> On 30 Mar 2015, at 17:11 , Gian Maria Niccolò Benucci
> wrote:
>
> Dear R-usrs,
>
> I am trying to perform a MANOVA on a data frame with 31 columns about soil
> parameters and 1 column containing the explanatory variable (Fraction) that
> have three levels.
> my code is the following:
>
> d
Dear all,
I am working with a likelihood function that requires the inverse of
many small covariance matrices for multivariate normal densities. When
the sample size is large, this calculation is really heavy. Those
matrices are independent but unfortunately I can hardly find a way to
vectori
> I am analyzing trend using Mann-kendall test for 31 independent sample, each
> sample have 34 years dataset. I supposed to find Kendall “tau” for each
> sample. The data is arranged in column wise (I attached the data).To find
> Kendall tau, I wrote R script as:
> ...
> Anyone can tell me ho
Greetings, I am working on a project where we are applying the
Kruskal-Wallace test to some factor data to evaluate their correlation with
existing grade data. I know that the grade data is nonnormal therefore we
cannot rely on ANOVA or a similar parametric test. What I would like to
find is a me
On 02/04/2015 9:31 AM, Feng Li wrote:
Dear all,
I am working with a likelihood function that requires the inverse of
many small covariance matrices for multivariate normal densities. When
the sample size is large, this calculation is really heavy. Those
matrices are independent but unfortunately
Thank you very much for the response. Then what does it mean? I am not a stat
person, but have to use it for my project. :(
Could you please recommend some readings about it? Thanks a lot!
On Wednesday, April 1, 2015 10:58 AM, Michael Dewey
wrote:
This is really a question about st
As a follow-up to Don's reply have a look at
https://github.com/hadley/devtools/wiki/Reproducibility and/or
ttp://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
for some useful suggestions on how to frame a question.
If you need to include sample data with your
This is a big topic. You might try looking for tutorials on "linear
models", with "rank" or "rank deficiency" as subtopics. One possible
book is:
http://www.amazon.com/Linear-Models-Chapman-Statistical-Science/dp/1439887330/ref=sr_1_5?s=books&ie=UTF8&qid=1427987551&sr=1-5&keywords=linear+models+in
> MannKendalltau<- numeric(nc) simply makes MannKendalltau a single
> integer equal to nc; that doesn't look sensible when the next thing you
> do is treat MannKendalltau as a vector.
No, numeric(nc) makes a "numeric" (double precision) vector of length nc
filled with zeros.
Perhaps you were thin
Please stop... you are acting like a broken record, and are also posting in
HTML format. Please read the Posting Guide and demonstrate that you have used a
search engine on this topic before posting again.
---
Jeff Newmiller
You cannot have columns which are linearly dependent.
Starting a project which uses statistics without having some sort of
local statistical backup seems unprofitable.
install.packages("fortunes") # if not already done
library(fortunes)
fortune(122)
On 02/04/2015 15:20, Fix Ace wrote:
Thank
I have a data frame 'mydata.final' (see below) that contains US
counties and a continuous numeric variable 'Mean.Wait' that ranges
from zero to 10 or so. I also created variable 'wait' that is based on
the 'Mean.Wait' and takes on discrete values from 1 (lowest values on
'Mean.Wait') to 5 (highest
On Apr 1, 2015, at 9:59 PM, Alaios wrote:
You are still replying with formatted email. This mailing list attempts to be
an HTML-free zone.
> Thanks.The code you gave me at the end works correctly.. I was wondering if
> there is more efficient way to access each element withouth this classic fo
On Apr 2, 2015, at 7:06 AM, S Ellison wrote:
>> I am analyzing trend using Mann-kendall test for 31 independent sample,
>> each
>> sample have 34 years dataset. I supposed to find Kendall “tau” for each
>> sample. The data is arranged in column wise (I attached the data).To find
>> Kendall
Hi there: I have a list of data frames with identical variable names. I’d
like to reverse scale the same variables in each data.frame.
I’d appreciate any one’s suggestions as to how to accomplish this. Right now,
I’m working with the code at the very bottom of my sample data.
Thanks, Simon K
Thank you for the explanation and comments. I managed to solve and your
comments are helpfully! thanks once again,Desta
On Thursday, April 2, 2015 7:19 PM, David Winsemius
wrote:
On Apr 2, 2015, at 7:06 AM, S Ellison wrote:
>> I am analyzing trend using Mann-kendall test for 3
Dimitri,
You could use colorRamp() and rgb() to get more continuous colors.
For example
newpal <- colorRamp(c("yellow", "red"))
missing <- is.na(mydata.final$Mean.Wait)
newcol <- ifelse(missing, "white",
rgb(newpal(mydat$Mean.Wait/max(mydat$Mean.Wait)), maxColorValue=255))
map('county', fill=TR
Thank you, Jean, but I think this newcol line is not working. I am running:
newcol <- ifelse(missing, "white",
rgb(newpal(mydata.final$Mean.Wait/max(mydata.final$Mean.Wait,
na.rm=T)),
maxColorValue=255))
# And I am getting:
Error in rgb(newpal(mydata.final$Mean.Wait/max(myda
Jean, I think I fixed it:
newpal <- colorRamp(c("yellow", "red"))
missing <- is.na(mydata.final$Mean.Wait)
newcol <- ifelse(missing, "white",
rgb(newpal(mydata.final$Mean.Wait[!is.na(mydata.final$Mean.Wait)]/
max(mydata.final$Mean.Wait,
na.rm=T)), maxColorValue=2
Hi Dimitri,
You can also try the color.scale function in plotrix, which allows you to
specify the NA color in the call.
newcol<-color.scale(mydata.final$Mean.Wait,extremes=c("yellow","red"),na.color="white")
Jim
On Fri, Apr 3, 2015 at 8:08 AM, Dimitri Liakhovitski <
dimitri.liakhovit...@gmail.c
Hi Simon,
How about this?
library(plotrix)
revlist<-grep("i",names(df),fixed=TRUE)
df[,revlist]<-sapply(df[,revlist],rescale,c(3,1))
Jim
On Fri, Apr 3, 2015 at 6:30 AM, Simon Kiss wrote:
> Hi there: I have a list of data frames with identical variable names.
> I’d like to reverse scale the s
Hi Collin,
Have a look at this:
http://stats.stackexchange.com/questions/70643/power-analysis-for-kruskal-wallis-or-mann-whitney-u-test-using-r
Although, thinking about it, this might have constituted your "perusal of
the literature".
Plus it always looks better when you spell the names properly
This is really cool, Jim - thanks a lot!
On Thu, Apr 2, 2015 at 6:18 PM, Jim Lemon wrote:
> Hi Dimitri,
> You can also try the color.scale function in plotrix, which allows you to
> specify the NA color in the call.
>
> newcol<-color.scale(mydata.final$Mean.Wait,extremes=c("yellow","red"),na.colo
I have been doing other things, but want again to learn R.
First some code:
#try 2 axes at bottom
x1 <- c(1,2,3,4,5,6,7,8,9)
x2 <- c(5,1,9,6,3,7,8,2,4)
plot(x1,x2,type="l") #scatter plot x1 on horizontal, x2 on vertical axis
axis(1,labels=TRUE,tick=TRUE,col="red",col.axis="red",lwd=4,line=2.5)
So
Thank you Jim, I did see those (though not my typo :) and am still
pondering the warning about post-hoc analyses.
The situation that I am in is that I have a set of individuals who
have been assigned a course grade. We have then clustered these
individuals into about 50 communities using standard
On 04/02/2015 10:40 PM, Duncan Murdoch wrote:
On 02/04/2015 9:31 AM, Feng Li wrote:
Dear all,
I am working with a likelihood function that requires the inverse of
many small covariance matrices for multivariate normal densities. When
the sample size is large, this calculation is really heavy. T
Hi Hurr,
Try this:
par(mar=c(7,4,4,2))
plot(x1,x2,type="l",xlab="")
require(plotrix)
fullaxis(1,at=pretty(x1),labels=round(1/pretty(x1),2),tick=TRUE,
col="red",col.axis="red",lwd=4,pos=-0.2)
mtext("x1",side=1,line=4.5)
Jim
On Fri, Apr 3, 2015 at 11:16 AM, Hurr wrote:
> I have been doing other
Thanks, That was good help; I tried this; I hope I didn't copy it wrong:
#try 2 axes at bottom
x1 <- c(1,2,3,4,5,6,7,8,9)
x2 <- c(5,1,9,6,3,7,8,2,4)
par(mar=c(7,4,4,2))
plot(x1,x2,type="l",xlab=””) #scatter plot x1 on horiz, x2 on vert axis
require(plotrix)
fullaxis(1,at=pretty(x1),labels=round(1/
Hi all,
I need to design an experiment very similar to the design described in
the AlgDesign vignette on page 44. When I run the code on the vignette,
I get the following error:
> library(AlgDesign)
> Rep<-gen.factorial(3,1,factor=1,varName="Rep")
> Block<-gen.factorial(3,1,factor=1,varName="Blo
Hi Hurr,
The problem is with the quotes in xlab="". In your version the quotes are
not ASCII double quotes but some sort of "smart" quotes. Either your email
client has substituted them or more likely, you are editing your code in
something like MS Word and cutting and pasting the result into R. Tr
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