The idea behind the treatment of NAs is that if you don't know what the
value is, how do you know if it's 1 or not (in your case)? Therefore, it's
normal that the result to your query is NA.
First, I guess your command was supposed to be:
y - ifelse(x==1, 'NO', 'YES')
Secondly, if you want to
Hi all,
I'm reviewing a paper and ran into a mystery (hopefully not to you) trying
to reproduce the stats. The authors uploaded their data to a repository.
I've created a similar (though random) data set for confidentiality reasons.
So the authors include one four-level treatment variable, but
Hi all,
I have a large number of measurements from which I select a large number of
unique vectors. For each vectors I would like to test which distribution might
be a candidate for fitting.
It is impossible to look on each vector separately but I can inside a for loop
test different models and
Dear lattice users,
I am trying to produce a lattice graph with two conditioning variables.
My problem is that I only want to show the strips for the levels of the second
conditioning variable.
I want to remove the strips for the levels of the first conditioning variable.
I tried with the strip
Hello,
Please provide a commented, minimal, self-contained, reproducible
code, as requested.
Regards,
Pascal
On 10 February 2014 19:48, Martin Ivanov tra...@abv.bg wrote:
Dear lattice users,
I am trying to produce a lattice graph with two conditioning variables.
My problem is that I only
Dear Pascal,
Thank You very much for Your reply. Here is a minimal working example:
library(lattice);
# this is with both strips plotted:
data - data.frame(x=1:5, y=6:10, f1=c(a1, a1, a1, a2, a3),
f2=c(b1, b2, b2, b1, b2));
p - xyplot(y ~ x | f1 + f2, data=data)
print(p);
# and this is my
Hi Martin
If you only had 1 strip then strip = FALSE would suffice.
With 2 I think you are getting into the realm of strip.default because of the
first 2 arguments
strip.default(which.given, which.panel, )
and having to modify it
An easy solution is
library(latticeExtra)
And for the lattice solution
see
? lattice::print.trellis
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet.com.au
-Original Message-
From: r-help-boun...@r-project.org
Hello,
I am relatively new to vegan and need some help with metaMDS. My R is
Version 3.0.2. I am analysing cuticular profiles of flys from different
locations an different gender. My data is in a 48 x 70 matrix. It is about
percentages and so, there are a lot of 0s included. If I run metaMDS, it
Hi,
I tried to reverse the data axis in stripchart
which should work according to docu with
xlim = c(Hi, Lo)
but it did not work.
Same effect in DotPlot from lessR.
Any there work arounds or hints?
Thanks
Hermann
example:
x - stats::rnorm(50)
stripchart(x)
# setting limits works
Dan,
That's a bug in setdiff. I've fixed it in the development version of
lubridate at http://github.com/hadley/lubridate. Thank you for finding it.
setdiff *does* intend to trap the two interval result and return an error.
The part of the visit where Chris is *not* there would be
I believe this is more a question for SO (stats.stackexchange.com).
There are many possible goodness of fit statistics that can easily be
calculated in R, but I think the fundamental question is: To what end?
First, there are probably several parametric distributions that give
(essentially)
Thanks everyone for the help. Dennis, the bquote version work great.
Thanks,
Doug
On 2/7/2014 7:08 PM, Dennis Murphy wrote:
Here's a bquote version:
x=c(1,2,3,4); y=c(1,2,3,4); z=c(1.25,1.5,2.5,3.5)
# first stats based on data, used to populate legend
wdt_n = 50; wdt_mbias = 0.58
wdt_mae =
Without more information, it is hard to say. You did not tell us much about the
data beyond the dimensions, but it looks like you have several different kinds
of measurements including location and gender (probably categorical) and
cuticular profiles (presumably numeric). Without the commands
You are correct, but I do not know why stripplot works
differently from plot. I can give you two ways to get around the
problem though. The easiest is to use the plot command to
emulate stripchart:
plot(x, rep(1, length(x)), xlim=c(2, -2), yaxt=n, ylab=,
pch=0)
Or using stripchart:
With data like the following, a frequency table in data frame form, I'd
like to fit a collection of loglm models
of independence of ~ attitude + memory for each combination of education
and age.
I can use apply() if I first convert the data to a 2 x 2 x 3 x 3 array,
but I can't figure out an
Hi,
set.seed(42)
mat1 - matrix(sample(40,5*3,replace=TRUE),ncol=3)
mat2 - apply(mat1,1,sort,decreasing=TRUE)
mat2[1,]- mat2[2,]
A.K.
Thank you.
And if I have a lot of rows? How can I apply this for row?
On Sunday, February 9, 2014 8:20 PM, arun smartpink...@yahoo.com wrote:
Based
Perhaps:
?tapply
and/or various wrappers like ?by .
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
H. Gilbert Welch
On Mon, Feb 10, 2014 at 8:05 AM, Michael Friendly
Hi,
I have a dataframe in R and would like to split the data (1900-1980) into two
sets.
For example, one dataframe should have data from 1900-1960 and the other from
1961-1980.
Thanks for your help.
AT.
[[alternative HTML version deleted]]
__
On Mon, 10 Feb 2014, Zilefac Elvis wrote:
I have a dataframe in R and would like to split the data (1900-1980) into
two sets. For example, one dataframe should have data from 1900-1960 and
the other from 1961-1980.
If you want two separate dataframes (while leaving the original), subset()
Suggestions:
Read the posting guide mentioned at the bottom of this email. Note request to
post in plain text and provide reproducible example. Knowing how your data are
constructed allows us to make more concrete suggestions as to how Rio solve
your problem.
Read help files: ?%in%, ?subset,
Thanks for the fast response!
My data is actually just the percentages of the substances found in the
cuticular layer of the flies. The quantitative and qualitative composition
of this layer differs in gender, locality and so on.
My 70 „species“ are the chemical compounds we identified and my 48
Hi:
I am running the following code to query Pubmed database, but getting
the error couldn't connect to host.
library(RCurl)
library(XML)
library(tm)
url - http://eutils.ncbi.nlm.nih.gov/entrez/eutils/esearch.fcgi?;
q - db=pubmedterm=saunders+nf[au]usehistory=y
esearch -
Hello,
I’m new to R and I’m trying to solve a simple problem for finding the radius
of a cylinder given the height and volume. I’ve included the data file where
the height in cm is column 2 and volume is in column 1. I know the equation
needed is r= √(V/H(3.14)). My question is how do I code this
On 2/10/2014 11:12 AM, Bert Gunter wrote:
Perhaps:
?tapply
and/or various wrappers like ?by .
Cheers,
Bert
Thanks, Bert
The examples for ?by gave the answer I was looking for:
# using by()
mods - with(Punishment,
by(Punishment, list(age, education),
function(x)
On 09 Feb 2014, at 10:56 , Paul Parsons pparsons...@gmail.com wrote:
Many thanks, Peter. Creating a wrapper function for integrand using
Vectorize, and then integrating the wrapper, does indeed solve the problem. I
tried your final suggestion, but the variable x still gets passed into
A reproducible example would help. Look at the example in metaMDS using the
dune data set. It consists of 20 observations (sites) with 30 species observed
(columns). You indicate that your compounds (species) are the rows and your
sites are the columns. So perhaps you are analyzing the
We have a policy of not doing your homework for you, but I can point you to a
valuable resource for learning R
http://cran.r-project.org/doc/contrib/Paradis-rdebuts_en.pdf. Your problem
involves simple arithmetic. There is no regression problem unless you are
trying to predict height from
Hi, I am trying to fit a SEM model (the ram specification of the model, along
with the estimated polycoric VCmatrix (CC32), is attacched), but I get the
following error message: Error in 1:m : NA/NaN argumentI did not find any
help on internet, so I tried to understend what is going wrong
My yearly Regression Modeling Strategies course is expanded to 4 days
this year to be able relax the pace a bit. Details are below.
Questions welcomed.
-
*RMS Short Course 2014*
Frank E. Harrell, Jr., Ph.D., Professor and Chair
We would like to announce the following statistics course:
Topic: Data exploration, linear regression, GLM GAM in R. With
introduction to R.
Where: Aston University. Birmingham, UK
When: 12-16 May, 2014
Info: http://www.highstat.com/statscourse.htm
Flyer:
Hi,
Try:
##Assuming X and Y are matrices (as stated)
Y[!Y %in% X,,drop=FALSE]
# numberall
#[1,] gen3
#[2,] gen4
#[3,] genx4
A.K.
I would like to exclude entire lines in matrix Y based in 2 collumns in
matrix X:
Matrix X:
number1 number2
gen1 genx1
gen1
Hello,
I am using the survey package for the first time to analyze a dataset that
has both weights and 200 BRR replication weights. When I try to run svyglm
on the output from svrepdesign, I get an error message that I do not know
how to interpret, and an extended period of time searching for
i.e.
remove.packaghttp://127.0.0.1:16060/help/library/utils/help/remove.packages
es('raster')
install.packages('raster')
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hey all, I uninstalled and then reinstalled raster, and the problem fixed
itself. Cheers
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
On Feb 10, 2014, at 6:34 AM, Bert Gunter wrote:
I believe this is more a question for SO (stats.stackexchange.com).
Actually it might get closed on SO since it is not an R programming question
per se but rather an advice for statistical approach. It's a better fit for
CrossValidated.com
I am trying to export the tree splitting results from a rpart object. I
can visualize the splits by printing the rpart object (shown below for my
rpart object, model.prune). However, this printed result is not a
dataframe and I am unable to export this information in .csv format.
I can view the
My mistake. I gave the right URL but referred to it as SO instead of CV.
Gotta get my nomenclature right!
Bert
On Feb 10, 2014, at 3:32 PM, David Winsemius dwinsem...@comcast.net wrote:
On Feb 10, 2014, at 6:34 AM, Bert Gunter wrote:
I believe this is more a question for SO
Hi All,
I¹m using the twang package to estimate ATE propensity scores. These are
being used (inverse probability weighting) in a Cox model estimated by cph
in the rms package with cluster(id) set. I set options to include a
datadist object before running cph.
I want to check for sensitivity to
Hi,
dat - read.table(text=A B
1 0
2 2
3 1
4 NA,sep=,header=TRUE)
within(dat,{C-A/B;C[!is.finite(C)]-0})
#
within(dat,{C-A/B;C[!is.finite(C)]-NA})
A.K.
Hi R users,
Just wondering is it possible to divide two columns with some NA or Zero
values? E.g.:
Hi,
I'm building a package in R and not able to find key information for a
couple of things.
1. I have a .png graphic that the R code uses in print-outs it creates.
What is the appropriate folder, or manner to incorporate a .png into the
package?
2. I have two .txt data files I want
Hi, This is My Dataset Requriment. If Possible To solve This Task. Please
send reply
I Have Take 3 Numeric Values
1.Is Age
2.Is Income
3.Is Cast
If We Have To Arranged In Order Is
Age Income Cast
15 15000 5000
18 12000 4500
20 16000 6000
25 22000 11000
28 11000 6000
30
Thank you for your help. This is not homework however and my difficulty is in
the fact that i cannot figure out how the declare a variable for the height
or the volume given a data set where the matrix is already formed. I have
looked on many sites for the is answer and it seems like it would be
Hi,
Try
dat1- as.matrix(read.table(text=V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11
6 6 6 6 6 0 0 0 0 0 0
5 5 5 5 5 5 5 5 5 5 5
1 0 0 0 0 0 0 0 0 0 0
5 5 5 0 0 0 0 0 0 0 0
6 8 11 0 9 0 0 0 0 0 0
2 0 0 2 0 0 0 0 0 2
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