Well, seems it's an assignment, so you should REALLY get them on your
own and not enquire the list. Foolish me...
M.
> -Original Message-
> From: einohr2...@web.de [mailto:einohr2...@web.de]
> Sent: Mittwoch, 20. Januar 2010 19:32
> To: Meyners,Michael,LAUSANNE,AppliedMathematics; r-help@
Peter Rote wrote:
>
>
> but i still have a problem to write to file. The problem is the slash in
> file names (Aerospace/Defense Products & Services ). If i want it to
> C:/ab/
> so "C:/ab/AdvertisingAgencies.txt" is ok but
> "C:/ab/Aerospace/Defense-MajorDiversified.txt" is not
>
>
As Rol
On Thu, Jan 21, 2010 at 8:47 AM, Roslina Zakaria wrote:
> Hi r-users,
>
> I try to draw histogram (in terms of probabilty) and superimpose with the
> gamma pdf. Using this code below I can get the plots BUT the y scale for the
> density is out of scale. How do I change the y-axis scale to max 1?
Hi r-users,
I try to draw histogram (in terms of probabilty) and superimpose with the gamma
pdf. Using this code below I can get the plots BUT the y scale for the density
is out of scale. How do I change the y-axis scale to max 1? Another thing,
how I do draw smooth line rather that points?
Dear R-helpers,
I have a question about giving index.
Suppose that I have a vector say, id=1:5, and each of them have some given
index but some of them may share the same index
Now, I have another vector, and I want to give it a index based what just
defined.
# 5 subjects
test1=1:5
# corresponding
Hi Jim & Gabor -
Apparently, it was most likely a hardware issue (shortly after
sending my last e-mail, the computer promptly died). After buying a
new system and restoring, the script runs fine. Thanks for your help!
On Tue, Jan 19, 2010 at 2:02 PM, jim holtman - jholt...@gmail.com
<+nabble+m
Dear R-helpers,
I have a question about giving index.
Suppose that I have a vector say, id=1:5, and each of them have some given
index but some of them may share the same index
Now, I have another vector, and I want to give it a index based what just
defined.
# 5 subjects
test1=1:5
# corresponding
Dear R-helpers,
I have a question about giving index.
Suppose that I have a vector say, id=1:5, and each of them have some given
index but some of them may share the same index
Now, I have another vector, and I want to give it a index based what just
defined.
# 5 subjects
test1=1:5
# corresponding
On Jan 20, 2010, at 11:10 PM, Erin Hodgess wrote:
Dear R People:
I'm sure that this is a very silly question, but I'm reading the
Software for Data analysis book by John Chambers, and I can't find the
"Declination" data that he refers to on page 15 in the book.
I did all of the usual stuff:
datatry2=transform(datatry,DIS=as.numeric(paste(datatry[,1],datatry[,2],sep="")))
It works like this !!
You understood perfectly well what I needed ;-)
Thanks to all that answered
Magali
--- En date de : Jeu 21.1.10, milton ruser a écrit :
De: milton ruser
Objet: Re: [R] concatenate 2 column
On Jan 20, 2010, at 11:14 PM, teurlai magali wrote:
Sorry I forgot the subject in my previous post
Here is an example of data : province and district code of 4 locations
datatry=matrix(c(8,12,3,3,1,1,16,1),4,2)
colnames(datatry)<-c("PROCODE","DISCODE")
> cbind( datatry, paste(datatry[, "PR
Hi Magali,
I confess that I still not figured out what you want,
but try this:
datatry2=transform(datatry,DIS=as.numeric(as.character(paste(datatry[,1],datatry[,2],sep=""
bests
milton
On Wed, Jan 20, 2010 at 11:14 PM, teurlai magali wrote:
> Sorry I forgot the subject in my previous post
On Jan 20, 2010, at 11:10 PM, Tena Sakai wrote:
Hi,
When I emulate what's suggested I get an error (shown below).
What does this mean?
Thank you.
Province <- c(1, 1, 13)
District <- c(1, 2, 2)
df$Distric_Unique <- with(df, paste(Province, District, sep=".") )
Error in eval(substitute(expr)
Sorry I forgot the subject in my previous post
Here is an example of data : province and district code of 4 locations
datatry=matrix(c(8,12,3,3,1,1,16,1),4,2)
colnames(datatry)<-c("PROCODE","DISCODE")
DISCODE is not unique for each location (there is a distrcit nb 1 in each
province)
I want to
Dear R People:
I'm sure that this is a very silly question, but I'm reading the
Software for Data analysis book by John Chambers, and I can't find the
"Declination" data that he refers to on page 15 in the book.
I did all of the usual stuff:
library(help=SoDA)
??Declination
looked for it onlin
Hi,
When I emulate what's suggested I get an error (shown below).
What does this mean?
Thank you.
> Province <- c(1, 1, 13)
> District <- c(1, 2, 2)
>
> df$Distric_Unique <- with(df, paste(Province, District, sep=".") )
Error in eval(substitute(expr), data, enclos = parent.frame()) :
invalid '
I'm not quite sure why, but reading in the *sorted* data (imported into
Excel, sorted, written to a text file) worked perfectly fine with
read.delim().
Thanks to those that replied!
-Robin
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btw, the re-order and swap make it much clearer, so thanks again!
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Oh, yes, my bad. Merge works! I guess the reordering automatically made by
Merge made me confused.
I wrote my code using Merge several months ago, and just didn't feel it
right today.
Thanks anyway!
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Hi,
don't worry about those setGenericS3()/setMethodS3(0 warnings - they
are just informative.
I am aware of that last warning:
Warning: package 'R.oo' claims to be built under R version 2.10.0 but
is missing some help files and needs to be re-installed
but it does not go away when you reinstal
On Jan 20, 2010, at 9:34 PM, teurlai magali wrote:
Hi
is there a way in R to create a new column vector with the numbers
of 2 others columns simply appended?
Example : I have a column with provinces codes (1 to 19)
I have a second column with districts codes (1 to ##, depending on
the pr
transform(DF, District_Unique = paste(Province, District, collapse = ''))
On Thu, Jan 21, 2010 at 12:34 AM, teurlai magali wrote:
> Hi
>
> is there a way in R to create a new column vector with the numbers of 2
> others columns simply appended?
>
> Example : I have a column with provinces codes
Hi
is there a way in R to create a new column vector with the numbers of 2 others
columns simply appended?
Example : I have a column with provinces codes (1 to 19)
I have a second column with districts codes (1 to ##, depending on the province)
I want to create a third column with a code that wo
A name such as ``Aerospace/Defense etc.'' is certainly not a legal
file name under unix-alike systems, and I suspect it would not
be even under Windoze. Even if it is, you shouldn't use it!
Change the name to ``Aerospace-Defense Products & Services''
or something like that, for goodness sake.
I have an very long, irregularly spaced time series (and I'm also new to
spectral analysis, so please be patient.) I want to use spec.ls as an
interpolator and then use the output to reconstruct the time series via
inverse fft. But so far I've been having difficulty doing this.
ts<-read.csv("t
by the way how do i change the output
"1016" "Advertising Agencies" "CMM"
"1803" "Advertising Agencies" "FMCN"
"2427" "Advertising Agencies" "IPG"
"3093" "Advertising Agencies" "MWW"
"3372" "Advertising Agencies" "OMC"
"4809" "Advertising Agencies" "VCLK"
"4832" "Advertising Agencies" "VISN"
"500
I have been through the help file archives a number of times, and still
cannot figure out what is wrong.
I have a tab-delimited text file. 76Mb, so while it's large.. it's not
-that- large. I'm running Win7 x64 w/4G RAM and R 2.10.1
When I open this data in Excel, i have 27 rows and 450932 rows, e
Thank you Dieter,
but i still have a problem to write to file. The problem is the slash in
file names (Aerospace/Defense Products & Services ). If i want it to C:/ab/
so "C:/ab/AdvertisingAgencies.txt" is ok but
"C:/ab/Aerospace/Defense-MajorDiversified.txt" is not
> head(AlexETF)
Jason Rupert wrote:
David,
Thank you very much for your response.
Is there any chance you can provide your example where mtext is used?
I also tried to use mtext, but I did not have any luck, so any help with the example below working with mtext or otherwise is greatly appreciated.
Thanks
Tena koe Jonathan
There are lots of ways - using base graphics or lattice or ggplot. As a
start, using base graphics:
plot(plothelpX, plothelpA2, col=factor(plothelpB))
points(plothelpX, plothelpA1, col=factor(plothelpB))
You'll also need the pch argument to plot and, presumably, xlab, ylab
and
On Wed, Jan 20, 2010 at 5:30 PM, Steve Lianoglou
wrote:
> Hi,
>
> On Wed, Jan 20, 2010 at 6:05 PM, Peng Yu wrote:
>> On Wed, Jan 20, 2010 at 5:04 PM, Peng Yu wrote:
>>> I don't find a tutorial on S3. "Bengtsson.pdf" cites MASS (1999
>>> edition). However, I don't think that MASS (2002 edition) c
One follow up question - the proposed solution was (notice - this time
I am introducing one NA in data frame "x")
x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,NA,10,20,100,200))
x$std.via.ave<-ave(x$values, x$factor, FUN=function(x)x/mean(x))
I compared the result to my own clumsy s
David,
Thank you very much for your response.
Is there any chance you can provide your example where mtext is used?
I also tried to use mtext, but I did not have any luck, so any help with the
example below working with mtext or otherwise is greatly appreciated.
Thanks again,
Jason
Does *cat(v[0])* produce what you need?
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Specifically, your setup of the data makes all the columns factors, so to
make Peter's code work you need to make the relevant columns numeric first.
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It may also be that path names with spaces in them are not supported.
Try it using a path that has no spaces in it.
On Wed, Jan 20, 2010 at 6:43 PM, Gabor Grothendieck
wrote:
> On Wed, Jan 20, 2010 at 6:01 PM, Jerry Floren
> wrote:
>> Next, I tried Gabor's code, which is so much more compact:
>
I'm trying to figure out the best way to do a scatterplot using the
following data as an example:
plothelp_x=1:10
plothelp_A1=sin(plothelp_x)
plothelp_A2=tan(plothelp_x)
plothelp_B=c("A","B","C","A","B","C","A","B","C","A")
(note that all 4 of these vectors have 10 entries each)
What I would l
Why not just use matrix(runif(9),nrow=3) ?
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On Wed, Jan 20, 2010 at 6:01 PM, Jerry Floren wrote:
> Next, I tried Gabor's code, which is so much more compact:
>
> ### Start of Gabor's code ###
> library(gdata)
> DF <- read.xls(U, pattern = "SAMPLE", as.is = TRUE)
> ## End of Gabor's code ##
>
> Is "U" the directory and path? Is "SAMPLE" the
On Jan 20, 2010, at 5:37 PM, hongwei wrote:
I have spent a whole afternoon searching for the solution, but got
nothing so
far. It should be a simple problem.
I have two datasets to merge. The first one just contains two ID
columns and
it looks like:
FromID ToID
1 2
1
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of hongwei
> Sent: Wednesday, January 20, 2010 2:38 PM
> To: r-help@r-project.org
> Subject: [R] Question about many-to-one merge
>
>
> I have spent a whole afternoon searching for
Thanks a lot for your helpful suggestion!
Dimitri
On Wed, Jan 20, 2010 at 5:56 PM, Chuck Cleland wrote:
> On 1/20/2010 5:37 PM, Dimitri Liakhovitski wrote:
>> Hello!
>>
>> I have a data frame with a factor and a numeric variable:
>>
>> x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,
Hi,
On Wed, Jan 20, 2010 at 6:05 PM, Peng Yu wrote:
> On Wed, Jan 20, 2010 at 5:04 PM, Peng Yu wrote:
>> I don't find a tutorial on S3. "Bengtsson.pdf" cites MASS (1999
>> edition). However, I don't think that MASS (2002 edition) clearly
>> explain what S3 is and help a user who knew very little
I have spent a whole afternoon searching for the solution, but got nothing so
far. It should be a simple problem.
I have two datasets to merge. The first one just contains two ID columns and
it looks like:
FromID ToID
1 2
1 3
2 1
2 3
3 1
3
Wow, this is as much hand-holding as one can expect, and for a R newbie it is
unbelievable and invaluable! Thanks again for all this help Marc!
Here is the screen print of my entire R session:
> .Machine$sizeof.pointer
[1] 8
>
> library(RODBC)
> conn <- odbcConnect(dsn='qdblocal', uid = "ciqxf
Hello, I am using te blpGetData() function to retrieve closing prices from
bloomberg on r. This is the code that I wrote:
library(RBloomberg)
conn=blpConnect
blpGetData(conn,"ANF UN Equity","PX_LAST","2009/09/01","2009/09/10")
and I get the following error:
Error in substring(paste("0", v$day, se
On Wed, Jan 20, 2010 at 5:04 PM, Peng Yu wrote:
> I don't find a tutorial on S3. "Bengtsson.pdf" cites MASS (1999
> edition). However, I don't think that MASS (2002 edition) clearly
> explain what S3 is and help a user who knew very little about S3 to
> quickly understand it. Could somebody let me
I don't find a tutorial on S3. "Bengtsson.pdf" cites MASS (1999
edition). However, I don't think that MASS (2002 edition) clearly
explain what S3 is and help a user who knew very little about S3 to
quickly understand it. Could somebody let me know if there are some
better learning materials to help
Sorry Gabor, but I am not quite there yet. Thanks David for your suggestion
on xlsReadWrite. Your message was posted while I was composing this one.
The Excel worksheet I want to read in is named "Paste Special"
I started with this code and thought it worked:
### Start of Code ###
library(RODB
On 1/20/2010 5:37 PM, Dimitri Liakhovitski wrote:
> Hello!
>
> I have a data frame with a factor and a numeric variable:
>
> x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200))
>
> For each level of "factor" - I would like to divide each value of
> "values" by the mean o
While I agree that this is someone's homework, it occur to me that we
can mess him up a little more.
He asked for 200 points *uniformly* distributed on (or in) a circle.
Well, he did NOT say "uniform random distribution".
So in fact the plot, "on" a circle would be
theta <-seq(0,2*pi, by =
Hi Peng,
On Wed, Jan 20, 2010 at 5:37 PM, Peng Yu wrote:
> R-lang.pdf has the following description in Section 3.1.1.
>
> """
> Any number typed directly at the prompt is a constant and is evaluated.
>> 1
> [1] 1
> Perhaps unexpectedly, the number returned from the expression 1 is a
> numeric. In
According to R-lang.pdf (Section 2):
Function mode gives information about the mode of an object in the
sense of Becker,
Chambers & Wilks (1988), and is more compatible with other
implementations of the S
language. Finally, the function storage.mode returns the storage mode
of its argument
in the
On Jan 20, 2010, at 1:36 PM, Eric Ma wrote:
>
> Thanks Marc for the quick reply. I confirmed the R binary I built is indeed
> 64-bit.
>
> sqlplus works fine, so is the odbcConnect() call.
>
> Any idea the error is thrown by RODBC or R?
>
> Eric
If you are getting the same error as in the or
Jerry Floren wrote:
Hi Gabor,
Thanks for your insights and suggestions. There was a post on the Wiki you
mentioned that makes me think this will work.
Unfortunately, the spreadsheet was designed to make it easy for lab staff to
enter their results, but not so easy for another program to read
R-lang.pdf has the following description in Section 3.1.1.
"""
Any number typed directly at the prompt is a constant and is evaluated.
> 1
[1] 1
Perhaps unexpectedly, the number returned from the expression 1 is a
numeric. In most
cases, the difference between an integer and a numeric value will b
Hello!
I have a data frame with a factor and a numeric variable:
x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200))
For each level of "factor" - I would like to divide each value of
"values" by the mean of "values" that corresponds to the level of
"factor"
In other word
Thanks a million Gabor. I was able to quickly import 21 test files. This will
save me hours, and should eliminate some errors.
Thanks,
Jerry Floren
Minnesota Department of Agriculture
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On Jan 20, 2010, at 4:55 PM, jshort wrote:
I'm attempting to generate matrices where the entries are randomly
generated
numbers of specified distribution.
The following code was an attempt to create a 3 by 3 matrix, where my
entries where randomly generated from a uniform (0,1) distributio
I'm attempting to generate matrices where the entries are randomly generated
numbers of specified distribution.
The following code was an attempt to create a 3 by 3 matrix, where my
entries where randomly generated from a uniform (0,1) distribution.
x = matrix(0,ncol = 3, byrow = T)
for(i in 1:
Jason Rupert wrote:
I have an instance where I need to include Greek letters on a plot title that is multiple lines.
I've searched the forums for an approach to do this, but most of the previous posts and replies seem to just address instances of single line examples and problems:, e.g.
http
Hi All,
I am new to python, R and rpy2. I am trying to print an Rvector say v and I
want to print 1st element so the statement will be
*print (v[0])* and this gives me *output [1] 876* but the real value I
should get is 876.
Can any one can help how to get rid of [1] while printing.
I am using
Tena koe Sean
I suspect the apply() and merge() functions are working, but they may
not be doing what you expect :-) You could try rbind() and aggregate():
> data.frame1$HAD <- as.numeric(NA)
> data.both <- rbind(data.frame1, data.frame2)
> aggregate(data.both[,-(1:3)], data.both[,1:3], sum, na.
Try this:
U <-
"http://www.mda.state.mn.us/en/sitecore/content/Global/MDADocs/licensing/map/mapreportform.aspx";
library(gdata)
DF <- read.xls(U, pattern = "SAMPLE", as.is = TRUE)
and now write an R program to create the desired data frame from DF.
On Wed, Jan 20, 2010 at 3:56 PM, Jerry Floren
Hi Gabor,
Thanks for your insights and suggestions. There was a post on the Wiki you
mentioned that makes me think this will work.
Unfortunately, the spreadsheet was designed to make it easy for lab staff to
enter their results, but not so easy for another program to read in the
data. That is w
Thank you - zoo did exactly what I needed...
On Jan 20, 12:20 pm, Henrique Dallazuanna wrote:
> Try the zoo package:
>
> plot(as.yearqtr(time.val), inc, col = 'red', type = 'l')
>
>
>
> On Wed, Jan 20, 2010 at 3:41 PM, mah wrote:
> > I am trying to generate a line graph with quarterly time bucke
Hi,
I'm looking to combine two data frames. Several of the columns are in
common while the others need to be summed up. The apply functions and
the merge functions don't seem to be working. I've included a basic
example of what I'm trying to do below. Thanks!
Sean
data.frame1<-as.data.
Thanks Marc for the quick reply. I confirmed the R binary I built is indeed
64-bit.
sqlplus works fine, so is the odbcConnect() call.
Any idea the error is thrown by RODBC or R?
Eric
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Is there a package for growth mixture modeling in R?
Rhoderick
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PLEASE do read the posting guide http://www.R-project.org/post
1. read.xls in gdata has the capability of
- reading an xls file starting from the first row with a given pattern
- converting an xls file to a csv file
- converting an xls file to a data frame
Would those capabilities be sufficient for the input side? See
examples in ?read.xls
2. If you do a g
One way to plot subsets of data identified by a grouping variable is to use
lapply() on a list of subsets. The approach is worth mentioning because
similar tactics are useful for many problems.
#List of unique values for grouping variable
#that is not necessarily a factor
names <- as.list(uniqu
Try this:
> t(apply(x, 1, function(x) (x == max(x)) - (x == min(x
[,1] [,2] [,3]
[1,] -101
[2,]1 -10
[3,]10 -1
You can avoid the transpose using plyr:
> library(plyr)
> aaply(x, 1, function(x) (x == max(x)) - (x == min(x)))
Var1 1 2 3
1 -1 0 1
2
Six times a year labs submit individual Excel spreadsheets to me. There are
usually around 60 labs and the spreadsheets have 12 columns with 20 rows.
Chapter 8 of the R Data Import/Export manual recommends converting .xls
files, to a text file. I have been manually converting the individual .xls
Hello, I need to retrieve datas from bloomberg.
I want to retrieve those datas in the fastest way as possible. I have two
options:
writing the datas from bbg to excel and reading from r the excel sheet or
directly
read the datas from from r with a Rbbg connection. Which connection is
faster?
Than
Use outer margins around the whole thing, see argument "oma" in ?par.
Uwe Ligges
On 20.01.2010 17:53, Lars Skjærven wrote:
Dear R users,
I'm making multiple plots within the same pdf page (par(mfcol = c(5,1)), and
want a legend for this at the bottom of all the plots. From previous mails
it ha
I don't understand what you are looking for.
The line
barplot(t(ld))
is placing the four treatments side by side, and is also stacking the
three values of typ.
Please indicate what you would like the figure to look like.
Rich
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On 21/01/2010, at 4:41 AM, omar kairan wrote:
Hi,
could someone help me with dilemma on the simulation of logistic
regressiondata with multicollinearity effect and high leverage point..
If that is the clearest way in which you can phrase your question then
I doubt that *anyone* can help you.
Got it.
Thanks
> CC: r-help@r-project.org
> From: dwinsem...@comcast.net
> To: aaron.fo...@students.tamuk.edu
> Subject: Re: [R] bootstrapping
> Date: Wed, 20 Jan 2010 10:34:20 -0500
>
>
> On Jan 20, 2010, at 10:23 AM,
> > wrote:
>
> >
> > Hello,
> >
> >
> >
> > I was able to boot
Ted, Alfredo, et al: Please stop doing this
jerk's homework for him!!!
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
I have an instance where I need to include Greek letters on a plot title that
is multiple lines.
I've searched the forums for an approach to do this, but most of the previous
posts and replies seem to just address instances of single line examples and
problems:, e.g.
https://stat.ethz.ch/p
On Wed, 2010-01-20 at 23:28 +0800, Bingzhang Chen wrote:
> Hi,
>
> I am stuck in one problem when doing nonmetric multidimensional
> scaling. I use the function 'metaMDS' in the package 'vegan' to work
> on the presence/absence community data. The problem is when two
> samples are identical (dissi
Meyners,Michael,LAUSANNE,AppliedMathematics schrieb:
Sorry, wrong button. Below a hopefully more helpful solution...
Etienne,
I don't see the point in avoiding some 'special' packages. If you are
willing to change your mind in this regard, try one of the following
solutions that work for me:
li
Try the zoo package:
plot(as.yearqtr(time.val), inc, col = 'red', type = 'l')
On Wed, Jan 20, 2010 at 3:41 PM, mah wrote:
> I am trying to generate a line graph with quarterly time buckets (with
> nice labels) on the x-axis. The first block of code below will
> generate the graph with nicely fo
I am trying to generate a line graph with quarterly time buckets (with
nice labels) on the x-axis. The first block of code below will
generate the graph with nicely formatted x-axis labels, but the
"type=" and "col=" options are not recognized when factors are used
for the x-axis.
The second bloc
You'll probably want to look at the 'by' function
d=data.frame(sex=rep(1:2,50),x=rnorm(100))
d$y=d$x+rnorm(100)
head(d)
cor(d)
by(d[,-1],d['sex'],function(df)cor(df))
You might also want to look at the doBy package
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Try googling "latticeExtra x.same" for some examples. Here's one:
http://www.mail-archive.com/r-help@r-project.org/msg39048.html
On Wed, Jan 20, 2010 at 9:44 AM, George Chen wrote:
> Hello,
>
> I would like to juxtapose two lattice graphs with common X axes such that
> the X axes line up. I am
Hi,
try c.trellis() from the latticeExtra package.
HTH,
baptiste
2010/1/20 George Chen :
> Hello,
>
> I would like to juxtapose two lattice graphs with common X axes such that the
> X axes line up. I am using plot right now but the edges are not neat and it
> would be nice if I could just d
Hello,
I would like to juxtapose two lattice graphs with common X axes such that the X
axes line up. I am using plot right now but the edges are not neat and it
would be nice if I could just draw 1 X axis and not both of them.
Here is my code:
upper<-bwplot(SignalUsed~as.fact
Try this:
sapply(as.data.frame.table(a), as.numeric)
On Wed, Jan 20, 2010 at 3:18 PM, rusers.sh wrote:
> Hi,
> See my example below.
> a<-array(1:12,c(2,3,2))
>> a
> , , 1
>
> [,1] [,2] [,3]
> [1,] 1 3 5
> [2,] 2 4 6
>
> , , 2
>
> [,1] [,2] [,3]
> [1,] 7 9 11
>
On Tue, 19 Jan 2010, jshort wrote:
Hi
How does one tell R that one is using an augmented matrix as appose to an
non-augmented matrix?
Explicitly creating an augmented matrix is unnecessary and
awkward if the object is to solve a system of linear equations.
See:
?solve
HTH,
Chu
On Jan 20, 2010, at 11:07 AM, BioStudent wrote:
Hi
I'm hoping someone can help me I am a relative newbie to R.
I have data that is in a similar format to this...
Experiment Score1 Score2
X -0.85 -0.02
X -1.21 -0.02
X 1.05 0.09
Y -1.12 -0.07
Y -0.27 -0.07
Y -0.93 -0.08
Z 1.1 -0.03
Z 2.4 0.
You can use this instead:
with(do.call(rbind, df.list), tapply(Score, list(Date, Show, Time), invisible))
On Wed, Jan 20, 2010 at 3:02 PM, Tony B wrote:
> Thank you for taking the time to reply Henrique. Although your
> solution does take away the zeroes and replaces them with NA's (which
> i pr
Thank you for taking the time to reply Henrique. Although your
solution does take away the zeroes and replaces them with NA's (which
i prefer), it unfortunately seems to reduce all of the other scores to
just '1':
> x <- with(do.call(rbind, df.list), tapply(Score, list(Date, Show, Time),
> length
Hi,
See my example below.
a<-array(1:12,c(2,3,2))
> a
, , 1
[,1] [,2] [,3]
[1,]135
[2,]246
, , 2
[,1] [,2] [,3]
[1,]79 11
[2,]8 10 12
I want to get a result something like
dim1 dim2 dim3 elements
111
121
1
Thanks so much for your assistance. In fact I did not read the documentation
carefully.
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___
Hi
I'm hoping someone can help me I am a relative newbie to R.
I have data that is in a similar format to this...
Experiment Score1 Score2
X -0.85 -0.02
X -1.21 -0.02
X 1.05 0.09
Y -1.12 -0.07
Y -0.27 -0.07
Y -0.93 -0.08
Z 1.1 -0.03
Z 2.4 0.09
Z -1.0 0.09
Now I can easily have a look at the
Hi,
could someone help me with dilemma on the simulation of logistic
regressiondata with multicollinearity effect and high leverage point..
Thank you
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I see now, thanks for explaining that. Would it be for you to add data.table
methods to ddply then, for this to happen? Or does a ddply method need to
be added to data.table?
"hadley wickham" wrote in message
news:f8e6ff051001200825q4009a122m122082a9df5fe...@mail.gmail.com...
> On Wed, Jan 20
Folks,
I've got a matrix x as follows:
> x <- matrix(c(1,2,3,5,3,4,3,2,1), ncol = 3, byrow = TRUE)
> x
[,1] [,2] [,3]
[1,]123
[2,]534
[3,]321
In each row of x, I want to replace the minimum value by -1, the maximum
value by +1 and all other values by 0.
Dear R users,
I'm making multiple plots within the same pdf page (par(mfcol = c(5,1)), and
want a legend for this at the bottom of all the plots. From previous mails
it has been suggested to use par(xpd=TRUE), increase the margin at the last
plot, and then draw the legend. However, when I do this,
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