Thank you Dr. Sarkar. yscale.components.log10.3 is pretty good choice.
John
- Original Message
From: Deepayan Sarkar
To: array chip
Cc: Don McKenzie ; Henrique Dallazuanna
; R-help Forum
Sent: Mon, September 27, 2010 8:07:15 PM
Subject: Re: [R] scientific vs. fixed notation in xypl
I have a dataset, example of the data is shown below:
Grouped Data: drain_irr ~ irr | crop
Year decades crop irrisystem drain_irrirr
1310 1995-96 1990s Citrusvarious 0.400 9.021
1311 1995-96 1990s Citrus drip 0.541 6.468
1312 1995-96 1990s Citrus overhea
If you type:
class(z.ex)
you'll see that your z-test produces an object with a class "htest."
Then type:
methods("plot")
you'll get a list of all the types of objects that the plot() function
knows how to make plots for. It doesn't look like plot has a sub-
method for an htest object,
You have provided no information as to what you mean by "my analysis
fails". Exactly what error message are you getting, what operation
system do you have, how much memory do you have, how much are you
using for all the other objects in your address space, etc..
Information like this would hel
On Mon, Sep 27, 2010 at 5:28 PM, array chip wrote:
> I found where the problem is with y-axis, the yscale.components.log10 should
> be
>
> yscale.components.log10 <- function(lim, ...) {
> ans <- yscale.components.default(lim = lim, ...)
> tick.at <- logTicks(10^lim,loc=1)
> ans$left$tic
I try to remember that timevar defines the wide columns you want and
v.names the values in the body of the table - but I always forget next
time I need to use reshape :)
On 28 September 2010 12:55, Toby Gass wrote:
> That does work, thank you. I didn't understand that the "fame"
> column would
Hi, I am a beginner in R and is trying to plot some dot plot from t test
result.
I was following the example
## Example from z.test -- requires TeachingDemos package
# library(TeachingDemos)
# z.ex <- z.test(rnorm(25,100,5),99,5)
# z.ex
# plot(z.ex)
but encounter this error for the last command
Hello Richard,
Since no one else has answered yet I'll venture a guess.
The following works on my little macbook...
x <- as.factor(sapply(letters[1:26], function(x) paste(rep(x, 10),
collapse="")))
So each of the 26 factor levels in x has a string representation of
100,000 chars. So I'm *g
Hi,
Teaching this year I was noticing that there seems to be an odd limitation in
the IQR function. It doesn't allow one to set the quantile type. It seems to
me an easy thing to fix. The current function is…
IQR <- function (x, na.rm = FALSE) diff(quantile(as.numeric(x), c(0.25, 0.75),
na
That does work, thank you. I didn't understand that the "fame"
column would be the time varying column.
Toby
On 28 Sep 2010 at 12:47, Michael Bedward wrote:
> Hi Toby,
>
> I think this should work...
>
> reshape(dat, v.names=c("weight"), idvar=c("valley", "plot", "trt"),
> timevar="fame", di
Hi Toby,
I think this should work...
reshape(dat, v.names=c("weight"), idvar=c("valley", "plot", "trt"),
timevar="fame", direction="wide")
Michael
On 28 September 2010 12:17, Toby Gass wrote:
> Hello, helpeRs,
>
> I've been trying, unsuccessfully, to change a dataframe from long to
> wide for
Hello again Ralf,
Using the instructions at that link (minus the bit about copying
config files from /usr/share), followed by...
install.packages("rimage", type="source")
...seems to have worked for me. I can now load the package and read /
display a jpeg file.
Michael
On 28 September 2010 10
Hello, helpeRs,
I've been trying, unsuccessfully, to change a dataframe from long to
wide format using reshape (the original). I would appreciate it if
someone could demonstrate the correct syntax. The script below will
create a toy example. The new wide data should have a column name
for
Dear all,
I am trying to look for a built in function that performs the cochran Q
test.
that is, cochranq.test(X)
where X is a contingency table (maybe a matrix or data.frame).
The output will naturally be the test statisitcs, p-value, etc.
A quick search on Google gives me the cochran.test in th
Hello, helpeRs,
I've been trying, unsuccessfully, to change a dataframe from long to
wide format using reshape (the original). I would appreciate it if
someone could demonstrate the correct syntax. The script below will
create a toy example. The new wide data should have a column name
for
I have a dataset, example of the data is shown below:
Grouped Data: drain_irr ~ irr | crop
Year decades crop irrisystem drain_irrirr
1310 1995-96 1990s Citrusvarious 0.400 9.021
1311 1995-96 1990s Citrus drip 0.541 6.468
1312 1995-96 1990s Citrus overhead
Hello,
You probably know this already but there are many textbooks on Group theory
in statistical processes.
Although this book may not help you, the authors may be able to lead you in
the right direction:
http://www.nd.com/products/nsbook.htm Neural Solutions
--
View this message in conte
Try something like this:
dfr <- read.table(textConnection("plate.id well.id Group HYB
rlt1
1 P1 A1 Control SKOV3hyb 0.190
2 P1 A2 Control SKOV3hyb 0.210
3 P1 A3 Control SKOV3hyb 0.205
4 P1 A4 Control SKOV3hyb 0.206
5 P
Hi Ralf,
Are you using Mac OSX ? There isn't a binary available for OSX but you
can apparently install the package from sources. However, it seems
that you have to manually install some system dependencies first:
http://www.r-cookbook.com/node/18
I'm going to try it myself shortly.
Michael
On
I found where the problem is with y-axis, the yscale.components.log10 should be
yscale.components.log10 <- function(lim, ...) {
ans <- yscale.components.default(lim = lim, ...)
tick.at <- logTicks(10^lim,loc=1)
ans$left$ticks$at <- log(tick.at, 10)
ans$left$labels$at <- log(tick.a
On 09/27/2010 08:09 PM, Paul Miller wrote:
> Hello Everyone,
>
> I'm trying to conduct a couple of power analyses and was hoping
> someone might be able to help. I want to estimate the sample size
> that would be necessary to adequately power a couple of
> non-inferiority tests. The first would b
Hello,
thank you very much for replying. The code yields an error message
Error in strptime(x, format, tz = tz) : invalid 'x' argument
But I can't see what's wrong with it, the Date/Time info is in the third
column, format is
"%Y-%m-%d %H:%M:%S". There is no time zone info in the data, could
Dear R-help list members,
I have the following question concerning the strucchange()-package: is
it possible to get the boundaries for one-sided (upper / lower) CUSUM
and MOSUM tests?
Thank you in advance.
Julia
__
R-help@r-project.org mailing l
On Mon, Sep 27, 2010 at 5:38 PM, Struve, Juliane
wrote:
> Hello,
>
> thank you very much for replying. The code yields an error message
>
> Error in strptime(x, format, tz = tz) : invalid 'x' argument
>
> But I can't see what's wrong with it, the Date/Time info is in the third
> column, format i
The expression 19:15:21:22 evaluates as follows:
19:15:21:22
[1] 19 20 21 22
Warning messages:
1: In 19:15:21 : numerical expression has 5 elements: only the first used
2: In 19:15:21:22 :
The expression 2:2:2:2 evaluates as follows:
2:2:2:2
[1] 2
I'm *guessing* you mean
c(19,15,21,22)
Hi,
I am still learning and find it very helpful in my research. I have searched
before posting this and could not figure out if there is a way to compare
matrices the way i am describing below:
I have a matrix of 64 cols and 64 rows which mostly has lots of zeros but has
other non-zero posit
Hi, I found an old thread which refered to 2 examples (Fig 8.4 & Fig 8.5) from
Dr. Sarkar's lattice book that address this problem using "xscale.components"
and "yscale.components":
http://lmdvr.r-forge.r-project.org/figures/figures.html
I am able to change the label of x-axis, but still can't
I think arima may do the trick. I have not used it, but this is the
functionality I have pulled from the ether.
On Mon, Sep 27, 2010 at 5:02 PM, Dr. Alireza Zolfaghari
wrote:
> Hi list,
> I have a set of data which I want to use time series analysis in R in order
> to forecast the value for futu
Thanks very much for your reply, Petr. That's a nice function.
Also, I found "gvarbrowser" function in package "gWidgest" is really a good
one. It's GUI and yet extendable.
library(gWidgets)
options(guiToolkit="RGtk2")
gvarbrowser(cont=gwindow("Object browser"))
I highly recommend to anyone who
Hi list,
I have a set of data which I want to use time series analysis in R in order
to forecast the value for future. I know there are some R functions, but not
sure how to use them. Would you please help me if you are familiar with time
series in R? I want to get value for Nov 2012 using the foll
Hello forum!
I am currently having trouble getting R to recognize one of my pch
characters. Here's what I have:
legend(120, 40, c("SAL-arterial", "SAL-portal", "GLP-1-arterial",
"GLP-1-portal"), bty="n", col=c("black"), pch=19:15:21:22, pt.bg="white",
pt.cex=2:2:2:2, lty=1, lwd=2, text.col="blac
Thanks Josh.
The variance of predictor should be var(beta_0+beta_1*newdata+epsilon). It
is actually the variance of dependent variable if we plug the concrete value
of independent variable into the model.
On Mon, Sep 27, 2010 at 2:09 PM, Joshua Wiley wrote:
> Hi,
>
> Try this:
>
> # using the
Hello
I have a data set like below:
plate.id well.id Group HYB rlt1
1 P1 A1 Control SKOV3hyb 0.190
2 P1 A2 Control SKOV3hyb 0.210
3 P1 A3 Control SKOV3hyb 0.205
4 P1 A4 Control SKOV3hyb 0.206
5 P1 A5 Control SKOV3hyb 0.
Hi,
Peter's suggestion is more general, but for just the weighted mean,
there is a built in function you can use (I do not know of any basic
weighted standard deviation or variance functions).
dat <- data.frame(age = 1:5, no = c(21, 31, 9, 12, 6))
weighted.mean(x = dat$age, w = dat$no)
Best rega
Hello-
After looking through ?spplot, I would expect that I could specify the
values of the cuts:
"...‘cuts’ number of cuts or the actual cuts to use..."
So in the following command,
spplot(lzm.krige.dir["var1.pred"], scales=list(draw=TRUE),
xlab="Easting",ylab="Northing",
cuts=seq(0.0,0.4,
Tena koe Eddie
One way:
eddie <- data.frame(grp=rep(c('small','medium','large','very large'), each=20),
wgt=rnorm(80, 100, 10))
with(eddie, plot(grp, wgt))
eddie$grp <- factor(eddie$grp, levels=c('small','medium','large','very large'))
with(eddie, plot(grp, wgt))
HTH ...
Peter Alspach
> -
Hi Eddie,
I've been on a role with the iris data, so I figure why stop.
Assuming that one variable is a factor, you can easily reverse it, and
if you want fine tuned control, then just reorder the levels. Here is
an example:
dat <- iris
boxplot(Sepal.Length ~ Species, data = dat)
boxplot(Sepal.L
Hi,
Try this:
# using the iris dataset
mydat <- iris
mymodel <- lm(Sepal.Length ~ Petal.Length + Species, data = mydat)
summary(mymodel)
newdat <- data.frame(Petal.Length = seq(1, 10, by = .1),
Species = factor(rep("virginica", 91)))
results <- predict(object = mymodel, newd
Hi
carolina plescia wrote:
Dear all,
in GADM map there are three levels (nation, province and precinct) for each
country of the world but for all of them you are never able to plot only one
part of a chosen country.
The "Spatial" object you get when you read in the shapefile has
convenient s
Hello All,
I noticed when I generated some boxplots, the data is presented in
alphabetical order along the x-axis (the data in this case was the four
quandrants of a sample area (NE,NW, SE, SW) that was my first column of
data). Is there a way to have R plot the data in a different order? I
i
This is quite elegant (thanks) and brings up a problem I could not
solve awhile back, although Dr. Sarkar did his best to help.
How do I do the same thing in a panel plot?
e.g., toy example
temp.df <- data.frame(X=seq(1,100,by=1),Y=seq(1,50.5,by=.5),class=rep
(c("A","B"),each=50))
xyplot(Y ~
Hi folks,
I use lm to run regression and I don't know how to predict dependent
variable based on the model.
I used predict.lm(model, newdata=80), but it gave me warnings.
Also, how can I get the variance of dependent variable based on model.
Thanks.
[[alternative HTML version deleted]]
I haven't done much with the type of data you're working with, but
here is a post that lists a few packages for doing sample size
calculations in R. Perhaps one of them will be helpful.
https://stat.ethz.ch/pipermail/r-help/2008-February/154223.html
Andrew Miles
On Sep 27, 2010, at 2:09 PM,
Thanks for the suggestion. But my example is just an example, I would prefer to
have some generalized solution, like what options(scipen=4) does in general
graphics, which usually gave pretty axis labels as well.
Any suggestions?
Thanks
John
From: Henrique Da
Mark -
Here's one way to get the percentages you want. Suppose your
data frame is called df:
correct = subset(as.data.frame(with(df,table(date,region,correct))),
correct=='yes')
all = as.data.frame(with(df,table(date,region)))
names(all)[3] = 'Total'
both = merge(correct,al
What I'm trying to do is to figure out how to create lattice charts of
%right by region, or alternatively, by date from a dataset of observations
that looks something like this:
date,location,region,correct
2010-09-10,a,r1,yes
2010-09-10,a,r1,yes
2010-09-10,a,r1,no
2010-09-11,a,r1,yes
2010-09-01,b
Try this:
xyplot(1:10~1:10, scales=list(log = T, labels = round(log(1:10), 4)))
On Mon, Sep 27, 2010 at 4:10 PM, array chip wrote:
> Hi I am using xyplot() to plot on the log scale by using scale=list(log=T)
> argument. For example:
>
> xyplot(1:10~1:10, scales=list(log=T))
>
> But the axis l
Hello Everyone,
I'm trying to conduct a couple of power analyses and was hoping someone might
be able to help. I want to estimate the sample size that would be necessary to
adequately power a couple of non-inferiority tests. The first would be a
log-rank test and the second would be a Wilcoxon
Hi I am using xyplot() to plot on the log scale by using scale=list(log=T)
argument. For example:
xyplot(1:10~1:10, scales=list(log=T))
But the axis labels are printed as scientific notation (10^0.0, etc), instead
of
fixed notation. How can I change that to fixed notation?
options(scipen=4) d
Hi Peter,
Thank you for your thoughtful reply. I am tweaking the setting print
settings you suggested. It looks like this is going to solve my problem.
Thanks very much for help.
Jonathan
On Sat, Sep 25, 2010 at 6:00 PM, Peter Ehlers wrote:
> On 2010-09-25 8:59, Jonathan Flowers wrote:
>
>>
I am learning R via a textbook that performs analysis with SPSS and SAS. In
trying to reproduce the results for an ordinal logit model, I get very
similar point estimates for my cut-off points, but the parameters for the
covariate q60 do not match. The estimate for q51 also matches. Is this
because
On Mon, Sep 27, 2010 at 5:27 AM, Ben Bolker wrote:
> Luis Felipe Parra quantil.com.co> writes:
>
>>
>> Hello, I am trying to unlist a list, which is attached, and I am having the
>> problem that when I unlist it the number of elements changes from 5065 to
>> 5084
>>
>> > x <- lapply(SumaPluvi,
On Mon, Sep 27, 2010 at 1:34 PM, Titus von der Malsburg
wrote:
> On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck
> wrote:
>> Try this zero width negative look behind expression:
>>
>>> gregexpr("(?!a+)(b+)", "abcdaabbc", perl = TRUE)
>> [[1]]
>> [1] 2 7
>> attr(,"match.length")
>> [1] 1 2
>
>
one way is the following:
col1 <- c(1,2,3,4,5,6)
col2 <- c(6,5,4,3,2,1)
m <- cbind(col1, col2)
col3 <- c(1,3,2,6)
col4 <- c(6,3,5,1)
n <- cbind(col3, col4)
ind.n <- do.call(paste, c(as.data.frame(n), sep = "\r"))
ind.m <- do.call(paste, c(as.data.frame(m), sep = "\r"))
ind.n %in% ind.m
I hope
I found a solution to my original question (see code below).
But I have a question about cosmetics, which I always find very challenging.
1. How can I make all dates appear on the X axis (rotated at 90
degrees vs. horizontal)?
2. How can I create vertical grid lines so that at each date there is
a
Hi everyone:
I have a kinda easy question but i do not know how to solve that in a simple
way.
I want to compare the rows of two matrices.
col1 <- c(1,2,3,4,5,6)
col2 <- c(6,5,4,3,2,1)
m <- cbind(col1, col2)
col3 <- c(1,3,2,6)
col4 <- c(6,3,5,1)
n <- cbind(col
http://www.amazon.com/Statistical-Design-George-Casella/dp/1441926143/ref=sr_1_1?s=gateway&ie=UTF8&qid=1285609902&sr=8-1
Hello Mohd. Yaseen,
Please check out the book by Dr. Casella and his website
www.stat.ufl.edu/~casella for the relevant R codes. Chapter 5 of this book
talks about Split Split P
*Hello,
I'm new to R and trying to do Split Split Plot Design analysis with aov
function in R. Sharing any worked example and suggestion will be highly
appreciated. Thanks
Regards!
*
--
*
Muhammad Yaseen
*
[[alternative HTML version deleted]]
__
You've tried:
gregexpr("b+", "abcdaabbc")
On Mon, Sep 27, 2010 at 12:48 PM, Titus von der Malsburg wrote:
> Dear list!
>
> > gregexpr("a+(b+)", "abcdaabbc")
> [[1]]
> [1] 1 5
> attr(,"match.length")
> [1] 2 4
>
> What I want is the offsets of the matches for the group (b+), i.e. 2
> and 7, not
I have 500,00 rows in my matrix and i was wondering whether there is any way to
get its SVD without breaking it to parts
because if R can only read about 1000 columns then to have a rectangular matrix
(diagonal i think they are called) I will need to have only 1000 rows
I want to know how i can
Hi All,
I would like to announce the release of Deducer 0.4-1 and JGR 1.7-2 to CRAN.
The updates should be propagating through the mirrors over the next few days.
On the Deducer side we have a number of nice improvements:
1. A new Text Field Widget for plug-ins is included, which is better suit
Fractional polynomials ("FPs") are an automatic way of fitting
non-linear, parametric effects. The R-package mfp implements a
frequentist inference approach for FP models. Recently, we have proposed
a Bayesian inference approach for normal FP models, which is based on
the quasi-default hyper-/
You could do this:
gregexpr("ab+", "abcdaabbcbb")[[1]] + 1
On Mon, Sep 27, 2010 at 2:25 PM, Titus von der Malsburg
wrote:
> On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna
> wrote:
> > You've tried:
> >
> > gregexpr("b+", "abcdaabbc")
>
> But this would match the third occurrence of b+ in
On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck
wrote:
> Try this zero width negative look behind expression:
>
>> gregexpr("(?!a+)(b+)", "abcdaabbc", perl = TRUE)
> [[1]]
> [1] 2 7
> attr(,"match.length")
> [1] 1 2
Thanks Gabor, but this gives me the same result as
gregexpr("b+", "abcdaab
On Mon, Sep 27, 2010 at 11:48 AM, Titus von der Malsburg
wrote:
> Dear list!
>
>> gregexpr("a+(b+)", "abcdaabbc")
> [[1]]
> [1] 1 5
> attr(,"match.length")
> [1] 2 4
>
> What I want is the offsets of the matches for the group (b+), i.e. 2
> and 7, not the offsets of the complete matches. Is there
On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna wrote:
> You've tried:
>
> gregexpr("b+", "abcdaabbc")
But this would match the third occurrence of b+ in "abcdaabbcbb". But
in this example I'm only interested in b+ if it's preceded by a+.
Titus
_
Thank you Jim, but just as the solution that I discussed, your
proposal involves deconstructing the pattern and searching several
times. I'm looking for a general and efficient solution. Internally,
the regexpr engine has all necessary information after one pass
through the string. What I need i
try this:
> x <- gregexpr("a+(b+)", "abcdaabbcaaacaaab")
> justA <- gregexpr("a+", "abcdaabbcaaacaaab")
> # find matches in 'x' for 'justA'
> indx <- which(justA[[1]] %in% x[[1]])
> # now determine where 'b' starts
> justA[[1]][indx] + attr(justA[[1]], 'match.length')[indx]
[1] 2 7 17
>
On M
On Mon, Sep 27, 2010 at 9:34 AM, Jonas Josefsson
wrote:
> I have a two-column table as follows where age is in the 1st column and the
> number of individuals is in the 2nd.
>
> age;no
> 1;21
> 2;31
> 3;9
> 4;12
> 5;6
You can use the following trick:
x = rep(age, no)
This repeats age[1] no[1]-ti
I have a two-column table as follows where age is in the 1st column and
the number of individuals is in the 2nd.
age;no
1;21
2;31
3;9
4;12
5;6
Can I use mean() and sd() to calculate the mean and standard deviation
from this or do I have to manually multiplicate 21*1+31*2 etc. / N?
_
Why do you want 2 pairs plots on the same device? There may be a better
approach to what you want to do.
You could use splom from the lattice package along with the print.trellis
function to put 2 on the same page.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthca
The rimage package appears to have been abandoned. One option is the EBImage
package from Bioconductor.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r
On 2010-09-27 4:54, Christophe Bouffioux wrote:
bwplot(v2 ~ v1 | z, data = ex3, layout=c(3,2),
pch = "|",
par.settings = list(
plot.symbol = list(alpha = 1, col = "transparent",cex = 1,pch = 20)),
panel = function(x, y){
panel.bwplot(x, y)
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jonas Sundberg
> Sent: Monday, September 27, 2010 1:43 AM
> To: r-help@r-project.org
> Subject: [R] make changes in existing vector with the apply function?
>
>
> Hi, I'm trying
You can use the scale function, just use the minimum instead of the mean and
the range instead of the standard deviation.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-proj
On Mon, Sep 27, 2010 at 8:48 AM, Justin Fincher wrote:
> Howdy,
> I have created a set of plots, but I wish to increase the dpi to 300
> (instead of the default 72). From the documentation, I thought that
> the "res" parameter to png should accomplish this, but it appears to
> greatly alter the
Dear list!
> gregexpr("a+(b+)", "abcdaabbc")
[[1]]
[1] 1 5
attr(,"match.length")
[1] 2 4
What I want is the offsets of the matches for the group (b+), i.e. 2
and 7, not the offsets of the complete matches. Is there a way in R
to get that?
I know about gsubgn and strapply, but they only give me
Dear R-ers!
Asking for your help with building the stacked area chart for the
following simple data (several variables - with date on the X axis):
### Creating a data set
my.data<-data.frame(date=c(20080301,20080402,20080503,20090301,20090402,20090503,20100301,20100402,20100503),
x=c(1.1,1.0,1.6,
Howdy,
I have created a set of plots, but I wish to increase the dpi to 300
(instead of the default 72). From the documentation, I thought that
the "res" parameter to png should accomplish this, but it appears to
greatly alter the appearance of my plot. (plot area becomes smaller,
plot lines bec
Howdy,
I have created a set of plots, but I wish to increase the dpi to 300
(instead of the default 72). From the documentation, I thought that
the "res" parameter to png should accomplish this, but it appears to
greatly alter the appearance of my plot. (plot area becomes smaller,
plot lines bec
On Mon, Sep 27, 2010 at 8:22 AM, Kennedy wrote:
>
> Hi,
>
> I want to perform a hierarchical clustering using the median as linkage
> metric. As I understand it the function hcluster in package amap have this
> option but it does not produce the results that I expect.
Also, if you have a large(r)
On Sun, 2010-09-26 at 09:41 -0700, Vik Rubenfeld wrote:
> I'm experienced in statistics, but I am a first-time R user. I would like to
> use R for correspondence analysis. I have installed R (Mac OSX). I have used
> the package installer to install the CA package. I have run the following
> l
On Mon, Sep 27, 2010 at 8:22 AM, Kennedy wrote:
>
> Hi,
>
> I want to perform a hierarchical clustering using the median as linkage
> metric. As I understand it the function hcluster in package amap have this
> option but it does not produce the results that I expect.
>
> In the example below M is
On 27/09/2010 11:11 AM, Czerminski, Ryszard wrote:
Is there an easy way to control smoothness of the contour lines?
In the plot I am working on due to the undersampling the contour
lines I am getting are jugged, but it is clear "by eye" these should
be basically straight lines.
Straight lines
Hi,
I want to perform a hierarchical clustering using the median as linkage
metric. As I understand it the function hcluster in package amap have this
option but it does not produce the results that I expect.
In the example below M is a matrix of similarities that is transformed into
a matrix of
Is there an easy way to control smoothness of the contour lines?
In the plot I am working on due to the undersampling the contour
lines I am getting are jugged, but it is clear "by eye" these should
be basically straight lines.
In maps package I found smooth.map function, but maybe there is a mor
Hi all,
I tried to install the rimage in order to get to the function
?read.jpeg. However, I get this error, independent what mirror I
choose:
install.packages("rimage")
--- Please select a CRAN mirror for use in this session ---
Warning message:
In getDependencies(pkgs, dependencies, available,
Hi,
set the 'vertex.label.dist' parameter:
g <- graph.ring(10)
tkplot(g, vertex.label.dist=1, layout=layout.circle)
See ?igraph.plotting for details.
Best,
Gabor
On Mon, Sep 27, 2010 at 11:18 AM, anderson nuel wrote:
> Dear r-help,
>
> I create a graph of my baysian network. I use the package
On Mon, Sep 27, 2010 at 7:49 AM, statquant2 wrote:
>
> thank you very much for this sql package, the thing is that thoses table I
> read are loaded into memory once and for all, and then we work with the
> data.frames...
> Do you think then that this is going to be quicker (as I would have thougth
Hi,
I'm not sure it's even possible (and if it is I don't know how, but I'm
no expert).
But I think it doesn't make much sense to have only one named column.
Just give it a vector:
vect_names <- c("myname1", "myname2", "myname3")
colnames(my_matrix) <- vect_names
HTH,
Ivan
Le 9/27/2010 10:
Hi,
It is because the column names do not exist. If you cast the matrix as a
data frame your code would work.
jon
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Lorenzo Cattarino
Sent: 27. september 2010 10:27
To: r-help@r-project.o
Hi Lorenzo,
The problem is that my_matrix does not have dimnames. See below.
my_matrix <- matrix (1:12,ncol=3)
str(my_matrix) ## does not have dimnames
dimnames(my_matrix) ## dimnames is NULL
colnames(my_matrix) <- "myname" # fails because you are trying to
alter the value of something that does
Is there an alternative to par(mfrow=c(2,1)) to get stacked scatterplot
matrixes generated with "pairs"?
I am using version 2.11.1 on Windows XP. The logic I am using follows, and
the second "pairs" plot replaces the first plot in the current graphics
device, which is not what I expected (or desi
Thank you all for those great links, I will look at those.
Thanks again
Colin
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Hi everybody,
using bwplot for producing panel boxplot with 3 dimensions
i want to add a mark on each boxplot representing one individual (on all its
dimensions)
till now, i didn't succeed getting the desired solution
I want as well to keep the median symbols as a line
Many thanks for your help
c
The functional form is given in chapter 4 of my book:
Wood S.N. (2006) Generalized Additive Models: An Introduction with R.
Chapman and Hall/CRC Press. (reserve your copy now for christmas)
... but note that by default mgcv reparameterises so that the
identifiability constraints on the smoo
thank you very much for this sql package, the thing is that thoses table I
read are loaded into memory once and for all, and then we work with the
data.frames...
Do you think then that this is going to be quicker (as I would have thougth
that building the SQL DB from the flat file would already be
Hi
Is there a maximum length for the character string representing a level
of a factor? I have a set of several million variables, each a factor
of length 19. Each factor level is a character string which in some
cases can be many thousands of characters long. I am trying to find out
why my
Hi, I'm trying to make some changes in a vector according to some conditions.
It takes too long time however with vector length > 10 and I guess a better
way would be using the apply function. I cannot sort out how, however.
As a for/if loop:
for (i in 1:length(PrH)) {
if (is.finite(PrH[i])
Hi R-users
I can not change the name of one column only of my matrix.
my_matrix <- matrix (1:12,ncol=3)
colnames(my_matrix)[1] <- 'myname'
Error in dimnames(x) <- dn :
length of 'dimnames' [2] not equal to array extent
thank you for your help
Lorenzo
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