwise the separate X term is confounded with s(z, by=X).
(gam detects such confounding and copes with it, but gamm can't).
Simon
On 17/03/11 17:47, I
I have a matrix say:
23 1
12 12
00
0 1
0 1
0 2
23 2
I want to count of number of distinct rows and the number of disinct element
in the second column and put these counts in a column. SO at the end of the
day I should have:
c(1, 1, 1, 2, 2, 1, 1) for the distinct rows and c(1, 1, 1
Did you post your data or hypothetical data?
Usually that helps make your problem more clear and more interesting
( likely to get a useful response to your post).
From: tintin...@hotmail.com
To: r-help@r-project.org
Date: Thu, 17 Mar 2011 17:38:14
Dear Jim,
17.03.2011 20:54, Jim Silverton wrote:
I have a matrix say:
23 1
12 12
00
0 1
0 1
0 2
23 2
I want to count of number of distinct rows and the number of disinct element
in the second column and put these counts in a column. SO at the end of the
day I should have:
c(1, 1
Hey everyone,
I'm having a little trouble with ggplot. I have two sets of y-values, one
whose range is contained in the other. Due to the nature of the y-values, I
wish to scale the y axis with a log base two transformation. Furthermore, I
wish to plot the two sets of y-values as boxplots in separ
Hey everyone,
Nevermind, I figured it out:
ggplot(data, aes(x1, y1))+geom_boxplot()+scale_y_log2(lim=c(...))
:)
Cheers,
Godwin
[[alternative HTML version deleted]]
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Hi,
thank you for your elaborate answer. I downloaded Prof. Dayton's pdf and
will read it tomorrow.
A friend also told me that our professor said you can actually compare
AICs for different distributions. Apparently it's not correct strictly
speaking, because of the two different likelihoods,
Hi all,
I would like to fit a gamm model of the form:
Y~X+X*f(z)
Where f is the smooth function and
With random effects on X and on the intercept.
So, I try to write it like this:
gam.lme<- gamm(Y~ s(z, by=X) +X, random=list(groups=pdDiag(~1+X)) )
but I get the error messag
On Thu, Mar 17, 2011 at 02:54:49PM -0400, Jim Silverton wrote:
> I have a matrix say:
>
> 23 1
> 12 12
> 00
> 0 1
> 0 1
> 0 2
> 23 2
>
> I want to count of number of distinct rows and the number of disinct element
> in the second column and put these counts in a column. SO at the en
So, I've been confused by this for a while. If I want to create functions in
an apply, it only uses the desired value for the variable if I create a new
local variable:
> lapply(1:5,function(h){k=h;function(){k}})[[1]]()
[1] 1
> lapply(1:5,function(k){function(){k}})[[1]]()
[1] 5
>
Normally, a
Hi,
I think it's a side effect of lazy evaluation, where you should
probably use the ?force like a jedi,
lapply(1:5,function(k){force(k) ; function(){k}})[[2]]()
HTH,
baptiste
On 18 March 2011 07:01, jamie.f.olson wrote:
> So, I've been confused by this for a while. If I want to create funct
Thanks Jim,
It turns out that the problem was that all columns had been converted
to factors. Once I converted them back to numeric variables the code
worked fine. If anybody is wondering how, you can do this with the
following:
mynumber <- as.numeric(levels(myfactor))[myfactor]
There are plen
Try this:
lapply(1:5,function(i){i;function()i})[[2]]()
or
lapply(1:5,function(i){i;function(j=i) j } )[[2]]()
## both give 2
Now try:
lapply(1:5,function(i){function(j=i) j } )[[2]]()
## gives 5 !
The problem is that if you do:
lapply(1:5,function(h){function(h)h)
what you get is a l
On Thu, Mar 17, 2011 at 10:32 AM, Joshua Ulrich wrote:
> On Wed, Mar 16, 2011 at 6:58 PM, jctoll wrote:
>> Hi,
>>
>> I'm struggling to figure out the way to change the name of a column
>> from within a loop. The problem is I can't refer to the object by its
>> actual variable name, since that wi
I attach the data (csv format). There are the 3 coordinates, (but as there are
not so many points I wanted two do 3 analysis in each of them collapsing one
variable).There are two variables to study I have posted the data as a ratio
between both states and as a percentage state between both sta
On Thu, Mar 17, 2011 at 10:23:33AM -0400, Gabor Grothendieck wrote:
> On Thu, Mar 17, 2011 at 9:38 AM, David Wolfskill wrote:
> ...
> > But the X-axis labels show up as "large integers" -- the POSIXct values
> > are apparently treated as numeric quantities for this purpose.
> ...
> Please cut this
Hi dear all,
It may be a simple question, i have a list output with different number of
elements as following;
[[1]]
[1] 0.86801402 -0.82974691 0.3974 -0.98566707 -4.96576856 -1.32056754
[7] -5.54093319 -0.07600462 -1.34457280 -1.04080125 1.62843297 -0.20473912
[13] 0.30659907 2.669081
Hey all!
I am working on my master thesis and I am desperate with my model.
It looks as following:
Y(t) = β1*X1(t) + β2*X2(t) + δ*(β1*((1+c)/(δ+c))+β2)*IE(t) -
β2*α*((1+c)/(δ+c))*(δ+g)* IE(t-1)
note: c and g is a constant value
The problem I encounter is that between IE(t) and IE(t-1) there is
Hello,
Thanks in advance for any help,
I have read a CSV file in which there is a column for an IP addr as in:
tmpInFile$V2
[1] "74.125.224.38" "74.125.224.38" "129.46.71.19" "129.46.71.19"
[5] "129.46.71.19" "129.46.71.19" "129.46.71.19" "129.46.71.19"
[9] "129.46.71.19" "129.46.71.19
Hi All,
I'm trying to plot data that is a time series of flows that are associated
with a specific level, and I would like each level to represent a colour
in a line plot. Here is some data that approximates what I'm using:
date=c(1:300)
flow=sin(2*pi/53*c(1:300))
levels=c(rep(c("
Dear List,
This is an embarrassing question, but I can seem to make this work
How do I
change the font size on the xlab and on the numbers shown in the x-axis on
the time series plot below. The arguments cex.lab and cex.axis do not seem
to be 'passing' to the plot function.
plot(ts(rnorm(100), s
Thank you for our reply. It's a pity, that 2 variables defined by different
formula have same name. If the variables had been named differently, I
wouldn't have problem at all and it looks like it's done on purpose.
Because I test a quality of data (performance of collecting data) not a
model which
?unlist
quantile(unlist(data))
> Hi dear all,
>
> It may be a simple question, i have a list output with different number of
> elements as following;
>
> [[1]]
> [1] 0.86801402 -0.82974691 0.3974 -0.98566707 -4.96576856
> -1.32056754
> [7] -5.54093319 -0.07600462 -1.34457280 -1.04080125
try this:
> x
[1] "74.125.224.38" "74.125.224.38" "129.46.71.19" "129.46.71.19"
"129.46.71.19" "129.46.71.19"
[7] "129.46.71.19" "129.46.71.19" "129.46.71.19" "129.46.71.19"
"129.46.71.19" "129.46.71.19"
> table(x)
x
129.46.71.19 74.125.224.38
10 2
> which.max(table
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