You can create an empty matrix (or even array) of list elements and
then assign your data frames to whichever element you want. Example:
# Allocate empty matrix...
> x <- matrix(list(), nrow=2, ncol=3);
# ...alternatively
> x <- array(list(), dim=c(2,3));
> print(x);
[,1] [,2] [,3]
[1,] NULL
On Tue, Sep 14, 2010 at 12:44 AM, David Winsemius
wrote:
> The second argument to mean is trim. I am not sure what mean(1, 3) is
> supposed to do but what it return is 1.
>
Thanks for the info. On this particular point I find the documentation
confusing. In ?mapply :
'‘mapply’ applies
‘FUN’ t
One more question, given that
plot(rnorm(1),rnorm(1), ylab=expression(a~b >= 3), cex.lab=1.2)
then sign >= seems to be smaller than the rest, seems like cex.lab=1.2
affects only the text in ylab.
1) Is there any way to alter its size to follow the size of the whole
expression? It works nice for
raje...@cse.iitm.ac.in wrote:
Hi,
I create several dataframes in a nested loop and would like to maintain them in
a matrix form with each dataframe represented by the row and the column. How
can I do this?
You can't, at least as you describe it.
However, you can add a column for "row ID"
Good day R-listers,
I'm working with a panel dataset, i've used many models ,
homogeneous (fixed effect, pooled ols and Driscoll and Kraay)
heterogeneous (swamy random coefficients) and would like to do a
post-estimation to select the model that best fit my regression. is
there any method, co
Hi,
I create several dataframes in a nested loop and would like to maintain them in
a matrix form with each dataframe represented by the row and the column. How
can I do this?
[[alternative HTML version deleted]]
__
R-help@r-project.org maili
Hi,
This is Krishan, I am working on r-project through vb.Net
While calling the r-project I got following error
System.Runtime.InteropServices.COMException (0x80040013): Exception from
HRESULT: 0x80040013
at Microsoft.VisualBasic.CompilerServices.LateBinding.InternalLateCall(Object
o, Type ob
Thanks for the hint. But I want to change the line thickness, mgp() concerns
the
the spacing. Also I think par() doesn't have any effect on lattice plots.
It seems to be an easy thing to make axis line thicker, anyone has nay
suggestions?
Thanks
John
- Original Message
From: Daisy
Thanks to Thomas Lumley and David Winsemius for their responses. I
had read a number of papers by Thomas and have ordered his book on
survey analysis, but I wanted to get some confirmation because I
wanted to get started before the book arrived. Thanks, again.
Dan
Daniel Nordlund
Bothell, WA US
David Winsemius comcast.net> writes:
> On Sep 13, 2010, at 8:27 PM, csiro.au> wrote:
>
> > I have site data with variables that vary across sites (like wind
> > speed, moisture content) and within sites I have experimental units
> > (logs) with associated variables (like decay class, suspen
check out ?par for all the details on plotting
‘mgp’ The margin line (in ‘mex’ units) for the axis title,
axis labels and axis line. Note that ‘mgp[1]’ affects
‘title’ whereas ‘mgp[2:3]’ affect ‘axis’. The
default is ‘c(3, 1, 0)’.
On Tue, Sep 14, 2010 at 8:56 AM,
Allow me to add to Michael's and Clifford's responses.
If you fit the same regression model for each group, then you are also
fitting a standard deviation parameter for each model. The solution
proposed by Michael and Clifford is a good one, but the solution assumes
that the standard deviation pa
On Mon, 13 Sep 2010, Daniel Nordlund wrote:
I have been asked to look at options for doing relative risk regression on
some survey data. I have a binary DV and several predictor / adjustment
variables. In R, would this be as "simple" as using the survey package to
set up an appropriate desig
Thanks for turning my half-baked suggestion into something that would
actually work Cliff :)
Michael
On 14 September 2010 12:27, Clifford Long wrote:
> If you'll allow me to throw in two cents ...
>
> Like Michael said, the dummy variable route is the way to go, but I believe
> that the coeffici
If you'll allow me to throw in two cents ...
Like Michael said, the dummy variable route is the way to go, but I believe
that the coefficients on the dummy variables test for equal intercepts. For
equality of slopes, do we need the interaction between the dummy variable
and the explanatory variab
Hello:
On 9/13/2010 3:44 PM, VojtÄch Zeisek wrote:
> Hello
>
> Dne Po 13. záÅà 2010 14:51:39 Tal Galili napsal(a):
> But here, I wish point out one issue, which can be fixed relatively
> easily: R would deserve much more better web running some good
> open-source CMS. I have very good exp
On Sep 13, 2010, at 8:27 PM, wrote:
I have site data with variables that vary across sites (like wind
speed, moisture content) and within sites I have experimental units
(logs) with associated variables (like decay class, suspension).
With normal response y one can use R to get the betwee
Thanks! This helps a lot!
On Mon, Sep 13, 2010 at 4:35 PM, Joshua Wiley wrote:
> Hi,
>
> Just to provide an example:
>
>
> #Sample data copied to variables "X", "Y", and "Z"
> Z <- Y <- X <- data.frame(A = 1:10)
>
> # A variable holding the names of the variables
> datasetname <- c("X", "Y", "Z")
Sure I tried that and it works but I was formerly using just 'proportion ' so I
find it strange that this has occurred.
-Original Message-
From: Phil Spector [mailto:spec...@stat.berkeley.edu]
Sent: Monday, September 13, 2010 4:08 PM
To: ROLL Josh F
Cc: r-help@r-project.org
Subject: Re:
I have site data with variables that vary across sites (like wind speed,
moisture content) and within sites I have experimental units (logs) with
associated variables (like decay class, suspension). With normal response y one
can use R to get the between and within site information using
aov(y~
I am trying this as you mentioned and getting an error which i cant fix
do you know where is the problem?
> df2[is.na(df2)] <- 0
> df2
X age no. age.1 no..1 age.2 no..2 age.3 no..3 age.4 no..4
1 center1 3 9 6 4 9 110 1 0 0
2 center2 5 3 9
On Sep 13, 2010, at 8:51 PM, Natasha Asar wrote:
> I am trying this as you mentioned and getting an error which i cant
> fix
> do you know where is the problem?
>
> > df2[is.na(df2)] <- 0
> > df2
> X age no. age.1 no..1 age.2 no..2 age.3 no..3 age.4 no..4
> 1 center1 3 9 6
Have a look at:
"Computing Thousands of Test Statistics Simultaneously in R" by Holger
Schwender and Tina Müller, in
http://stat-computing.org/newsletter/issues/scgn-18-1.pdf
Hadley
On Mon, Sep 13, 2010 at 4:26 PM, Alexey Ush wrote:
> Hello,
>
> I have a question regarding how to speed up the t
Hello Doug,
Perhaps it would just be easier to keep your data together and have a
single regression with a term for the grouping variable (a factor with
3 levels). If the groups give identical results the coefficients for
the two non-reference grouping variable levels will include 0 in their
confi
I have been asked to look at options for doing relative risk regression
on some survey data. I have a binary DV and several predictor /
adjustment variables. In R, would this be as "simple" as using the
survey package to set up an appropriate design object and then running
svyglm with family=
Thanks David. It almost does what I wanted, except it's plotting the point
characters 3 time for each line (left, middle and right): o---o---o
I can live with that if there is no way to get rid of the point characters at
the ends.
Thanks very much!
John
- Original Message
From: D
Maybe prop.table ?
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
On Sep 13, 2010, at 6:25 PM, array chip wrote:
Hi all, I
When I plot both lines and points using type=c('l', 'p') in
xyplot(), if I want
to include in legend both of them using keys=list(lines=list(col=1:3),
points=list(pch=1:3)), the lines and points are plotted side by side
in legend.
On 09/13/2010 08:41 PM, Sunny Srivastava wrote:
> Dear R-Helpers,
> I have a list l1 like:
>
> l1[[1]]
> a b c
>
> l1[[2]]
> d
>
> l1[[3]]
> e f
>
> I want an output res like:
>
> res[[1]]
> 1 1 1
>
> res[[2]]
> 2
>
> res[[3]]
> 3 3
>
> Essentially, I want to replicate each index equal to t
Hi ,
SO i have been on a role of asking simple questions lately. So much for
feeling like im getting this R business.
I wrote a script 2 weeks ago that utilized "proportion" to turn values in a
table (from "table") into proportions to then graph. I now get an error
that proportion is not a
Hello,
I have a question regarding how to speed up the t.test on large dataset. For
example, I have a table "tab" which looks like:
a b c d e f g h
1
2
3
4
5
...
10
dim(tab) is 10 x 100
I need to do the t.test for each ro
Thank you very much list!
On Mon, Sep 13, 2010 at 5:04 PM, Phil Spector wrote:
> Sunny -
> I don't think mapply is needed:
>
> lapply(1:length(mylist),function(x)rep(x,length(mylist[[x]])))
>>
> [[1]]
> [1] 1 1 1
>
> [[2]]
> [1] 2
>
> [[3]]
> [1] 3 3
>
>
Hello,
We've got a dataset with several variables, one of which we're using
to split the data into 3 smaller subsets. (as the variable takes 1 of
3 possible values).
There are several more variables too, many of which we're using to fit
regression models using lm. So I have 3 models fitted (one
Try this:
relist(rep(1:length(l1), sapply(l1, length)), l1)
On Mon, Sep 13, 2010 at 3:41 PM, Sunny Srivastava
wrote:
> Dear R-Helpers,
> I have a list l1 like:
>
> l1[[1]]
> a b c
>
> l1[[2]]
> d
>
> l1[[3]]
> e f
>
> I want an output res like:
>
> res[[1]]
> 1 1 1
>
> res[[2]]
> 2
>
> res[[3]]
Hi, another question: is there any argument that controls the line width of
axis
box of xyplot()? I tried lwd=2 or lwd.axis=2 in xyplot() or within
scales=list()
argument, without success.
Thanks
John
__
R-help@r-project.org mailing list
https://st
Hello
Dne Po 13. září 2010 14:51:39 Tal Galili napsal(a):
Hello all,
There is currently a (very !) lively discussions happening around the
web, surrounding the following topics:
1) Is R efficient? (scripting wise, and performance wise)
2) Should R be written from scratch?
3) What should be the
Hi all, I
When I plot both lines and points using type=c('l', 'p') in xyplot(), if I want
to include in legend both of them using keys=list(lines=list(col=1:3),
points=list(pch=1:3)), the lines and points are plotted side by side in legend.
Is there anyway to plot the points in the middle of th
On 2010-09-13 16:11, Kevin Burnham wrote:
The line I have now is this:
plot(alldata$haa_Haa, pch=alldata$Subject.Group)
This more or less gives me what I want, but instead of using the actual
group number as the plotting point it converts it to the pch character
corresponding to that number. I
Hi all,
I have a package that contains a function foo that calls a function
.fooInternal via match.fun('.fooInternal'). This step is necessary
because I want to give the user an option to override .fooInternal
with a custom function. The .fooInternal function name is not
exported. The function foo
The line I have now is this:
plot(alldata$haa_Haa, pch=alldata$Subject.Group)
This more or less gives me what I want, but instead of using the actual
group number as the plotting point it converts it to the pch character
corresponding to that number. Is there a way that the above line could be
m
Have you considered something like the following:
install.packages('sos') # if it's not already installed
library(sos)
si <- findFn('spline interpolation')
# 125 matches
summary(si)
# in 64 packages
si
# opens the results as a table in a web browser
# sorted to put first the package with the mo
On Sep 13, 2010, at 5:48 PM, Kevin Burnham wrote:
Hi All,
I am trying to plot per cent correct scores (column name =PerCorr)
for each
of about 40 subjects. I would like the character representing each
score to
be a number between 0 and 4 depending on the subject's group (from the
column S
On Sep 13, 2010, at 5:43 PM, Mestat wrote:
Hi listers,
If I would like to check if a variable contains certain value, I would
write:
if (10 %in% x)
And If I would like to check the opposite, that 10 is not into x.
How would
be?
Not sure about your terminology, "into x"?. why wouldn't it
Hi All,
I am trying to plot per cent correct scores (column name =PerCorr) for each
of about 40 subjects. I would like the character representing each score to
be a number between 0 and 4 depending on the subject's group (from the
column Subject.Group).
Also, I would ideally be able to order the d
!10 %in% x
(or !(10 %in% x) if you don't believe in R's precedence rules.
- Phil Spector
Statistical Computing Facility
Department of Statistics
On Sep 13, 2010, at 5:20 PM, Liviu Andronic wrote:
Hello
On Sun, Sep 12, 2010 at 5:37 PM, Sebastian Gibb > wrote:
Hello,
thanks for your answer.
mapply fits to my needs.
One thing that seems strange is that if you use
tree[[1]]$node$values <- 1:10
tree[[2]]$node$values <- 3:12
you still g
Hi listers,
If I would like to check if a variable contains certain value, I would
write:
if (10 %in% x)
And If I would like to check the opposite, that 10 is not into x. How would
be?
Thanks in advance,
Marcio
--
View this message in context:
http://r.789695.n4.nabble.com/Condition-in-tp2538110
Hello
On Sun, Sep 12, 2010 at 5:37 PM, Sebastian Gibb wrote:
> Hello,
>
> thanks for your answer.
> mapply fits to my needs.
>
One thing that seems strange is that if you use
tree[[1]]$node$values <- 1:10
tree[[2]]$node$values <- 3:12
you still get
> mapply(mean, tree[[1]]$node$values, tree[[2]]
There was an award session at the Vancouver JSM where both Ross Ihaka
and Robert Gentleman spoke. The simply-start-over post sounds very
much like Ross's JSM talk. Robert's response was much more positive -
coming from the perspective of developing BioConductor I think. I
don't recall the detail
Although not an R solution, I would highly recommend the generic mapping tools
GMT for this type of work.
http://gmt.soest.hawaii.edu/
Cheers,
Dylan
On Monday 13 September 2010, Craig Stanton wrote:
> Hello all,
> I'm very new to R and am having some trouble with the results of the
> inte
Sunny -
I don't think mapply is needed:
lapply(1:length(mylist),function(x)rep(x,length(mylist[[x]])))
[[1]]
[1] 1 1 1
[[2]]
[1] 2
[[3]]
[1] 3 3
- Phil Spector
Statistical Computing Facility
Hello all,
I'm very new to R and am having some trouble with the results of the
interp function. I'm trying to produce a chart roughly akin to a weather map
with natural looking filled contours over a large region of the south pacific.
I've got a list of points and values to be mapped to
Thanks all for the excellent help!
Kind regards,
Alej
2010/9/13 Dennis Murphy :
> Hi:
>
> demo(graphics) has a lot of nice examples of how to use various features -
> one of them is font selection.
>
> font = 3 => italic font = 4 => bold italic
>
> You need to select the correct as
Hi,
Just to provide an example:
#Sample data copied to variables "X", "Y", and "Z"
Z <- Y <- X <- data.frame(A = 1:10)
# A variable holding the names of the variables
datasetname <- c("X", "Y", "Z")
# Use mget() to collect all the variables in a list
# get() only gets one variable at a t ime,
Hello,
I am currently using the polar.plot function in the plotrix package to graph
data. Unfortunately, it seems that the default for the labels is to have a
background color that is covering the line representing my data, making it
difficult to read. Is there a way to make this label backgrou
Se ?get function.
On Mon, Sep 13, 2010 at 3:40 PM, Jie Li wrote:
> All,
>
> For example, I have a dataset named "ABC" loaded into R
> > ABC
> [,1] [,2] [,3]
> [1,]147
> [2,]258
> [3,]369
>
> and I also have a variable datasetname
> > datasetname
> [1] "ABC
Dear R-Helpers,
I have a list l1 like:
l1[[1]]
a b c
l1[[2]]
d
l1[[3]]
e f
I want an output res like:
res[[1]]
1 1 1
res[[2]]
2
res[[3]]
3 3
Essentially, I want to replicate each index equal to the number of elements
present in that index.
Below is what I do to accomplish this:
l1 <- list
All,
For example, I have a dataset named "ABC" loaded into R
> ABC
[,1] [,2] [,3]
[1,]147
[2,]258
[3,]369
and I also have a variable datasetname
> datasetname
[1] "ABC"
and I want to add this "ABC" dataset to an existing list "alldata"
> alldata <-NULL
Hello all,
There is currently a (very !) lively discussions happening around the
web, surrounding the following topics:
1) Is R efficient? (scripting wise, and performance wise)
2) Should R be written from scratch?
3) What should be the license of R (if it was made a new)?
Very serious people hav
Hi Uwe,
The problem is most likely because the original poster doesn't have
the latest version of plyr. I correctly declare this dependency in
the DESCRIPTION
(http://cran.r-project.org/web/packages/reshape2/index.html), but
unfortunately R doesn't seem to use this information at run time,
genera
LCOG1 lcog.org> writes:
[snip]
>
> I need to do the same for a fueltype where each record has a character
> representing the data instead of a numeral(as in year). No reproducible
> code or data because i think this is pretty straight forward. I could do
> this using a series of loops but i t
Dears,
does anyone know whether there is any complete list with explanations for the
several Trellis parameters (i.e. the ones you one gets with trellis.par.set() )
available? Many of them are self explanatory or easy to guess but there is such
a vast amount of them that this is not possible f
CRAN has a significant update to rms. Windows and unix/linux versions
are available and I expect the Mac version to be available soon.
The most significant improvement is addition of latex=TRUE
arguments to model fitting print methods, made especially for use
with Sweave.
Here is a summary o
Hello,
I'm attempting to use the ratetable argument to
survexp in the survival package. I use
the example from the ?survexp help page below,
and then slightly modify it to produce an error.
library(survival)
data(pbc)
#fit a model without any factors
pfit1 <- coxph(Surv(time, status > 0) ~ tr
Try this:
format(dt, '%Y-%m-%d'), if you want Date class:
as.Date(format(dt, '%Y-%m-%d'))
On Mon, Sep 13, 2010 at 2:24 PM, Andrew Yee
wrote:
Thanks David, now I wonder how you can have as.Date() render the
date using
local time rather than UTC.
Andrew
On Mon, Sep 13, 2010 at 12:08 PM
Hi guys,
Cant seem to find a solution for this. I am looking for a substitute for
cut that can transform character vectors. So as cut would be used below to
transform YearCat-> YearCat2 base on Year and vector
Ag<-c("00-'70","'71-'75","'76-'85","'86-'09") using 'cut '
FleetData$YearCat<-cut(F
Henrique, thanks for your suggestion. For my applications, character would
have been sufficient, so your suggestion of using format() works fine too.
Perhaps I should submit a feature request for as.Date() to let you specify
local time.
Thanks,
Andrew
On Mon, Sep 13, 2010 at 2:06 PM, Henrique D
On Sep 13, 2010, at 1:24 PM, Andrew Yee wrote:
> Thanks David, now I wonder how you can have as.Date() render the
> date using local time rather than UTC.
>
Since we are both in the EDT TZ at the moment, our times are UTC-4
(hours = 60*60 seconds)
> dt <- as.POSIXct("2010-08-22 23:14:52")
You can try this also:
Vectorize(sample, 'size')(y, 1:3)
On Mon, Sep 13, 2010 at 12:30 PM, James Hudson wrote:
> Id like to sample the vector y repeatedly. In this dummy dataset, Id
> like to sample (and store) it 1, 2, and 3 times.
>
>
>
> Is there a straightforward way to do this without
thanks for your help
I am trying to work around this in R but i have the feeling that this is going
to but the center and age next to each other which is not what i need...
it might be me not being able to find my head around this but...
i need a table with age as columns and center as rows
if
Thank you Marc - the first scenario.
On Mon, Sep 13, 2010 at 2:04 PM, Marc Schwartz wrote:
>
> On Sep 13, 2010, at 10:30 AM, James Hudson wrote:
>
> > Iâd like to sample the vector âyâ° repeatedly. In this dummy dataset,
> > Iâd
> > like to sample (and store) it 1, 2, and 3 times.
>
This is to announce that we plan to release R version 2.12.0 on Friday,
October 15, 2010.
Those directly involved should review the generic schedule at
http://developer.r-project.org/release-checklist.html
The source tarballs will be made available daily (barring build
troubles) via
http://cran.
Thanks David, now I wonder how you can have as.Date() render the date using
local time rather than UTC.
Andrew
On Mon, Sep 13, 2010 at 12:08 PM, David Winsemius wrote:
>
> On Sep 13, 2010, at 11:56 AM, Andrew Yee wrote:
>
> I'm trying to understand why as.Date() is converting a the modified dat
This is to announce that we plan to release R version 2.12.0 on Friday,
October, 2010.
Those directly involved should review the generic schedule at
http://developer.r-project.org/release-checklist.html
The source tarballs will be made available daily (barring build
troubles) via
http://cran.r-p
On Sep 13, 2010, at 10:30 AM, James Hudson wrote:
> I‚d like to sample the vector „y‰ repeatedly. In this dummy dataset, I‚d
> like to sample (and store) it 1, 2, and 3 times.
>
> Is there a straightforward way to do this without using a „for‰ loop?
>
> x <- c(1 :3)
>
> y <- c(1:10)
>
> (run.
Without the square term you can just use the rule for addition in sines:
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
So a regression of y= a + b* sin(2*pi/360*x + c) can be fit as:
lm( y~ sin( 2*pi/360*x) + cos( 2*pi/360/x ) )
If you need the actual values of b and c then you will need to do a littl
On Sep 13, 2010, at 5:24 AM, tuggi wrote:
hello,
can i calculate a sum to infinity in R.
i want to do something like this:
\sum_{i=0}^\infty
\frac{2^{-d-1}}{\Gamma(\frac{d-1}{2})}\left(\frac{\Gamma(2d-3)(2-
d)_{i}\Gamma(i+1,-z/2)2^{i+1}}{\Gamma(d-1)(4-2d)_{i}i!}\right)+
\\
\sum_{i=0}^\inft
?dev.copy
--
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-help-boun...@r-
> project.org] On Behalf Of Joel
> Sent: Monday, September 13, 2010 7:13 AM
> T
The zoomplot or updateusr functions in the TeachingDemos package may help.
But be cautious, the thing that grabs the most attention is where 2 (or more)
lines cross, when you plot multiple lines on different scales the crossing is
meaningless, but that meaningless point is what draws the eyes.
On Sep 13, 2010, at 11:56 AM, Andrew Yee wrote:
I'm trying to understand why as.Date() is converting a the modified
date of
a file from August 22 to August 23.
foo <- file.info(file.to.process)
str(foo)
'data.frame': 1 obs. of 10 variables:
$ size : num 5.37e+09
$ isdir : logi FALSE
$
Is this a recent version of R? If so, please report to the maintainer.
Otherwise, please also report that it does not work with your version of
R so that the maintainer can add a version dependency.
Best,
Uwe Ligges
On 13.09.2010 17:42, Paul Metzner wrote:
Hi!
I updated to reshape2 yesterday
On Mon, Sep 13, 2010 at 8:02 AM, Marc Schwartz wrote:
>
> On Sep 13, 2010, at 8:59 AM, Joshua Wiley wrote:
>
>> If it's not possible to use their particular algorithms, does anyone
>> think it would be helpful/practical to try to write a general scoring
>> system? I imagine a function with argume
I'm trying to understand why as.Date() is converting a the modified date of
a file from August 22 to August 23.
> foo <- file.info(file.to.process)
> str(foo)
'data.frame': 1 obs. of 10 variables:
$ size : num 5.37e+09
$ isdir : logi FALSE
$ mode :Class 'octmode' int 436
$ mtime : POSIX
Thank you sir.
Regards,
Phil
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Id like to sample the vector y repeatedly. In this dummy dataset, Id
like to sample (and store) it 1, 2, and 3 times.
Is there a straightforward way to do this without using a for loop?
x <- c(1 :3)
y <- c(1:10)
(run.sample <- sample (y, x))
Thanks very much,
James Hudson
Hi!
I updated to reshape2 yesterday and tried to make it work. Unfortunately, it
mainly throws error messages at me (good thing it's reshape2 1.0 and not
reshape 2.0). The most recent is:
Error in match.fun(FUN) : object 'id' not found
When I manually create an object 'id', it says:
Error in
Hello all,
I want to specify a minimum number of valid arguments for the mean
function--I have 5 variables but I want the mean only of cases that have at
least 3 valid answers. What is the best way to do that?
Thank you very much!
Luana
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Thanks Gábor!
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On Sep 13, 2010, at 15:59 , Joshua Wiley wrote:
> If it's not possible to use their particular algorithms, does anyone
> think it would be helpful/practical to try to write a general scoring
> system? I imagine a function with arguments for column names, a list
> where each element is a vector t
Hi:
demo(graphics) has a lot of nice examples of how to use various features -
one of them is font selection.
font = 3 => italic font = 4 => bold italic
You need to select the correct aspect of the plot, however: font.main,
font.sub, font.lab or font.axis (from par()).
HTH,
Dennis
On Sep 13, 2010, at 8:59 AM, Joshua Wiley wrote:
> If it's not possible to use their particular algorithms, does anyone
> think it would be helpful/practical to try to write a general scoring
> system? I imagine a function with arguments for column names, a list
> where each element is a vector
I'm still relatively new to R, so I tried the first of you two solutions:
.First <- function(){
source("Friedman-Test-with-Post-Hoc.r.txt")
}
Thanks very much for that, it works perfectly
Cheers
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On Sep 13, 2010, at 10:34 AM, Filoche wrote:
Hi again everyone.
Anyone know if there's any limitation with gap.plot concerning the
fill
color of plotted markers? I would like to fill the circles with a
color :
library(plotrix);
gap.plot(c(1,2,3,4,10), c(1,2,3,4,10), c(5,9), pch = 21, co
Hi:
Here's a ggplot2 example:
p <- ggplot(mtcars, aes(x = wt, y = mpg))
p + geom_rect(xmin = 2, xmax = 3, ymin = 0, ymax = Inf, fill = 'green',
alpha = 0.2) + geom_point() + theme_bw()
I don't know why the alpha transparency doesn't work in this example, but
plotting the points after shadin
check out this link
http://finzi.psych.upenn.edu/R/library/seqinr/html/col2alpha.html
On Mon, Sep 13, 2010 at 9:58 AM, threshold wrote:
>
> Thanks for replying, indeed works.
> I forgot to mention that I am looking for a 'transparent shading' s.t. the
> plot symbols are still visible on the shad
Hi again everyone.
Anyone know if there's any limitation with gap.plot concerning the fill
color of plotted markers? I would like to fill the circles with a color :
library(plotrix);
gap.plot(c(1,2,3,4,10), c(1,2,3,4,10), c(5,9), pch = 21, col = "red");
However, it only change the color of the
On Sat, 2010-09-11 at 14:41 -0700, Peng, C wrote:
> Is this something you want to have (based on a simulated dataset)?
>
> counts <- c(18,17,15,20,10,20,25,13,12)
> #risk <- round(rexp(9,0.5),3)
> risk<- c(2.242, 0.113, 1.480, 0.913, 5.795, 0.170, 0.846, 5.240, 0.648)
> gm <- glm(counts ~ risk, f
Did you check if the data in "da" has any NA in the dependent or the
independent data?
Remember that your function llk.mar is going to evaluate dnorm for each pair.
If any of those
pairs has an NA value, your function will return an NA at the end
(sum(c(NA,1,2,3)) = NA)
I would check if the ll
On Sep 13, 2010, at 8:50 AM, Alejo C.S. wrote:
Dear list, I making some box-and-whisker plots in R with vertebrate
data. The x axis are species names that must be in italics. I tried
with the "axis" function but no luck, and it seems that affects both
axes.
Any tip?
In bwplot just add:
...,
If it's not possible to use their particular algorithms, does anyone
think it would be helpful/practical to try to write a general scoring
system? I imagine a function with arguments for column names, a list
where each element is a vector that indicates the numbers that
correspond to various subsc
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