On 13/07/2012 21:37, Greg Snow wrote:
A permutation test may be appropriate:
Yes, it may, but precisely which one is unclear. You are testing
whether the two samples have an identical distribution, whereas I took
the question to be a test of differences in dispersion, with differences
in lo
On 7/13/2012 8:37 PM, darnold wrote:
Hi,
I hope that folks can give me some simple approaches to taking the data set
below, which is accumulated in two columns called "long" and "group", then
arrange the data is the "long" column into a data frame containing five
variables: "Group 1", "Group 2",
Hi,
I hope that folks can give me some simple approaches to taking the data set
below, which is accumulated in two columns called "long" and "group", then
arrange the data is the "long" column into a data frame containing five
variables: "Group 1", "Group 2", "Group 3", "Group 4", and "Group 5".
Very nice suggestion. I am getting some very kind help here.
x1 <- round(rnorm(10,60,3))
x2 <- round(rnorm(10,65,3))
x3 <- round(rnorm(10,70,3))
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method="stack",
pch=4,
offset=1/2,
col="blue",
set.seed(42)
d <- array(as.integer(round(runif(125)*10, 0)), dim=c(5, 5, 5))
data_int <- apply(d, c(1,2), sum)
-
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77840-4352
- Original Message -
From: "Sam B
On 2012-07-13 11:37, Rui Barradas wrote:
Hello,
Or maybe the argument 'pos' of axis().
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method="stack",
pch=4,
offset=1/2,
col="blue",
lwd=2,
las=1,
xlim
On 2012-07-13 15:55, Katharine Miller wrote:
I have a function that requires a distance matrix of class dist with
species names as row names. For the life of me, I cannot figure out how to
get dist() to include species names.
I am sure this must be easy, because a lot of packages and functions
I have a function that requires a distance matrix of class dist with
species names as row names. For the life of me, I cannot figure out how to
get dist() to include species names.
I am sure this must be easy, because a lot of packages and functions out
there require dist objects to have row name
Sorry, I meant Kolmogorov-Smirnov test.
Thanks Peter for correction.
Weidong
On Fri, Jul 13, 2012 at 4:56 PM, Peter Ehlers wrote:
> On 2012-07-13 13:33, Weidong Gu wrote:
>>
>> Hi Joerg,
>>
>> Seems Mann-Whitney-Wilcoxon test (ks.test in R) would do the work
>> which tests differences anywhere
On 2012-07-13 06:07, Chris82 wrote:
Hello,
hiere is a small reproducible example.
All z.i which are NA should be transparent at the plot, but they are white
colored.
### Example image.plot regular x,y grid ###
x <- seq(2,2.9,0.1)
y <- seq(42,42.9,0.1)
z <- matrix(seq(-5,4.9,0.1),nrow=10)
i
I just want to integrate a 3D data set along one dimension to obtain a
2D data set. Something like:
(given array "d" with dim nx,ny,nz ...)
data_int<-array(dim=c(nx,ny))
for (n in 1:ny) {
for (m in 1:nx) {
data_int[m,n]<-sum(d[m,n,])
}
}
The thing is, given R's facility w
On 2012-07-13 13:33, Weidong Gu wrote:
Hi Joerg,
Seems Mann-Whitney-Wilcoxon test (ks.test in R) would do the work
which tests differences anywhere in two distributions, e.g. tails,
interquartiles and center.
The ks.test() function refers to the Kolmogorov-Smirnov test,
not the Wilcoxon test.
If not linear, then perhaps nlrob() in package robustbase.
-
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77840-4352
- Original Message -
From: "Stephen Sefick"
To: "Lauren Vogric" , r-help@r-project
A permutation test may be appropriate:
1. compute the ratio of the 2 IQR values (or other comparison of interest)
2. combine the data from the 2 samples into 1 pool, then randomly
split into 2 groups (matching sample sizes of original) and compute
the ratio of the IQR values for the 2 new samples.
Hi Joerg,
Seems Mann-Whitney-Wilcoxon test (ks.test in R) would do the work
which tests differences anywhere in two distributions, e.g. tails,
interquartiles and center.
Weidong Gu
On Fri, Jul 13, 2012 at 7:32 AM, Schaber, Jörg
wrote:
> Hi,
>
> I have two non-normal distributions and use interq
For something like this the best (and possibly only reasonable) option
is to use simulation. I have posted on the general steps for using
simulation for power studies in this list and elsewhere before, but
probably never with coxph.
The general steps still hold, but the complicated part here will
They are due to measurement error, sample of a different population, or
... ? What is the unusual event? Does it explain something important
about the system that you are working on? I am not telling you not to
do what you are doing, but just writing things that I consider when I am
doing re
S+ and, I assume, R compute r^2 when there is no intercept as
sum(fitted(M1)^2) / sum(y^2)
where M1 is the fitted model and y the response.
See, for example,
http://web.ist.utl.pt/~ist11038/compute/errtheory/,regression/regrthroughorigin.pdf
for the derivation of this formula.
Bill Dunlap
How do I read/write liblinear models to files?
E.g., if I train a model using the command line interface, I might want
to load it into R to look the histogram of the weights.
Or I might want to train a model in R and then apply it using a command
line interface.
--
Sam Steingold (http://sds.podval
Hi Petr,
Yes, that's really very helpful.
Petr : Using this interpretation, AB occurs at lines 1,3,4 and not 1,3,5.
Is this correct?
Vineet : Yes , thats right sorry for the typo.
Petr: If some sequence contains several ocurrences of a pattern, for
example,
the sequence
A, B, A, B
contai
Do you have a good reason to throw these points out?
On Fri, Jul 13, 2012 at 2:17 PM, David L Carlson wrote:
> I didn't actually see any question in this posting, but instead of removing
> the outliers consider using a robust linear model.
>
> library(MASS)
> ?rlm
>
> The TeachingDemos package h
I didn't actually see any question in this posting, but instead of removing the
outliers consider using a robust linear model.
library(MASS)
?rlm
The TeachingDemos package has a data set called outliers to show what can
happen when you iteratively remove "outliers" in the way you suggest.
Hello,
Or maybe the argument 'pos' of axis().
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method="stack",
pch=4,
offset=1/2,
col="blue",
lwd=2,
las=1,
xlim=c(53, 77),
xaxt="n")
axis(1, at = seq(55, 75
What I'm trying to do is create best fit line in R for a set of data points
and then remove all the outliers to re-create a best fit. I can't use IQR
because the outliers I have in mind are easily within the range, but way out of
line for the best fit, which is ruining the fit. I'd rather thro
On Jul 13, 2012, at 18:54 , darnold wrote:
> OK, got this far:
>
> x1 <- round(rnorm(10,60,3))
> x2 <- round(rnorm(10,65,3))
> x3 <- round(rnorm(10,70,3))
> stripchart(list(sample1=x1,sample2=x2,sample3=x3),
> method="stack",
> pch=4,
> offset=1/2,
> col="
try something like
abline(h=1.9)
John Kane
Kingston ON Canada
> -Original Message-
> From: dwarnol...@suddenlink.net
> Sent: Fri, 13 Jul 2012 09:54:35 -0700 (PDT)
> To: r-help@r-project.org
> Subject: Re: [R] Side by side strip charts
>
> OK, got this far:
>
> x1 <- round(rnorm(10,60,
On Jul 13, 2012, at 17:59 , arun wrote:
> Hi Peter,
>
> I copied the data from your email and run it again.
>
> dat1<-read.table(text="
> 2.5 3.6 7.1 7.9
> 100 3 4 23
> 200 3.1 4 3 3
> 300 2.2 3.3 24
> ",sep="",header=TRUE)
>
> dat1
> X2.5 X3
While excluding the intercept may make sense, your formula for r^2 assumes
that there was an intercept (that is why mean(y) is in your expression for
sqtot).
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From: Pamela Krone-Davis [mailto:pkrone-da...@csumb.edu]
Sent: Friday, July 13, 201
Pamela
R squared with a non-zero, and with a zero intercept can be very different as
the regression line that you get with and without a zero intercept can be very
different. Have you plotted your data plot(k[,2],k[,1]) to see if a zero
intercept is reasonable for your data? Have you drawn the r
Hello,
I am using the locfit to fit a non parametric glm model to data with a gamma
distributed response variable. In the parametric glm regression the diagnostics
were based on the study of the standardized deviance or pearson residuals. How
can I estimate the the standardized Pearson residua
You might want to look at
http://support.microsoft.com/kb/214230
entitled
Incorrect output is returned when you use the Linear Regression (LINEST)
function in Excel
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [m
What does Excel give for the following data, where the by-hand formula
you gave is obviously wrong?
> x <- c(1, 2, 3)
> y <- c(13.1, 11.9, 11.0)
> M1 <- lm(y~x+0)
> sqerr <- (y- predict(M1)) ^ 2
> sqtot <- (y - mean(y)) ^ 2
> 1 - sum(sqerr)/sum(sqtot)
[1] -37.38707
Bill Dunlap
On 13/07/2012 9:46 AM, Abhi Raghavan wrote:
Hi,
I'm trying to run an R script from the command line and I do it in the
following way:
R CMD BATCH -q *
*I write the R script from inside a Perl program (on UNIX of course!)
and execute the shell command using the function "system".**What
intrigu
OK, got this far:
x1 <- round(rnorm(10,60,3))
x2 <- round(rnorm(10,65,3))
x3 <- round(rnorm(10,70,3))
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method="stack",
pch=4,
offset=1/2,
col="blue",
lwd=2,
las=1)
Any ideas on how
First of all, although the question uses the term "list", the object named
'a' is clearly not a list as R defines the term. It is a vector.
Thus a better example answer is
a <- c('abc', 'def')
paste(a , collapse=' ')
and this works for however many elements there are in the vector a.
The ex
On 13.07.2012 03:18, elong zhang wrote:
Dear All:
Could anybody help me figure out why I get the Error message below while I
running the example code of bugs() function in R2OpenBUGS packages? I have
tried the code both in Win 7 and Ubuntu 12.04, but they show the same
message. My R version is
On 13.07.2012 12:51, Tom Porteus wrote:
It seems to not be recognising a function you are calling within your model -
taking a quick look you might want to check that table() is a function in
OpenBUGS.
g <- length(table(G))
[This time also CCing the OP]
Additional hint:
For debugging purpo
On 13.07.2012 12:51, Tom Porteus wrote:
It seems to not be recognising a function you are calling within your model -
taking a quick look you might want to check that table() is a function in
OpenBUGS.
g <- length(table(G))
Additional hint:
For debugging purposes, it is advisable to open t
Thanks Dennis and Mike... I'm getting it!!!
Sent from my Android
Rui Barradas wrote:
>Hello,
>
>To know why, just evaluate the condition, with 't1$' before 'version_1':
>
>which(as.character(t1$version_1) %in% a) != 0
>[1] TRUE TRUE
>
>
>It allways evaluates to TRUE, therefore, subset() returns
> -Original Message-
> I have question concerning box plot and it's whiskers. As I
> understood from the description of the boxplot() function, if
> the range value is positive the plot whiskers extend out from
> the box to the most extreme data points defined by the values
> of the
?plot.default
## documents the frame.plot arg and
?boxplot
## documents miscellaneous graphical parameters that go into ...
I would agree that the documentation is relatively poor, a legacy of the
original graphics system, which splits up parameter documentation among
par(), plot(), and specific t
Hello,
To know why, just evaluate the condition, with 't1$' before 'version_1':
which(as.character(t1$version_1) %in% a) != 0
[1] TRUE TRUE
It allways evaluates to TRUE, therefore, subset() returns all rows.
See if this isn't simpler than both of your forms.
v2 <- subset(t1, version_1 %in% a
Hi,
I have been using lm in R to do a linear regression and find the slope
coefficients and value for R-squared. The R-squared value reported by R
(R^2 = 0.9558) is very different than the R-squared value when I use the
same equation in Exce (R^2 = 0.328). I manually computed R-squared and the
E
Hi,
I have two non-normal distributions and use interquartile ranges as a
dispersion measure.
Now I am looking for a test, which tests whether the interquartile ranges from
the two distributions are significantly different.
Any idea?
Thanks,
joerg
[[alternative HTML version deleted]]
On Fri, Jul 13, 2012 at 9:40 AM, mdvaan wrote:
> Thanks, I see that it is working in the sample data. My data, however, gives
> me an error message:
>
> data <- strapplyc(text, batch[[l]])
> Error in structure(.External("dotTcl", ..., PACKAGE = "tcltk"), class =
> "tclObj") :
> [tcl] couldn't co
The whiskers will not be symmetrical if 1.5*IQR extends beyond the maximum or
minimum values. In that case, the whisker stops at the maximum or minimum.
-
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77840-4352
Apologies -- replied to the wrong message.
luke
On Fri, 13 Jul 2012, luke-tier...@uiowa.edu wrote:
They look fine to me.
luke
On Fri, 13 Jul 2012, Joshua Wiley wrote:
Dear John,
Thanks very much for the reply. Looking at the optimizers, I had
thought that the objectiveML did what I wante
They look fine to me.
luke
On Fri, 13 Jul 2012, Joshua Wiley wrote:
Dear John,
Thanks very much for the reply. Looking at the optimizers, I had
thought that the objectiveML did what I wanted. I appreciate the
clarification.
I think that multiple imputation is more flexible in some ways bec
Hi,
I'm trying to run an R script from the command line and I do it in the
following way:
R CMD BATCH -q *
*I write the R script from inside a Perl program (on UNIX of course!)
and execute the shell command using the function "system".**What
intrigues me is the strange error that I see in th
Thanks, I see that it is working in the sample data. My data, however, gives
me an error message:
data <- strapplyc(text, batch[[l]])
Error in structure(.External("dotTcl", ..., PACKAGE = "tcltk"), class =
"tclObj") :
[tcl] couldn't compile regular expression pattern: parentheses () not
balan
Thanks all you guys' help!
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Thank you a ton! This is perfect!
From: Jean V Adams [via R] [mailto:ml-node+s789695n4636370...@n4.nabble.com]
Sent: Thursday, July 12, 2012 4:59 PM
To: Lauren Vogric
Subject: Re: Grabbing Indexes of a certain standard deviation
I wrote a little function called first() to help with situations lik
Dear R users,
I have question concerning box plot and it's whiskers. As I understood from
the description of the boxplot() function, if the range value is positive
the plot whiskers extend out from the box to the most extreme data points
defined by the values of the IQR times range (default 1.5)
It seems to not be recognising a function you are calling within your model -
taking a quick look you might want to check that table() is a function in
OpenBUGS.
g <- length(table(G))
TP
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Dear community,
I'm using rpart and would like to extract 95% prediction interval at each
terminal node from a regression tree (built with rpart, method= "anova") Is
that possible?
Thanks in advance, u...@host.com as crossp...@host.com
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Why does the subset not work in the which() version below?
Thank you
v1 <- subset(t1,
version_1==as.character("100-1")
| version_1==as.character("100-2"))
a<-c("100-1", "100-2")
v1 <- subset(t1, which(a==as.character(version_1)) != 0)
[[alternative HTML version deleted]]
___
Dear John,
Thanks very much for the reply. Looking at the optimizers, I had
thought that the objectiveML did what I wanted. I appreciate the
clarification.
I think that multiple imputation is more flexible in some ways because
you can easy create different models for every variable. At the sam
> BTW, is there an R command to read just the first line of the file?
scan() or readLines() will read as many lines of the file as you want.
Use the file() function to open a "file connection" so a subsequent
read.table() will start where scan() or readLines() finished. E.g.,
> tfile <- tempfi
Try this:
> x <- read.table(text = " 2.5 3.6 7.1 7.9
+ 100 3 4 2 3
+ 200 3.1 4 3 3
+ 300 2.2 3.3 2 4", header = TRUE, check.names = FALSE)
>
> x
2.5 3.6 7.1 7.9
100 3.0 4.0 2 3
200 3.1 4.0 3 3
300 2.2 3.3 2 4
> names(x)
[1] "2.5"
Hello All,
Does anyone know where I can find information about how to do a power analysis
for Cox regression with a time-varying covariate using R or some other readily
available software? I've done some searching online but haven't found anything.
Thanks,
Paul
_
On 13/07/2012 9:50 AM, Bert Gunter wrote:
Jessica:
On Fri, Jul 13, 2012 at 1:35 AM, Jessica Streicher wrote:
> two things:
>
> - R always counts from 1, not from 0
> - listmembers are accessed by using [[ ]] , not [ ]
>
FALSE! -- or at least not clearly stated:
> x <- list(a=letters[1:3],b
On Jul 13, 2012, at 04:27 , arun wrote:
> Hello,
>
> I saw your reply in nabble. Sorry about that. I thought the dataset had
> only few columns.
>
> #You can read first line of a file using:
> readLines("foo.txt",n=1)[1]
>
>
> #The more generic colname substitution
> dat1<-read.table(text=
Jessica:
On Fri, Jul 13, 2012 at 1:35 AM, Jessica Streicher wrote:
> two things:
>
> - R always counts from 1, not from 0
> - listmembers are accessed by using [[ ]] , not [ ]
>
FALSE! -- or at least not clearly stated:
> x <- list(a=letters[1:3],b=1:4)
> x[[2]]
[1] 1 2 3 4
> x[2]
$b
[1] 1 2
Hi,
I need to generate HTML index file from Rd file.
example
Module Class Function
Description
Csvfunction csvoperations Module Description
http://test Read CSV Function
Descripti
Hello,
hiere is a small reproducible example.
All z.i which are NA should be transparent at the plot, but they are white
colored.
### Example image.plot regular x,y grid ###
x <- seq(2,2.9,0.1)
y <- seq(42,42.9,0.1)
z <- matrix(seq(-5,4.9,0.1),nrow=10)
image.plot(x,y,z)
### overplotting b
Someone pointed me to this paper:
http://www.validlab.com/goldberg/paper.pdf
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Thank you for the explanation. Good to know about the issue how double values
are "constructed" by a bit system. This makes me handling double values with
care in using it in R or aother languages control structures etc.
Thank you also for the hint concerning the Null vs. Zero vs.. issue. Yes,
th
On Jul 13, 2012, at 12:35 AM, sanchez ana wrote:
> Dear All,
>
> I am using the function vuong from pscl package to compare 2 non nested
> models NB1
> (negative binomial I ) and Zero-inflated model.
>
>
> NB1 <- glm(, , family = quasipoisson), it is an
> object of class: "glm" "lm"
> zinb <
On Thu, Jul 12, 2012 at 8:17 AM, Bharat Warule wrote:
> Hello,
>
> I am using read.csv.sql first time for reading the large data file.If I am
> ran this code that showns warning “closing unused connection”.
>
> Is it I am missing any argument from my command or how to comeout from this
> warning?.
Hello,
Check the structure of what you have, df and newdf. You will see that in
df dateTime is of class POSIXlt and in newDf newDateTime is of class
POSIXct.
Solution:
[...]
df$dateTime <- strptime(df$dateTime,"%m/%d/%Y %H:%M")
df$dateTime <- as.POSIXct(df$dateTime)
[...]
Hope this helps,
thanks!
On Fri, Jul 13, 2012 at 4:35 AM, Jessica Streicher wrote:
> two things:
>
> - R always counts from 1, not from 0
> - listmembers are accessed by using [[ ]] , not [ ]
>
> try
>
> t1[t==ll[[1]], "v"] <- 99
>
> greetings Jessi
>
>
> On 11.07.2012, at 15:47, Charles Stangor wrote:
>
> > I c
Hello,
Within my department I would like to share the latest version(s) of my R
scripts to my colleagues using subversion. This repository is password
protected (WebDAV) as it should not be accessed by external persons.
In the majority of my scripts I load scripts using the source() function, b
Dear all, I am analyzing data from a survey question where I have many "Don'k
know" responses. Respondents could either say "Don't know" or give a rating
from 0 to 10. I could analyse the "Don't know" responses separately using a
logistic regression to see who is the "typical" person who doesn
I have the following dataframe with the first column being of type datetime:
dateTime <- c("10/01/2005 0:00",
"10/01/2005 0:20",
"10/01/2005 0:40",
"10/01/2005 1:00",
"10/01/2005 1:20")
var1 <- c(1,2,3,4,5)
var2 <- c(10,20,30,40,50)
df <- dat
Hi,
here i have a Max and Min values
Min <-3
Max <-6
and also a matrix like this,
ABCXYZ PQR
-- ------
2 43
5 48
7 13
In this i need to check each particular
Hi, I'm currently working on the below codes however whenever I run it in
openbugs it gives an error message saying: unknown type of logical function
error pos 76. Any help would be appreciated.
## bugs code
library(R2OpenBUGS)
sink("C:/Users/CCF/Documents/Suzie Work/PTY Project/Waterhole
Correc
Thanks to both of you, you are probably right that memory is the limiting
factor. I have no knowledge about the available memory on the lab machines,
but I will find out and make sure that this is the explanation.
Best,
Ulrike
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Hello,
This information is new. Are you able to import data from access?
A.K.
- Original Message -
From: imnew
To: r-help@r-project.org
Cc:
Sent: Thursday, July 12, 2012 9:52 PM
Subject: Re: [R] plot graph by first letter
Hi, for my dataset is actually retrieve from a access file
Thank you for this useful code!
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h
On Fri, Jul 13, 2012 at 12:21 PM, Martin Ivanov wrote:
> Dear R users,
>
> I am struggling with the colorkey on a levelplot lattice graphic.
> I want that no ticks are printed on the colorkey. That is, I want their size
> tck=0.
> Merely setting tck=0 t in the colorkey parameter does not work. S
You could probably make them numeric, like
> v<-c("a","a","b","c")
> f<-factor(v)
> as.numeric(f)
[1] 1 1 2 3
to get a numeric "rock_id", but i wouldn't per se recommend it.
You should ask someone who knows more about the scientific side of this method
to tell you how factorial data is proper
On Thu, Jul 12, 2012 at 03:51:54PM -0500, Vineet Shukla wrote:
> I have independent event sequences for example as follows :
>
> Independent event sequence 1 : A , B , C , D
> Independent event sequence 2 : A, C , B
> Independent event sequence 3 :D, A, B, X,Y, Z
> Independent event sequence
two things:
- R always counts from 1, not from 0
- listmembers are accessed by using [[ ]] , not [ ]
try
t1[t==ll[[1]], "v"] <- 99
greetings Jessi
On 11.07.2012, at 15:47, Charles Stangor wrote:
> I can't seem to determine how to get the name of a list member to
> substitute:
>
> ll <- lis
To use variables in mathematical expressions, bquote can be used:
#Term in in .( ) is evaluated
plot(1:10,main=bquote(frac(alpha,.(2+2
--
GnuPG Key: 0x7340821E
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PLEA
Hello, Ivan,
you'll find argument frame (actually frame.plot) explained in
?plot.default
Regards -- Gerrit
Hi Peter,
I had never heard of this 'frame' argument and it's a breakthrough for me to
be finally able to get rid of this frame!
But where is this argument explained? I couldn't
Dear R users,
I need to add minor axis ticks to my graph. In traditional R this is easily
achievable by simply
adding a second axis with the minor ticks. But how to do that in trellis? I am
already out of ideas.
Any suggestions will be appreciated.
Best regards,
Martin
-
Hi Peter,
I had never heard of this 'frame' argument and it's a breakthrough for
me to be finally able to get rid of this frame!
But where is this argument explained? I couldn't find it in plot(),
boxplot(), bxp() or par().
Thank you for your answer :)
Ivan
--
Ivan CALANDRA
Université de B
Hello,
Works unchanged with me.
Yesterday it could have worked for some other reason, like having other
variables in my environment, which I had, but this time I have started
anew. Try including
tUnitsmall <- tUnitsort[, cols]
and then use this data.frame to see what happens.
Rui Barradas
Hello Veronica,
what makes you think that this is an error? It is a warning that your specified
SVAR-model is **just** identified and hence an over-identification test cannot
be conducted. You can suppress this warning by not asking for an
over-identification in the first place, by setting lrte
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