Dear r Users,
I am new in r. I am trying to estimate regression quantiles in complex
surveys.I used these commands.
mydesign
-svydesign(ids=~IDSCHOOL,strata=~IDSTRATE,data=TUNISIA,nest=TRUE,weights=~TOTWGT)
bootdesign - as.svrepdesign(mydesign,type=auto,replicates=150)
I need to create a color vector with 4000 colors, and I have another
variable, it contains 4000 element. If my variable is number of bed, and I
want the color corresponds to the number of bed in this variable. How should
I do?
--
View this message in context:
I answered this question which was part of the online course correctly by
executing some commands and guessing.
But I didn't get the gist of this approach though my R code works.
I have a training and test dataset.
nrow(training)
[1] 251
nrow(testing)
[1] 82
head(training1)
diagnosis
What's going on is very simple. I apologize for not getting my thoughts
organized from the start.
Confidence bands for mu_y (the expected value of y) will have straight
line edges if and only if there is a *single predictor* in the model.
If there are two or more predictors the edges will
Hi,
The following code adapted from Michael post
(https://stat.ethz.ch/pipermail/r-help/2012-March/306069.html) works just fine
on Linux Debian, but not on Windows 7 (no points on plots 2 and 3). More
surprisingly, if the
first plot is a boxplot, it works on both OS... and if I do a pdf (using
First off, please post in plain text not HTML. It is a forum
requirement. HTML posts are often ignored, especially if they look
messy in plain text, which is all that the mailing list software
distributes. It strips off the HTML portion entirely.
Your question is vague in that you didn't indicate
On Tue, 16 Sep 2014 09:34:39 PM wszsdmjj wrote:
I need to create a color vector with 4000 colors, and I have another
variable, it contains 4000 element. If my variable is number of bed,
and I
want the color corresponds to the number of bed in this variable. How
should
I do?
Hi wszsdmjj,
I
Dear all,
I am having some difficulties trying to control color palette and legend of a
filled.countour plot.
Basically, I am plotting volumetric soil moisture which ranges from 0 to 1
(although the data excerpt I'm providing here ranges from 0 to 0.4, my complete
dataset ranges from 0 to
Hello,
I'm wondering if this is expected?
cut(structure(1, class=Date), structure(c(11100,1),
class=Date))
[1] NA
Levels: 2000-05-23
The help page says that for ‘Date’ objects, only ‘day’, ‘week’,
‘month’, ‘quarter’ and ‘year’ are allowed [for the 'breaks'
argument]. Though
On 17/09/2014 12:04, Benjamin Tyner wrote:
Hello,
I'm wondering if this is expected?
It is as documented!
cut(structure(1, class=Date), structure(c(11100,1),
class=Date))
[1] NA
Levels: 2000-05-23
The help page says that for ‘Date’ objects, only ‘day’, ‘week’,
Dear useRs,
I have a data frame y starting from 1961 to 2010 in the following manner
(where A,B,C .., I are station names and the values uder these are
discharge values.)
dput(y)
structure(c(1961, 1961, 1961, 1961, 1, 1, 1, 1, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13, 14, 15,
Use the 'tidyr' package: your 'month' does not match your desired output -
x - structure(c(1961, 1961, 1961, 1961, 1, 1, 1, 1, 1, 2, 3
+ , 4, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
+ , 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27
+ , 28, 29, 30, 31, 32, 33,
You probably told R to write out the file as a single long line with fields
separated alternately by 380 TABs and one newline � that�s what the ncol
argument does (write is just a small wrapper around cat()).
cat() doesn�t print lines that are longer than 2 GiB, so it will insert an
extra \n
Thankyou very much jim. you always have something new for the R-ookies.
Eliza
Date: Wed, 17 Sep 2014 09:26:07 -0400
Subject: Re: [R] column names to row names
From: jholt...@gmail.com
To: eliza_bo...@hotmail.com
CC: r-help@r-project.org
Use the 'tidyr' package: your 'month' does not
Hello,
I am looking for ANN ARIMA or ANN ES ( Artificial Neural Networks Hybrid with
ARIMA or Exponential Smoothing ) R examples or packages ?
as referenced in
http://cs.uni-muenster.de/Professoren/Lippe/diplomarbeiten/html/eisenbach/Untersuchte%20Artikel/Zhan03.pdf
regards
Here's another approach using stack():
y - data.frame(y)
E - with(y, data.frame(year, month, day,
stack(data.frame(y), select=4:12)))
colnames(E)[4:5] - c(discharge, station)
But there are some differences. For my E:
str(E)
'data.frame': 36 obs. of 5 variables:
$ year : num
On 17 Sep 2014, at 05:58 , isabe...@ghement.ca wrote:
The example is based on a real data set (please keep it
confidential), which is attached to this e-mail as mod.data.csv.
Don't worry, it will stay between you two and the Internet
(Actually, the list software seems to have
https://github.com/hadley/devtools/wiki/Reproducibility
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
John Kane
Kingston ON Canada
-Original Message-
From: aparna.ec...@gmail.com
Sent: Mon, 15 Sep 2014 14:40:37 -0500
To: r-help@r-project.org
Hi Dennis,
Why is there that warning ? I think my syntax is
right. Isn't it not? So the warning can be ignored ?
Thanks,
Mohan
On Wed, Sep 17, 2014 at 9:48 PM, Dennis Murphy djmu...@gmail.com wrote:
No reproducible example (i.e., no data) supplied, but the following
Hello,
I am trying to interface in my teaching some elementary probability with Monte
Carlo ideas. In sampling from a finite population, the number of distinct
samples of size 'k' from a population of size 'n' , when individuals are
selected with replacement and the selection order does not
Hello,
Try function ?sample. Something like, if 'x' is a vector of size n,
sample(x, k, replace = TRUE)
If you want indices into 'x', try instead
sample(n, k, replace = TRUE)
Hope this helps,
Rui Barradas
Em 17-09-2014 19:25, Giovanni Petris escreveu:
Hello,
I am trying to interface in
You have not shown all of your code and it is difficult to diagnose the
issue.
I assume that you are using the data from:
library(AppliedPredictiveModeling)
data(AlzheimerDisease)
If so, there is example code to analyze these data in that package. See
?scriptLocation.
We have no idea how
Thank you, Rui, but I have the feeling that in this way you are drawing
uniformly *ordered* samples, and you get a bias if you consider them to be
unordered.
Best,
Giovanni
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: Wednesday, September 17, 2014
On 17/09/2014 2:25 PM, Giovanni Petris wrote:
Hello,
I am trying to interface in my teaching some elementary probability with Monte
Carlo ideas. In sampling from a finite population, the number of distinct
samples of size 'k' from a population of size 'n' , when individuals are
selected with
On 17/09/2014 3:07 PM, Duncan Murdoch wrote:
On 17/09/2014 2:25 PM, Giovanni Petris wrote:
Hello,
I am trying to interface in my teaching some elementary probability with
Monte Carlo ideas. In sampling from a finite population, the number of distinct
samples of size 'k' from a population of
Hi Duncan,
You are right. The idea of the derivation consists in 'throwing' k placeholders
(* in the example below) in the list of the individuals of the population.
For example, if the population is letters[1:6], and the sample size is 4, the
following code generates uniformly a 'sample'.
On 17/09/2014 3:46 PM, Giovanni Petris wrote:
Hi Duncan,
You are right. The idea of the derivation consists in 'throwing' k placeholders
(* in the example below) in the list of the individuals of the population.
For example, if the population is letters[1:6], and the sample size is 4, the
I am fundamentally not understanding something about how this is set up,
but after a few hours of googling I am going to ask and I apologize if its
quite basic. I have sanger data that I am reading into ape with read.dna.
x-read.dna(/Volumes/Storage/file.phy, format = interleaved)
It has IUPAC
Hi,
I am running quantile regressions with replicates, but I don't know how
to calculate the Pseudo R squared for quantile regression with
replicates. I have used the following commands:
rho - function(u,tau=.5)u*(tau - (u 0))
V - sum(rho(fit$resid, fit$tau))
where fit is my
In the following example, I am trying to plot a response on a log scale,
and add one or more smoothed
curves, e.g., lowess and abline. In base graphics, I'd like to do this
using log=y, so that the Y axis is
spaced on the log scale, but labeled as the actual response values.
Using ggplot2, I'm
Hi all,
I would like to apply the vcov function from the survey package for the
variables api00 and api99 grouped by the stype variable which can assume H,
M and E categories (see below)
I have tried svyby but as far as I know covariant matrices don't work with
this function
In the code
Hello Tim,
Sorry about the slow response, I just found this message.
On Sep 4, 2014, at 9:23 AM, Tim Richter-Heitmann trich...@uni-bremen.de wrote:
Hi there,
I have two questions about heatmap.2 in gplot.
My input is a simple square matrix with numeric values between 75 and 100 (it
is a
Thanyou David!!!
:)
From: dcarl...@tamu.edu
To: jholt...@gmail.com; eliza_bo...@hotmail.com
CC: r-help@r-project.org
Subject: RE: [R] column names to row names
Date: Wed, 17 Sep 2014 14:04:52 +
Here's another approach using stack():
y - data.frame(y)
E - with(y, data.frame(year,
Thanks Brian!
The confusion was due to my failure to notice that these two functions have
opposite defaults for 'right':
args(cut.default)
function (x, breaks, labels = NULL, include.lowest = FALSE, right = TRUE,
dig.lab = 3L, ordered_result = FALSE, ...)
args(cut.Date)
Thank you!
That does exactly what I was looking for.
Best,
Giovanni
From: Duncan Murdoch [murdoch.dun...@gmail.com]
Sent: Wednesday, September 17, 2014 15:02
To: Giovanni Petris; r-help@R-project.org
Subject: Re: [R] Generating unordered, with
On Sep 17, 2014, at 4:34 PM, Michael Friendly wrote:
In the following example, I am trying to plot a response on a log
scale, and add one or more smoothed
curves, e.g., lowess and abline. In base graphics, I'd like to do
this using log=y, so that the Y axis is
spaced on the log scale, but
Oops, I forgot to mention that an bug was preventing RowSideColors from
working properly. It is fixed in version 2.14.2 of gplots which I've just
uploaded to CRAN and am attaching to this email.
-Greg
On Wed, Sep 17, 2014 at 5:12 PM, Gregory R. Warnes g...@warnes.net wrote:
Hello Tim,
Sorry
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