dear R-community,
Is it possible to have one pie chart of a large radius and then a pie
chart with smaller radius placed on top of the larger so the centers of
the two pie charts coincide?
I managed to place a filled circle with smaller radius in the piechart,
see below.
Thanks in advance,
Hi All,
Will try and keep brief, excuse the poor graphics skills, first time using
GIMP. I would like to make the first one look (somewhat) like the second if
possible.
From: http://d.imagehost.org/0519/Screen_shot_2010-12-08_at_3_50_23_PM.png
To here:
Hi,
The easiest way to get the wide curly braces in your plot might be the
tikzDevice package. In its vignette you'll find an example of placing
an arbitrary tikz element in a plot, featuring a curly bracket. Sadly,
the internal coordinate system used by ggplot2 might make the
positioning a
so it could be zero in both cases when given column has 0 set in it
--
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Sent from the R help mailing list archive at Nabble.com.
Hi,
I have used Vegan to construct an NMDS ordination plot. I plotted sites of
three forest types with the site number in it. My reviewer has asked me to
use different symbols for each of the forest types.
Can anyone send me how I can do this in R in simple steps. I have used the
options like
Hi Paul,
I am using Sweave and MiKTeX and the results are really impressive, but
it's often quite complicated (or impossible) to share the rnw-files with
my colleagues/clients. So it depends with/for whom you are working.
Perhaps as an alternative you could use a simpler markup format e.g.
Hello,
You could also be interested by the ascii package, which allows to use
Sweave with more simple markup languages such as asciidoc, txt2tages
or restructuredtext. An asciidoc based Rnw file is in my opinion more
readable and sharable than a latex based.
Best.
On Wednesday, December 8,
Two more cents: pandoc support is in my todo list... But I don't have
the time actually.
On Wednesday, December 8, 2010, Patrick Hausmann
patrick.hausm...@uni-bremen.de wrote:
Hi Paul,
I am using Sweave and MiKTeX and the results are really impressive, but it's
often quite complicated (or
Hi,
Still another way would be to count the levels of the factors (if it is
indeed a factor):
length(levels(example$V1))
Ivan
Le 12/8/2010 06:08, Jorge Ivan Velez a écrit :
Hi,
One way would be
example- read.table(textConnection(V1 V2
+ x y
+ y x
+ z b
+ a c
+ b j
+ d l
+ c
Thanks Jim!
--- On Tue, 12/7/10, jim holtman jholt...@gmail.com wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] Help on loops
To: Anup Nandialath anup_nandial...@yahoo.com
Cc: r-help@r-project.org
Date: Tuesday, December 7, 2010, 7:47 PM
use split and lapply to make it easier.
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 12/08/2010 12:29 AM, Paul Miller wrote:
Hello Everyone,
Hi Paul,
Been learning R over the past several months. Read several books and
have learned a great deal about data manipulation, statistical
analysis, and graphics.
Now I want to
Hi
x-rnorm(20)
max(x[x=0])
[1] -0.1028260
x-c(x,0)
max(x[x=0])
min(x[x=0])
AFAIK there is no other number fulfilling both criteria other than zero.
So if you want both criteria to be true (as your and suggests), just
check if there is zero.
Regards
Petr
r-help-boun...@r-project.org napsal
hi lexi
indeed, for a week or so i can also plot the diagnostics - same script, no
change in data... maybe an update of a package, i don't know.
katharina
--
dr. katharina manderscheid
soziologisches seminar
universität luzern
kasernenplatz 3
6000
Hi
Use par(new=TRUE)
before calling new plot
pie(rep(1,12), col=rainbow(24), radius=0.9)
par(new=TRUE)
pie(rep(1,12), col=topo.colors(24), radius=0.5)
But use piechearts only when you want to deceive your audience.
Regards
Petr
r-help-boun...@r-project.org napsal dne 08.12.2010 09:08:13:
On Wed, 2010-12-08 at 13:54 +0530, Sinu P wrote:
Hi,
I have used Vegan to construct an NMDS ordination plot. I plotted sites of
three forest types with the site number in it. My reviewer has asked me to
use different symbols for each of the forest types.
Can anyone send me how I can do this
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
I would like to have a data.frame, where one column contains functions,
and another one lists. i.e.:
FUN - function(l) {return(l$a+l$b+l$c}
LIST - list(a=1, b=2, c=3)
d - data.frame(fun=FUN, no=LIST, value=2, b=TRUE)
FUN - function(l)
I have got a list with 3 colums x,y,z, now I want do delete whole rows where
column z has NA values
I am not intereted in x and y columns if there are also NA values or not,
moreover I do not want to touch them because if there would be NA that would
mean a more serious error for me, and I have
It sounds like you want to use a list instead of a dataframe,
especially if the elements are a different length.
d - list() # initialize
d[[length(d) + 1]] - list() # extend
d[[length(d)]]$fun - sin # add a function
d[[length(d) + 1]] - list() # extend again
d[[length(d)]]$fun - cos #
Are you using a 'dataframe' instead of a 'list?
If it is a dataframe, then the following will work:
df - df[!is.na(df$z), ] # only keep rows without z == NA
On Wed, Dec 8, 2010 at 5:16 AM, madr madra...@interia.pl wrote:
I have got a list with 3 colums x,y,z, now I want do delete whole rows
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
rgl crashes my R session, when resizing the rgl graphic window.
I am using Ubuntu Maversick, with dual monitor setup. If I disconnect
one monitor, I can resize it a little bit, but it still craches if I
enlarge it to much.
I assume that the
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 12/08/2010 11:22 AM, jim holtman wrote:
It sounds like you want to use a list instead of a dataframe,
No - I would like to have a data,frame. I am aware of the differences,
but as far as I understand, each column in a data.frame can have a
On 12/08/2010 07:08 PM, Sybille Wendel (Udata) wrote:
dear R-community,
Is it possible to have one pie chart of a large radius and then a pie
chart with smaller radius placed on top of the larger so the centers of
the two pie charts coincide?
I managed to place a filled circle with smaller
A good time to learn how to debug R scripts. There are several
references and you might start with the 'debug' package and setting
options(error=utils::recover)
so that when the error occurs, you get placed in the 'browser' and can
look around to see what the problem is. I would assume that
Rainer M Krug wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
rgl crashes my R session, when resizing the rgl graphic window.
I am using Ubuntu Maversick, with dual monitor setup. If I disconnect
one monitor, I can resize it a little bit, but it still craches if I
enlarge it to much.
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 12/08/2010 12:05 PM, Duncan Murdoch wrote:
Rainer M Krug wrote:
Hi
rgl crashes my R session, when resizing the rgl graphic window.
I am using Ubuntu Maversick, with dual monitor setup. If I disconnect
one monitor, I can resize it a little
Dear list,
When playing around with the [ method for S4 classes I noticed that it gets
called twice in my example.
setClass(testClass,
representation(a=character))
setMethod([, signature(x = testClass, i = ANY, j=ANY),
function (x, i, j, ..., drop){
Hello everyone,
I would like to find out if there are already implemented function for legendre
polynomials. I tried google but returns nothing. How do you suggest me to
search
for that?
Regards
Alex
[[alternative HTML version deleted]]
On 08/12/2010 6:07 AM, Rainer M Krug wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 12/08/2010 12:05 PM, Duncan Murdoch wrote:
Rainer M Krug wrote:
Hi
rgl crashes my R session, when resizing the rgl graphic window.
I am using Ubuntu Maversick, with dual monitor setup. If I
Hi Alaios
Have a look at the package gsl. There is a family of Legendre functions,
with additional functions available from the ALF package
(http://www.gnu.org/software/gsl/).
All the best
Chris Campbell
MANGOSOLUTIONS, R Training and Consulting
T: +44 (0)1249 767700 Ext: 233
F: +44 (0)1249
Hi,
Try the package orthopolynom on CRAN.
HTH,
baptiste
On 8 December 2010 12:51, Alaios ala...@yahoo.com wrote:
Hello everyone,
I would like to find out if there are already implemented function for
legendre
polynomials. I tried google but returns nothing. How do you suggest me to
Hi all.
I am currently attempting to build a list of sparse matrixes. That I have
already achieved, by
vmat - list()
for (i in 1:n) {
vmat - c(vmat, sparseMatrix(i,j,x=data)
}
How I am trying to select those elements from the list where the column e.g.
999 is not null. I can do this for
Hello Germán.
You probably want something like:
sapply(vmat, function(curMat){
curMat[,999] != 0
})
Or if you want the indices, just surround this with a which.
HTH.
Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653,
I don't think it is being called twice; you are seeing the return
value printed out:
setClass(testClass,
+ representation(a=character))
[1] testClass
setMethod([, signature(x = testClass, i = ANY, j=ANY),
+function (x, i, j, ..., drop){
+print(void function)
uups,
of course you are right. Thanks!
like this it becomes obvious:
setMethod([, signature(x = testClass, i = ANY, j=ANY),
function (x, i, j, ..., drop){
print(void function)
NULL
}
)
Am 08.12.2010 um 13:39 schrieb jim holtman:
I don't think it is
Hello
I hope my question makes sense. It is possible to specify the shape paramenters
in a glm model with family Gamma?
Thanks in advance
Rosario
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Thanks Nick! That's exactly what I wanted!
The surrouding with a which was not exactly what I wanted, but
which(apply(sapply(vmat, function(currMat){
currMat[,999] != 0
}),2,any))
did do the trick.
Thank you very much!!
Cheers,
Germán
2010/12/8 Nick Sabbe nick.sa...@ugent.be
Hello
Hi is there any command or setting that allows you to do the plot command but
it dose not print the plot on screen?
So when you are saveing an plot in a function you dont want it to display
the plot just save it.
--
View this message in context:
Hello,
I am interested in Figure 2 in
http://www-stat.stanford.edu/~tibs/ftp/cus.pdf
Can anyone tell please how to create this plot?
Many thanks
Samuel
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
The create a file with the graphics:
?pdf
?jpg
?postscript
The question is if you are generating it, what do you want to do with
it. If you don't say otherwise, it will go to the default graphics
which is probably a window on your console. You will have to direct
it where you want it.
On Wed,
On 08/12/2010 8:14 AM, Joel wrote:
Hi is there any command or setting that allows you to do the plot command but
it dose not print the plot on screen?
So when you are saveing an plot in a function you dont want it to display
the plot just save it.
Yes, that's the normal behaviour when
Date: Tue, 7 Dec 2010 17:22:57 -0800
From: ryan.steven.gar...@gmail.com
To: r-help@r-project.org
Subject: [R] Parallel Scan of Large File
Is it possible to parallel scan a large file into a character vector in 1M
chunks using scan() with the
For (at least) boxplot() and hist(), you can set plot=FALSE
But since we don't know what type of plot you want, difficult to be sure.
Ivan
Le 12/8/2010 14:14, Joel a écrit :
Hi is there any command or setting that allows you to do the plot command but
it dose not print the plot on screen?
So
Do a little reading on how to use the graphics commands. I would look at
plot
lines
segments
a combination of those will let you easily create the output. It
would seem you need some data object that has the dimensions of the
bar lengths you want to create, but given that, it is not much of a
Rosario Garcia Gil M.Rosario.Garcia at slu.se writes:
Hello
I hope my question makes sense. It is possible to specify the shape
paramenters in a glm model with family Gamma?
It doesn't look that way, but you can do a Gamma model with
a specified shape parameter (albeit much less
Thank you for your reply. My question was how to create Fig. 2? I am using
UniCox; the aa=uniCoxCV gives list of aa$ devcvm aa$ncallcvm aa$se.devcvm,
aa$devcv, aa$ ncallcv and aa$ folds, which data I should plot here to create
the graph in Fig. 2?
Many thanks,
Sam
--- On Wed, 8/12/10, jim
You cannot, plotMap() decides things for you. For more flexibility than:
polys - data.frame(PID=rep(1,4), POS=1:4, X=c(0,1,1,0), Y=c(0,0,1,1))
plotMap(polys, xlim=c(-.5,1.5), ylim=c(-.5,1.5), projection=LL)
from the help page, consider trying:
library(sp)
library(maptools)
sp_polys -
On 12/7/2010 9:35 PM, Yihui Xie wrote:
shell(paste(yap, C:/WINDOWS/TEMP/Rtmpz0QkT8/file311f289a.dvi))
I can confirm that using shell() directly on the .dvi file generated by
latex() works, while system() does not -- it hangs
as before.
However, Yihui's patch, in this form still hangs, so
On 2010-12-07 12:01, kv wrote:
hello list,
i'm a bit puzzled by the error message i get when i copy past this in R:
Hello everybody,
I would like to calculate the median for each factor combination, with only one
value per factor combination as an output.
Could anybody help me?
For example:
# make table
g-1:2
group-rep(g, each=5)
session-c(1,1,2,2,2,1,1,1,2,2)
rt-seq(length=10,300, 800)
rt-round(rt,
Thanks Peters, what i suspected (i.e. ties). For what it's worth i would add
that the problem happends when length(data)100 (i.e. the second condition
in the mc.default() function).
Best,
--
View this message in context:
Hi Marianne,
Please consider the following:
with(table, aggregate(rt, list(group, session), FUN = median))
HTH,
Jorge
On Wed, Dec 8, 2010 at 9:29 AM, Marianne Stephan wrote:
Hello everybody,
I would like to calculate the median for each factor combination, with only
one value per factor
Might Wayland fix it in Narwhal ?
Duncan Murdoch murdoch.dun...@gmail.com wrote in message
news:4cff7177.7030...@gmail.com...
On 08/12/2010 6:07 AM, Rainer M Krug wrote:
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On 12/08/2010 12:05 PM, Duncan Murdoch wrote:
Rainer M Krug wrote:
Hi
Thank you.
I tried this but not sure if I have implemented this correctly ... basically
if function1 hangs .. I need the timeout to be triggered and then the
process moves to the next function call.
I have this:
function1 - function(x){
setTimeLimit(elapsed = 5*60, transient = FALSE)
step 1
Thank you, Peter.
(see inline)
On Wed, Dec 8, 2010 at 14:58, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2010-12-07 12:01, kv wrote:
hello list,
i'm a bit puzzled by the error message i get when i copy past this in R:
[.]
in my computer adjbox says:
maximal number of iterations
Dear R Group Users:
If there is a resource or easy way to calculate the distance
between zipcodes, for example, if I have the zipcode for 200 patients,
and the zipcode of a hospital, how to calculate the travel distance of
these individual patient to the hospital based on the zipcode. Your
Prof. Maechler , Koller,
Thank you very much, problem solved.
Best,
Seasons greetings.
--
View this message in context:
http://r.789695.n4.nabble.com/robustbase-problem-bug-in-adjbox-function-tp3077134p3078429.html
Sent from the R help mailing list archive at Nabble.com.
On Wed, Dec 8, 2010 at 10:41 AM, XINLI LI lihaw...@gmail.com wrote:
Dear R Group Users:
If there is a resource or easy way to calculate the distance
between zipcodes, for example, if I have the zipcode for 200 patients,
and the zipcode of a hospital, how to calculate the travel distance of
Thank you Jorge, this is exactly what I needed!
Another solution that works too, suggested by Steve:group.med - tapply( rt,
list(group, session), median)
From: jorgeivanve...@gmail.com
Date: Wed, 8 Dec 2010 09:43:47 -0500
Subject: Re: [R] How can I calculate the median for each factor
for example I have matrix with two columns
x,y
1,0.56
2,9.55
2,7.56
5,2.55
5,0.56
3,0.55
2,0.56
2,1.56
so I need to take average from y values placed where x==2
--
View this message in context:
Thanks Jim,
Ave does what I wanted.
It is simpler and probably more efficient
than unlisting Sn as I tried.
Still I remain puzzled with the structure
of the by() or tapply() output and how
to access the individual cumsums.
Yes the split command is useful for checking
the result.
Gerrit.
Op
On 2010-12-08 06:39, kv wrote:
Thanks Peters, what i suspected (i.e. ties). For what it's worth i would add
that the problem happends when length(data)100 (i.e. the second condition
in the mc.default() function).
Best,
I don't think that the length of 100 is special (other
than in the
try tapply() if you want the values for all levels of x, or calculate your
mean after a subset()
Check the documentation on both of these.
Mike
On Wed, Dec 8, 2010 at 9:54 AM, madr madra...@interia.pl wrote:
for example I have matrix with two columns
x,y
1,0.56
2,9.55
2,7.56
5,2.55
My apologies to Martin.
I should have known how prompt he would be and I should have
been more on the ball in checking incoming mail.
The new version of robustbase on R-Forge does indeed fix this.
Thanks to the robustbase team.
Peter
On 2010-12-08 07:57, Peter Ehlers wrote:
On 2010-12-08
Upcoming R Programming Techniques Courses
http://www.xlsolutions-corp.com/Rcourses
*** R Fundamentals and Programming Techniques
Washington DC, Dec 16-17, 2010
San Francisco, Dec 16-17
More on website
http://www.xlsolutions-corp.com/Rcourses
Ask for group discount and reserve
Hi all,
I try to interpolate a data set in the form:
timeErg
0.0048.65
1.5056.08
3.0038.33
4.5049.65
6.0061.39
7.5051.25
9.0050.45
10.50 55.11
12.00
I think
?table
is what you want.
-- Bert
On Wed, Dec 8, 2010 at 1:19 AM, Ivan Calandra
ivan.calan...@uni-hamburg.de wrote:
Hi,
Still another way would be to count the levels of the factors (if it is
indeed a factor):
length(levels(example$V1))
Ivan
Le 12/8/2010 06:08, Jorge Ivan Velez
Hi all,
I am trying to fit a logistic regression for a bivariate response using five
independent variables in a stepwise procedure. My outputs look okay but does
any one know (or is there any literature on) how the confidence intervals
are calculated for the reported odds ratios..?
Thanks!
I am trying to understand the function acf
stats:::acf shows me the function
I am having trouble understanding the usage $acf in the following
acf - array(.C(R_acf, as.double(x), as.integer(sampleT),
as.integer(nser), as.integer(lag.max), as.integer(type ==
correlation),
See McCullagh and Nelder's GLM book for details -- and also probably
VR's MASS for a concise summary, although I don't have it at hand and
can't be sure it's there. Really, practically any book on GLM should
have details.
**HOWEVER** You should realize that all these references are wrong
in the
On Dec 8, 2010, at 11:59 AM, S.M. Raghavan wrote:
Hi all,
I am trying to fit a logistic regression for a bivariate response
using five
independent variables in a stepwise procedure. My outputs look okay
but does
any one know (or is there any literature on) how the confidence
intervals
Provide at least a subset of the data you want to plot (e.g., 10 data
points). I assume that the data provides enough information to determine
how long each of the branches/cross ties are for the elements that you want
to plot. I don't use UniCox, nor do I know that the list of items that it
From: ggrothendi...@gmail.com
Date: Wed, 8 Dec 2010 10:50:40 -0500
To: lihaw...@gmail.com
CC: r-help@r-project.org
Subject: Re: [R] GIS Help: distance calculation based on ZIP Code
On Wed, Dec 8, 2010 at 10:41 AM, XINLI LI wrote:
Dear
Hi all,
I want to estimate parameters from a VARMA(p,q)-Modell.
The equations of the model or the model structures is given by:
Xt=beta1+beta2*Xt-1+beta3*Yt-1+epsilon1
Yt=beta4+beta5*Yt-1+espilon2
epsilon1 and espilon2 are white noise.
Xt is given by a vector of n elements e.g. (2,
Is anyone aware of a way to seasonally adjust time series data using X-12 ARIMA
and TRAMO/SEATS from within R? I know that that one can seasonally adjust data
with gretl, which I understand offers some level of R integration. However, all
the examples I've seen of gretl/R integration involve
Hi here is the code as example
lars is in package lars
x-matrix(rnorm(20*5,0,1),20,5)
bs-matrix(sample(seq(1:10),5),5,1)
er-rnorm(20,0,1)
y-x%*%bs+er
lobj-lars(x,y,type=lasso)
names(lobj)
[1] call type df lambda R2
[6] RSSCp actionsentry
Thanks for the three great answers! For those who are curious, I timed the
three approaches:
nr - 15812
nc - 64636
mymat - matrix(nrow=nr, ncol=nc)
mymat[1,1] - 1 # see note below
# mydf is created elsewhere
dim(mydf)
# 109102633
colnames(mydf)
# x y a
# approach 1:
# mymat[ mydf$x +
You might be better off partitioning the file before processing with
R. If you are planning on using skip = n to skip over records
before processing, then the last thread you would start would have to
read through 7/8 of the file before starting. The actual I/O, plus
the looking for the line
Dear community,
I have now taken my R-file from lectures and intend to use it at home, but
have a problem
reading the Data from the file. I have installed and loaded the Package
xlsReadWrite so far.
I have also Changed directory.
This is what I have entered
daten=read.xls(Daten A2)
This is my
On Wed, 2010-12-08 at 08:48 -0800, anna_m wrote:
Hi all,
I try to interpolate a data set in the form:
time Erg
0.00 48.65
1.50 56.08
3.00 38.33
4.50 49.65
6.00 61.39
7.50 51.25
9.00 50.45
Matthew Dowle wrote:
Might Wayland fix it in Narwhal ?
I hope those names mean something to Rainer, because they mean nothing
to me.
Duncan Murdoch
Duncan Murdoch murdoch.dun...@gmail.com wrote in message
news:4cff7177.7030...@gmail.com...
On 08/12/2010 6:07 AM, Rainer M Krug wrote:
Thank you very much, I will look into it.
Best,
xing
On 12/8/10, Mike Marchywka marchy...@hotmail.com wrote:
From: ggrothendi...@gmail.com
Date: Wed, 8 Dec 2010 10:50:40 -0500
To: lihaw...@gmail.com
CC: r-help@r-project.org
Subject: Re:
HI:
Is there a way to display a dataset on a dialog window? I am creating an
application
with Visual Basic and R, and I want the user to be able to see the dataset used
print out
on a dialog window. Not sure if there is a better way to do this, but basically
when the user
click a button on a VB
On Wed, 8 Dec 2010, S.M. Raghavan wrote:
Hi all,
I am trying to fit a logistic regression for a bivariate response using five
independent variables in a stepwise procedure. My outputs look okay but does
any one know (or is there any literature on) how the confidence intervals
are calculated
I have simple scatterplot of temperature data taken daily from many years so
at particular date there are usually more than one measurement, but not all
data is complete.
Now I want to draw line that will show me probability of finding data
between defined bands of temperatures, for example +5
Hello,
I've been looking through ?phantom and ?expression and this forum for
examples of how I might be able to manipulate some of the names that appear
on the y-axis of the barplot below. For example, the gw in ECgw would
appear as a subscript...or qr would be the theta symbol followed by
Hi!
I have a large dataset where I need to recompute a value in each row based
on z-sores from another look-up table. The look-up table is arranged by raw
score in the first column and age in the first row, eg:
20 30 40
1 .3 .5 .7
2 .2 .3 .4
3 .1 .4 .7
the main matrix will have
On Wed, Dec 8, 2010 at 6:18 PM, XINLI LI lihaw...@gmail.com wrote:
Thank you very much, I will look into it.
If it really is only 200 then I'd start typing them into Google's
direction finder and typing in the distance/time given. Even if you
only do one per minute they'll all be done in half a
On 2010-12-08 09:16, Dick Knox wrote:
I am trying to understand the function acf
stats:::acf shows me the function
I am having trouble understanding the usage $acf in the following
acf- array(.C(R_acf, as.double(x), as.integer(sampleT),
as.integer(nser), as.integer(lag.max),
Dear R People:
Suppose I have the following in a file:
1 1.171504 1.010070
2 -0.9317064 1.860900
3 -0.06522837 0.6561147
4 -1.817026 0.02619137
5 1.426983 0.5995691
6 -0.2844911 1.155561
7 -0.6920972 0.7633124
8 0.3129615 5.121108
I want to use read.table to bring it in and I want the first
Hi All,
How do I add these axis labels?
###
p=seq(0,1,length.out=500)
p=p[-c(1,length(p))]
g1=log(p/(1-p))
g2=qnorm(p)
g3=log(-log(1-p))
g4=-log(-log(p))
plot(p,g1,
'n',ylim=c(-5,5),las=1,
bty='n',
xaxt='n',yaxt='n',
Try this modification of your code.
-tgs
ynames -
base.dat.sel2$base.dat.Covariate[order(base.dat.sel2$base.dat.US.Num.Obs.to.Achieve.Starting.Residual,decreasing=F)]
ynames - as.character(ynames)
ynames[12]-expression(theta[r])
ynames[13]-expression(EC[gw])
Yes, just set the colClasses argument to read.table (this will also tend to
speed up the reading, though only noticeable for really big files).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From:
On Dec 8, 2010, at 2:40 PM, Erin Hodgess wrote:
Dear R People:
Suppose I have the following in a file:
1 1.171504 1.010070
2 -0.9317064 1.860900
3 -0.06522837 0.6561147
4 -1.817026 0.02619137
5 1.426983 0.5995691
6 -0.2844911 1.155561
7 -0.6920972 0.7633124
8 0.3129615 5.121108
I want to
It isn't clear to me what you want to do. Do you want the axes to show? Do
you want labels for the lines? Do you want a legend? What is your desired
output?
-tgs
On Wed, Dec 8, 2010 at 2:42 PM, casperyc caspe...@hotmail.co.uk wrote:
Hi All,
How do I add these axis labels?
Hi all,
How can one evaluate NAs in a numeric dataframe column? For example, I have
a dataframe (demo) with a column of numbers and several NAs. If I write
demo.df = 10, numerals will return TRUE or FALSE, but if the value is
NA, NA is returned. But if I write demo.df == NA, it returns as NA
Hi All,
I am new to the R program (only my 2nd day trying it out!) and this may seem
like a really stupid question but I was wondering if someone could help. I have
managed to calculate the pearson correlation coefficient for my data and now I
have a correlation coefficient matrix with a whole
Thomas Stewart wrote:
It isn't clear to me what you want to do. Do you want the axes to show?
Do
you want labels for the lines? Do you want a legend? What is your
desired
output?
-tgs
On Wed, Dec 8, 2010 at 2:42 PM, casperyc caspe...@hotmail.co.uk wrote:
Hi All,
How do I
On 2010-12-08 12:10, Wade Wall wrote:
Hi all,
How can one evaluate NAs in a numeric dataframe column? For example, I have
a dataframe (demo) with a column of numbers and several NAs. If I write
demo.df= 10, numerals will return TRUE or FALSE, but if the value is
NA, NA is returned. But if I
Date: Wed, 8 Dec 2010 19:25:09 +
Subject: Re: [R] GIS Help: distance calculation based on ZIP Code
From: b.rowling...@lancaster.ac.uk
To: lihaw...@gmail.com
CC: marchy...@hotmail.com; r-help@r-project.org
On Wed, Dec 8, 2010 at 6:18 PM,
Hi!
How can one evaluate NAs in a numeric dataframe column? For example, I have
a dataframe (demo) with a column of numbers and several NAs. If I write
demo.df = 10, numerals will return TRUE or FALSE, but if the value is
NA, NA is returned. But if I write demo.df == NA, it returns
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