On Tue, 20 Jan 2009, Stavros Macrakis wrote:
I'm rather confused by the semantics of factors.
snip actual confusion
It is all very confusing. Of course, most of this behavior is
documented and is easily determined by experimentation, but it would
be easier to learn and teach the language
Hi there,
I have a list of dataframes (generated by reading multiple files) and all
dataframes are comparable in dimension and column names. They also have a
common column, which, I'd like to use for merging. To give a simple example of
what I have:
df1 - data.frame(c(LETTERS[1:5]),
As a follow-up, I don't see any reason why rle() shouldn't work on factors.
There's no ambiguity about what the result should be, and the current
implementation in rle() would work on factors if they could get past the
pre-test.
-thomas
On Wed, 21 Jan 2009, Thomas Lumley wrote:
On
Hi:
I am a newbie using grid and lattice.
I want to update a xyplot without redrawing and I was wondering if there is
a way of adding a point to a pointsGrob ?
For example, take the example from the R Graphics Book page 218-219:
grid.newpage()
angle - seq(0, 2*pi, length=21)[-21]
x - cos(angle)
What version of R are you using? I get this:
do.call(cbind, mylist)
df1.pos df1.data df2.pos df2.data df3.pos df3.data
1 A2 A6 A9
2 B6 B2 B3
3 C3 C9 C6
4 D
Gabor Grothendieck schrieb:
What version of R are you using?
R version 2.8.1 (2008-12-22) (running Windows)
I get this:
do.call(cbind, mylist)
df1.pos df1.data df2.pos df2.data df3.pos df3.data
1 A2 A6 A9
2 B6 B2
Hi Daniel,
Is there a way I can add a point to pg (or GRID.points.755) to update
the
xyplot without redrawing all the points ?
I think you want to look at ?grid.edit to do this.
grid.edit(yourGROB, redraw = FALSE)
HTH, Mark.
Daniel Kornhauser-3 wrote:
Hi:
I am a newbie using grid
Hello,
I'm trying to analyze the behavior of a series of numbers.
Let's say I have the following data: 0 0 0 20 20 20 50 50 53 56 23 24 21 10
0 4 129 159 30 0 0 0 (for example) - these numbers are some measurements
that are taken on equal interval of times (but the specific moment in time
(e.g.
Try:
library(forecast)
plot(forecast(auto.arima(x)))
On Wed, Jan 21, 2009 at 4:01 AM, Yana Mileva yanamil...@googlemail.com wrote:
Hello,
I'm trying to analyze the behavior of a series of numbers.
Let's say I have the following data: 0 0 0 20 20 20 50 50 53 56 23 24 21 10
0 4 129 159 30 0 0
Hello.
Does anyone know, if there is a function in R to compute the vector
autocorrelations?
Thank you in advance.
Regards,
Andreas.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
Hello, everyone
I wonder if R has something similar to Excel function
EDATE(start_date; months) which returns a serial number of the date
that is the indicated number of months before of after the start date.
Example (the second column EDATE(first_column; -6)):
01.01.1999 01.07.1998
See ?julian
julian(Sys.Date(), Sys.Date() - 10) # 10
[1] 10
attr(,origin)
[1] 2009-01-11
and R News 4/1.
On Wed, Jan 21, 2009 at 4:37 AM, Sergey Goriatchev serg...@gmail.com wrote:
Hello, everyone
I wonder if R has something similar to Excel function
EDATE(start_date; months) which returns
I created the graph at the bottom using xyplot in the lattice package. I
added a title using the main=Title command in xyplot, however it is
plotted too close to the legend for my liking. To remedy this I
increased
the upper margin of the plot using plot(data, position = c(0,0,1,.9))
and
Hi
data-read.table(data.txt, header=T, sep='\t')
Error in file(file, r) : cannot open the connection
In addition: Warning message:
In file(file, r) : cannot open file 'data.txt': No such file or
directory
gives me an error so it is not possible to reproduce your function.
However it looks
Hi
It depends if you have Excel available
write.excel-function(tab, ...) write.table( tab, clipboard, sep=\t,
row.names=F)
this function I use for copying object through clipboard to opened Excel
file.
just
write.excel(someobject)
open excel list
Ctrl-V
puts an object into a list.
Regards
Hi
From the R for OS X FAQ page: (http://cran.r-project.org/bin/macosx/RMacOSX-FAQ.html
)
4.4.6 Editor (internal and external): Using AppleScript it is easy to
implement Command-E and Command-Return like functionality.
How?
Regards, Gregor.
__
Good morning to all,
I should evaluate a function for every cell of a given
matrix with n rows and n columns.
This function, named fun(), has got two
arguments: the number of the row and the number of the column which
characterized every single cell.
So the result should be
fun(1,1)
Hi
?outer
regards
Petr
r-help-boun...@r-project.org napsal dne 21.01.2009 11:47:03:
Good morning to all,
I should evaluate a function for every cell of a given
matrix with n rows and n columns.
This function, named fun(), has got two
arguments: the number of the row and the number
Here's a simulated example (although really for this model structure, one
might as well fit seperate models for each factor level).
## Simulate some data with factor dependent smooths
n - 400
x - runif(n, 0, 1)
f1 - 2 * sin(pi * x)
f2 - exp(2 * x) - 3.75887
f3 - 0.2 * x^11 * (10 * (1 - x))^6 +
Good morning to all,
I should evaluate a function for every cell of a given
matrix with n rows and n columns.
This function, named fun(), has got two
arguments: the number of the row and the number of the column which
characterized every single cell.
So the result should be
fun(1,1)
Dear Gabor,
Thanks for that!
Still, it is not really similar to how EDATE works.
With julian(Sys.Date(), Sys.Date() - 10) one moves 10 days back.
The problem is that I need to move by months, not by days, as months
have different number of days.
I need to come to the same day when I move
1) Please update your R as requested in the posting guide, and provide
the 'at a mimimum' information requested.
2) Try reading ?postscript and setting the encoding: I suspect yours
is not one of the cases that are guessed correctly.
3) If all else fails, give a 'commented, minimal,
Team,
I have a time series data with lot of seasonality and trends. I am familiar
with some forecasting techniques, but for this set, I believe STL would be
the best. Please let me know any tips to use STL using R.
Best,
Kishore
[[alternative HTML version deleted]]
On Wed, 21 Jan 2009, Sergey Goriatchev wrote:
Dear Gabor,
Thanks for that!
Still, it is not really similar to how EDATE works.
With julian(Sys.Date(), Sys.Date() - 10) one moves 10 days back.
The problem is that I need to move by months, not by days, as months
have different number of days.
I
enrico.fosco...@libero.it enrico.fosco...@libero.it napsal dne
21.01.2009 11:54:49:
I can't use the function outer because my function fun() doesn't take
vectors
as arguments.
If you can not
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Dear Prof. Ripley,
Thank you for help.
Yes, that is an interesting question you pose. I already thought
myself how February should be handled, as EDATE(31.08.2008; -6)
returns 29.02.2008.
In Excel it is not a problem, since this nonexisting date is then used
in VLOOKUP function where one can have
On Wed, 21 Jan 2009, Sergey Goriatchev wrote:
Dear Prof. Ripley,
Thank you for help.
Yes, that is an interesting question you pose. I already thought
myself how February should be handled, as EDATE(31.08.2008; -6)
returns 29.02.2008.
In Excel it is not a problem, since this nonexisting date is
Thomas Lumley wrote:
On Tue, 20 Jan 2009, Stavros Macrakis wrote:
I'm rather confused by the semantics of factors.
snip actual confusion
It is all very confusing. Of course, most of this behavior is
documented and is easily determined by experimentation, but it would
be easier to learn
This will give you 6 months after today. Use negative
numbers to move backwards:
d - Sys.Date()
seq(d, length = 2, by = paste(6, months))[2]
[1] 2009-07-21
On Wed, Jan 21, 2009 at 6:27 AM, Sergey Goriatchev serg...@gmail.com wrote:
Dear Gabor,
Thanks for that!
Still, it is not really
Li, Hua wrote:
Dear R helpers:
Let's say I have some data X,
X - runif(1000, 1, 100)
pdf('X.pdf', width=100,height=5)
hist(X, breaks=1000)
dev.off()
I find that, on x-axis the coordinates are 0e+00, 2e+09, 4e+09, 6e+09,
8e+09, 1e+10. Only five
Try this also:
cbind(pos = mylist$df1$pos, data.frame(mylist)[grep(data,
names(data.frame(mylist)))])
On Wed, Jan 21, 2009 at 6:19 AM, Antje niederlein-rs...@yahoo.de wrote:
Hi there,
I have a list of dataframes (generated by reading multiple files) and all
dataframes are comparable in
Have you tried specifying an absolute path in prefix.string instead of a
relative one?
HTH,
Thierry
Good point! This actually solves the problem, I just wanted to avoid
this as I wanted to send it to someone else... so I just reordered the
files and it works.
Solutions given by
Henrique Dallazuanna schrieb:
Try this also:
cbind(pos = mylist$df1$pos, data.frame(mylist)[grep(data,
names(data.frame(mylist)))])
Hi Henrique,
cool solution - that's seems to be the easiest way!
though I thought there should be some possibiliy of multiple merge
Anyway, this will do it
Hi Antje,
Try this:
merge(merge(mylist[[1]], mylist[[2]], by = pos), mylist[[3]])
On Wed, Jan 21, 2009 at 11:19 AM, Antje niederlein-rs...@yahoo.de wrote:
Henrique Dallazuanna schrieb:
Try this also:
cbind(pos = mylist$df1$pos, data.frame(mylist)[grep(data,
names(data.frame(mylist)))])
merge.zoo can do a multiple merge. We create a list of
zoo objects, mylist.z, and then perform the merge:
library(zoo)
mylist.z - lapply(mylist, function(x) zoo(x$data, as.character(x$pos)))
do.call(merge.zoo, mylist.z)
df1 df2 df3
A 2 6 9
B 6 2 3
C 3 9 6
D 1 7 2
E
Thank you all for your help!
Sergey
On Wed, Jan 21, 2009 at 13:25, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
This will give you 6 months after today. Use negative
numbers to move backwards:
d - Sys.Date()
seq(d, length = 2, by = paste(6, months))[2]
[1] 2009-07-21
On Wed, Jan
Dear Juliet,
I found out that the issue of matching more than two vectors can be avoided by
pasting vectors into one using paste(). Also, it is not actually necessary for
my problem to use match(). I was looking for a method to match points by
coordinates, this can be done be using equal
Hello everybody!
I have a list of length 5000 whose components are (mostly) ts objects,
but some these components are intentionally left empty, ie, they are NULL
components of the list. My question is how can I get the position of these
null components in a effective way (I'm trying to avoid a
Dear all,
I have a zoo object that has following structure:
str(bldata)
zoo [1:5219, 1:12] 91.9 91.8 91.7 91.8 91.7 ...
- attr(*, index)=Classes 'dates', 'times' atomic [1:5219] 7305
7306 7307 7308 7309 ...
.. ..- attr(*, format)= chr m/d/y
.. ..- attr(*, origin)= Named num [1:3] 1 1
Try this:
x - list(NULL, 1:10, 25, NULL, more)
x
[[1]]
NULL
[[2]]
[1] 1 2 3 4 5 6 7 8 9 10
[[3]]
[1] 25
[[4]]
NULL
[[5]]
[1] more
which(sapply(x, is.null))
[1] 1 4
On Wed, Jan 21, 2009 at 7:42 AM, diego Diego dhab...@gmail.com wrote:
Hello everybody!
I have a list of
Please reduce your examples down to small amounts of data and use dput so
that they are reproducible.
On Wed, Jan 21, 2009 at 9:32 AM, Sergey Goriatchev serg...@gmail.com wrote:
Dear all,
I have a zoo object that has following structure:
str(bldata)
zoo [1:5219, 1:12] 91.9 91.8 91.7 91.8
Another possibility is Reduce:
Reduce(function(x,y,by='pos')merge(x,y,by='pos'),mylist)
pos data.x data.y data
1 A 2 69
2 B 6 23
3 C 3 96
4 D 1 72
5 E 9 51
- Phil Spector
Thanks a lot for all the mails!
With all the different examples I now have got it working. The problem
finally after I had worked out the newd2 as supplied by Gavin was that I
found the plot so strange looking. I have altered my newd2 a bit, and after
your e-mails Simon I figured out how to
Hi,
I do have a data set with some missing values that appear as blanks. I want
to fill these blanks with an NA. How can this be done? Thanks for your help
--
View this message in context:
http://www.nabble.com/filling-blanks-with-NA-tp21584278p21584278.html
Sent from the R help mailing list
Hello everybody!
We have a problem with a linux server that crashes when we try to read large
datasets in R.
The R code is as followed:
komplett - read.spss(komplett2003aar.sav, to.data.frame =TRUE, reencode
=Latin1)
The information about the linux server is:
Linux version 2.6.24-19-generic
Hello,
I've a time series, T, for an irregular sequence of days (data for bank
holidays weekends have been removed). I would like to aggregate this series
by into weeks using
aggregate(x, by=weeks, FUN=mean).
To do this, I think that x needs to be a timeSeries object, and when I try
to call
Hi,
kayj wrote:
Hi,
I do have a data set with some missing values that appear as blanks. I want
to fill these blanks with an NA. How can this be done? Thanks for your help
Something like this?
my.data-data.frame(var=c(1,2,5,,66,4,3,,67,5,3,2,1,4,32,56,23),
stringsAsFactors=F)
Dear R-users,
Has anyone written a function for multifractal detrended
fluctuation analysis? The fractal package does mono-fractal DFA,
but not multifractal as far as I can tell. The MF-DFA approach is
presented in:
J. W. Kantelhardt, S. Zschiegner, E. Koscielny-Bunde, S. Havlin, A.
Hello everybody!
I have a problem when I try to perform a forecast of an ARIMA model
produced by an auto.arima function. Here is what I'm doing:
c-auto.arima(fil[[1]],start.p=0,start.q=0,start.P=0,start.Q=0,stepwise=TRUE,stationary=FALSE,trace=TRUE)
# fil[[1]] is time series of monthly data
In addition to this, is it in anyway possible to get the variance in the
prediction graph as a shade similar to the GAM plots?
2009/1/21 Robbert Langenberg mcre...@gmail.com
Thanks a lot for all the mails!
With all the different examples I now have got it working. The problem
finally after I
Try this also:
is.na(my.data$var) - which(my.data$var == )
On Wed, Jan 21, 2009 at 12:33 PM, kayj kjaj...@yahoo.com wrote:
Hi,
I do have a data set with some missing values that appear as blanks. I want
to fill these blanks with an NA. How can this be done? Thanks for your
help
--
Dear Dr. Grothendieck,
First of all, I realized I did not load zoo package before I tried the
first str(bldata). If I load zoo and then do str(bldata) I get the
following:
'zoo' series from 7305 to 14609
Data: num [1:5219, 1:12] 91.9 91.8 91.7 91.8 91.7 ...
- attr(*, dimnames)=List of 2
..$
Dear list,
I'm posting the solution to my problem in case others may find this useful.
This code was sent to me by Phil Spector. With a bit of cleaning, it can easily
be converted to a usable format. Thanks to Gabor Grothendieck, David winsemius
and Martin Morgan for also sending possible
Hi All,
I have an script in R which accepts user inputs for certain parameters,
particularly dates, which the user inputs as character strings.
eg:
date1 - 2009-01-21
The script later parses the input via the as.Date function:
as.Date(date1)
However, as.Date encounters an error when the
Dear Dr. Grothendieck,
My purpose in this case it to have the structure of the object on
non-Bloomberg machine the same as that of the object on the Bloomberg
machine.
That way I am sure that whatever code I write away from Bloomberg
machine will work on it when I copy the code to it.
Also, I am
I do have a data set with some missing values that appear as blanks. I
want
to fill these blanks with an NA. How can this be done? Thanks for your
help
Please help us to help you. What form are the data in? Are they in a
text file, or are they in R already? What do you mean by 'blanks'?
Dear list members,
I am using a cross validation of a generalised linear model (glm). The cv.glm
function (from boot package) returns an error as so-called „delta“ value. I
would like to get to a (cross-validated) squared q, because I want to directly
compare it to the squared correlation
Hi,
I need to rbind two data frames. Each one has a header . after the rbind I
would like to keep the header for each and have the two data frames
separated by a line. Is this possible to do in R?
For example
weight_meanweight_sd.dev
F 14.3 4.932883
M 34.7 10.692677
What do you want to do with it? Is this just for printing out? What
other types of transformations are you intending to do? Why not just
put them in a 'list' and then write your own specialized print
routine.
On Wed, Jan 21, 2009 at 10:30 AM, SNN s.nan...@yahoo.com wrote:
Hi,
I need to
?try
On Wed, Jan 21, 2009 at 10:55 AM, Brigid Mooney bkmoo...@gmail.com wrote:
Hi All,
I have an script in R which accepts user inputs for certain parameters,
particularly dates, which the user inputs as character strings.
eg:
date1 - 2009-01-21
The script later parses the input via the
So far we have dput(a). Does that represent what you have?
what you want? Can you provide both and a description in words.
See the last line to every message to r-help.
On Wed, Jan 21, 2009 at 11:13 AM, Sergey Goriatchev serg...@gmail.com wrote:
Dear Dr. Grothendieck,
My purpose in this case
Gregor,
Section 6.1 of the FAQ provides further examples. It does depend on
the editor you are using. Several editors come with add-ons that
support R.
Recently, on the Mac specific r-sig-mac mailing list ( r-sig-...@stat.math.ethz.ch
), Smultron came up.
Regards,
Rob
On Jan 21, 2009,
HI:
Could someone help me with the seq function? I have a range of values starting
from 1 to 52 but I want seq to start at 27 by=2, but when it reaches 51 start
with with number 1 to 25. is this possible. I can do the basics of seq() but I
can't figure how to do this one. This is how I want my
G'day Felipe,
On Wed, 21 Jan 2009 09:29:03 -0800 (PST)
Felipe Carrillo mazatlanmex...@yahoo.com wrote:
Could someone help me with the seq function? I have a range of values
starting from 1 to 52 but I want seq to start at 27 by=2, but when it
reaches 51 start with with number 1 to 25. is this
Is this what you want:
x - seq(1, 52, 2)
x
[1] 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43
45 47 49 51
(x + 26) %% 52
[1] 27 29 31 33 35 37 39 41 43 45 47 49 51 1 3 5 7 9 11 13 15 17
19 21 23 25
On Wed, Jan 21, 2009 at 12:29 PM, Felipe Carrillo
Hi Felipe,
concatenate two sequences using c():
c(seq(from=27,to=51,by=2),seq(from=1,to=25,by=2))
HTH,
Stephan
Felipe Carrillo schrieb:
HI:
Could someone help me with the seq function? I have a range of values starting
from 1 to 52 but I want seq to start at 27 by=2, but when it reaches 51
On Wed, 2009-01-21 at 09:29 -0800, Felipe Carrillo wrote:
HI:
Could someone help me with the seq function? I have a range of values
starting from 1 to 52 but I want seq to start at 27 by=2, but when it reaches
51 start with with number 1 to 25. is this possible. I can do the basics of
Hi,
I'm trying to use metaMDS with a dissimilarity matrix of angles, not
Bray-Curtis, and I wanted to know if there is an in-built function to
produce a plot of stress values against dimensions, that could be used to
determine the 'true' dimension of the solution. The number of objects is
only a
Hello,
i'm tring to use a cox's model for a survival analysis. I have a dataset,
this is a part:
VOD SESSO fonte_sct donor RT_CGY STATOBMT TEMPO morto
1 0 F midrelated 1200
CP651
2 0 M mid
It works like a charm,thank you all for your help.
--- On Wed, 1/21/09, jim holtman jholt...@gmail.com wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] seq()
To: mazatlanmex...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Wednesday, January 21, 2009, 9:40 AM
Is this what you
the heatmapCol function of package MKmisc might help ... try:
library(MKmisc)
example(heatmapCol)
Best
Matthias
Liu, Hao [CNTUS] wrote:
Dear All:
I tried to use heatmap.2 to generate hierarchical clustering using the
following command:
heatmap.2(datamatrix, scale=row, trace=none,
Hi,
This is just for print out so it looks nice. what I have is a loop that
loops over my variables and calculates the mean and the sd for these
variables. Then I need to rbind them so I can stack them up in one file. the
problem is that only the fist header appears and the rest do not. this is
Are you sure that you want to loop over variables for each subset, and do
this within each subset? A simple way to generate summary statistics
(across variables and within categories) is the summaryBy function in the
doBy package:
Hi,
I have a data.frame SAMPLES with columns:
Site Site# Season Day1 Day2 Day3
Day1, Day2, Day3 are class Date, the other columns are numeric or
factor.
I have a date mydate that may or may not be listed in my data.frame
and I need to find that out.
If mydate is there, I
Dear Gonçalo,
If DF is your data set, something like this should do the job:
which(DF==yourdate,arr.ind=TRUE)
# Example
Dates-c('01/18/2009','01/19/2009','01/20/2009','01/21/2009')
DF-data.frame(
+Site=sample(4),
+ Site#=sample(4),
+ Season=sample(4),
+
Hi,
I have some code with a bunch of apply/sapply/lapply calls to
different functions. I am trying to profile it using Rprof, however
the resulting summary looks like this (output from 'R CMD Rprof'):
% total % self
totalseconds selfsecondsname
100.00
Dear All,
I have interval data (for Mon-Sun, 00-24h) of an activity and would like
to visually plot them in a matrix-like plot, where color A would be
assigned to the activity, and color X to unspecified time usage. Note
that the activities are not in standardised units (hours or so), but
I am working with a list of dates and I would like to replace each date with
the one that comes after, ie. 1/1/07 will become 1/5/07, 1/5/07 will become
1/7/07, etc. The number of days between my dates always varies, so I can't
just increase each one by 5 days or so. Does anyone know of a way I
My bad. I used system(test -r blah) to see if a file was readable,
forgetting that not all Windows installations have the test program (from
Cygwin) installed. I've changed this to use the R function file.access() in
version 2.5, which I've just submitted to CRAN. The Windows binary should be
Hello R Users,
Suppose I have data with the structure below:
Group_Name Pre_Test Post_Test
Grp_A xxx xxx
Grp_A xxx xxx
Grp_A xxx xxx
...
Grp_B xxx xxx
Grp_B xxx xxx
...
Grp_Z
Hello,
i'm tring to use a cox's model for a survival analysis. I have a dataset,
this is a part:
VOD SESSO fonte_sct donor RT_CGY STATOBMT TEMPO morto
1 0 F midrelated 1200
CP651
2 0 M mid
Hello,
I'm writing a function like this:
f-function(x,y,...) {
...
assign(x,y,envir=?)
}
I need the caller (of f) 's environment for the ? so that the
assignment is done at the right place. To be specific, when the code
f(x,1) appears in environment A, I need the assignment of 1 to x
happen in
de Jong, S. (1993) SIMPLS: an alternative approach to partial least squares
regression. Chemometrics and Intelligent Laboratory Systems, 18, 251263
Thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Try this:
f - function(env = parent.frame()) env$x
g - function(x=1) f()
x - 2
g() # 1
On Wed, Jan 21, 2009 at 5:45 PM, Yi Zhang yizhan...@gmail.com wrote:
Hello,
I'm writing a function like this:
f-function(x,y,...) {
...
assign(x,y,envir=?)
}
I need the caller (of f) 's environment
Following the recent NYT article about R, I thought this group is not
only ready for R but ready to take it one step further.
Got models in R? Deploy and score them in ADAPA in minutes on the
Amazon EC2 cloud computing infrastructure!
Zementis ( http://www.zementis.com ) has been working with
In ?title I see the
plot(cars, main = )
title(main = list(Stopping Distance versus Speed, cex=1.5, col=red,
font=3))
I can't seem to generalize this to use several colors in a single title.
What I'd like is
in latex-ish
\red{Hair color} \black{ and } \blue{Eye color}
to serve also as an
Dear All,
I was wondering if it is possible to generate a regression summary (it does
not matter at this stage if from an lm or for example a glm estimate) in
which to obtain the joint significance of a set of regressors?
Examples could be looking at the joint significance level of a
2009/1/21 Michael Friendly frien...@yorku.ca:
In ?title I see the
plot(cars, main = )
title(main = list(Stopping Distance versus Speed, cex=1.5, col=red,
font=3))
I can't seem to generalize this to use several colors in a single title.
Solution from
Michael Friendly wrote:
In ?title I see the
plot(cars, main = )
title(main = list(Stopping Distance versus Speed, cex=1.5, col=red,
font=3))
I can't seem to generalize this to use several colors in a single title.
What I'd like is
in latex-ish
\red{Hair color} \black{ and } \blue{Eye
Hi,
is there a way to define, that a line drawn via abline() should only
go from for example -2 to 1 on the x-axis (with something working
similiar to xlim()) ?
thanks for any help!
__
R-help@r-project.org mailing list
I use this function (a lot):
ablinepiece - function(a=NULL,b=NULL,reg=NULL,from,to,...){
# Borrowed from abline
if (!is.null(reg)) a - reg
if (!is.null(a) is.list(a)) {
temp - as.vector(coefficients(a))
if (length(temp) == 1) {
a - 0
b -
Thanks, that's great!
Am 22.01.2009 um 01:18 schrieb Remko Duursma:
I use this function (a lot):
ablinepiece - function(a=NULL,b=NULL,reg=NULL,from,to,...){
# Borrowed from abline
if (!is.null(reg)) a - reg
if (!is.null(a) is.list(a)) {
temp -
Greetings all,
Sorry for posting what is primarily not an R question
(though it will have an R target in due course).
Let F(x) be the CDF of the Normal -- i.e. pnorm(x)
Let f(x) be the density function -- i.e. dnorm(x)
Define G(psi) = Integral[-inf,inf] F(x)*f(x)*exp(x*psi) dx
Is G(psi) a
On Wed, Jan 21, 2009 at 5:57 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this:
f - function(env = parent.frame()) env$x
Thanks. What if the x in env$x is an argument passed in? e.g. f -
function(x, env=parent.frame()) { #assign to env$x ? }
g - function(x=1) f()
x - 2
g() # 1
Hi all,
Is there a way to study the lead and lag relation of two time series?
Let's say I have two time series, At and Bt. Is there a systematic way
of concluding whether it's A leading B or B leading A and by how much?
Thanks!
__
try
install.packages('car')
?car::linear.hypothesis
hth,
Kingsford Jones
On Wed, Jan 21, 2009 at 4:20 PM, Roberto Patuelli
roberto.patue...@lu.unisi.ch wrote:
Dear All,
I was wondering if it is possible to generate a regression summary (it does
not matter at this stage if from an lm or for
Try a search on
cross correlation time series
HTH,
Jim Porzak
TGN.com
San Francisco, CA
http://www.linkedin.com/in/jimporzak
use R! Group SF: http://ia.meetup.com/67/
On Wed, Jan 21, 2009 at 5:17 PM, Michael comtech@gmail.com wrote:
Hi all,
Is there a way to study the lead and lag
One of:
?by
?aggregate
?ave
Next time include the package name where these functions come from AND
code that creates an example data situation and you will increase your
probability of getting a more prompt and complete reply.
--
David Winsemius
On Jan 21, 2009, at 4:54 PM, Ferry wrote:
I am using the following script to make .jpg files.
jpeg('plotx.jpg')
ddat -read.table(file,header=T)
attach(ddat)
tdat-read.table(file1)
plot(xx1,yy1,type='p',pch=1,col=blue,cex=0.2,xlim=c(0,3.5),ylim=c(-75,75))
points(tdat,col=green,pch=1,cex=0.2)
dev.off()
The problem is that I want the
Try this:
myassign - function(x, val, env = parent.frame())
assign(deparse(substitute(x)), val, env)
myassign(x, 3)
x # 3
On Wed, Jan 21, 2009 at 8:03 PM, Yi Zhang yizhan...@gmail.com wrote:
On Wed, Jan 21, 2009 at 5:57 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this:
f -
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