Hi:
A univariate t-distribution with 1 df is equivalent to the standard Cauchy
distribution, for
which it is well established that the population mean does not exist. You're
basically
simulating a vector version of the Cauchy distribution (an IID 8-dimensional
version),
so the same problem is
Shant Ch wrote:
Warning messages:
1: In if (freq) x$counts else { :
the condition has length 1 and only the first element will be used
...
hist(W,breaks=20,probability=T)
A variable called T lying around with length(T) != 1 ?
Try probability=TRUE instead.
--
Peter Dalgaard
Center for
Dear ALL R Experts,
I have a time series and I want to detect presence of seasonality in this
time series. I know the method of plotting the Autocorrelation function(acf)
and if there are significant lags after period s,2s,3s,... then s is the
period for the time series. But my problem is
Hi there,
I checked google for aov. usually one uses summary to see whether the p-value
is small.
but I want to put aov in my script. how can I get the p-value, (0.1115, 0.6665,
0.6665 in the following example)?
thanks
YU
Hi:
str() is usually helpful:
str(summary(aov.ex2))
List of 1
$ :Classes anova and 'data.frame': 4 obs. of 5 variables:
..$ Df : num [1:4] 1 1 1 12
..$ Sum Sq : num [1:4] 76.5625 5.0625 0.0625 311.25
..$ Mean Sq: num [1:4] 76.5625 5.0625 0.0625 25.9375
..$ F value: num [1:4]
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Thank you very much for that.
What is then if I have unpaired and unbalanced samples?
Best regards,
Cheba
2010/5/7 Bert Gunter gunter.ber...@gene.com
Perhaps this might help clarify:
sample A: 10 15 20
sample B: 12 15 22
Median of sample A = 15; median of sample B = 15.
On 09.05.2010 18:37, bbslover wrote:
a-1:5
b-2:6
plot(a,b)
Error in function (width, height, pointsize, record, rescale, xpinch, :
Graphics API version mismatch
before, R 2.10 , plot() is ok. Now, R 2.11.0 does not work
YYou have probably an old version of the graphics package
Hey guys,
I have a doubt here , It is something simple I guess, what am I missing out
here ??
f - function(y) function() y
tmp - vector(list, 5)
for (i in 1:5) tmp[[i]] - f(i)
tmp[[1]]() # returns 5;
z - f(6)
tmp[[1]]() # still returns 5; it should return 6 ideally right ???
Even if I dont
See ?colSums
On Mon, May 10, 2010 at 12:44 AM, vincent.deluard
vincent.delu...@trimtabs.com wrote:
Hi R users,
I have a matrix m of the type:
m
X4.20.2010 X4.19.2010 X4.16.2010
[1,] 0.008319468 0. -0.008250825
[2,] 0.005574136 0.01816118 0.073081608
[3,] -0.047830688
The hash-2.0.0 has been released to CRAN.
This package implements a data structure similar to hashes in Perl and
dictionaries in Python but with a purposefully R flavor. For objects of
appreciable size, access using hashes outperforms native named lists and
vectors.
This version accounts for
You are missing 'force'.
See 'The R Inferno' page 90.
In this case you can define:
f - function(y) { force(y); function() y}
On 10/05/2010 11:06, sayan dasgupta wrote:
Hey guys,
I have a doubt here , It is something simple I guess, what am I missing out
here ??
f- function(y) function()
Version 2.2.0 of package bcp is now available. It replaces the suggests of
NetWorkSpaces (previously used for optional parallel MCMC) with the
dependency on package foreach, giving greater flexibility and supporting a
wider range of parallel backends (see doSNOW, doMC, etc...).
For those
Hey
thanks for your help ,
But thats not exactly the problem I have
See I am fine with
tmp[[1]]() being = 5 and not 1; but then
for (i in 1:5) tmp[[i]] - f(i)
z - f(6)
tmp[[1]]() ## should give 6 right ? Because f(6) was last evaluate so in
parent.frame() y should be 6 ???
On Mon, May 10,
sayan dasgupta wrote:
Hey guys,
I have a doubt here , It is something simple I guess, what am I missing out
here ??
f - function(y) function() y
tmp - vector(list, 5)
for (i in 1:5) tmp[[i]] - f(i)
tmp[[1]]() # returns 5;
z - f(6)
tmp[[1]]() # still returns 5; it should return 6 ideally
When you call a function R passes a promise to it for each argument.
A promise consists of the unevaluated variable together with the
environment in which it should evaluate the variable when time comes
to evaluate it. Thus tmp[[1]] contains function(y) y and in the
environment of function(y) y
Dear all,
when trying to replicate John M. Quick's example for correlations between
multiple variables posted on:
http://rtutorialseries.blogspot.com/2009/11/r-tutorial-series-zero-order.html
with R 2.11.0 (GUI 1.33) using my MacBook Pro with OX X 10.5.8 I got the
following error message
On 10/05/2010 7:36 AM, Ruben Garcia Berasategui wrote:
Dear all,
when trying to replicate John M. Quick's example for correlations between
multiple variables posted on:
http://rtutorialseries.blogspot.com/2009/11/r-tutorial-series-zero-order.html
with R 2.11.0 (GUI 1.33) using my MacBook Pro
Hello everybody,
I'm trying to install a package I have built. This package contains three
scripts with various functions (S3 as well as S4 classes)
I run at first the package.skeleton command with:
package.skeleton(affyAnalysis, namespace=TRUE, code_files =c(defS3.R,
defS4.R, qc.R))
See the replace argument in ?randomForest.
Andy
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri
Liakhovitski
Sent: Friday, May 07, 2010 12:21 PM
To: r-help
Subject: [R] randomForest: sampling with replacement?
I've not seen the book myself, but Graham Williams (author of the rattle
package) has been working on a book that perhaps may fit your need.
http://datamining.togaware.com/survivor/index.html
Andy
From: Wensui Liu
good question!
if there is such a book, i'd also like to read as well.
On
Hello,
I've tried to install randomForest on a Ubuntu 8.04 Hardy Heron system.
I've repeatedly rec'd the error:
install.packages(randomForest, dependencies = TRUE)
ERROR: compiliation failed for package 'randomForest'
** Removing '/home/admuser/R/i486-pc-linux-gnu-library/2.6/randomForest'
I don't know much about how permissions are managed under ubuntu, but
you can try a couple of things:
- Dirk had worked very hard at automating building of CRAN packages for
Debian (from which Ubuntu is based). I believe randomForest is among
one of those available. I'm not sure if you need to
On 10/05/2010 7:43 AM, Assa Yeroslaviz wrote:
Hello everybody,
I'm trying to install a package I have built. This package contains three
scripts with various functions (S3 as well as S4 classes)
I run at first the package.skeleton command with:
package.skeleton(affyAnalysis, namespace=TRUE,
Would you please confirm that it is a bootstrap sample with
replacement?
Someone should note that the definition of a bootstrap sample is a
sample with replacement (usually of size n).
I've read quite a few papers where they claim to be using the
bootstrap. Upon further review (sometimes to
On 9 May 2010 15:25, Ky Mathews ky.math...@sydney.edu.au wrote:
Hi,
I want to customise the segments on an xyplot. Below is a simple example
of what I'm trying to do...
#Example dataset
x - c(-0.25, 0.25, 0.8)
y - c(-0.5, 0, 0.75)
gp - c(A, I, C)
my.data -
On 05/10/2010 12:32 AM, bbslover wrote:
many thanks . I can try to use test set with 100 samples.
anther question is that how can I rationally split my data to training set
and test set? (training set with 108 samples, and test set with 100 samples)
as I know, the test set should the same
Steve,
The list r-sig-debian is a more appropriate forum for this question as it is
dedicated to R on Debian / Ubuntu.
On 10 May 2010 at 08:24, steve_fried...@nps.gov wrote:
|
| Hello,
|
| I've tried to install randomForest on a Ubuntu 8.04 Hardy Heron system.
|
| I've repeatedly rec'd the
The Mantel-Byar test is a simple modification to the logrank test. Can
R perform this test or any other similar test?
Raphael
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PLEASE do read the posting guide
Dear all,
I'm trying to create a dot plot with error bands with
Dotplot(labels ~ Cbind(estimate, lower, upper), data=For.plot)
where estimate, lower and upper are numerical vectors, and labels is a
character vector that contains labels.
The problem is that labels are automatically sorted
Hi,
In the locfit package, could someone please let me know the automatic selection
of smoothing parameter if Gauss kernel density function is used as weight
function?
thanks
Fir
[[alternative HTML version deleted]]
__
Hi,
I have a dataset with many variables and observations.
The variable Group has two levels: C and P,
the Month variable has four levels: 0, 1, 2 and 3.
I want to extract a subset of the variable Weight,
considering only 1 and 3 levels for Months of the Group
variable.
I tried the command
now. it is ok. I uninstall R2.11.0, then delete an packages in the library,
and install again R2.11.0. ok, it does works.
thank you!
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Dear r-help,
Could you help me to find the function which create an empty matrix.
I use matrix(), but it gives *a single value that is NA and length of this
matrix is 1.*
**
*Best Regards*
[[alternative HTML version deleted]]
__
Dear all,
I'm trying to create a dot plot with error bands with
Dotplot(labels ~ Cbind(estimate, lower, upper), data=For.plot)
where estimate, lower and upper are numerical vectors, and labels is a
character vector that contains labels.
The problem is that labels are automatically sorted
Hi:
This is more a statistical question than an R question (apologies!).
I have some income data as follows:
$5000 : 598
$5000-$1 : 2586
$65001-$7 : 202
$70001+ : 446
I.e an open ended income class for incomes $70k.
What would be the best way to estimate mean income?
Something
On 10.05.2010 14:35, Silvano wrote:
Hi,
I have a dataset with many variables and observations.
The variable Group has two levels: C and P,
the Month variable has four levels: 0, 1, 2 and 3.
I want to extract a subset of the variable Weight, considering only 1
and 3 levels for Months of the
On 10.05.2010 13:15, anderson nuel wrote:
Dear r-help,
Could you help me to find the function which create an empty matrix.
Help is:
?matrix
Uwe Ligges
I use matrix(), but it gives *a single value that is NA and length of this
matrix is 1.*
**
*Best Regards*
[[alternative
Hi,
My first suggestion would be to supply a sample data (maybe using the
function dput) showing what you have, what you want to do, and what
you've tried (you say that subset() didn't work but we don't know how
you've typed it).
Then, we'll see!
Ivan
Le 5/10/2010 14:35, Silvano a écrit :
On 10/05/2010 7:15 AM, anderson nuel wrote:
Dear r-help,
Could you help me to find the function which create an empty matrix.
I use matrix(), but it gives *a single value that is NA and length of this
matrix is 1.*
Not sure what you mean by an empty matrix, but here's one interpretation:
I think matrix(nrow=0, ncol=0) will do it.
-Ista
On Mon, May 10, 2010 at 7:15 AM, anderson nuel anderson@gmail.comwrote:
Dear r-help,
Could you help me to find the function which create an empty matrix.
I use matrix(), but it gives *a single value that is NA and length of
this
On May 10, 2010, at 10:00 AM, Alexey Bessudnov wrote:
Dear all,
I'm trying to create a dot plot with error bands with
Dotplot(labels ~ Cbind(estimate, lower, upper), data=For.plot)
where estimate, lower and upper are numerical vectors, and labels is
a character vector that contains
See the kern argument in ?locfit.raw.
Andy
From: FMH
Hi,
In the locfit package, could someone please let me know the
automatic selection of smoothing parameter if Gauss kernel
density function is used as weight function?
thanks
Fir
[[alternative HTML version
Hi,
I used
#--
# Análise do PESO -
#--
pesom1 = cbind(PESO[MÊS==1])
pesom3 = cbind(PESO[MÊS==3])
#- Grupo Placebo -
pesom1p = pesom1[GRUPO=='P']
pesom3p = pesom3[GRUPO=='P']
t.test(pesom1p, pesom3p, paired=T)
#- Grupo Cogumelo -
R experts -
I'm using John Fox's sem package to analyze a simple path model (two correlated
predictor variables directly influencing a single criterion variable):
Predictor1 - Criterion
Predictor2 - Criterion
Predictor1 - Predictor2
I'm giving a presentation on this material next week, and
Dear Elisabeth,
I'm not sure if I have understood your question -- are you trying
to use a different logarithmic base?
Kind regards,
Xianwen
On Sun, May 9, 2010 at 8:05 PM,
(Near) non-identifiability (especially in nonlinear models, which include
linear mixed effects models, Bayesian hierarchical models, etc.) is
typically a strong clue; usually indicated by software complaints (e.g.
convergence failures, running up against iteration limits, etc.).
However this is
no one?
any pointers would really be greatly appreciated!
thanks,
kay
-
Kay Cichini
Postgraduate student
Institute of Botany
Univ. of Innsbruck
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Loader's book is the ultimate reference (other than the source code...). I
don't have it with me right now, so can't tell you for sure. However, it seems
like an odd thing to do to use a different kernel at the selection step than
the final estimation step. I should think if you specify the
Hello!
Thank you for answering!
What I am trying to do is to plot my raw values (biomass of different
species) on a logaritmic y-axis with the base of e. When I type log=y,
the axis transforms into a logaritmic axis with the base of 10.
Best regards,
Elisabeth
Dear Elisabeth,
I'm not
Sums of correlated increments have the same correlation as the original
variables...
library(mvtnorm)
X- matrix(0,nrow=1000,ncol=2)
for(i in 1:1000){
Y - rmvnorm(1000,mean=mu,sigma=S)
X[i,] - apply(Y,2,sum)
}
cor(Y)
[,1] [,2]
[1,] 1.000 0.4909281
[2,] 0.4909281 1.000
Hello:
I am using the ggplot2 package on R 2.10.1. I am plotting points using
geom_pointrange. Is there a way to overlay hashmarks on the points,
specifically the median and the min and max of the range?
Cheers,
Michael
[[alternative HTML version deleted]]
Thank you Andy. I'm aware of that but my question is about the way the locfit
package use to automatically determine the bandwidth/smoothing parameter if
Gauss kernel density function is used as the weight function. Does the
generalized cross validation criteria is imposed here or does the
Dear R users,
I recently developed a plotting function in R and introduced it to my
coworkers. The function is designed to make plotting easier and more efficient,
which will in turn be more cost-effective for the company. The reviews for the
function have been positive thus far, except for
I'm searching for an r command that will notify me if I create a time that
does not exist due to Daylight Savings Time. For example, if I run the
following command on a windows machine
ISOdatetime(2010,03,14,2,10,0, tz = ) # My system time is set to the
United States Central Time Zone
[1] NA
R
Dear Elisabeth,
log(X) shall return logarithm of X, with the base of e.
If you are not sure, you can specify e to be the base, e.g.:
log(X,exp(1))
.
Snow in Tromsø is melting, so will your problem.
King regards,
Dear R users and specially Albyn and Giovanni,
thanks for your answers, but in fact I supposed the same at the
beginning of my problem. However, when I generate the data seldom I
obtain the expected correlation. For example using this code:
fz-function(n,t,rho){
f-NULL
for(i in 1:n){
On May 10, 2010, at 2:55 PM, Sergio Andrés Estay Cabrera wrote:
Dear R users and specially Albyn and Giovanni,
thanks for your answers, but in fact I supposed the same at the
beginning of my problem. However, when I generate the data seldom I
obtain the expected correlation. For example
I think your problem is that HL07 is not a data frame but a list. read.spss()
produces a list rather than a data frame by default, and svydesign() requires a
data frame. You can use as.data.frame() to turn HL07 into a data.frame.
The weights also look a bit strange -- you have sampling
Lets say I have a generated data frame with variables that follow a
naming convention:
title,a1,a2,b1,b2,b3,c1,c2,c3,c4...
I am plotting every column (starting from a1) as a line in a plot.
That works. However my diagram becomes very unorganized. Creating
legends is nice, but trying out
Hi,
Lattice and ggplot2 are both ideally suited for this task. Consider
this example,
library(ggplot2)
d = data.frame(x=1:10, a1=rnorm(10), b1=rnorm(10))
m = melt(d, id =x) # reshape into long format
qplot(x, value, data=m, geom=path, colour=variable)
library(lattice)
xyplot(value~x, data=m,
When running DEMA(data, 5) on a vector 'data' of length 5, my R engine
stops. Is this function or the R environment facing a bug here or am I
doing something wrong? DEMA should work if the smoothing window size
is the same size as the the data length, right?
(I am working with Eclipse 3.5. and
Sigh; that's because I forget to include it in the 'substitute' list.
Here is a version that works (and has been tested...).
ourtitle - function(title, footnotes=NULL) {
if (length(footnotes) 0) {
fn - paste(footnotes, collapse=' ')
title - eval(substitute(expression(bold(paste(title,
Hi,
I use R with a screen reader to do statistical and financial analysis.
There are a couple of things that your student can do.
R was created on Linux, and then ported to Windows. So she can use R under
Linux where the CLI is perfectly accessible.
There are 2 main screen readers for Linux:
Many thanks for this suggestion. Indeed, labels turned out to be a
factor, and after reordering the levels I got the plot I wanted.
Alexey
David Winsemius wrote:
On May 10, 2010, at 10:00 AM, Alexey Bessudnov wrote:
Dear all,
I'm trying to create a dot plot with error bands with
I am writing to ask if R has a build- in function to calculate this
polylogarithm Li_n(z) function , also known as the Jonquière's function
defined as
Li_n(z)=sum_(k=1)^infty(z^k)/(k^n)
Thanks
Andy
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R-help@r-project.org mailing list
R experts,
I am working with large multivariable data frames ( 50 variables)
and I would like to scroll horizontally across my output
to view each data frame rather than having to scroll down vertically-
wrapped data frames.I have been using R Commander as a programming
interface. If I assign
I tried writing that function before, but the expression() command is not
preserved when we incorporate it into a function(). The first example you gave
me, if run in R, produced a title that reads title instead of stuff. It
seems that in this case the expression() can't read the character
Hi All
I need some help with plotting a step function, currently I'm using
sfun - stepfun(c(1, 2, 5,10, 20), c(0, 11, 22, 33, 44, 0), f=0)
plot(sfun, pch=NA, main=, xlim=c(1,20))
which I working fine, but my data is in the following format:
Min Max Value
1 2 11
2 5 22
510 33
10 20 44
1. To
Hi Mike,
You can set options(width = x), where x equals the number of columns.
-Ista
On Monday 10 May 2010 16:06:54 Michael H wrote:
R experts,
I am working with large multivariable data frames ( 50 variables)
and I would like to scroll horizontally across my output
to view each data frame
I installed the lattice package, and got an error that R was not able
to remove the previous version of lattice. Now my installation seems
to be currupt, even affecting other packages. I am getting this error
when loading TTR:
library(TTR)
Loading required package: xts
Loading required package:
Dear all:
I would like to create a landscape of environmental values that follow a
uniform frequency distribution and also have spatial autocorrelation in the
landscape. I was wondering if there is an algorithm and/or package out there
that creates autocorrelation of values that are distributed
Perhaps even ?View would be useful here. I've never used R Commander,
so don't know it would be useful in that environment.
Michael H wrote:
R experts,
I am working with large multivariable data frames ( 50 variables)
and I would like to scroll horizontally across my output
to view each
Hi: Thanks to Dennis and Fernando for your help reordering the levels.
Now I have a different issue:
I am trying to get the cumulative weekly values using cumsum and it
appears to output the wrong values. Here's my dataset:
winter - structure(list(week = c(26L, 27L, 28L, 29L, 30L, 31L, 32L,
On May 10, 2010, at 4:46 PM, Ralf B wrote:
I installed the lattice package, and got an error that R was not able
to remove the previous version of lattice. Now my installation seems
to be currupt, even affecting other packages. I am getting this error
when loading TTR:
library(TTR)
Loading
It is working correctly, You have NAs in your data, so you get NAs in the
cumsum.
On Mon, May 10, 2010 at 5:09 PM, Felipe Carrillo
mazatlanmex...@yahoo.comwrote:
Hi: Thanks to Dennis and Fernando for your help reordering the levels.
Now I have a different issue:
I am trying to get the
I realized that after I hit the send button...I was looking at the wrong
dataset,Brr
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
From: jim holtman jholt...@gmail.com
To: Felipe Carrillo mazatlanmex...@yahoo.com
Elisabeth, question to you: How is it that you recognise that it is
a logaritmic axis with the base of 10, as opposed to any other base?
Ted.
On 10-May-10 17:15:04, Elisabeth Bjerke Rastad wrote:
Hello!
Thank you for answering!
What I am trying to do is to plot my raw values (biomass of
Hi,
I'm using maxlik with functions specified (L, his gradient hessian).
Now I would like determine some robust standard errors of my estimators.
So I 'm try to use vcovHC, or hccm or robcov for example
but in use one of them with my result of maxlik, I've a the following
error message :
under Unix/X11 (and IIRC under Windows), edit(some.data.frame) will
invoke a smallish spreadsheet-like data editor, which will allow you to
move around your data with no fuss. I probably wouldn't use it for any
serious data entry, but found damn useful in a lot of data-debugging
situations (you
I'm trying to use cex.axis or something like this to augment the size
labels' on levelplot axis, but it doesn't work.
Could someone help me?
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Is there a predict method/syntax which I could use to generate
predictions (and other output from predict() methods) if I have the
model parameter estimates from a training dataset but not the data used
to generate the original model (the models were generated by a
collaborator using STATA and for
Are there functions/packages which support plots (bar and/or line) where
I provide the point estimate and some error measure rather than the raw
data?
I often have to summarize/present data from multiple sources where the
original data is unavailable and do not yet have a good solution for
this.
On 10/05/2010 6:13 PM, Chaudhari, Bimal wrote:
Is there a predict method/syntax which I could use to generate
predictions (and other output from predict() methods) if I have the
model parameter estimates from a training dataset but not the data used
to generate the original model (the models
On May 10, 2010, at 4:29 PM, vjaneiro wrote:
I'm trying to use cex.axis or something like this to augment the size
labels' on levelplot axis, but it doesn't work.
Could someone help me?
Look at the thread Re: [R] Increasing the font size on axes in
trellis from two days ago.
--
On May 10, 2010, at 6:17 PM, Chaudhari, Bimal wrote:
Are there functions/packages which support plots (bar and/or line)
where
I provide the point estimate and some error measure rather than the
raw
data?
I often have to summarize/present data from multiple sources where the
original data
On Mon, 10 May 2010, RATIARISON Eric wrote:
Hi,
I'm using maxlik with functions specified (L, his gradient hessian).
Now I would like determine some robust standard errors of my estimators.
So I 'm try to use vcovHC, or hccm or robcov for example
but in use one of them with my result of
Perhaps you should be using the chron package. It has no time zones
in the first place.
On Mon, May 10, 2010 at 2:26 PM, Garrett Grolemund g...@rice.edu wrote:
I'm searching for an r command that will notify me if I create a time that
does not exist due to Daylight Savings Time. For example,
On May 10, 2010, at 4:54 PM, Laura S wrote:
Dear all:
I would like to create a landscape of environmental values that
follow a
uniform frequency distribution and also have spatial autocorrelation
in the
landscape.
Perhaps:
X - distribution on one dimension
Y - distribution on another
I'm learning ggplot and am a little confused. Sometimes discrete scales work
like I'd expect, and sometimes they don't. For example...
This works exactly like one would expect:
df-data.frame(names=c(Bob,Mary,Joe,Bob,Bob))
ggplot(df,aes(names))+geom_histogram()
But this yields an error:
On May 10, 2010, at 7:36 PM, John Rauser wrote:
I'm learning ggplot and am a little confused. Sometimes discrete
scales work
like I'd expect, and sometimes they don't. For example...
This works exactly like one would expect:
df-data.frame(names=c(Bob,Mary,Joe,Bob,Bob))
May I ask for further illumination of my dumbness?
I had been merrily plotting ..density.. and using xlim() to constrain the
x-axis like this and everything was working fine:
df-data.frame(names=c(Bob,Mary,Joe,Bob,Bob))
ggplot(df,aes(names,..density..,group=1))+geom_histogram()+xlim(Bob,Mary)
Hi - a newbie question, if someone can please help
I want to change X1, X2,,.to X.1 X.2 etc in the names below. I am using
the Principal Component Regression function (pcr) and it seems to want it
this way
datap3.pcr - pcr(water ~ X, 10, data = datap3, Validation =cv)
Error in
Is this what you want:
x - paste(X, 1:10, sep='')
x
[1] X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
sub(X, X., x)
[1] X.1 X.2 X.3 X.4 X.5 X.6 X.7 X.8 X.9 X.10
On Mon, May 10, 2010 at 8:53 PM, Ravi Ramaswamy raram...@gmail.com wrote:
Hi - a newbie question, if someone can please
Jim - thanks.
It worked, however pcr program is still not accepting it there is a
sample yarn data loaded and I am comparing. My data looks like this
X1 X2 X3 X4 X5 X6 X7
X8 X9 X10 X11 X12 X13 X14 X15 X16 X17
You need to provide an 'str(yarn)' so we can see what the structure is. I
don't know what the 'pcr' program is expecting, but it looks like from what
you have provided that 'yarn' might be a dataframe and X is a vector.
Look at the documentation for pcr and see what it expects.
On Mon, May 10,
There is definitely a difference. The pcr program is Principal Component
Regression, but document says
The formula argument should be a symbolic formula of the form response ~
terms, where response is the name of the response vector or matrix (for
multi-response models) and terms is the name of
Hi r-sers,
I have a data of relative frequencies for the interval of 0-20,
20-40,...380-400. I would like the two data on the same graph using the same
x-axis label. My question is how to get a smooth curve using kernel density
code if it possible for this data.
cbind(rel_obs,rel_gen)
Dear r-help list members,
I am quite new to R, and hope to seek advice from you about a problem I have
been cracking my head over. Apologies if this seems like a simple problem.
I have essentially two tables. The first (Table A) is a standard patient
clinicopathological data table, where rows
Rhelpers:
I'd like to modify this RSQLite statement:
rs_stations-dbSendQuery(con_stations, select * from stations)
so that stations is actually an R variable, e.g.:
stations=c(stationA,stationB)
How would I modify the above statement to query from stations[[1]]
(aka stationA)?
--j
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