The posting guide (a link at the bottom of every e-mail from the list)
suggests including a small example of what you are trying to do.
Without an example, we have to guess at what you are trying to do.
I am still guessing since you still did not include an example. It
appears that part of the
text to the margins).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Greg Snow
Sent: Friday, September 07, 2007 9:01 AM
To: Yves
?options
Look at the entry on 'digits'. Does that fix the problem? If not, give
a little more detail on what you are doing.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL
Not all of us are familiar with lisp (I have done a little, but not
enough to really understand what you are asking). If you tell us what
find, member, cond, and loop do, or what functionality you are looking
for, then we will have a better chance of telling you how to do the same
in R.
Just
It works for me:
boxplot( split(state.x77[,'Frost'], state.region) )
text( 1:4, -5, rep('test',4), col='green' )
Show us what you tried and maybe we can be of more help,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
You may also want to look at the cnvrt.coords function in the TeachingDemos
package. It may be a bit simpler than mixing grid and base.
-Original Message-
From: Sébastien [EMAIL PROTECTED]
To: Prof Brian Ripley [EMAIL PROTECTED]
Cc: R-help r-help@stat.math.ethz.ch
Sent: 9/3/07 7:46 PM
Sorry, I did not think about the nested design (did not read carefully enough).
Another thing to try (untested) is to create a column of 1's and include that
specifically and exclude the default intercept, then your column of 1's acts as
the intercept, but a p-value is returned from it.
Note
?mapply
mapply('+', a, b, SIMPLIFY=FALSE)
colSums(mapply('+', a, b))
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On
You could make multiple calls to the rug function using a different
color for each call. Using tapply or by may automate this for you.
Another option would be to create your own rug using the segments
function (which will plot multiple colors).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
If your main goal is to do a loess fit, then make predictions from that,
then using the 'get' function may do what you want:
tmp.var - get(ORDINATE)
lo - loess(percent ~ ncms * tmp.var, d, ...
grid - expand.grid(tmp.var=MINVAL:MAXVAL, ncms=MINCMS:MAXCMS)
predict(lo, grid)
Here you stick with
Take a look at the my.symbols function in the TeachingDemos package.
The last example shows how to create a hexagonal grid and the 2nd to
last example shows how to plot several small line plots onto a larger
plot. Combining these 2 examples should give you what you want.
Hope this helps,
--
The best option is to use a bar chart or dot chart instead of a pie chart.
-Original Message-
From: Adam Green [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 9/4/07 10:21 AM
Subject: [R] Pie Chart Labels
I am having trouble finding out how to adjust the
A couple of things that may help:
1. you can create your boxplots by using boxplot with plot=FALSE, then
plotting that info using the bxp function. The bxp function returns the
locations of the boxes as part of the output which could then be used
for placing labels or refrence lines.
2. The
What is happening is that before the regex engine can look at your
pattern, the R string parsing routines first process your input as a
string. In the string processing there are certain things represented
using a backslash. Try this code in R:
cat('here\tthere\n')
The \t is made into a tab
Try this code (with the mydf that you generate below):
library(TeachingDemos)
plot( c(0,5), c(0,1), xlab='State', ylab='ylab', axes=FALSE, type='n' )
axis(1, at= (1:5) - 0.5, labels=paste('state',1:5))
box()
for(i in 1:5){
with( subset(mydf, State==i),
subplot( plot(Position,
Here is the approach that I would take.
Use a different plot for each day but line them all up like so:
x - c(1, 2, 10, 12)
y - c(100, -20, 50, 25)
day - c(1,1,2,2)
my.df - data.frame(x=x,y=y,day=day)
par(mfrow=c(1,2), oma=c(5,4,4,2)+0.1, mar=c(0,0,0,0))
tmp.yr - range(my.df$y)
for (i in
generally works better).
Erich Neuwirth wrote:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Erich Neuwirth
Sent: Wednesday, August 29, 2007 4:46 PM
To: r-help
Subject: Re: [R] Excel
Greg Snow wrote:
Or do you trust all of your clients to know
Matt Austin wrote:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Austin, Matt
Sent: Thursday, August 30, 2007 1:25 PM
To: r-help
Subject: Re: [R] Excel
Ah . . . the hammer analogy. In a conversation like this
it's not a question of will
Try calling summary.lm on your object (if it is an aov object then summary
calls summary.aov which does not show the intercept, but calling summary.lm
directly does give info on the intercept).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
This is my understanding of what is happening.
1. Standardize all the x variables to have mean 0 and variance 1 (possibly y as
well).
2. Compute the unconstrained least squares regression.
3. Sum the abs values of the b's.
That sum is the scaling factor. A bound of 1 means the sum above (and
Erich Neuwirth said:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Erich Neuwirth
Sent: Wednesday, August 29, 2007 12:43 PM
To: r-help
Subject: Re: [R] Excel
Excel bashing can be fun but also can be dangerous because
you are makeing your
You need to save the output in order to do something with it (the
default if you don't save the output is to call the print method, so
what you are seeing is the results of calling print.htest on the return
value from t.test).
Save the output by doing something like:
out - t.test(c(1:5,7:11),
From: Birgit Lemcke [mailto:[EMAIL PROTECTED]
Sent: Friday, August 24, 2007 2:13 AM
To: Greg Snow
Cc: R Hilfe
Subject: Re: [R] ordered factors in data.frame
Thanks a lot Greg for your help
The important element in the fit object is named coefficients not coef.
Sometimes the partial matching can cause confusion when it tries to help. In
your case it creates a copy of the coefficients, changes the 2nd value, then
creates a new component to fit called coef with these values (not
See the wtd.quantile function in the Hmisc package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Seung Jun
Sent: Tuesday,
Deb,
Others have answered your question, I just want to give you something
else to think about.
Stacked bar plots may not be the best tool to use for this situation,
they are ok for comparing the bottom and top groups, but all the other
groups don't line up in a stacked bar plot making those
Does the following code do something like what you want (but with the
built in data):
plot(as.integer(state.region), state.x77[,'Murder'],
+ xaxt='n', xlim=c(0,5), col=as.integer(state.region))
axis(1, at=1:4, labels=levels(state.region))
oo - order(order( as.integer(state.region),
The attach function only attachs a copy of the data (changes to the data don't
show up in the attached copy). Also you need to tell R what to do with the
result of as.ordered (where to save it).
Try something like:
table$V3 - as.ordered(table$V3)
Or
table - transform(table,
You can set up the 3 plots per page by using:
par(mfrow=c(3,1))
Then there are a couple of options for skipping the top graphics
position if the graph fails. If you know that the graph failed then you
can just use plot.new() (or frame()) to skip the top plot and plot the
next one in the 2nd
Oops, I read further down in your original post and see that you already
knew about par(mfg=c(2,1)). To get it to advance to page 2 for the 4th
plot try calling plot.new() which should move you to the next page, then
doing par(mfg=c(1,1)) should cause the next graph to be at the top.
Hope this
Here are a couple of options that you could look at:
The biglm package also has the bigglm function which you only call once
(no update), you just need to give it a function that reads the data in
chunks for you. Using bigglm with a gaussian family is equivalent to
lm.
You could also write your
Ted Harding wrote:
[snip]
So you if you want the density plot, you would need to calculate
this for yourself. E.g.
H1$density - counts/sum(counts)
plot.histogram(H1,freq=FALSE)
shouldn't that be:
H1$density - counts/sum(counts)/diff(brkpts)
?
I think at least part of your problem is that in the lm example you use
data=x (correct), but in the t.test example you use data=warpbreaks, so
in that case it is uing the full data set, not just the portion passed
by the 'by' function. Try the t.test example again with data=x and see
what
Jim Lemon Wrote:
I also greatly enjoyed Ted's rebuttal of the Bar charts are
evil and must be banned argument. If bar charts are
appropriate for the audience, give 'em bar charts. One great
way to turn off your customers is to tell them what they can
and can't do with your product.
I
?
Quoting Greg Snow [EMAIL PROTECTED]:
My original intent was to get the original posters out of
the mode of
thinking they want to match what the spreadsheet does and into
thinking about what message they are trying to get across. To get
them (and possibly others) thinking I made
Another couple of things to think about:
You could use the layout function to set up your multiple plots and
include an extra plotting area at the bottom to place the legend in.
If you stick with the solution below then the cnvrt.coords function from
the TeachingDemos package may be useful (will
with John Kane on Wed
1 August and again today on similar lines, and I think it's
time an alternative point of view was presented, to
counteract (I hope usefully) what seems to be a draconianly
prescriptive approach to the presentation of information.
On 07-Aug-07 21:37:50, Greg Snow
think that Greg's points are valid too, and in this
particular
case, bar plots are a bad choice and adding numbers at variable
heights causes a perception error as I wrote previously.
Thanks for your elaboration on this important subject.
Frank
On 07-Aug-07 21:37:50, Greg
30 is not 30% of 300 (it is 10%), so your prop.test below is testing
something different from your hand calculations. Try:
prop.test(c(.30,.23)*300,c(300,300), correct=FALSE)
2-sample test for equality of proportions without continuity
correction
data: c(0.3, 0.23) * 300 out
Does this do what you want?
x2 - rbinom( 200, 1, ifelse(x, .95, p1/.6) )
y2 - rbinom( 300, 1, ifelse(y, .8, p1/.6) )
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From:
You can call par('usr') to find the current coordinates, then use the
text function (the pos and offset arguments may be helpful as well)
The cnvrt.coords function from the TeachingDemos package may also be of
help for finding coordinates for the text function.
Hope this helps,
--
Gregory
Try:
data - expand.grid( a= seq(0,1,.1), b= seq(0,1,.1) )
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of esh
Generally adding the numbers to a graph accomplishes 2 things:
1) it acts as an admission that your graph is a failure
2) it converts the graph into a poorly laid out table (with a colorful
and distracting background)
In general it is better to find an appropriate graph that does convey
the
Why do you feel the need to add the numbers to your barplot?
Occasionally this can be useful (and others have answered the question
on how to do it), but often when you feel the need to add numbers to a
graph you are really doing 2 things:
1. admitting the graph has failed
2. replacing the graph
Be aware that the effects of calls to par usually only last for the
duration of the graphics device, not the R session. If you put a call
to par in your startup script, then it will open a graphics device and
set the option, but if you close that graphics device and do another
plot then a new
Look at the power.examp and run.power.examp functions in the
TeachingDemos package. Do these do what you want? If not you can look
at the code in them to see how to fill part of the area under the curve.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain
It's not clear exactly what you want, but you may want to look at the
layout function, or the subplot function in the TeachingDemos package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From:
?bxp
-Original Message-
From: Pietrzykowski, Matthew (GE, Research) [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 8/2/07 6:04 AM
Subject: [R] boxplot hinge customization
Hello R users wiser than I -
I am trying to produce a boxplot with quantiles defined
The persp function has a col argument that you can use to specify the colors.
-Original Message-
From: Juliane Willert [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 7/26/07 12:55 AM
Subject: [R] colored heights in 3D plot (persp)
Hello everybody,
I have a
show
outliers that may be of interest.
From: Dong GUO ?? [mailto:[EMAIL PROTECTED]
Sent: Sat 7/28/2007 10:19 PM
To: Greg Snow
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] plot
Thanks again, Greg, It really helps.
Would you please let me know more reference
Check out the biglm package for some tools that may be useful.
-Original Message-
From: Eric Doviak [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 7/30/07 9:54 AM
Subject: [R] the large dataset problem
Dear useRs,
I recently began a job at a very large and
Marc gave some good general advice, here are a couple more things that are more
specific to your problem.
Remember that most R functions return information, sometimes invisibly, but it
is good to save the results.
This includes the summary function (all the numbers that get printed out are
Graphs that rely on 3-d effects tend to distort the data rather than enlighten
the viewer. If your goal is to distort the data (which I doubt), then most of
us don't want to help. On the other hand, if you really do want to enlighten
the viewer (even if that is just you), then tell us what
Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: Dong GUO 郭东 [mailto:[EMAIL PROTECTED]
Sent: Friday, July 27, 2007 12:09 PM
To: Greg Snow
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] plot
Many thanks, Greg and Justin.
The matrix
You may want to look at the interaction function (a quick way to make the
single factor with 4 levels that you mention).
You can create your own sets of contrasts and set them using the C or contrasts
functions, then use the split argument to summary.aov to look at the individual
degrees of
You may want to look at the R2HTML package as one approach (others have
already told you about sink and cat).
Another approach is to use the variations on sweave. Here you set up a
template file with the code you want run as well as any explanitory text
(you can even write an entire report),
One of the nice things about the R Graph Gallery is that if you click on
the R logo underneath the graph (may need to scroll down a bit) it will
show you the code used to create that particular graph.
You may also want to look at the subplot function in the TeachingDemos
package for another way
Start with something like this:
pairs(iris, panel= function(x,y,...){
points(x,y);
abline(lm(y~x))})
Then substitute in your data and extra args and any other enhancements
you want.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain
Under windows you could use a graphics window for the stimulus and
capture the keypress using the getGraphicsEvent function (this does not
require hitting enter, but the graphics window needs to be the active
window through the experiment). The playSudoku function in the sudoku
package shows one
Look at the commandArgs function to see if that does what you want.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Aydemir, Zava
The rug function will take line and position arguments for the location
of the rug plot, so you can do something like:
rug( x1, line=0 )
rug( x2, line=1 )
rug( x3, line=2 )
Where line=0 means right on the x axis, line=1 means 1 textline height
below the x axis (-1 would be the same distance
There are many options that can help, but which to use depends a lot on
what your research question is. What are you really looking for in
these relationships?
Here are a couple of tools that may help.
If my.data is a matrix or data.frame with just the variables of interest
then
look at
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: Ken Williams [mailto:[EMAIL PROTECTED]
Sent: Thursday, July 12, 2007 3:50 PM
To: Greg Snow; R-help@stat.math.ethz.ch
Subject: Re: [R] Compute rank within factor groups
On 7/12/07 4:28 PM, Greg Snow
One approach would be to use a permutation test.
Lets start with a simpler case: x1 and y1 are 2 variables measured
under condition 1 and x2 and y2 are the same variables measured under
condition 2. The null hypothesis is that conditions 1 and 2 make no
differences on the measurements (same
Not a strict proof, but think of it this way:
The liklihood of getting a particular value of x has 2 parts. 1st x has
to be generated from h, the liklihood of this happening is h(x), 2nd the
point has to be accepted with conditional probability f(x)/(c*h(x)). If
we multiply we get h(x) * f(x)/
Use xx[[gene]] instead of xx$gene (the $ is a shorthand for [[ with some
extra magic to be more convenient, the magic is getting in your way, so
go back to the [[ syntax (make sure you double the braces)).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain
Try something like:
mytable - table(known, modelout)
prop.table( mytable, 1 )
Also look at ?addmargins and the CrossTable function in the gmodels
package.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
Look at ?ave and try something like:
wc$rank - ave( wc$score, wc$report, FUN=rank )
This works even if the dataframe is not pre sorted.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original
PROTECTED]
(801) 408-8111
-Original Message-
From: Ken Williams [mailto:[EMAIL PROTECTED]
Sent: Thursday, July 12, 2007 2:50 PM
To: Greg Snow; R-help@stat.math.ethz.ch
Subject: Re: [R] Compute rank within factor groups
On 7/12/07 3:42 PM, Ken Williams [EMAIL PROTECTED] wrote
The built in power functions are for the fairly straight forward
situations. Yours does not appear to fit into any of those. You need
to think through your problem a bit more before starting to think about
power.
What do you mean by effect size of 1.5 (is that 1.5 standard deviations?
Or raw
I nominate the following 2 pieces from Bill's reply for fortunes
(probably 2 separate fortunes):
All this becomes even more glaring if you take the unusal
step of plotting the data.
and
What sort of editor would overlook this clear and
demonstrable message leaping out from the data in
I don't know what is causing your problem, But if you goal is to
produce html then you may want to look at the R2HTML package. It may do
what you want without using sink.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801)
One approach is to use the fact that vectors are automatically
replicated to the correct length when subscripting, so you can do
something like:
my.matrix[ c(FALSE,TRUE,FALSE), 3 ]
To get every 3rd element starting at the 2nd element, and the 3rd
column.
Hope this helps,
--
Gregory (Greg)
Use rbind instead of c:
df - rbind(df1,df2)
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Aydemir, Zava
The R2HTML package can be used to create and html transcript of your
session (you do need to start, end, and tell it to include graphs). I
use this when doing an R demonstration (teaching), so that the students
can have a transcript that shows the commands I typed along with their
output
The first question you should ask yourself (and really think about the
answer) is Why do I want to do this?
Fancy colors and shadings can detract from a plot rather than enhance
it. The wrong choice of colors can make bars look bigger or smaller
than they really are. Reading Tufte's The visual
The sample function has a prob argument that determines the
probabilities of each element being sampled, put your proportion of
women in there and see if that works for you.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
Are you looking for pmax? (look at the help ?pmax and the examples and
see if that does what you want).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL
I don't know of a current package that does this (others may), but if
you know what you expect your data to look like you can simulate it and
calculate power that way.
Basically, write a function that will simulate data with the level of
measurement error that you expect in the real data (or have
Here is one approach:
tmp - scan()
1: 1 127 1 261 1 142 1 183 1 234 1 162 2 173 2 321 2 168 2
20: 197 2 213 2 261 3 198 3 126 3 167 3 154 3 134 3 187 3 109 3 210
41:
Read 40 items
my.df - as.data.frame( matrix(tmp, ncol=2, byrow=TRUE) )
names(my.df) - c('Category','Variable')
For a), if you really want the plots overlayed, it is best to use
something like matplot, or use the points or lines function to add the
later plots to the first. You can also get the overlay effect using
par(new=TRUE), but then you need to be careful with axis labels and
scales and the plot may
You can look at the power.examp function in the TeachingDemos package to
see if it gives you the graph you want, or look at the code to see how
you could modify it to give you the plot you want.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
There are 2 approaches that should work for you (actually there are probably
more, but these 2 are what I would suggest).
The first is to wrap barplot in your own function that also adjusts the
parameters and adds the axis. The other is to use the output from subplot to
go back and annotate
Try:
plot(.., bty='l')
Does that do what you want? (see the bty parameter in ?par for details)
If you don't want the lines extending beyond the axes on the right and
top then you could do something more like:
plot(5:10, 5:10, bty='n')
library(TeachingDemos)
lines(cnvrt.coords( c(0,0,.5),
?invisible
-Original Message-
From: Jason Q McClintic [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 6/19/07 2:04 PM
Subject: [R] A question about plots and lists in functions
R-helpers:
I tried googling and couldn't find anything.
I have a function I am
The triplot function in the TeachingDemos package (I don't know about
the one in klaR, or the others mentioned) honors the type='l' argument
and passes it on to points. So if you know where you want the contours
drawn, you can use triplot to draw the lines (it also has an add
argument that could
-Original Message-
From: Kuhn, Max [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 14, 2007 3:11 PM
To: Greg Snow; [EMAIL PROTECTED]; r-help@stat.math.ethz.ch
Subject: RE: [R] R vs. Splus in Pharma/Devices Industry
Greg,
Thanks for the kind words about odfWeave
There is the TkBrush function in the TeachingDemos package for R. It is not
web based.
-Original Message-
From: Roy Mendelssohn [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 6/12/07 10:26 AM
Subject: [R] [OT]Web-Based Data Brushing
I apologize for the
when the function is called.
Stone.
Sundar Dorai-Raj wrote:
Hi, Greg,
type = 'b' won't work according to ?locator. Try type = 'o'.
HTH,x
--sundar
Greg Snow said the following on 6/13/2007 7:27 AM:
Does
locator(type='l')
(or type ='b')
Work for you
But sweave is expanding. There is a driver for HTML sweaving in the
R2HTML package and the odfWeave package allows you to sweave with open
office docs (which can be converted to/from MS word). I personally like
using LaTeX and the original sweave, but I work with people who want
everything in MS
Try:
lm( formula( paste( Ytext, '~ Xvar' ) ), data=X)
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Pedro Mardones
Sent:
In my case, the MS word users know just enough about statistics to know
that they need a statistician (me or one of my group), so it is usually
me that sets up the template. This is generally for a set of
graphs/tables that will be included in a paper or presentation. They do
most of the initial
Does
locator(type='l')
(or type ='b')
Work for you?
-Original Message-
From: ryestone [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 6/8/07 10:19 AM
Subject: [R] ievent.wait
I am working on a plot and would be like to click on a few points and then
Look at the subplot function in the TeachingDemos package.
-Original Message-
From: Héctor Villalobos [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch r-help@stat.math.ethz.ch
Sent: 6/11/07 5:48 PM
Subject: [R] barplot and map overlay
Hi,
I wonder if it is possible with the graphics
Peter Lercher wrote:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Peter Lercher
Sent: Friday, June 08, 2007 3:07 AM
To: r-help@stat.math.ethz.ch
Subject: [R] overplots - fixing scientific vs normal notation
in output
Moving from S-plus to
Try this:
plot(runif(10), ylab=, xlab=Black, standard?)
mtext('Red, Bold', side=2, line=3, col='red', font=2)
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL
You need to do 1 of 2 things (but not both).
Either register your database file with your odbc driver (done outside
of R)
Or
Use odbcConnectAccess in place of odbcConnect
The 2nd is simpler if you just want to import that file, the first may
be better in the long run if you are going to be
Tim,
First, I personnally am a big fan of LaTeX, Emacs, and ESS and I think
that in the long run you would benefit from learning all of them
(probably start with Emacs, then ESS, then LaTeX once you already have a
knowledge of Emacs and how it can help).
Since you asked about the simplest way to
To decide on which way to do a one tailed test, you should really ask a
few questions (before looking at the data) either of yourself, your
client, or other expert in the field.
I would start by asking the 3 questions:
1. What will I/you do if A is less than B?
2. What will I/you do if A is
1. Yes you need a shape file with that information.
2. One place to look is:
http://www.census.gov/geo/www/cob/bdy_files.html There are shapefiles
(the .shp ones) for counties and zip codes (as well as many other
things). These work pretty directly with the tools in the maptools and
sp packages.
1 - 100 of 317 matches
Mail list logo
