Hello R-users,
I am trying to obtain Type III SS for an ANOVA with subsampling. My design
is slightly unbalanced with either 3 or 4 subsamples per replicate.
The basic aov model would be:
fit - aov(y~x+Error(subsample))
But this gives Type I SS and not Type III.
But, using the drop() option:
How to generate all possible samples of size 5 for the two variables in one
analysis.
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Hi,
I have two groups of data of different size:
group A: x1, x2, , x_n;
group B: y1, y2, , y_m; (m is not equal to n)
The two groups are independent but observations within each group are
not independent,
i.e., x1, x2, ..., x_n are not independent; but x's are independent from
Dear Users,
I wish to know at what confidence level is the confidence interval
provided in the Spectrum function (plot.spec) plots.
The only information provided in the help regarding this is : a confidence
interval will be plotted by plot.spec: this is asymmetric, and the width of
the
Am 21.05.2011 06:30, schrieb Sharma D:
How to generate all possible samples of size 5 for the two variables in one
analysis.
Like this?
x - 1:5
y - 6:10
expand.grid(x,y)
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Thomas,
your example works with R-2.13.0 and Apache running on
Gentoo. You may try to use http://127.0.0.1:5984/path/fn.R;,
just to exclude that this problem has anything to do with
name resolution.
Best
Hugo
Dear List,
I have problems with the function source() using a
url of the kind:
Person A is working on the file on their computer the path to the data would
be (Mac OSX) /Users/PersonA/Dropbox/Project/data.csv However, to Person B
the path would be /Users/PersonB/Dropbox/data.csv I'm looking for a way to
keep the path to data.csv universal and independent of who is
Hi,
ia have similar problem you had.Did you manage to find out what that error
meant?
thanks, m
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Hi all,
I have tried to create a DocumentTermMatrix with a tm package, but i get this
error :
Error in tolower(txt) :
invalid input 'PROD Z LAHKO GNETNO MELJNO GLINO, ... in 'utf8towcs'
I tried doing this as it is showed in :
http://www.r-project.org/doc/Rnews/Rnews_2008-2.pdf
Let's say I have the data frame 'dd' below. I'd like to select one
column from this data frame (say 'a') and keep its name in the
resulting data frame. That can be done as in #2. However, what if I
want to make my selection based on a vector of names (and again keep
those names in the resulting
dd[[ myname]]
Sent from my iPad
On May 21, 2011, at 7:37, Lars Bishop lars...@gmail.com wrote:
Let's say I have the data frame 'dd' below. I'd like to select one
column from this data frame (say 'a') and keep its name in the
resulting data frame. That can be done as in #2. However, what if I
Or
dd[,myname]
should work too.
If you are worried about getting multiple columns, you can just make myname a
vector of column names using c() before you use either Jim's list indexing or
the above matrix indexing syntax.
Are you looking for:
dd[, a, drop=FALSE]
On 21/05/2011 12:37, Lars Bishop wrote:
Let's say I have the data frame 'dd' below. I'd like to select one
column from this data frame (say 'a') and keep its name in the
resulting data frame. That can be done as in #2. However, what if I
want to make my
Got it...the problem was with Slovenian characters. Once i replaced them with
normal characters it works fine.
Tnx anyway, m
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Matevž Pavlič
Sent: Saturday, May 21, 2011 1:27 PM
To:
Got it...the problem was with Slovenian characters. Once i replaced them with
normal characters it works fine.
Tnx anyway, m
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of mpavlic
Sent: Saturday, May 21, 2011 1:06 PM
To:
Hello!
I've tried for a while - but can't figure it out. I have data frame x:
y=c(a,b,c,d,e)
z=c(m,n,o,p,r)
a=c(0,0,1,0,0)
b=c(2,0,0,0,0)
c=c(0,0,0,4,0)
x-data.frame(y,z,a,b,c,stringsAsFactors=F)
str(x)
Some of the values in columns a,b, and c are 0:
I need to write a loop through all the cells
Dear Sam,
I don't think that you can use the Anova() function in the car package for
this design because there would be no way to specify the error term.
Sorry,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
Hello everybody,
I need an help because I don´t know if the command for the ANOVA analysis I am
performing in R is correct. Indeed using the function aov I get the following
error:In aov (..) Error() model is singular
The structure of my table is the following: subject, stimulus,
On May 21, 2011, at 9:12 AM, Dimitri Liakhovitski wrote:
Hello!
I've tried for a while - but can't figure it out. I have data frame x:
y=c(a,b,c,d,e)
z=c(m,n,o,p,r)
a=c(0,0,1,0,0)
b=c(2,0,0,0,0)
c=c(0,0,0,4,0)
x-data.frame(y,z,a,b,c,stringsAsFactors=F)
str(x)
Some of the values in columns
Dmitri:
1. I did not read your whole missive. I prefer mystery novels. ;-)
2. I suggest you banish Excel language (cells) from your vocabulary
and think in R's terms of whole objects that one indexes into.
3. If I understand correctly, you can't combine results into a data
frame, because they
Dimitri Liakhovitski-2 wrote:
Hello!
I've tried for a while - but can't figure it out. I have data frame x:
y=c(a,b,c,d,e)
z=c(m,n,o,p,r)
a=c(0,0,1,0,0)
b=c(2,0,0,0,0)
c=c(0,0,0,4,0)
x-data.frame(y,z,a,b,c,stringsAsFactors=F)
str(x)
Some of the values in columns a,b, and c are 0:
I've got a gls formula that's a mix of continuous and ordered variables.
I wanted to use gls because I wanted to use the varIdent structure.
Anyway, attempts to use predict.gls choke with the error that the
levels I use are not allowed for one of them -- the first one
alphabetically, so I'd
Mitchell Maltenfort mmalten at gmail.com writes:
I've got a gls formula that's a mix of continuous and ordered variables.
I wanted to use gls because I wanted to use the varIdent structure.
Anyway, attempts to use predict.gls choke with the error that the
levels I use are not allowed
Alexander Kapeller mail at alexander-kapeller.de writes:
Hello,
I like to fit data against a negative binominal distribution
[snip]
I am wondering how to translate the size and the mu parameter into the usual
number of successes (n) and the contiuous probability (p) . To me it is not
On May 21, 2011, at 7:37 AM, Lars Bishop wrote:
Let's say I have the data frame 'dd' below. I'd like to select one
column from this data frame (say 'a') and keep its name in the
resulting data frame. That can be done as in #2. However, what if I
want to make my selection based on a vector of
Is there a way to use tm with the Slovene characters? I ask,
because I was hoping to use tm with languages like Arabic, Urdu,
Farsi, and Hebrew. If you need to translate Slovene characters, it
could create problems with using the software for the desired purpose in
many languages,
Correction, this is not an issue with multilevel, rather a quirk with
aggregate.
Sill looking for help, anyone?
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Hi everyone,
i am trying to group close numbers in a vector.
For example i have a vector x = [1 2 4 7 9 10 15].
I want the code to pick 1 2 4 (max difference between successive numbers is
2) and assign them to variable a, then pick 7 9 10 and assign them to b and
15 to c. But since i do not know
On May 21, 2011, at 11:56 AM, eeecon wrote:
Correction, this is not an issue with multilevel, rather a quirk with
aggregate.
So change you subject.
Sill looking for help, anyone?
About what?
--
David Winsemius, MD
West Hartford, CT
__
On May 20, 2011, at 3:54 PM, eeecon wrote:
Hi,
My code indicates there may be a bug in multilevel.
I doubt this is actually the case, can anyone tell me what is wrong
with my
code?
The data file for this code can be downloaded here:
http://cameron.econ.ucdavis.edu/mmabook/mma15p4gev.asc
Hi everyone,
i am trying to group close numbers in a vector.
For example i have a vector x = [1 2 4 7 9 10 15].
I want the code to pick 1 2 4 (max difference between successive numbers
is
2) and assign them to variable a, then pick 7 9 10 and assign them to b
and
15 to c. But since i do not
Hello,
I am newbie to R and I want to do this:
for(i in 1:6)
{
ds[i] - list(df=data.frame(oilDF[,1],oilDF[,i+2]),
df2=data.frame(oilDF2[,1],oilDF2[,i+2]))
}
#oilDF and oilDF2 are 2 data frames with several columns. They have different
number of rows
#I want to have for example
Dear members,
I apologize for the relatively simple request, but I couldn't find exactly
what I was looking for. I have a binary vector [1,0] representing
presence/absence at 1 second intervals over length(N). I would like to
convert this to a new time series vector with N/60 elements that
Dear R gurus,
I'm trying to solve what I assume is a fairly simple problem, but I'm having
trouble finding the proper approach. I have a matrix where each column is some
object (e.g. a car) and each row is a numeric measurement of a feature of said
object (e.g. horse power, top speed, etc.).
Hi mac,
Try
N - 6000
x - sample(1:0, N, TRUE)
tapply(x, rep(1:(N/60), each = 60), sum)
HTH,
Jorge
On Sat, May 21, 2011 at 1:59 PM, andyjmac wrote:
Dear members,
I apologize for the relatively simple request, but I couldn't find exactly
what I was looking for. I have a binary vector
Hi Rui,
Here is one option:
ds - vector(list, 6)
for(i in 1:6) ds[[i]] - list(df = mtcars[, c(i, i + 2)], df2 =
mtcars[, c(i, i + 2)] + 10)
another could be:
altds - lapply(1:6, function(x) {
list(df = mtcars[, c(x, x + 2)], df2 = mtcars[, c(x, x + 2)] + 10)
})
all.equal(ds, altds)
For
Hi Mark,
Is there a reason you cannot simply include the make of the car along
with all the other data? Using your example:
cbind(as.data.frame(t(x)), carmake)
then instead of applying across columns, apply across rows, and have
your custom function decide what to do based on the column named
What does *strip=function *and *strip.default *means
for eg: *strip=function(bg='white', ... )
strip.default(bg='white', ...)
*Ram
[[alternative HTML version deleted]]
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Thanks for the tip Josh! I thought about combining the data, but wondered if
there was a way to pass in the information as a separate argument. But I went
ahead with your solution and it worked perfectly. Thanks!
On May 21, 2011, at 12:57 PM, Joshua Wiley wrote:
Hi Mark,
Is there a reason
On May 21, 2011, at 3:15 PM, Ramnath R wrote:
What does *strip=function *and *strip.default *means
for eg: *strip=function(bg='white', ... )
strip.default(bg='white', ...)
I am guessing from my experience that you are working on a plotting
problem and trying to understand some
I am unsure what is being returned, and what is supposed to be
returned, when using 'predict' with type='expected' for an aftreg
survival model. The code below first generates a weibull model, then
uses predict to create a vector of the linear predictors, then
attempts to create the 'expected'
Amen to David's comments (I'm sorry to say ...)
In addition: Learn to use the Help system. ?help (at the command line)
and An Introduction to R manual provide info.
To answer your question:
library(lattice)
?strip.default
-- Bert
On Sat, May 21, 2011 at 1:27 PM, David Winsemius
Mike,
On Sat, May 21, 2011 at 10:15 PM, Mike Harwood harwood...@gmail.com wrote:
I am unsure what is being returned, and what is supposed to be
returned, when using 'predict' with type='expected' for an aftreg
survival model.
I'm sorry, but there is no 'predict.aftreg'. The generic 'predict'
Hi,
I don't think the final verdict has been spoken. Peter's posts have hinted at
ill-conditioning as the crux of the problem. So, I decided to try a couple of
more things: (1) standardizing the covariates, (2) exact gradient, and (3) both
(1) and (2).
I compute the exact gradient using a
Hi:
Does this work for the first problem?
library(reshape2)
subset(melt(x, id = c('y', 'z')), value 0)
y z variable value
3 c oa 1
6 a mb 2
14 d pc 4
The second problem is so convoluted I don't even know where to start...
HTH,
Dennis
On Sat, May 21,
Hi Robert,
thanks for your reply. is there a way to store them in separate vectors?
and when i try it with a different example i got different result. For
example if x = [1 2 8 9]
i want the result to be x1 = [1 2] and x2 = [8 9].
thanks
On Sat, May 21, 2011 at 7:16 PM, Robert Baer rb...@atsu.edu
Is this what you are after:
x = c(1 ,2 ,4 ,7 ,9 ,10 ,15)
# partition if the difference is 2)
breaks - cumsum(c(0, diff(x) 2))
# partition into different lists
split(x, breaks)
$`0`
[1] 1 2 4
$`1`
[1] 7 9 10
$`2`
[1] 15
On Sat, May 21, 2011 at 6:03 PM, Salih Tuna saliht...@gmail.com
Hi,
I would appreciate some help with a very basic problem in using pgfSweave. That
is, dealing with the sweave.sty file. After some googling, I did the following
:
(1) copied the sweave.sty file to my trial folder and (2) added the line
\usepackage{Sweave} to the Rnw file. Are there more
On 21/05/11 15:15, Sam Chand wrote:
Hello R-users,
I am trying to obtain Type III SS for an ANOVA with subsampling. My design
is slightly unbalanced with either 3 or 4 subsamples per replicate.
The basic aov model would be:
fit- aov(y~x+Error(subsample))
But this gives Type I SS and not
Yes this is exactly what i want, thanks Jim.
One last question, (i am sure this is a very simple question but i am still
learning) how can i write the output to a txt file seperately?
vector 1 to one file and vector 2 to another and etc?
thanks
salih
On Sat, May 21, 2011 at 11:41 PM, jim
From: rvarad...@jhmi.edu
To: marchy...@hotmail.com; pda...@gmail.com; alex.ols...@gmail.com
CC: r-help@r-project.org
Date: Sat, 21 May 2011 17:26:29 -0400
Subject: RE: [R] maximum likelihood convergence reproducing Anderson Blundell
1982
--- On Fri, 5/20/11, David Winsemius dwinsem...@comcast.net wrote:
From: David Winsemius dwinsem...@comcast.net
Subject: Re: [R] Adding a numeric to all values in the dataframe
To: John Kane jrkrid...@yahoo.ca
Cc: bill.venab...@csiro.au, ramya.vict...@gmail.com, r-help@r-project.org
I have a simple system of linear equations to solve for X, aX=b:
a
[,1] [,2] [,3] [,4]
[1,]1211
[2,]3004
[3,]1 -4 -2 -2
[4,]0000
b
[,1]
[1,]0
[2,]2
[3,]2
[4,]0
(This is ex Ch1, 2.2 of Artin, Algebra).
So, 3 eqs
Hi I want to analyse my data (using nonlinear regression ) but I can not load
and install the package. this package (nls) not available in my R. how get
it.
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nls is part of the stats 'package'.
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Please let me know how to get this package. Thanks
On Sat, May 21, 2011 at 9:50 PM, dslowik [via R]
ml-node+3541502-716551039-239...@n4.nabble.com wrote:
nls is part of the stats 'package'.
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If you reply to this email, your message will be added to the
On May 21, 2011, at 3:27 AM, Zablone Owiti wrote:
Dear Users,
I wish to know at what confidence level is the confidence interval
provided in the Spectrum function (plot.spec) plots.
The default level is clearly indicated in the arguments of the Usage
section on the help page of
On May 21, 2011, at 9:33 PM, magushi wrote:
Hi I want to analyse my data (using nonlinear regression ) but I can
not load
and install the package. this package (nls) not available in my R.
how get
it.
nls() is a function that is part of the stats package which is
installed
solve() only works for nonsingular systems of equations.
Use a generalized inverse for singular systems:
A- matrix(c(1,2,1,1, 3,0,0,4, 1,-4,-2,-2, 0,0,0,0), ncol=4, byrow=TRUE)
A
[,1] [,2] [,3] [,4]
[1,]1211
[2,]3004
[3,]1 -4 -2 -2
[4,]00
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