Hi Lazarus,
It would be more simple with mdat as a matrix (before coercing to a
data.frame). It might be a simpler way to compare a matrix with a vector but I
don't find it for the moment; in any case, this works:
mdatT <- matrix(mdat %in% c(1, 11, 20), ncol=3)
> mdat[!apply(mdatT, 1, any), ]
On Wed, 4 Mar 2009, Vadlamani, Satish {FLNA} wrote:
Hi:
Sorry if this is a double post. I posted the same thing this morning and did
not see it.
I just started using R and am asking the following questions so that I can plan
for the future when I may have to analyze volume data.
1) What are
Rolf Turner wrote:
>
> Sports scores are random variables. You don't know a priori what the
> scores are
> going to be, do you? (Well, if you do, you must be able to make a
> *lot* of money
> betting on games!) After the game is over they aren't random any
> more; they're
> just numbers. But th
Dear R-help team,
I am getting addicted to using R but keep on getting many challenges on the way
especially on data management (data cleaning).
I have been wanting to drop all the rows if there values are `NA' or have
specific values like 1 or 2 or 3.
mdat <- matrix(1:21, nrow = 7, ncol=3,
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Kris
> Sent: Wednesday, March 04, 2009 8:28 PM
> To: r-help@r-project.org
> Subject: [R] new r user
>
> When adding a trend line to a scatterplot (e.g. abline
> (90,4,col="red")
When adding a trend line to a scatterplot (e.g. abline (90,4,col="red"), I
believe the "90" is the intercept and "4" is the slope. How do I determine
the intercept and slope for the abline command?
Kristopher R. Deininger
Management Strategy & Entrepreneurship
PhD Student 2012
Robert H. Smith
try
?abline
everything should be there for you
stephen sefick
On Wed, Mar 4, 2009 at 11:30 PM, kris deininger
wrote:
>
>
>
> When adding a trend line to a scatterplot (e.g. abline
> (90,4,col=”red”), I believe the “90” is the intercept and “4” is the
> slope. How do I determine the intercept a
On Wed, Mar 4, 2009 at 5:22 PM, oren cheyette wrote:
> Trying to use dates in their R-native form (e.g., POSIXct) rather than
> turning them into character strings, I've encountered the following problem.
> I create a data frame where one column is dates. Then I use "by()" to do a
> calculation on
Hi Kingsford - this is exactly what I am looking for...
Many thanks!!
Kingsford Jones wrote:
>
> I'm guessing you processed a data frame with the 'by' function.
> Rather than restructuring the by output, try using a different
> function on your data frame. For example
>
>> #install.packages(d
When adding a trend line to a scatterplot (e.g. abline
(90,4,col=red), I believe the 90 is the intercept and 4 is the
slope. How do I determine the intercept and slope for the abline command?
_
eet.
[[alternative
I'm guessing you processed a data frame with the 'by' function.
Rather than restructuring the by output, try using a different
function on your data frame. For example
> #install.packages(doBy)
> summaryBy(breaks ~ tension + wool, data=warpbreaks, FUN=sum)
tension wool breaks.sum
1 LA
On Wed, Mar 4, 2009 at 8:40 PM, Fuchs Ira wrote:
> So functions in the base package are all written in C?
No, a large proportion are written in R and the code can be seen in
the console by typing the function name. C is generally used where
speed is a concern.
Kingsford
> Thanks.
> On Mar 4,
So functions in the base package are all written in C?
Thanks.
On Mar 4, 2009, at 10:26 PM, Kingsford Jones wrote:
> see
>
> https://stat.ethz.ch/pipermail/r-help/2008-January/151694.html
>
> hth,
> Kingsford Jones
>
> On Wed, Mar 4, 2009 at 7:30 PM, Fuchs Ira wrote:
> > How can I print the def
Many thanks, Hadley! It was really helpful.
Cheers,
Swaroop
-Original Message-
From: hadley wickham [mailto:h.wick...@gmail.com]
Sent: Wednesday, March 04, 2009 9:56 AM
To: Vedula, Satyanarayana
Cc: r-help@r-project.org
Subject: Re: [R] Selecting one row or multiple rows per ID
On
see
https://stat.ethz.ch/pipermail/r-help/2008-January/151694.html
hth,
Kingsford Jones
On Wed, Mar 4, 2009 at 7:30 PM, Fuchs Ira wrote:
> How can I print the definition of a function that is in the base package?
>
> for example, if I type:
>
> which.min
>
> I get
>
> function (x)
> .Internal(w
Hi R users,
I have an R object with the following attributes:
> str(sales.bykey1)
'by' int [1:3, 1:2, 1:52] 268 79 118 359 87 147 453 130 81 483 ...
- attr(*, "dimnames")=List of 3
..$ GROUP: chr [1:3] "III" "II" "I"
..$ year : chr [1:2] "2006" "2007"
..$ week : chr [
Well, it seems like I may need to use a few different correlation coefficient
tests:
(1) For the Nominal scale to Interval Scale, I may need to be using the
Point-biserial correlation coefficients (rpb). It turns out that the ltm
Package calculates that correlation coefficient. Will be trying
How can I print the definition of a function that is in the base
package?
for example, if I type:
which.min
I get
function (x)
.Internal(which.min(x))
How can I see the definition of this function?
Thanks.
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h
On 5/03/2009, at 3:06 PM, David Winsemius wrote:
I mostly agree with you, Rolf (and Gunter). I would challenge your
joint use of the term "scientists". My quibble arises not regarding
biomedical practitioners (who may be irredeemable as a group) but
rather regarding physicists. At least in tha
Sueli Rodrigues esalq.usp.br> writes:
>
> Olá. Tenho um arquivo que a cada 6 linhas corresponde uma amostra da qual
> preciso dos coeficientes da regressão linear. Como faço para que o
> programa distinga a cada 6 linhas como uma amostra e não calcule como um
> todo?
> Estou usando a função: mode
I mostly agree with you, Rolf (and Gunter). I would challenge your
joint use of the term "scientists". My quibble arises not regarding
biomedical practitioners (who may be irredeemable as a group) but
rather regarding physicists. At least in that domain, I believe those
domain experts are
Olá. Tenho um arquivo que a cada 6 linhas corresponde uma amostra da qual
preciso dos coeficientes da regressão linear. Como faço para que o
programa distinga a cada 6 linhas como uma amostra e não calcule como um
todo?
Estou usando a função: model=lm(y ~ x)
Sueli Rodrigues
Eng. Agrônoma - UNES
Dear Uwe,
Thank you very much for your email. I think I have worked out that the problem
was related to the coordinates of the legend that are manually specified in the
leg_loc command. However, I'm not exactly sure what was wrong with exporting
the picture of the plot...
To avoid the proble
Hi,
is there a way to place text at the upper left corner (or another
corner) of the plot?
I want to place it really at the upper left corner of the whole plot
(the file I get),
not at the upper left corner of the plot-region.
I tried text() and mtext(), and corner.label() of the plotri
Dear Oren,
Try this:
> x <- data.frame(A= c("X", "Y", "X", "X", "Y", "Y", "Z" ), D =
+ as.POSIXct(c("2008-11-03","2008-11-03","2008-11-03", "2008-11-04",
+ "2009-01-13", "2009-01-13", "2009-01-13")), Z = 1:7)
>
> m<-with(x,tapply(Z, list(A,D), sum))
> m[rownames(m)==x$A[1],]
2008-11-03 2008-11-04
On 5/03/2009, at 12:13 PM, Bert Gunter wrote:
"The purpose of the subject or discipline ``statistics'' is in essence
to answer the question ``could the phenomenon we observed have arisen
simply by chance?'', or to quantify the *uncertainty* in any estimate
that we make of a quantity."
May I
Trying to use dates in their R-native form (e.g., POSIXct) rather than
turning them into character strings, I've encountered the following problem.
I create a data frame where one column is dates. Then I use "by()" to do a
calculation on grouped subsets of the data. When I try to extract values
fro
"The purpose of the subject or discipline ``statistics'' is in essence
to answer the question ``could the phenomenon we observed have arisen
simply by chance?'', or to quantify the *uncertainty* in any estimate
that we make of a quantity."
May I take strong issue with this characterization? It i
Hi:
Sorry if this is a double post. I posted the same thing this morning and did
not see it.
I just started using R and am asking the following questions so that I can plan
for the future when I may have to analyze volume data.
1) What are the limitations of R when it comes to handling large da
Hi:
This is a test mail. Thanks.
Satish
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reprodu
Well, you're the one who offered code without designating what
libraries were loaded or required. Here's my sessionInfo, ... what's
yours?
> sessionInfo()
R version 2.8.1 Patched (2009-01-07 r47515)
i386-apple-darwin9.6.0
locale:
en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached
Hi list,
I dont know why the top of map gets truncated: I appreciate if any one give
me a solution.
filename="C:\\temp\\test.pdf"
pdf(file=filename, width=15, height=10)
library(maps)
require("mapproj")
longlatLimit<-c(-106.65, -93.53 , 25.93 , 36.49)
par(plt=c(0,1,0,1),cex=1,cex.main=1) #Set
When I run this code i get the following error messages
Error in mapgetg(database, gon, as.polygon, xlim, ylim) :
NA/NaN/Inf in foreign function call (arg 6)
In addition: Warning messages:
1: In min(x, na.rm = na.rm) :
no non-missing arguments to min; returning Inf
2: In max(x, na.rm = na.rm)
The example on the help page would seem to be completely on point if I
understand your desire to be plotting text at particular long,lat
coordinates:
?map
text(long, lat, "text")
#
data(ozone)
map("state", xlim = range(ozone$x), ylim = range(ozone$y))
text(ozone$x, ozone$y, ozone$median)
bo
I am trying to overlay a data frame with lat and longitude(which refer to
zip codes) on the map of US that I get by using map ("states").
Is there anyway to do this or do I have to resort to using maptools?
thank you
[[alternative HTML version deleted]]
__
You can also change the column names to something else en mass:
colnames(dat) <- paste("X",1:100,sep="")
I next tried constructing the X names inside data.frame, but failed
using the paste function. The help page for data.frame has a paragraph
that begins "How the names of the data frame are
Type unique(density)
How many different unique values does density take? The wiggliness of the
smooth term consumes degrees of freedom. That is, the more wiggly your
smooth term, the more DFs it consumes. If you get the error you got, you
have to reduce the degrees of freedom alloted to the smooth
Could that be extended to generate a population data set with known skew and
kurtosis?
If so, how?
Thanks in advance for suggested solutions.
Harold
Daniel Nordlund-2 wrote:
>
>
>
> Something like this may help get you started.
>
> std.pop <- function(x,mu,stdev){
> ((x-mean(x))/sd(x
Have you considered using a 'list'? much easier to manage than a lot
of individual objects.
mylist <- lapply(1:100, runif)
On Wed, Mar 4, 2009 at 4:34 PM, Manli Yan wrote:
> Hi:
> I need to create many variables at one time,how to do this in R?
> for eg ,X1,X2...X100?
>
> Thanks~
>
>
Does cex.axis not work in that it reduces the size for both x and y
axes? If that's the case try calling plot with axes=FALSE, and then
add axes seperately with the axis function.
Kingsford Jones
On Wed, Mar 4, 2009 at 1:59 PM, Tiffany Vidal
wrote:
> I am trying to reduce the font size for y-ax
Why use text()?
There is a function called "mtext" for that task.
hth.
batho...@googlemail.com schrieb:
Hi,
is there a way to place text out of the plot region with text() ?
thanks!
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Dear Manli,
Do you mean the names of the variables? If so, something like this should
work:
paste('X',1:100,sep="")
HTH,
Jorge
On Wed, Mar 4, 2009 at 4:34 PM, Manli Yan wrote:
> Hi:
> I need to create many variables at one time,how to do this in R?
> for eg ,X1,X2...X100?
>
> Thank
Hi Manli,
you may consider structuring your data in some appropriate form like
data.frame or list. Its often not the best way holding information
separated in many variables.
But if you *really* want to create 100 separate variables, something like
for (i in 1:100) assign(paste("X",i,sep=""),
Hi,
is there a way to place text out of the plot region with text() ?
thanks!
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and provide
On Wed, Mar 4, 2009 at 2:34 PM, Manli Yan wrote:
> Hi:
> I need to create many variables at one time,how to do this in R?
> for eg ,X1,X2...X100?
It depends what you want. If you want 100 random normal variables of
length 10, stored in a data.frame with names V1, V2, ..., V100 try
dat <-
The help page for ?"for" says that:
The index seq in a for loop is evaluated at the start of the loop;
changing
it subsequently does not affect the loop. The variable var has the
same type
as seq, and is read-only: assigning to it does not alter seq.
The help file is not right when se
Thanks,
Here is my partial solution, from what you suggested me:
library(TeachingDemos)
z<-colors()
zz<-col2grey(z)
#index sorted
zzz<-sort(zz,index.return = TRUE)$ix
x<-z # colors in order or their greyscale
y<-z # greyscale sorted in gradient
for (i in 1:length(z)){
x[i]<-z[zzz[i]]
y[
I am trying to reduce the font size for y-axis labels, not ylab, but the
actual categorical names. I have tried cex, cex.axis, cex.lab, font,
but none seem to do the trick. Any ideas? thank you.
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Hi:
I need to create many variables at one time,how to do this in R?
for eg ,X1,X2...X100?
Thanks~
[[alternative HTML version deleted]]
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PLEASE do read
I'll reply to my own post - to make sure no one wastes his/her time on that.
I was able to solve the problem only after I modified the original
function interaction.plot (see below). All I did I added one line
before the final } - asking it to return the means on the numeric
(dependent) variable.
A
Hi Ken,
The help page for ?"for" says that:
The index seq in a for loop is evaluated at the start of the loop; changing
it subsequently does not affect the loop. The variable var has the same type
as seq, and is read-only: assigning to it does not alter seq.
So you cannot do what you want to do
It's in the R-FAQ. I can't remember it's 7.20 or 7.35 but it's in that
general area.
--
David Winsemius
On Mar 4, 2009, at 3:38 PM, Bob Roberts wrote:
Hi,
I formed a 49 by 3 data frame by reading in a text file using
read.table(), and combining it with a matrix that I made by using
un
Hi all,
I need a little help with flow control in R. What I'd like to do is to
advance a for loop by changing its counter. However, what seems obvious
to me does not yield the proper results. An example of my problem is
for (i in seq(1, some_number, some_increment)){
if (some
Thank you, David, however, I am not sure this approach works.
Let's try it again - I slightly modifed d to make it more clear:
d=data.frame(xx=c(1,1,1,1,2,2,2,2,3,3,3,3),yy=c(3,3,4,4,3,3,4,4,3,3,4,4),zz=c(-1.1,-1.3,0,0.6,-0.5,1,3.3,-1.3,4.4,3.5,5.1,3.5))
d[[1]]<-as.factor(d[[1]])
d[[2]]<-as.factor
## you may need to
## install.packages("HH")
library(HH)
tmp <- data.frame(y=rnorm(500),
g=rep.int(c("A", "B", "C", "D"), 125),
a=factor(rbinom(500, 1, .5)))
bwplot(y ~ g | a, data=tmp)
bwplot(y ~ a | g, data=tmp)
tmp$ga <- with(tmp, interaction(a, g))
posi
Hi,
I formed a 49 by 3 data frame by reading in a text file using read.table(),
and combining it with a matrix that I made by using unlist() on a list of
character strings. I would like to do some simple arithmetic operations on the
elements in the data frame columns (e.g. column 3/column2) b
Hi, every body!
I am a new comer for R, so my question would unavoidablely sounds stupid. Sorry!
in my experiment, there are two type of soil ( soil F and soil D), each half of
them were subjected to steam sterilize (result in FS and DS soil). A equal
volume of soil from two of the four soil typ
On 4/03/2009, at 10:06 PM, Marc Vinyes wrote:
Dear Carlos and Kjetil,
Thanks for your answer.
I do not think that is the way to go. If you believe that your
algorithm
is better than the existing one, talk to the author of the package
and
discuss the improvement. The whole community will
On 5/03/2009, at 4:54 AM, Michael A. Miller wrote:
"Rolf" == Rolf Turner writes:
On 4/03/2009, at 11:50 AM, Michael A. Miller wrote:
Sports scores are not statistics, they are measurements
(counts) of the number of times each team scores. There
is no sampling and vanishingly small possi
This does not fully answer your question, but there is a function col2grey (or
col2gray) in the TeachingDemos package that will help you see what a given
color plot will look like (approximately) when printed/photocopied in grayscale.
For your example you would do something like:
> plot(1:6,col
Q1:
See if this seems any better. I took the liberty of reconstruction
your initial example in a longer dataframe:
dta <- data.frame(val = sample(t,1000), g = gl(4, 250, labels=c("A",
"B", "C", "D")) , G2 = gl(2,1, labels=c("XX", "YY")))
#arguments to data.frame are recycled so one does no
William Dunlap wrote:
> I entered the same x into Excel 2003 and used the formulae
> =percentile(A1:10,0),
> =percentile(A1:A10,.125), ..., =percentile(A1:A10,1) and got the results
>1, 1.125, 2.25, 3, 4, 6.875, 8, 8.875, 10
> This matches only R's type 7, the default.
>
hurray! in this
On Wed, Mar 4, 2009 at 11:58 AM, Christian Pilger
wrote:
>
> Dear R-experts,
>
> recently, I started to discover the world of R. I came across a problem,
> that I was unable to solve by myself (including searches in R-help, etc.)
>
> I have a flat table similar to
>
> key1 key2 value1
>
> ab
Hello,
First, sorry for sending HTML emails earlier. This should now be in
plain-text mode.
I have two time series (ts) objects, 1 is yearly (population) and the
other is quarterly (bankruptcy statistics). I would like to produce a
quarterly time series object that consists of bankruptcy/populati
On Wed, Mar 4, 2009 at 11:58 AM, Christian Pilger
wrote:
>
> Dear R-experts,
>
> recently, I started to discover the world of R. I came across a problem,
> that I was unable to solve by myself (including searches in R-help, etc.)
>
> I have a flat table similar to
>
> key1 key2 value1
>
> ab
See if this helps. After your code, submit this to R:
with(d, text(xx[xx==3],zz[xx==3],paste("3, ",zz[xx==3])))
After that has convinced you that xx and zz are being used properly,
you can try the more general approach:
with(d, text(xx,zz,paste(xx, " , ", zz)))
I would have used ZZ rather
Hi Anuj,
Take a look at ?nls. It might be useful in this case.
HTH,
Jorge
On Wed, Mar 4, 2009 at 1:22 PM, anujgoel wrote:
>
> Dear R Community,
> I am plotting this simple x-y plot (raw data & plot attached).
> I cant fit a linear regression line to it. I have to figure out what is the
> best
Excel 2003's help for percentile just says it interpolates
between the quantiles in the data:
Array is the array or range of data that defines relative standing.
K is the percentile value in the range 0..1, inclusive.
If array is empty or contains more than 8,191 data points,
Pls forgive me heavy-handed data generation -- newby ;-)
### start ###
# example data
g <- rep.int(c("A", "B", "C", "D"), 125)
t <- rnorm(5000)
a <- sample(t, 500, replace=TRUE)
b <- sample(t, 500, replace=TRUE)
# what I actually want to have:
boxplot(a | b ~ g)
# but that does obviously not pr
On Mar 4, 2009, at 1:22 PM, anujgoel wrote:
Dear R Community,
I am plotting this simple x-y plot (raw data & plot attached).
I cant fit a linear regression line to it. I have to figure out what
is the
best fit for this graph.
That is virtually impossible to define rigorously. The "best fi
Hello!
I asked in this forum about what kind of seasonality the function arima() from
stats implements. Now that I have been answered that it implements the
Box-Jenkins multiplicative seasonality, I would like to ask whether there is in
R possibility to model ARIMA with additive seasonality. I m
Hello,
I have two time series objects, 1 is yearly (population) and the other is
quarterly (bankruptcy statistics). I would like to produce a quarterly time
series object that consists of bankruptcy/population. Is there a pre-built
function to intelligently divide these time series:
br.ts = ts(da
anujgoel wrote:
Dear R Community,
I am plotting this simple x-y plot (raw data & plot attached).
I cant fit a linear regression line to it. I have to figure out what is the
best fit for this graph. Is there a way to tell which regression to use for
this kind of data?
Also, after selecting the
On Wed, 4 Mar 2009, thibert wrote:
Hi,
I am looking for a colormap (in color) that look like a gradient in gray
scale. It is to allow people without color printer to print the color graph
and have something meaningful in gray scale.
It can be something like this
plot(1:6,col=c(1,7,5,3,2,4),p
At 12:19 AM 3/4/2009, you wrote:
plot(x,rho,pch=id)
Or this.
Tim
> dat
id x rho
1 A 1 0.1
2 B 20 0.5
3 C 2 0.9
> labels<-dat$id
> labels
[1] "A" "B" "C"
> plot(dat$x,dat$rho,pch=labels)
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See ?reshape
Uwe Ligges
Christian Pilger wrote:
Dear R-experts,
recently, I started to discover the world of R. I came across a problem,
that I was unable to solve by myself (including searches in R-help, etc.)
I have a flat table similar to
key1key2value1
abcd_1 BP 10
abcd_1
On Wed, Mar 4, 2009 at 10:52 AM, Dimitri Liakhovitski wrote:
> Hello - and sorry for what might look like a simple graphics question.
>
> I am building an interaction plot for d:
>
> d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.4,3.5,3.3,-1.1,-1.3))
> d[[1]]<-as.factor(d[[1]])
> d[[
Hi,
I am looking for a colormap (in color) that look like a gradient in gray
scale. It is to allow people without color printer to print the color graph
and have something meaningful in gray scale.
It can be something like this
plot(1:6,col=c(1,7,5,3,2,4),pch=c(1,20,20,20,20,20))
but with an
Dear R-experts,
recently, I started to discover the world of R. I came across a problem,
that I was unable to solve by myself (including searches in R-help, etc.)
I have a flat table similar to
key1key2value1
abcd_1 BP 10
abcd_1 BSMP1A
abcd_1 PD 25
abcd_2 BP 20
a
Dear R Community,
I am plotting this simple x-y plot (raw data & plot attached).
I cant fit a linear regression line to it. I have to figure out what is the
best fit for this graph. Is there a way to tell which regression to use for
this kind of data?
Also, after selecting the best fit model, I n
Hi Jim,
How are you? I saw your posting. I am trying to do clustering for co
authorship.What I have is undirected graph .I want to have clusters for 393
nodes.
I am attaching the file along with this mail.If you move to the section
Cluster I am looking to do something like that.Is it somethin
Hi
I'm trying to do a GAM analysis and have the following codes entered
into R (density is = sample number, alive are the successes)
density<-as.real(density)
y<-cbind(alive,density-alive)
library(mgcv)
m1<-gam(y~s(density),binomial)
at which point I get the following error message
Error in
For the record
residuals(model)
1 2 3 4 5
5.55860143 -0.00073852 2.49255235 -1.41987341 -0.00042425
6 7 8
-0.94389158 2.72987046 -1.15760836
residuals(model, "pearson")
1 2 3
Greg Snow wrote:
> Here is another approach that still uses strspit if you want to stay with
> that:
>
>
>> tmp <- '(-0.791,-0.263].(-38,-1.24].(0.96,2.43]'
>> strsplit(tmp, '\\.(?=\\()', perl=TRUE)
>>
> [[1]]
> [1] "(-0.791,-0.263]" "(-38,-1.24]" "(0.96,2.43]"
>
> This uses the Per
Hello,
I have two time series objects, 1 is yearly (population) and the other is
quarterly (bankruptcy statistics). I would like to produce a quarterly time
series object that consists of bankruptcy/population. Is there a pre-built
function to intelligently divide these time series.
The series I
:-) works!
Sundar Dorai-Raj schrieb:
(Sorry for the repeat. Forgot to copy R-help)
Try,
test = data.frame(expand.grid(c(1:10), c(1:10)))
z = test[,1] + test[,2]
test = cbind(test, z)
names(test) = c("x", "y", "z")
require(lattice)
wireframe(z ~ x*y, data = test,
par.settings = list(axis.line
Here is another approach that still uses strspit if you want to stay with that:
> tmp <- '(-0.791,-0.263].(-38,-1.24].(0.96,2.43]'
> strsplit(tmp, '\\.(?=\\()', perl=TRUE)
[[1]]
[1] "(-0.791,-0.263]" "(-38,-1.24]" "(0.96,2.43]"
This uses the Perl 'look-ahead' indicator to say only match on
On 04-Mar-09 16:56:14, Wacek Kusnierczyk wrote:
> (Ted Harding) wrote:
>
>> So, with reference to your original question
>> "Excel has percentile() function. R function quantile() does the
>> same thing. Is there any significant difference btw percentile
>> and quantile?"
>> the answer is tha
Thank you for finding this. Yes in some cases the parameter settings need to
be updated by a call to plot.new for the calculations to be correct (if you
carried out your example 2 more times you would see that the 3rd plot is also
incorrect since it is still using the dimensions of the 2nd plot
(Sorry for the repeat. Forgot to copy R-help)
Try,
test = data.frame(expand.grid(c(1:10), c(1:10)))
z = test[,1] + test[,2]
test = cbind(test, z)
names(test) = c("x", "y", "z")
require(lattice)
wireframe(z ~ x*y, data = test,
par.settings = list(axis.line = list(col = "transparent")),
par.box =
Hi All,
This is a slightly arcane question, but I'm wondering if anyone else
uses vi mode with R? On my platform, across several versions, there is
some broken behaviour. When executing commands like 'df)' (to delete
up to the next bracket) the cursor moves to the next ), but nothing is
d
(Ted Harding) wrote:
> So, with reference to your original question
> "Excel has percentile() function. R function quantile() does the
> same thing. Is there any significant difference btw percentile
> and quantile?"
> the answer is that they in effect give the same results, though
> differ
Hello - and sorry for what might look like a simple graphics question.
I am building an interaction plot for d:
d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.4,3.5,3.3,-1.1,-1.3))
d[[1]]<-as.factor(d[[1]])
d[[2]]<-as.factor(d[[2]])
print(d)
interaction.plot(d$xx, d$yy, d$zz,
type
Hi guys I have been using R for a few months now and have come across an
error that I have been trying to fix for a week or so now.I am trying to
build a classifer that will classify the wine dataset using Naive Bayes.
My code is as follows
library (e1071)
wine<- read.csv("C:\\Rproject\\Wine\\
On 04-Mar-09 16:10:29, megh wrote:
> Yes, I aware of those definitions. However I wanted to know the
> difference btw the words "Percentile" and "quantile", if any.
> Secondly your link navigates to some non-english site, which I could
> not understand.
"Percentile" and "quantile" are in effect t
#Hi,
#
#somebody knows how to remove the outer box around a wireframe and
reduce the height
#
#
test = data.frame(expand.grid(c(1:10), c(1:10)))
z = test[,1] + test[,2]
test = cbind(test, z)
names(test) = c("x", "y", "z")
require(lattice)
wireframe(z ~ x*y, data = test, par.box = c(col = "tran
Yes, I aware of those definitions. However I wanted to know the difference
btw the words "Percentile" and "quantile", if any. Secondly your link
navigates to some non-english site, which I could not understand.
Dieter Menne wrote:
>
> megh yahoo.com> writes:
>
>
>
>> To calculate Percenti
> -Original Message-
> From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Sent: Wednesday, March 04, 2009 3:17 AM
> To: Daniel Nordlund
> Cc: r-help@r-project.org
> Subject: Re: [R] How to generate fake population (ie. not
> sample) data?
>
> On Wed, Mar 4, 2009 at 2:48 AM, Dani
Dear Ajay,
just to deny the implicit statement 'corporate user'='moron' surfacing
here and there in this interesting thread :^). This might be a
statistical regularity but should by no means be considered a theorem,
as there are counter-examples available. You can find people willing to
learn both
> "Rolf" == Rolf Turner writes:
> On 4/03/2009, at 11:50 AM, Michael A. Miller wrote:
>> Sports scores are not statistics, they are measurements
>> (counts) of the number of times each team scores. There
>> is no sampling and vanishingly small possibility of
>> systemati
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