[EMAIL PROTECTED] wrote:
Dear all,
I'm using friedman rank test in R. Need to know how to calculate rank
sum
in friedman test (R options to get rank sum in friedman test). Kindly help me.
Hi jeevitesh,
The sums of ranks in the Friedman test are simply the column sums of the
I've successfully import my synteny data to R by using scan
command. Below show my results. My major problem with my data is how
am i going to combine the column names with the data( splt) where i
have tried on cbind but a warning message occur. I have realized
that the splt data only
I'm using rainbow function to generate 10 colors for the plot and it is
difficult to tell the neighboring colors from each other. How can I make
the colors more differently.
Using 10 colours is always going to be difficult, but take a look at
ColorBrewer
Hi,
I have a collection of .txt documents in my working folder for which I want to
do some text mining. If I run TextDocCol from the tm package, R crashes with
some memory issues. Does anyone has any idea if this is related to R itself or
to the tm package?
Below you can find what is
Dear expeRt,
I occasionally use help.search() function to find appropriate
functions in R.
Whenever committing help.search() after rerunning R, it takes so much
time to find help (around 2 minutes).
Is there any way to make it faster in help.search()?
I think indexing help database is one of
Wang, Zhaoming (NIH/NCI) [C] wrote:
Hello
I'm using rainbow function to generate 10 colors for the plot and it is
difficult to tell the neighboring colors from each other. How can I make
the colors more differently.
If all you want is for neighbouring colours to be distinguishable you
hello
which graph can I use for categoric var. vs categoric var. ?
regards
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
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R-help@r-project.org mailing list
On Mon, 7 Jan 2008, Barry Rowlingson wrote:
rainbow() is great - for drawing rainbows - but the palettes from the
RColorBrewer package are much better for statistical plots as someone
else suggested. When I write code for plots I try and use RColorBrewer
if it's there:
...additionally, there
Nasser Abbasi wrote:
Duncan Murdoch [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Why did you change the parameters? If you used the same ones as above,
you get
sghyper(a=-1, k=-1, N=5)
$title
[1] Generalized Hypergeometric
$Mean
[1] 0.2
Dear Celine,
Please excuse the delay in responding over the holiday period. Here
are my answer to your questions. If you are having further
difficulties or these answers don't seem right do let me know.
(1) Yes, I recommend splitting your data by chromosome. If you perform
a genomewide analysis,
hello
I want to have subset of my dataset that is equal to 0. But,
I got this error message.
hs2=hs[hs$hey60==0 ,]
Error in hs[hs$hey60 == 0, ] : (subscript) logical subscript too long
regards
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
http://www.nabble.com/file/p14664173/at-modality.png
I created the above image with R and I have one problem left:
Some of the labels of the axes do not show up, probably because there's not
enough space.
I use the following code to create the plot:
modality - read.table(results.table,
Hi Michael,
a solution would be not to draw axes directly, but afterwards with the
axis-command:
plot(length, col=color, axes=F)
axis(2,at=your_positions, labels=your_labels, cex.axis=0.6, las=1);
# left axis
axis(1,at=your_positions, labels=your_labels, cex.axis=0.6, las=1);
# bottom axis
tom soyer tom.soyer at gmail.com writes:
Hi,
I just discovered decompose() and stl(), both are very nice! I am wondering
if R also has a function that calculates the seasonal index, or make the
seasonal adjustment directly using the results generated from either
decompose() or stl(). It
Can you either provide a subset of data or at least do 'str(hs)' so
that we know what the structure of you data looks like. It could be a
matrix, dataframe, list, Each makes difference in how it is
handled.
On Jan 7, 2008 8:47 AM, temiz [EMAIL PROTECTED] wrote:
hello
I want to have
On Mon, 7 Jan 2008, Dong-hyun Oh wrote:
Dear expeRt,
I occasionally use help.search() function to find appropriate
functions in R.
Whenever committing help.search() after rerunning R, it takes so much
time to find help (around 2 minutes).
Is there any way to make it faster in
Hello everybody
Is there a way to output data into a file like the unformatted, direct
fortran files? That's the kind that can be opened with:
open(30, file='file', format='unformatted', access='direct', rec=1000,
convert='big endian')
Thank you in advance
Kostas Douvis
Dear John,
The first issue of R News had a very nice article by Brian Ripley on
accessing databases from R; you'll find it at
http://cran.r-project.org/doc/Rnews/Rnews_2001-1.pdf. There's also a
section on databases in the R Data Import/Export manual, which comes with R.
BTW, I don't generally
http://www.nabble.com/file/p14664173/at-modality.png
I created the above image with R and I have one problem left:
Some of the labels of the axes do not show up, probably because there's
not
enough space.
Try
par(las=1)
then draw the plot
plot(length, col=colour)
Regards,
Richie.
R can talk to MySQL via the RMySQL and RODBC packages: which is easier
will depend on your infrastructure, including your OS.
You haven't even told us your OS. But the 'R Data Import/Export Manual'
and 'An Introduction to R', both of which ship with R, are the manuals to
start with, as well as
On Mon, 7 Jan 2008, Costas Douvis wrote:
Hello everybody
Is there a way to output data into a file like the unformatted, direct
fortran files? That's the kind that can be opened with:
open(30, file='file', format='unformatted', access='direct', rec=1000,
convert='big endian')
Fortran file
Is there a way to output data into a file like the unformatted, direct
fortran files? That's the kind that can be opened with:
It sounds like you want to use cat(), e.g.
cat(x = , 1:10, file=test.txt)
See also write for outputting matices to file; and write.table and
write.csv for data
xlsReadWritePro natively reads and writes Excelfiles (v97 - 2003) on
Windows. I have uploaded a small bugfix update thereof. Fixes:
error with dimnames in 1-dimensional arrays
misleading error (assertion) messages with wrong from/to arguments.
Download package:
temiz [EMAIL PROTECTED] wrote in news:478225E1.5010706
@deprem.gov.tr:
hello
which graph can I use for categoric var. vs categoric var. ?
plot() works if there is a method for the class of object. Here is an
example with contingency table given to plot() that appeared on the
list last
Dear All,
I hope I am not asking a FAQ. I am dealing with a problem of graph
theory [connected components in a non-directed graph] and I do not
want to rediscover the wheel.
I saw a large number of R packages dealing for instance with the
k-means method or hierarchical clustering for spatially
See ?rollmean in the zoo package.
On Jan 7, 2008 3:31 PM, Abu Naser [EMAIL PROTECTED] wrote:
Hi all R users,
Can anyone please let me know how to do the moving average with R?
With regards,
Abu
_
Free games, great
Lorenzo, why can't you actually generate the graph to find the
connection components? With the 'igraph' package this is something like:
g - graph.adjacency( DIST 0.5, mode=undirected )
g - simplify(g)
no.clusters(g)
assuming you have your distance matrix in 'DIST'. If N is too big
then you
Have a look at ?staxlab library(plotrix)or
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-create-rotated-axis-labels_003f
--- mika03 [EMAIL PROTECTED] wrote:
http://www.nabble.com/file/p14664173/at-modality.png
I created the above image with R and I have one
problem left:
Achaz von Hardenberg fauna at pngp.it writes:
Dear all,
I am performing a binomial glmm analysis using the lmer function in
the lme4 package (last release, just downloaded). I am using the
Laplace method.
However, I am not sure about what I should do to test for the
significance
Dear Lorenzo,
if I understand your posting correctly, this is exactly what Single
Linkage clustering does if you cut the dendrogram tree at your threshold
distance.
Therefore you can use hclust with method=single (which produces the full
dendrogram; you have to generate the Euclidean
hi, this may be trivial, but we can't seem to find anything adequate,
(although there is a work around with match() ). We are looking for something
along the lines of
plot(table1[table1$var2==or(a,b,c,d),var1])
would be handy, with the potential or() function leading to what
Hi all R users,
Can anyone please let me know how to do the moving average with R?
With regards,
Abu
_
Free games, great prizes - get gaming at Gamesbox.
__
Try this:
plot(table1[table1$var2 %in% c(a,b,c,d),var1])
On 07/01/2008, Sebastian Leuzinger [EMAIL PROTECTED] wrote:
hi, this may be trivial, but we can't seem to find anything adequate,
(although there is a work around with match() ). We are looking for something
along the lines of
I wonder how one in R can fit a 3rd degree polynomial to some data?
Say the data is:
y - c(15.51, 12.44, 31.5, 21.5, 17.89, 27.09, 15.02, 13.43, 18.18, 11.32)
x - seq(3.75, 6, 0.25)
And resulting degrees of polynomial are:
5.8007 -91.6339 472.1726 -774.2584
THanks in advance!
--
Jonas
look at ?%in%; I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
Thanks you for your answers.
I finally managed to have odfweave functionning well by downloading Rtools
available at http://www.murdoch-sutherland.com/Rtools/ and selecting the
option that allows you to update the variable PATH; I re-start my computer and
miracle !! it works
Sylvie
try this:
y - c(15.51, 12.44, 31.5, 21.5, 17.89, 27.09, 15.02, 13.43, 18.18,
11.32)
x - seq(3.75, 6, 0.25)
coef(lm(y ~ x + I(x^2) + I(x^3)))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Hello,
I am planning to implement Genetic Algorithms in solving classification
problems. As far as I can see there is only genalg package that I can use. I
can see that there are gafit and rgenoud packages that can be used for
regression and optimisation problems but not for classification.
Hi Jerad,
Thanks for your quick response. I have installed the zoo using
R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
wondering why?
With regards,
Abu
_
[[replacing trailing spam]]
I'm assuming the package installed correctly (I've only ever installed
packages from within R)
Have you loaded the zoo package before calling rollmean?
ie.
library(zoo)
On Jan 7, 2008 4:31 PM, Abu Naser [EMAIL PROTECTED] wrote:
Hi Jerad,
Thanks for your quick response. I have
Hi Jerad,
Thanks for your quick response. I have installed the zoo using
R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
wondering why?
With regards,
Abu
_
Free games, great prizes - get gaming at
Abu Naser wrote:
Hi Jerad,
Thanks for your quick response. I have installed the zoo using
R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
wondering why?
In R, don't forget to load zoo by
library(zoo)
Uwe Ligges
With regards,
Abu
Hii all, may I introduce myself. my name is Putro Nugroho i'am astudent of
Gadjah Mada University, i'am doing on my final exam about Parametric Regression
Model with gamma and log-gamma regression model. I have a problem about the
syntax program on R because it doesn't work with gamma
http://www.nabble.com/file/p14668788/paragraphs.png
Hi,
R is is world full of wonders... I created the attached plot, and I think
it's exactly what I need! Well, actually I think it is more that wht I
need...
I wanted R to show the mean values of the categories on the x-axis and maybe
the
You may also wish to consider runmed(), which is a more robust running
median, and is automatically available as part of the stats package.
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Jared O'Connell
Aha... I forgot to load the library.
Thanks everyone who responded my quaries.
With regards,
Abu
_
Fancy some celeb spotting?
__
R-help@r-project.org mailing list
Hi Jerad,
Thanks for your quick response. I have installed the zoo using
R CMD INSTALL zoo_1.4-1.tar.gz but the R could not recognize rollmean-
wondering why?
With regards,
Abu
_
Fancy some celeb spotting?
Dimitris Rizopoulos wrote:
try this:
y - c(15.51, 12.44, 31.5, 21.5, 17.89, 27.09, 15.02, 13.43, 18.18,
11.32)
x - seq(3.75, 6, 0.25)
coef(lm(y ~ x + I(x^2) + I(x^3)))
Or use the 'poly' function:
coef(lm(x~poly(y,3)))
(Intercept) poly(y, 3)1 poly(y, 3)2 poly(y, 3)3
4.875
Thanks for both the replies.
I am now giving a try to the suggestion by Gabor since it looks easier
(for me) to implement.
I am testing it, but so far it does what I have in mind.
I am going now through the documentation of the igraph package. I can
count the cluster number, but I also want to
You might want to google box and whisker plot.
mika03 wrote:
http://www.nabble.com/file/p14668788/paragraphs.png
Hi,
R is is world full of wonders... I created the attached plot, and I think
it's exactly what I need! Well, actually I think it is more that wht I
need...
I
Hi Paul,
Your problem statement does not make much sense to me. You say that an
analytical solution can be found easily. I don't see how.
This is a variational calculus type problem, where you maximize a
functional. Your constraint dx/dt=u(t) means that there exists a solution
(the
mika03 [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
http://www.nabble.com/file/p14668788/paragraphs.png
R is is world full of wonders... I created the attached plot, and I think
it's exactly what I need! Well, actually I think it is more that wht I
need...
I wanted R to show
On Mon, Jan 07, 2008 at 05:25:44PM +0100, Lorenzo Isella wrote:
Thanks for both the replies.
I am now giving a try to the suggestion by Gabor since it looks easier
(for me) to implement.
I am testing it, but so far it does what I have in mind.
I am going now through the documentation of the
Talbot Katz wrote:
Hi.
Suppose I have a vector that I partition into disjoint, contiguous
subvectors. For example, let v = c(1,4,2,6,7,5), partition it into
three subvectors, v1 = v[1:3], v2 = v[4], v3 = v[5:6]. I want to
find the maximum element of each subvector. In this example,
Just looking at this again what I meant was that unless you can linearize it
you can't use linear programming. You still could use general nonlinear
optimization subject to constraints.
Define new variables z[i] = x[i] - x[i-1]
Then x[i] = z[0] + ... + z[i]
so maximize (we dropped the first and
If your main goal is to find 10 colors that are easy to tell apart and
look good in a graph, then look at the RColorBrewer package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL
Hi all,
I'm to inexperienced to come up with the matrix solution elusively appearing
on the horizon for the following problem and would appreciate if you could
give me a nudge ...
I have two vectors a, and b and need to find the closest match for each
value of a in b.
How to do that efficiently?
On Jan 7, 2008 4:32 PM, Ravi Varadhan [EMAIL PROTECTED] wrote:
Your problem statement does not make much sense to me. You say that an
analytical solution can be found easily. I don't see how.
This is a variational calculus type problem, where you maximize a
functional. Your constraint
Jonas,
In statistical sense polynomial is a linear regression fit. The function
that handles linear fitting is called lm. Here is how you can reproduce
your results:
lm(y ~ x + I(x^2) + I(x^3))
Unless you are really after the polynomial coefficients it is probably
better to use orthogonal
Vectors are numerics. I'd like to know for each a[i], which b[j] delivers
the smallest abs(a[i]-b[j]).
Thanks for your time.
Joh
Bert Gunter wrote:
Vectors of what? How is closest defined? See e.g. ?dist
Bert Gunter
Genentch Nonclinical Statistics
-Original Message-
From:
Try this:
b[which.min(a-b)]
On 07/01/2008, Johannes Graumann [EMAIL PROTECTED] wrote:
Vectors are numerics. I'd like to know for each a[i], which b[j] delivers
the smallest abs(a[i]-b[j]).
Thanks for your time.
Joh
Bert Gunter wrote:
Vectors of what? How is closest defined? See e.g.
Try:
sapply(paste(Data$x, 1:3, sep=), function(x)eval(parse(text=x)))
On 07/01/2008, Gregory Gentlemen [EMAIL PROTECTED] wrote:
Dear R users,
I'd like to evaluate a vector of characters. For example, say I have a data
frame called Data including the field names x1, x2, x3, and I'd like to a
I think what he meant is that x is a function of t so if dx/dt is regarded
to be a function of t, which we shall call u(t), then u(t)'s absolute value
at each value of t is less than or equal to 1 (as opposed to u(t) being
known).
On Jan 7, 2008 11:32 AM, Ravi Varadhan [EMAIL PROTECTED] wrote:
Yes, Gabor, that is the interpretation that I meant. Thanks for the
clarification.
Paul
On Jan 7, 2008 5:59 PM, Gabor Grothendieck [EMAIL PROTECTED] wrote:
I think what he meant is that x is a function of t so if dx/dt is regarded
to be a function of t, which we shall call u(t), then u(t)'s
This is a fortune(106) situation (with a little fortune(77) in there as well).
I am unsure what exactly the original poster is wanting to do, but accessing
the columns of a data frame (or list) by character strings can be done using [[
]] rather than $ and should be able to work for this
I have a dataframe DF with 4 columns (variables) A, B, C, and D, and
want to create a new dataframe DF2 by keeping B and C in DF but
counting the frequency of D while collapsing A. I tried
by(DF$D, list(DF$B, DF$C), FUN=summary)
but this is not exactly what I want. What is a good way to do
Thanks a lot! This is exactly what I wanted.
Gang
On Jan 5, 2008, at 2:20 PM, David Winsemius wrote:
Gang Chen [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Suppose I have a two-way table of nominal category (party
affiliation) X ordinal category (political ideology):
party
Talbot,
Try this:
PartMax - function(x, breaks)
{
breaks - c(breaks, length(x) + 1)
sapply(seq(length(breaks) - 1),
function(i) max(x[breaks[i]:(breaks[i + 1] - 1)],
na.rm = TRUE))
}
PartMax(c(1,4,2,6,7,5),c(1,4,5))
[1] 4 6 7
PartMax(6:1,c(1,4,5))
Hi all,
I am trying to name list objects, but am having trouble doing so. At the
moment I have list alninc that has 3 objects and I can refer to them as
alninc[[1]] . . .alninc[[3]]. I would like to refer to them as alninc$F,
alninc$V and alninc$G. How do I do this? This appears to be an
On Mon, 7 Jan 2008, [EMAIL PROTECTED] wrote:
Jonas,
In statistical sense polynomial is a linear regression fit. The function
that handles linear fitting is called lm. Here is how you can reproduce
your results:
lm(y ~ x + I(x^2) + I(x^3))
Unless you are really after the polynomial
names(alninc) - c(F,V,G)
See ?names
Gabor
On Mon, Jan 07, 2008 at 01:35:59PM -0500, Wade Wall wrote:
Hi all,
I am trying to name list objects, but am having trouble doing so. At the
moment I have list alninc that has 3 objects and I can refer to them as
alninc[[1]] . . .alninc[[3]]. I
Try testing the performance of transforming your series to
one in which the values of each partition are larger than all
prior partitions and the untransforming back:
# test data
myseq - c(1, 4, 2, 6, 7, 5)
part - c(1, 4, 5)
M - max(myseq)
# transform
myseq2 - myseq + M * cumsum(replace(0 *
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf
Of Wade Wall
Sent: Monday, January 07, 2008 10:36 AM
To: [EMAIL PROTECTED]
Subject: [R] Naming list objects
Hi all,
I am trying to name list objects, but am having trouble doing so. At the
moment I
Henrique Dallazuanna wrote:
Try: sapply(paste(Data$x, 1:3, sep=),
function(x)eval(parse(text=x)))
On 07/01/2008, Gregory Gentlemen [EMAIL PROTECTED] wrote:
Dear R users,
I'd like to evaluate a vector of characters. For example, say I
have a data frame called Data including the field
On 1/7/2008 1:28 PM, Gang Chen wrote:
I have a dataframe DF with 4 columns (variables) A, B, C, and D, and
want to create a new dataframe DF2 by keeping B and C in DF but
counting the frequency of D while collapsing A. I tried
by(DF$D, list(DF$B, DF$C), FUN=summary)
but this is not
Thanks Gabor! This is the sort of thing I was trying to devise, but I ran up
against my extensively documented brain power limitations ;-) Per your
remarks, it remains to be seen how it performs compared to a good
implementation with one of the apply functions.
-- TMK --
212-460-5430
Johannes Graumann wrote:
Hi all,
I'm to inexperienced to come up with the matrix solution elusively
appearing on the horizon for the following problem and would appreciate if
you could give me a nudge ...
I have two vectors a, and b and need to find the closest match for each
value of a
Yes, I misstated it when I said that I would keep B and C. I want to
collapse column A, but count the frequency of D as a new column in
the new dataframe DF2 while collapsing A. The rows of columns B, C,
and D of course would be reduced because of A collapsing.
For example, if dataframe DF
Sorry the new column in DF2 should be called FreqD instead of FreqA.
How can I get DF2 with aggregate?
Gang
On Jan 7, 2008, at 2:38 PM, Gang Chen wrote:
Yes, I misstated it when I said that I would keep B and C. I want to
collapse column A, but count the frequency of D as a new column in
Hello everyone,
I have an overlay plot it's nice but you can't see all the data. I would
like to know if there is a way to get a plot that gives a side by side
plot so that each plot would be next to each other. The two plots have
the same data are of different species. At the moment this is the
SAS programming is easy if everything you want to do fits easily into the
row-at-a-time DATA step paradigm. If it doesn't, you have to write macros,
which are an abomination. DATA step statements and macros are entirely
different programming languages, with one doing evaluations at compile time,
[EMAIL PROTECTED] said the following on 1/7/2008 2:59 PM:
Hello everyone,
I have an overlay plot it's nice but you can't see all the data. I would
like to know if there is a way to get a plot that gives a side by side
plot so that each plot would be next to each other. The two plots have
On Monday 07 January 2008, [EMAIL PROTECTED] wrote:
Hello everyone,
I have an overlay plot it's nice but you can't see all the data. I would
like to know if there is a way to get a plot that gives a side by side
plot so that each plot would be next to each other. The two plots have
the same
[EMAIL PROTECTED] wrote:
Hello everyone,
I have an overlay plot it's nice but you can't see all the data. I would
like to know if there is a way to get a plot that gives a side by side
plot so that each plot would be next to each other. The two plots have
the same data are of different
Ido M. Tamir wrote:
matchpt
Thanks for this hint. It is exactly what I'm looking for.
Cheers, Joh
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Dear R community,
I am plotting a histogram and would wish to display another variable of
the same dataset in a very narrow heatmap just below the x-axis. Never mind
the specifics of my task: How can I draw a second graph/image just below a
first graph/image?
Thank you!
Georg.
***
?layout
check:
example(layout) to see its features.
b
On Jan 7, 2008, at 5:06 PM, Georg Ehret wrote:
Dear R community,
I am plotting a histogram and would wish to display another
variable of
the same dataset in a very narrow heatmap just below the x-axis.
Never mind
the specifics of
Try use args. 'position' and 'more'
h1-hist()
h2-hist()
print(h1,position=c(0,0,1,1),more=T)
print(h2,position=c(,,,)
Weidong Gu,
University of Alabama, Birmingham
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Georg Ehret
Sent: Monday, January 07,
Hi Mika03,
It would be useful to know what function you used to create your plot.
Assuming you used boxplot, do this:
?boxplot
?boxplot.stats
Julian
mika03 wrote:
http://www.nabble.com/file/p14668788/paragraphs.png
Hi,
R is is world full of wonders... I created the attached plot,
I fear I risk being viewed as something of a curmudgeon, but the truth must be
stated. S-Plus, R, SAS, etc. are all similar in that they are all tools to an
end and not an end in themselves. Any one of the three can do most statistical
analyses one might want to do. I could point out the
John Sorkin wrote:
I fear I risk being viewed as something of a curmudgeon, but the truth must
be stated. S-Plus, R, SAS, etc. are all similar in that they are all tools to
an end and not an end in themselves. Any one of the three can do most
statistical analyses one might want to do. I
?par and have a look at mfrow
par() controls a vast amount of plotting options for
regular graphics
Example
aa - rnorm(100, 25, 4)
bb - aa/2
par(mfrow=c(2,1))
hist(aa)
plot(bb)
--- Georg Ehret [EMAIL PROTECTED] wrote:
Dear R community,
I am plotting a histogram and would wish to
?par and have a look at mfcol or mfrow.
Example
exp-cbind(abs(round(rnorm(10),2)*10), seq(100, 200,
by=10))
ref-cbind(abs(round(rnorm(10),2)*10), seq(100, 200,
by=10))
op - par(mfrow=c(1,2))
plot(ref, col=red)
plot(exp, col=blue)
par(op)
--- [EMAIL PROTECTED] wrote:
Hello everyone,
I have
I'm looking for a way to improve code that's proven to be inefficient.
Suppose that a data source generates the following table every minute:
Index Count
0 234
1 120
7 11
30 1
I save the tables in the following CSV format:
time,index,count
names(alninc) - c(F, V, G)
--- Wade Wall [EMAIL PROTECTED] wrote:
Hi all,
I am trying to name list objects, but am having
trouble doing so. At the
moment I have list alninc that has 3 objects and I
can refer to them as
alninc[[1]] . . .alninc[[3]]. I would like to
refer to them as
Does this do what you want?
x - c(1,4,2,6,7,5)
x.i - c(1,4,5)
partiMax - function(vec, index){
+ # create a vector to identify the subvectors
+ x.b - diff(c(index, length(vec) + 1))
+ # split up the vector
+ x.s - split(vec, rep(seq_along(index), times=x.b))
+
Frank,
I believe you are proving my point. The difference is not so much the language
as the end users. I use SAS, R, and SPlus on a regular basis. For some
analyses, SAS is easiest to use, for some R (or SPlus). I can be just as
dangerous using SAS and I can be with R if I don't think about
Dear all,
I would like to do a goodness-of-fit test on my data to see if they follow a
mixture of 2 poisson distributions. I have small numbers for observed values.
Most of them 5. The chisq.test gives warning message: Chi-squared
approximation may be incorrect in: chisq.test(x , p = prob).
Hi all:
I have a directory of files as in - bunchafiles -
list.files(path=/data/2.3/2006, pattern=returns,
full.names=T,recursive=T) Each file is a bunch of returns for a
particular date (unique). There are like 252 files or so.
With a custom function myread (below), I define a
One thing to do is to use Rprof() on your script so that you can
determine where time is being spent. My guess it that most of the
time is in the wtd.quantile function. If your Counts don't get too
big, another way is to use 'quantile' directly:
Index - c(0,1,7,30)
Count - c(234,120,11,1)
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