Enrico R. Crema enryu_cr...@yahoo.it 28/10/2010 12:08:55
Well...thanks to everybody for suggestions and reading...I guess I
have to study more!
...
On 28/10/10 12:23, Enrico R. Crema wrote:
I can't provide you a link for more details,
The canonical link is R FAQ 7.31 in the
When I use
Plot(series1)
lines(series2)
The graph will use the y axis scaling for series 1 so some of series
2 is
cut off. How do I control the y axis scaling?
Plot them together as a multivariate series as in the examples
already
pointed to.
...or try
plot(series1, type=l,
A simple alternative is to use at to contrl plot locations:
boxplot( ..., at=rev(1:nlevels(depthM)))
which just rearranges where they are plotted.
Example:
set.seed(1023)
x - gl(5, 5)
y-rnorm(25)
boxplot(y~x, horizontal=TRUE)
boxplot(y~x, at=rev(1:nlevels(x)), , horizontal=TRUE)
Steve E
If you usre filled.contour, then use contour() as a part of the function
supplied as the axis parameter, you can correctly overlay contours on
the colour contour plot.
Steve e
-Original Message-
From: Fiona Berryman f.berry...@wlv.ac.uk
To: r-help@r-project.org
Sent: 11/25/2010
John Fox j...@mcmaster.ca 14/07/2010 15:49:21
the development version of the car package on R-Forge, which can be
installed via install.packages(car,
repos=http://R-Forge.R-project.org;
(and which has an argument named center, not location).
Would you consider a plaintive request to
At the risk of repeating a post I haven't read, the two constructs are
different because 1:3 returns a vector of integers (class integer) and
c(1, 2, 3) is a bit more conservative about what '1' and the like mean
and returns a vector of class numeric.
lapply(df1, class)
lapply(df2, class)
On
check the robustbase package and rlm or hubers in MASS
You'll need to know which m-estimator you want, but both those packages
include at least some m-estimators.
Strictly, so does the base package; the mean is an m-estimator. Just
not a very robust one!
S Ellison
Iasonas Lamprianou lampria
Neat. But why assign the functions to separate variables at all?
mdlChooser - function(type=c(one,two)) {
type - match.arg(type)
m - switch(type,
one=function(x,N0,r) N0*exp(x*r) ,
two=function(x,N0,r,K) (N0*K)/(N0+(K-N0)*exp(-x*r))
)
m
}
also works without appearing to assign
Frank Harrell f.harr...@vanderbilt.edu 11/08/2010 17:02:03
This problem seems to cry out for one of the many available robust
regression methods in R.
Not sure that would be much more appropriate, although it would
_appear_ to work. The PB method is a sort of nonparametric/robust
approach
I use savePlot() with type=emf (remember to include the right file
extension - although savePlot will use the type as default extension, it
will only do so if there are no '.'s in the filename, so filenames like
'../plots/aplot' won't get the extension automatically. Really should
have made that
try
lapply(df, class)
Steve E
On Thu, Aug 26, 2010 at 4:31 PM, Daniel Brewer
daniel.bre...@icr.ac.uk wrote:
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
John Sorkin jsor...@grecc.umaryland.edu 10/09/2010 13:21:09
What is the difference (or differences) between lme and lmer? Both
appear to perform mixed effects regression analyses.
From a user's point of view:
- lme only accepts nested random effect; lmer handles crossed random
effects
- lme has
See ?lm and, more usefully, ?formula
Gundala Viswanath gunda...@gmail.com 29/09/2010 15:51
I am refering to a function call like this:
data(iris)
x - svmlight(Species ~ ., data = iris)
I tried to see the content of it by typing:
Species ~ .
but it gives nothing. How can I see it's
They probably look different because the y-axis distance is to the
bottom of the number and the x-axis to the top; character adjustment is
putting the actual locations in the 'same' place but with opposing
orientation.
Try fooling about with the mgp argument in axis():
par(mfrow=c(1,1), cex.axis
The gamlss package, by Mikis Stasinopoulos and available at
http://www.gamlss.com/ as well as from CRAN, is also very flexible,
allowing shape and scale adjustment.
Steve E
David Winsemius dwinsem...@comcast.net 28/10/2009 14:18
You might want to take a look at this article by WEI, PERE,
Lathouri, Maria m.lathour...@imperial.ac.uk 10/28/09 6:02 PM
I want to have as label in y-axis Temperature (oC).
First, look at ?plotmath and find the 'degree' symbol...
Then look at the symbol for spacing.
Then try
ylab=expression(Temperature~degree*C)
Hmm...
set.seed(17*11)
d-data.frame(africa=sample(50, 10),
europe= sample(50, 10),
n.america= sample(50, 10),
s.america= sample(50, 10),
antarctica= sample((1:50)/20, 10)
)
#Get three top from each row
R OO is documented for S3 classes under section 5 (Object-oriented
programming) in the R language definition.
I guess the issue is somewhat philosophial as to how you use it.
R philosophy _mostly_ separates data from operations on data, so the OO
model provides classes for data and essentially
There's a note on this issue at
http://www.nag.co.uk/numeric/RunderWindows.asp
which claims it can 'safely be ignored'.
Your mileage may vary, I presume.
I had the same error message under a previous version (but with a
different library and during installation, not loading, some packages)
plot(1:8, f, axes=FALSE)
axis(1, at=1:8, labels=time)
axis(2)
ANJAN PURKAYASTHA anjan.purkayas...@gmail.com 21/05/2010 17:15:48
Hi,
I need to plot $time on the x-axis and $f on the y-axis for the
following
data:
timef
0h0.00
0.5h0.54
1h1.15
2h2.33
4h1.57
6h2.19
18h
I'm writing a package, and would appreciate advice on controlling the
help documentation cross-references for a redefined generic.
I wanted to define a cbind equivalent for an object that mostly behaves
like a data frame. base::cbind dispatches to a data frame method if
_any_ parameter is a data
cut() might work if you want a character version...
set.seed(221)
x-rnorm(50)
y-rnorm(1)
xh-hist(x)
cut(y, breaks=xh$breaks)
#or
as.character(cut(y, breaks=xh$breaks))
#or for a 'level number' version (a bit like which())
as.numeric(cut(y, breaks=xh$breaks))
Jim Lemon j...@bitwrit.com.au
See ?.Machine
Saji Ren saji@gmail.com 24/12/2009 06:07:27
Hello,everyone:
I met this notation when I read the original code of function
quantile.There is one sentence as below:
eps - 100 * .Machine$double.eps
when I input .Machine$double.eps in R, it returns [1]
2.220446e-16.
Can anyone
If you're interested in handling outliers in analytical proficiency
testing read current and past IUPAC guidance on PT (see
http://www.iupac.org/objID/Article/pac7801x0145) and particularly
references 1, 4, 5, and 11 therein, for starters.
Although one might reasonably ask whether outliers are
If you set them a problem that has them doing the same sort of thing
five times and compare the time it takes with code pasted from an editor
(eg Tinn-R) and the time it takes via menius, you may have more luck
convincing them.
A command line sequence is harder than menus the first two times but
The ?plotmath page and the example ## How to combine math and numeric
variables : should be helpful...?
Thomas Bschorr bsch...@phys.ethz.ch 05/06/2010 10:36:42
Hi,
I desperately try to do s.th. like
m=1.23455
sig=0.84321
plot(1,1)
text(0.8,1,sprintf(Sigma=%1.2f±%1.2f,m,sig))
where
Hadley Wickham had...@rice.edu 02/07/2010 14:59:53
Where is %in% documented within R? I'm pretty sure it's a different
action to ?%in%, and it's not mentioned in ?formula
?formula in R 2.9.2 says in para 2:
The %in% operator indicates that the terms on its left are nested
within those on
Duncan Murdoch murd...@stats.uwo.ca 12/01/2010 18:07:46
I need to get the p values for a table with 15000 entries of t
values.
...
Put the t values into a vector, then use pt() in an appropriate way
... and don't forget any necessary correction for multiple comparisons;
see
?p.adjust
I think you should rather look at the origin= parameter in barchart. See
?panel.barchart for a discussion of this exact problem:
origin: the origin for the bars. For grouped displays with 'stack
=
TRUE', this argument is ignored and the origin set to 0.
Otherwise, defaults
Thomas Lumley tlum...@u.washington.edu 15/01/2010 16:07
Which should I use or does it matter, please?
I would say to use = if you are teaching people familiar with C or
Java, and to use - otherwise.
Nothing like an option to induce polarisation!
'=' is used in at least two contexts in R,
name toString(20)
is from Excel or OpenOffice; means 'logical and' in R, not string
concatenation.
paste() is simpler; sprintf() is more precise as to decimal places and
format.
anna lippelann...@hotmail.com 26/01/2010 14:09:15
Hello there, I want to create a string from strings and
If you know the likely range, uniroot would do it.
f.ytm-function(ytm) 100 / (1+ytm) +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010
uniroot(f.ytm, interval=c(0,25))
#$root has the answer
Craig P. Pyrame crap...@gmail.com 01/02/2010 10:19
Madhavi Bhave wrote:
Dear R helpers
I am
Might this be a firewall-like issue and nothing to do with html or R?
If I understand the 2.10 help system, it operates by starting R as an
http server - effectively a web server, operating on the local machine
(127.0.0.1) with an unusual IP port (which doesn;t seem to be consistent
from one run
I would like to ask how to extract the p-value for the whole model
from
summary(lm).
If you mean the p-value given at the end of the summary() printout, it
isn;t held in the summary object. But information to get it is. Using
the ?lm example:
ctl -
Try
shell('\file.xls\')
where 'file.xls' is the excel filename.
The escaped quotes (\) are not strictly necessary if the filename contains no
spaces, but they are essential if it does.
Sergey Goriatchev serg...@gmail.com 08/02/2010 12:48:20
Hello, everyone
I wonder if it is possible to
jim holtman jholt...@gmail.com 08/02/2010 14:09:52
Typically R does not have macros;
I know exactly why Jim Holtman said that; R doesn't have a separate
'macro' construct with separate 'macro variables'.
But it is perhaps a bit misleading to say that R doesn't have macros
without saying a bit
You could try aggregate:
If we call your data frame df:
aggregate(df[2], by=df[1], FUN=min)
will get you what you asked for (if not necessarily what you need ;-)
)
Switching the columns around is easy enough if you need to; proceeding
stepwise:
df.new-aggregate(df[2], by=df[1], FUN=min)
Try
f - function(nbr){
y-rnorm(nbr)
y1 - mean(y)
plot(y)
invisible( y1)
}
That will return y1 invisibly, so
f(100)
plots but returns nothing visible but
w-f(100)
plots and places the return value in w
Dennis Murphy djmu...@gmail.com 02/19/10 9:33 PM
Hi:
Perhaps you want
test$Y-NULL
test[Z]-NULL
Todd Ogden to...@columbia.edu 23/02/2010 16:02:01
There might be more elegant ways, but this will do it:
test-test[-match(Y,names(test))]
On Feb 23, 2010, at 7:04 AM, Knut Krueger wrote:
Hi to all,
test - data.frame(X=c(1:4),Y=c(5:8),Z=c(8:11))
test - test[,-2]
Ted Harding [EMAIL PROTECTED] 14/09/2007 10:59:47
On the contrary! It adds to our collective wisdom.
We all have to suck eggs, and usually can successfully perform the act.
Ted,
Thanks for the kind remarks.
But we'll have to get off the egg topic, or we'll all end up as acknowledged
Sorry!
paste(x*100,%)
Marc Schwartz [EMAIL PROTECTED] 27/09/2007 15:56:05
On Thu, 2007-09-27 at 14:36 +0100, stat stat wrote:
I am wondering if there is any procedure to write a particular value
in Percentage format, still maintaining it's numeric character. for
example I want to write
Is there any way to password-protect script files (either within R or
otherwise)?
The question seemed to me to be more about password protection against
modification, rather than encryption.
I'd have thought this was something a decent OS could take care of. It seems as
daft to try to get a
I think there is reason to be surprised, I am, too. ...
What am I missing?
Read the formula and ?summary.lm more closely. The denominator,
Sum((y[i]- y*)^2)
is very large if the mean value of y is substantially nonzero and y*
set to 0 as the calculation implies for a forced zero intercept. In
Ermmm... from ?integrate
Description:
Adaptive quadrature of functions of one variable over a finite or
infinite interval.
R maps infinite intervals to a finite interval before numerical integration,
provided that you tell it that the limits are infinite.
Using integrate() over
Divaker,
Thanks for the data.
For me,
summary(aov(Purity~Supplier/Batch, process))
gives exactly the same results for mean squares as
aov(Purity~Supplier+Error(Supplier/Batch), process)
except that the latter gives no p-values (because Supplier appears as
both error term and fixed effect,
I'm no guru, but can you not use do.call here?
i.e. does
foo-function(...) #Not x(...)
{
arglist-list(...)
#remove bad characters from arglist here.
# then
do.call(bar,arglist)
}
do what you want?
David Keegan [EMAIL PROTECTED] 01/02/2008 11:00:15
Hi,
foo-x(...)
{ # Need
Stas Kolenikov [EMAIL PROTECTED] 11/02/2008 18:54
I would think the FDA regulations could go as far as
specific SAS syntax, or at least to specify SAS PROCs to be used.
This is unnecessary caution. FDA (like the MHRA in the UK, where I come
from) should not endorse a single supplier, even
Also i have read in Quinn and Keough 2002, design and analysis of
experiments for
biologists, that a variance component analysis should only be conducted
after a rejection
of the null hypothesis of no variance at that level.
Hmmm...
This does rather assume that 'no significant result' means
Devred, Emmanuel [EMAIL PROTECTED] 13/03/2008 18:08:45
Dear R-users,
I haven't found a way in the searchable archive to overplot a contour
(lines) over a surface.
I got round this by using the plot.axis parameter, which accepts a set
of plotting commands and isn't fussy about what. Since
R on windows includes some useful functions for windows font handling.
Start with
?windowsFonts
windowsFonts()
should show Arial mapped to sans. (Arial is the default in Windows,
at least in my locale; ?windows tells you that too)
Then try
par(family=sans, font=2) #specifies Arial bold as
Duncan Murdoch murd...@stats.uwo.ca 15/07/2009 15:04:29
* R has a new predictive punishment module. It punishes you for
things
it knows you will do later.
Dang! Does that mean it'll return errors for the statistical idiocy I
haven't yet got around to committing?
;-)
Steve E
Since you;re already specifying the axis tick locations for the x-axis
any ylim for y, one way of dealing with it is to give the axes an extra
tick at each end: Instead of
axis(side=1, at=c(1,2,3,4), labels=gut)
use
axis(side=1, at=0:5, labels=gut)
Similarly for axis(2), use
axis(2,
I agree; the stripchart defaults place things too close to the edges for
'neat' layout.
Use at= coupled with ylim (or xlim, if vertical) to place the rows
explicitly and leave room at the margins.
Extending the last example in ?stripchart:
stripchart(decrease ~ treatment,
main =
You could look at ?lmList in package lme ...
Idgarad idga...@gmail.com 21/07/2009 17:27:52
How can I get the results of lm() into a list so I can loop through the
results?
e.g.
myResults[1] - lm(...)
myResults[2] - lm(...)
myResults[3] - lm(...)
...
myResults[15] - lm(...)
myResults[16] -
You probably missed the bit in the lattice documentation which says that
few if any of the standard par() parameters work on lattice. lattice
uses its own system.
Look at the xyplot help page and seek out the scales argument. That
tells you that scales is a list, optionally with x and y
Liviu Andronic landronim...@gmail.com 07/24/09 7:06 PM
.. use save.image() to save R's workspace and re-load it when
re-opening R via load().
Or you could just use save.image in your default directory for R and R
will then open it automatically. I'd use that sparingly though; it is
not always
boxplot itself is hardwired to produce the boxplot.stats list, and that
is not easy to change.
To get a different set of stats, you would need to do things in rwo
stages:
i) create a boxplot object of the type returned by boxplot, but using
your own stats
ii) call bxp on that object.
That's kind
If you use loess instead of lowess, you can get standard errors and
hence approximate confidence bands by using predict.
Example:
plot(cars)
plx-predict(loess(dist~speed, data=cars), se=T)
lines(cars$speed,plx$fit)
lines(cars$speed,plx$fit+2*plx$s, lty=2) #rough ready CI
Stephan Kolassa [EMAIL PROTECTED] 08/10/08 8:41 PM
I like RSeek:
http://www.rseek.org/
The burning question is why an audio compay would be running a web
search engine for an open source statistics package...??!
***
This email and
giov [EMAIL PROTECTED] 13/08/2008 10:59:32
just a question...I don't know
what is the distribution of my data (normal, T, etc...). So, how can I
set
the type parameter?
You must assume an underlying distribution or you can't do an outlier
test.
Outliers are just unusually extreme data
The help page on binary operators (see ?==) confirms that binary
representation of fractional representation is not catered for and
points to all.equal as a more suitable test method for those cases.
Steve E
Thomas Lumley [EMAIL PROTECTED] 13/08/2008 16:47
Integers (up to a fairly high limit)
giov,
It sounds like you have approximately symmetric distributions. If that
is so, and particularly if the standard deviation is less than about 20%
of the mean, I'll stick my neck out and say I would assume underlying
normality for outlier testing purposes unless there's a reason to do
An r hel search picks up
http://tolstoy.newcastle.edu.au/R/help/01a/1162.html
from 2001, which points towards the ts package.
Öhagen Patrik [EMAIL PROTECTED] 19/08/2008 09:00:12
Dear List,
I have used all my resources (i.e. help.search) and I still havn't been able
to figure out if there
see ?gsub and loook at the last examle
suman Duvvuru [EMAIL PROTECTED] 20/08/2008 05:19
I would like to know how to convert a string with characters to all
uppercase or all lowercase? If anyone could let me know if there exists
a
function in R for the conversion, that will be very helpful.
Strictly, you need to type the extension _if your filename or path
include a period (.)_ but not otherwise.
The filename test alone will be saved as paste(test,type). So will,
for example, /plots/test. But filenames such as
test.it or ../plots/test will not include the extension
automatically.
Duncan Murdoch [EMAIL PROTECTED] 26/08/2008 16:17:34
If this is indeed the case, switch; the expected gain is
positive because _you already have the information that you hold the
median value of the three possibilities_. The tendency when
presented
with the problem is to reason as if this is
Sherri;
The boxplot stats include outlier values in $out and their group number
in $group.
Let me assume that you want to put the boxes at integer positions on
the x-axis (the default is the same). The we have
y-rnorm(60)
g-gl(3,20)
labels=paste(Value, 1:length(y))
bx-boxplot(y~g)
bx.at-1:3
You can use the 'symbol' element in an expression. For example
plot(1:5)
text(2,3,expression(symbol(\xD1)))
S.
Nuno Prista [EMAIL PROTECTED] 06/19/08 9:55 PM
Dear colleagues,
Can anyone of you tell me how to write a nabla symbol in an R graph?
Thanks in advance,
Nuno
Two thoughts:
i) If you have a narrow distribution, the density can be higher than 1. The
area comes out at 1 for density, and n for the frequency.
ii) hist() will not show the same frequencies as density() unless hist has unit
bin sizes. density*length is showing number per unit change in
Lord Yo [EMAIL PROTECTED] 07/08/08 9:00 AM
I am trying to add a percent sign to my labels in a hist() plot.
?hist says labels: logical or character. This should be a clue;
labels could be a character vector.
Try
x-rlnorm(128, 1)
h-hist(x, plot=F)
plot(h,
Since the standard normal distribution goes to infinity in both
directions, you can't have random normal constrained to +-1.5.
You can have _truncated_ standard normal, though, if that's really what
you want.
+-1.5 is/are the normal quantiles at pnorm(c(-1.5,1.5)). So if we
generate 500
Sandy,
You can re-order a factor with
df$Eyeball-factor(df$Eyeball, levels=c(Normal, Mild, Moderate,
Severe), ordered=T)
(assuming df is your data frame and that you want an _ordered_ factor;
the latter is not essential to your plots)
Incidentally, NULL isn't a particularly friendly item to
Paul;
You asked
using ...
optFederov(~.,dat,6)
... does the job with good efficiency.
I would be interested to know what your objection to this is S
I have no issue with AlgDesign in principle, but the question was
specifically about _fractional_ factorials, so I answered that.
As to
What exactly is on line 1743?
Genuinely empty cells will probably read as NA anyway. Excel errors, prefixed #
in excel, will be read as comment markers and cause the remainder of the line
to be ignored. This often causes this kind of error.
Steve E
Monna Nygård [EMAIL PROTECTED] 30/04/2008
The factor order defaults to alphabetical, and boxplot follows that.
Re-ordering the factor to the order of interest is probably the best way
of handling it.
However, if you don;t want to do that, you could perhaps also use the
at= parameter in boxplot. For example
x-rnorm(50)
See
?chull
juli pausas [EMAIL PROTECTED] 05/05/08 1:56 PM
Hi,
Is there a way in R to plot an envelope line from a cloud of points
(x, y data) ? That is, a smooth line that include all points, where
the points do not follow a strait linear pattern. Could somebody
redirect me to some package or
Guru S [EMAIL PROTECTED] 09/05/2008 08:18
I fitted tree growth data with Chapman-Richards growth function using
nls.
When I try to run the anova() function I get this:
Error in anova.nls(fit.nls) : anova is only defined for sequences of
nls objects
There seem to be two problems.
The first is
Jason Lee-14 wrote:
I want to plot V1 against V2 with the condition that I
want to mark(color) the point(row 5) on the graph. This is so i could see
some triangle shape thing on the graph for row 5.
You could try something crude like
plot(V1,V2, col=ifelse( (1:length(V1)) == 5,
The low R2 says the model does not explain much of the variance.
But the high significance arises from the very large number of degrees
of freedom.
This is not an 'incompatibility'; just what happens with large
dispersion, small effects and a very large number of observations.
But you clearly
A bit of fiddling suggests that the problem is less to do with rpart
(which will quite happily acept weights defined inside a function) and
more to do with the formula argument to the test function.
This is probably because defining the formula in the global environment
(as you implicitly do by
Birgitle [EMAIL PROTECTED] 05/06/2008 10:33:20
I think you should specify your grouping factor:
Yes, the message says the grouping factor is missing.
But xx should not be a matrix.
Your x data should be in a (1-dimensional) vector of length 276, and so should
your grouping factor.
Toy
Lawrence,
use hist(..., axes=F)
then put your own axis on with axis(1,...)
Example:
y-rnorm(200)
hist(y,axes=F)
axis(2)
axis(1, at=seq(-3,3,1))
Steve E
Lawrence Hanser [EMAIL PROTECTED] 06/09/08 7:02 AM
Dear Friends,
I am doing a rather simple histogram on a vector of data, MR. I set
breaks
? How one cand do this in R?
Many thanks
--- S Ellison s.elli...@lgc.co.uk schrieb am Di, 28.4.2009:
Von: S Ellison s.elli...@lgc.co.uk
Betreff: Re: [R] duplicate 'row.names' are not allowed
An: r-h...@stat.math.ethz.ch, amor Gandhi amorigan...@yahoo.de
Datum: Dienstag, 28. April 2009, 14:17
Apologies; my earlier reply preceded this extra info.
If you are reading a large file with duplicate row names, try reading
it in as is, then simply converting the numeric parts to a matrix.
For example, if you had the same kind of data frame as before
x1 - rnorm(11,5,1)
x2 - runif(11,0,1)
nam -
Well, it's clearly not pseudoreplication if it's not replication!
But the observations within animal could well be associated. You seem
to have a straightforward experiment with multiple treatment
combinations on multiple subjects.
You could do several things. The most obvious is probably to
It's sometimes called a pyramid plot.
The plotrix package has one. See ?pyramid.plot
dxc13 dx...@health.state.ny.us 05/04/09 7:58 PM
Hi all,
I cannot think of the technical name of this plot, but I want to create
a
plot with the data below that looks like two back-to-back horizontal bar
The problem is in the code.
When you say
mysummary-tapply(myfactor,mydata,length)
mysummary
you have used mydata as a factor and myfactor as the data.
tapply has (correctly) used the ordered labels in the grouping factor
(mydata) to label its output.
If you did what you probably intended:
The usual way of doing this in R functions is via deparse, as in the
examples in ?deparse:
fd-function(x) plot(x, main=deparse(substitute(x)))
q-1:10
fd(q)
as.character works for simple cases but is probably not the best
option. Compare
fd(log(q))
with
fc-function(x) plot(x,
lehe timlee...@yahoo.com 08/05/2009 09:58:40
1. How to plot several lines in a figure?
See ?lines
2. How to open another figure window?
see ?windows
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This email and any attachments are confidential. Any
Assuming a constant bin width, you need to multiply the density by
n*binwidth, where the bin width is (obviously!) the width of the
histogram bins.
Jacques Wagnor jacques.wag...@gmail.com 05/09/09 5:10 PM
Dear List,
When I plot a histogram with 'freq=FALSE' and overlay the histogram
with a
Probabilities can only be even approximately linearly related to a
continuous predictor variable for a limited range, otherwise the model
will predict probabilities below 0 or above 1.
At some point, they have to tail off... unless you are modelling
something trivial like 'probability of being
Also check the asp= parameter in plot.default and plot.window; this sets
the aspect ratio so that 1 unit in x is the same physicla length as 1
unit in y. I don;t know whether it is respeced by your particular
biplot, though.
Bernardo Rangel Tura [EMAIL PROTECTED] 20/11/2007 10:51:12
On Mon,
Within the base package collection, see
?colorRamp and ?colorRampPalette
Then
or.to.blue-colorRampPalette(c(orange,green,blue)) #green
included to avoid muddy blues
plot(df$X, df$Y,type=n)
text(df$X, df$Y, labels=df$Z,
col=or.to.blue(length(df$Z))[rank(df$Z)])
You don't need to order by Z to do
Ted,
I don't think you can include the axis ticks in plot(), but you can
specify them in axis() using
plot(x,y,xlim=c(0,1000), axes=F)
box()
axis(1,at=seq(0,1000,50),labels=F)
axis(1,at=seq(0,1000,200),labels=T)
Obviously, if you want this to be variable for different xlim etc, your
simplest
--- Kapoor, Bharat [EMAIL PROTECTED] wrote:
1.I could not print the data in ROF column below
the bar charts, How to get full labels for each bar
insted I createda new coumn ReasonCode just to fit
below the bars.
You could plot the barchart horizontally, and increase the margins to
hadley wickham [EMAIL PROTECTED] 11/12/2007 17:02
Could someone recommend a good book on regular expressions with focus
on
applications/use as it might relate to R. I remember there was a
mention of
such a reference book recently, but I could not locate that message
on the
archive.
It's
?jitter is simpler:
x-rep(1:10,10)
y-x
plot(x,y) #100 points, only 10 show
plot(jitter(x),jitter(y)) #overlap removed.
Milton Cezar Ribeiro [EMAIL PROTECTED] 18/12/2007 04:36
Hi Wayne,
I have two suggestion to you.
1. You add some random noise on both x and y data or
2. You
Hiding in the windows faq is the observation that R's computation is
single-threaded, and so it cannot use more than one CPU. So multi-core
should make no difference other than allowing R to run with less
interruption from other tasks. That is often a significant advantage,
though.
Andrew
I am looking at MM-estimates for some interlab comparison work. The
usual situation in this particular context is a modest number of results
from very expensive methods with abnormally well-characterised
performance, so for once we have good variance estimates (which can
differ substantially for
.. and don't forget that 6 - x works but 6 = x won't ...
Gabor Grothendieck [EMAIL PROTECTED] 01/13/08 10:50 PM
No.
f - function(a = 3, b = 4) a-b
f(b = 10)
[1] -7
f(b - 10)
[1] 6
but if you only replace it in the context:
x - ...
then it should be ok.
On Jan 13, 2008 5:41 PM, Nasser
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