Ups, I mean library(reshape) not plyr, sorry
# Example datasets
# Input
propsum <- data.frame(coverClass=c("C", "G", "L", "O", "S"),
R209120812=c(NA, 0.49, 0.38, 0.04, 0.09),
R209122212=c(0.05, 0.35, 0.41, 0.09, 0.10))
library(reshape)
xpropsum <- melt
Hi:
How about this? Calling your matrix a,
a[order(rowSums(a)), ]
[,1] [,2] [,3]
[1,]912
[2,]2 111
[3,]34 13
HTH,
Dennis
On Fri, Apr 23, 2010 at 1:10 PM, Jonathan wrote:
> Hi r-help community,
>This question isn't so much a syntax/coding one, but here
Thank you Christopher, that is actually very helpful. =)
Carol
On Sat, Apr 24, 2010 at 12:13 PM, Christopher W Ryan
wrote:
> You should be aware that if you have read the data from your csv file
> into an object in R, generally nothing you do with that object in R
> will change your original csv
You should be aware that if you have read the data from your csv file
into an object in R, generally nothing you do with that object in R
will change your original csv file (this is generally a good thing),
until/unless, you save or write something from R to that csv file
(which is sometimes not a
Hi,
I've got a csv format of file. There are seven columns of variables all
together, and I've used the diff() function to take the differences between
the variables in one column. Now I want to add the differences as the 8th
column variable to the csv file. Is there any way that I can do this?
T
On Apr 23, 2010, at 8:06 PM, Muhammad Rahiz wrote:
Thanks David & Peter,
The locator() works but not practical as I have to repeat the
process many times.
Does the code works on linear regression only?
Should work for any process that can produce a function.
When i tried to find the int
On Apr 23, 2010, at 6:07 PM, Ryan Garner wrote:
>
> How do I pass a filename as an argument to the system command to count the
> number of records in that file? I only know how to do it by hardcoding it.
>
> HARDCODING EXAMPLE
>> test <- matrix(1:20,ncol=5)
>> write(x = test,file = "test.txt")
>
How do I pass a filename as an argument to the system command to count the
number of records in that file? I only know how to do it by hardcoding it.
HARDCODING EXAMPLE
> test <- matrix(1:20,ncol=5)
> write(x = test,file = "test.txt")
> records <- as.numeric(system("cat test.txt | wc -l",intern =
I have pretty big data sizes, like matrices of .5 to 1.5GB so once i need to
juggle several of them i am in need of disk cache. I am trying to use
bigmemory package but getting problems that are hard to understand. I am
getting seg faults and machine just hanging. I work by the way on Red Hat
Linu
That's perfect, thanks!
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__
did you look at the survSplit() function in the survival library? you may
have to recode your date variable to a zero point, i.e. Jan 1 1900 = 0, and
measure your time like that. The survSplit function will split your
survival duration into specified cut points. I hope this helps
Corey
-
C
Thanks David & Peter,
The locator() works but not practical as I have to repeat the process
many times.
Does the code works on linear regression only? When i tried to find the
intersection at a non-linear curve, i get the following error
Error in optimize(f = function(x) abs(xyf(ds) - n), c(
On Apr 23, 2010, at 6:24 PM, David Winsemius wrote:
On Apr 23, 2010, at 6:09 PM, tamas barjak wrote:
Thank you!
But, I'm sorry, cannot be understood what I wrote...(I'm sorry, I
use a machine translation program)...
Well:
I share out onto 4 parts the graphical area:
>layout(matrix(1:4
On Apr 23, 2010, at 6:09 PM, tamas barjak wrote:
Thank you!
But, I'm sorry, cannot be understood what I wrote...(I'm sorry, I
use a machine translation program)...
Well:
I share out onto 4 parts the graphical area:
>layout(matrix(1:4,ncol=2, byrow=T))
I would like to choose onto which I
Hi all,
The error I got was this:
Error in normalizePath(path) :
path[1]="C:\Program Files\R\library/gdata": The system cannot find the
file specified
(after some other packages updated fine)
For running:
update.packages(checkBuilt=TRUE, ask=FALSE)
My sessionInfo ():
R version 2.11.0 (2010-
Thank you!
But, I'm sorry, cannot be understood what I wrote...(I'm sorry, I use a
machine translation program)...
Well:
I share out onto 4 parts the graphical area:
>layout(matrix(1:4,ncol=2, byrow=T))
I would like to choose onto which I draw
example:
>plot(x, y, ...) <--- I would like 1 No
The reshape() function would be used to reorganize your event history
data set so that the Discussion and Agreement dates are one above the
other in the same column, presumably with another variable to flag
which is which (although you don't seem to need that).
Then, separately, you read your
Try this:
xtabs(values ~ ind + cover, cbind(cover = propsum$coverClass,
stack(propsum)))
On Fri, Apr 23, 2010 at 1:43 AM, wrote:
> Greetings all,
>
> I am having difficulty transposing, reshaping, flipping (not sure which) a
> data frame which is read from a DBF file. I have tried using t(), r
On Apr 23, 2010, at 4:57 PM, tamas barjak wrote:
Hi All!
I have 2 plain questions:
1.) I know that very primitive question, but that to grant it, that
the drawing
on the screen divided up onto which part draw
for example:
layout(matrix(1:4,ncol=2, byrow=T))
plot(x, y, ...) <--- 1. scree
While the OP turned out to want to transpose his data, for those who are
reading and would like some (ugly) code to flip a tensor about an
arbitrary axis, here goes:
*_* Carl
#my own cheap matrix flipflopper
flip<-function(x,flipdim=1) {
#axis is index, so 1 is rows, 2 is cols (for ma
There might be some package.
But you can also do something like:
results<-tapply(test$val,test$ty,function(x){
out1<-as.data.frame(length(x))
out2<-as.data.frame(mean(x))
out3<-as.data.frame(median(x))
out4<-as.data.frame(sd(x))
out<-cbind(out1,out2,out3,out
karena wrote:
> I need to create 10 matrices. say matrix 1-10.
>
> matrix_1 is 1 by 1
> matrix_2 is 2 by 2
> matrix_3 is 3 by 3
>.
>.
>.
> matrix_10 is 10 by 10
>
> I am just wondering if there are some functions in R that are similar to the
> macro variables in SAS. so I can create
Hi All!
I have 2 plain questions:
1.)
I know that very primitive question, but that to grant it, that the drawing
on the screen divided up onto which part draw
for example:
layout(matrix(1:4,ncol=2, byrow=T))
plot(x, y, ...) <--- 1. screen
plot(y, z, ...) <--- 3. screen
plot(z, x, ...) <---
Hi Dimitris,
This code is very helpful, although I note that the columns are not
preserved, ie if
> M <- matrix(sample(13, 9, TRUE), 3, 3)
produces:
> M
[,1] [,2] [,3]
[1,] 1392
[2,] 101 12
[3,] 1167
then
> n <- ncol(M)
> ind <- apply(M, 2, which.max)
> maxs
David, I'm not entirely sure at first glance (guess this is part of
the problem!), but after a little searching around, it does look like
this can be addressed using the "Hungarian algorithm," which reorders
the rows of a (square) matrix to minimize the sum of the elements
along the diagonal (the t
Serdal,
You still don't want explicit dummy variables. You have instead indicated
that you need two
factors, one for industryGroup, and one for industry nested within
industryGroup.
The model you suggested
lnQ~lnC+lnM+lnL+lnE+eco_inno+inno+(sum)ind_3d
will be easier to read and interpret as
one way is:
M <- matrix(sample(13, 9, TRUE), 3, 3)
n <- ncol(M)
ind <- apply(M, 2, which.max)
maxs <- M[cbind(ind, 1:n)]
diags <- diag(M)
diag(M) <- M[cbind(ind, 1:n)]
M[cbind(ind, 1:n)] <- diags
M
I hope it helps.
Best,
Dimitris
On 4/23/2010 10:10 PM, Jonathan wrote:
Hi r-help community,
Package tm does not have a function TermDocMatrix.
Where did you get the idea that it does?
There _is_ a function TermDocumentMatrix, however.
-Peter Ehlers
On 2010-04-23 10:21, Ignacio mas data wrote:
Hi List
I have the next code and the error. I have try with other codes and I have
the sa
Can you specify how you would re-order this matrix:
matrix(1:16, 4,4)
> apply(mtx, 2, which.max)
[1] 4 4 4 4
--
David.
On Apr 23, 2010, at 4:10 PM, Jonathan wrote:
Hi r-help community,
This question isn't so much a syntax/coding one, but here goes:
Let's say I have matrix of arbitrary dim
a...@ufl.edu (Allen S. Rout) writes:
> Greetings.
>
[ ... ]
D'oh, that was supposed to be aimed at R-devel.
- Allen S. Rout
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://w
Hi r-help community,
This question isn't so much a syntax/coding one, but here goes:
Let's say I have matrix of arbitrary dimensions and I'd like to
reorder the rows in such a way that I could maximize the sum of the
entries along the diagonal.
For example, for this 3x3 matrix:
[,1] [,
On Apr 23, 2010, at 3:56 PM, Jon Zadra wrote:
I took a look at xYplot() but it is beyond over-complex, at least
for my level, and the documentation isn't too clear.
Is there a simple way to make functions like plotCI() and errbar()
use the lattice reference rather than base graphics?
Don
Greetings.
There's a coalescing group which is working to imitate Dirk's fine
translation of CRAN to APT, in the Fedora/Red-Hat flavored land.
One of the things we're trying to do is implement as much as is
possible in R directly, and use as much of the existing package
management infrastruct
I took a look at xYplot() but it is beyond over-complex, at least for my
level, and the documentation isn't too clear.
Is there a simple way to make functions like plotCI() and errbar() use
the lattice reference rather than base graphics?
It seems like creating a plot of two lines across thre
Hi,
I have a very simple question, but I'm obviously not able to solve the
problem on my own.
I have a data.frame like
sample(c("A","B","C"),size=20,replace = T)->type
rnorm(20)->value
data.frame(ty=type,val=value)->test
There must be some built in functions, that will do some descriptive
On Apr 23, 2010, at 3:26 PM, Juliet Hannah wrote:
I should have mentioned that I also tried:
install.packages("Hmisc")
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
package ‘Hmisc’ is not available
If you have the necessary tools and install.packages("Hmisc",
Hi Dimitri,
Thanks for help,
i knew that way that you have suggested but in fact my case is a little bit
complicated than this,
therefore, i am writing a better explanation now,
my model is trying to see the effect of eco-innovative industries on total
output, thats why i am trying to e
I should have mentioned that I also tried:
> install.packages("Hmisc")
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
package ‘Hmisc’ is not available
On Fri, Apr 23, 2010 at 3:15 PM, David Winsemius wrote:
>
> On Apr 23, 2010, at 3:09 PM, Juliet Hannah wrote:
>
>>
Hi Hiro,
Here is a suggestion:
y <- rep(c("ACAM2343", "ACAM3838", "ACPP2598", "ACPA8355", "DEAM234",
"DEPA38485"), each = 10)
y
x <- y
x[grepl("PA|PP", x)] <- "WIN"
x[substr(y, 3, 4) == 'AM'] <- "LOSE"
x
Another way would be using regular expressions directly. See ?regex for more
informati
On Apr 23, 2010, at 3:09 PM, Juliet Hannah wrote:
I installed R 2.11.0, and I don't think I can load the doBy package
now. Any suggestions?
library("doBy")
Loading required package: survival
Loading required package: splines
Error in loadNamespace(i[[1L]], c(lib.loc, .libPaths())) :
there i
I installed R 2.11.0, and I don't think I can load the doBy package
now. Any suggestions?
> library("doBy")
Loading required package: survival
Loading required package: splines
Error in loadNamespace(i[[1L]], c(lib.loc, .libPaths())) :
there is no package called 'Hmisc'
Error: package/namespace
On Apr 23, 2010, at 11:51 AM, Uwe Ligges wrote:
>
>
> On 23.04.2010 18:10, Juliet Hannah wrote:
>> Hi List,
>>
>> I have a question about uninstalling and installing R on linux, which
>> I am new to.
>>
>>> sessionInfo()
>> R version 2.10.1 (2009-12-14)
>> x86_64-unknown-linux-gnu
>>
>> local
Hi weix1,
Try either "v:\\work\\gene" or "v:/work/gene". For more information see
"Basic Regular Expressions" under ?regex as well as the Windows FAQ.
HTH,
Jorge
On Fri, Apr 23, 2010 at 12:37 PM, weix1 <> wrote:
>
> I have string as follows:"v:\work\gene"
> however, i can not write this as c
On Apr 23, 2010, at 2:30 PM, Dimitri Liakhovitski wrote:
Your question is unclear.
Maybe not.
> mtx <- matrix(1:16, 4)
> nmtx <- matrix(0, 4,4)
> nmtx[1:4,1] <- mtx[row(mtx)==col(mtx)]
> nmtx[1:3,2] <- mtx[row(mtx)==col(mtx)-1]
> nmtx[1:2,3] <- mtx[row(mtx)==col(mtx)-2]
> nmtx[1:1,4] <-
Hi Cláudio,
Try this:
require(gtools)
combinations(10, 2)
combinations(10, 2, repeats = TRUE)
HTH,
Jorge
2010/4/23 Cláudio Sá <>
> Hi!
>
> Is there any easy/fast way to combine vectors with restrictions?
>
> Example:
>
> a=c(1,2,3), b=c(4,5,6), c=c(7,8,9)
>
> I want all combinations of this
Please see the FAQ on R for Windows2.16 where the distinction between the
standard file
separator "/" and the usual Windows file separator "\" is discussed.
On Fri, Apr 23, 2010 at 12:37 PM, weix1 wrote:
>
> I have string as follows:"v:\work\gene"
> however, i can not write this as character st
On Apr 23, 2010, at 1:30 PM, anderson nuel wrote:
Hello,
Could you help me to find the function which gives a vector that
indicate
the element in a vector A that are not in a vector B.
?"%in%"
... and look at the examples
Best Regards
[[alternative HTML version deleted]]
__
Hello,
Within the development of a package, I would need to build a specific method
for the "pmin" function.
I first make "pmin" generic
pmin <- function (..., na.rm = FALSE) UseMethod("pmin")
pmin.default <- base::pmin
Now, within my new method, I would like to change the argumen
attached is the data set which generates the rgui.exe cpu loop...
here are commands :
library(rpart)
train<-read.csv("traindata.csv",header=T)
y<-as.numeric(train[,18])
x<-train[,1:3]
fit<-rpart(y~.,x)
On Thu, Apr 22, 2010 at 9:25 AM, Terry Therneau wrote:
> --- Begin included message
>
> I h
Hi!
Is there any easy/fast way to combine vectors with restrictions?
Example:
a=c(1,2,3), b=c(4,5,6), c=c(7,8,9)
I want all combinations of this 3 vectors with length=2.
Like this:
1,4
1,5
1,6
1,7
1,8
1,9
2,4
2,5
2,6
2,7
... and so on.
Thanks in advance.
[[alternative HTML version d
I have string as follows:"v:\work\gene"
however, i can not write this as character string in R
what i did is:
path<- as.character("v:\work\gene")
but it seems that \ is eliminated:
Warning messages:
1: '\w' is an unrecognized escape in a character string
2: '\g' is an unrecognized escape in a c
Hi all,
Here is what I am trying to do. It doesn't seem like a complex command but I
am a beginner at R.
Lets say I have this variable:
test$v1 = rep(c("ACAM2343", "ACAM3838", "ACPP2598", "ACPA8355", "DEAM234",
"DEPA38485"), each = 10)
1. if test$v1 character string contains "PP" or "PA" set t
Thanks for all your reply. I am a biologist and only start to use R
recently. Yes, It is my fault. I assume that perm.test is a very popular
method. Thanks for Dennis to point out that it is in exactRankTests package.
I did not realize that there are so many R package. I guess I did not ask
the ri
Thanks Josh,
But I am not sure if the reshape function can create new rows based on
the meeting variable. For the second act there are three meetings, i.e.
one meeting between discussion and agreement, and this should be entered
as a separate row.
Best, Thomas
On Fri, 2010-04-23 at 07:45 -0700,
Hello,
Could you help me to find the function which gives a vector that indicate
the element in a vector A that are not in a vector B.
Best Regards
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch
Hi List
I have the next code and the error. I have try with other codes and I have
the same problem.
> reut21578 <- system.file("texts", "crude", package = "tm")
> (r <- Corpus(DirSource(reut21578), readerControl = list(reader =
> readReut21578XMLasPlain)))
A corpus with 20 text documents
> (r
On 2010-04-23 11:46, David Winsemius wrote:
On Apr 23, 2010, at 1:06 PM, Muhammad Rahiz wrote:
Does anyone know of a method that I can get the intersection where the
red and blue curves meet i.e. the value on the x-axis?
x <- 1:10
y <- 10:1
plot(x,y)
abline(lm(y~x),col="blue")
abline(h=2.5,co
Your question is unclear.
On Fri, Apr 23, 2010 at 5:45 AM, Usman Munir
wrote:
> Hi
>
> Suppose I have a matrix (cohort are rows and years are columns)
>
>
>
> [2000] [2001] [2002] [2003]
>
> [C1] 0.01 0.03 0.02 0.09
>
> [C2] 0.06 0.05 0.07 0.11
>
> [C3] 0.1
On Apr 23, 2010, at 1:54 PM, Ondřej Mikula wrote:
Dear R-helpers,
I need two simple functions, I guess they exist in R, but I am unable
to find them.
The first function should cyclically rotate elements in a vector.
("guyrot" in the package "wavethresh" should do it, but there's some
problem w
Serdal,
I think what David is saying: just take those 2 variables you have and
specify them as factors. Assuming your data frame is called "MyData":
MyData$Ind_1<-as.factor(MyData$Ind_1)
MyData$Ind_2<-as.factor(MyData$Ind_2)
This way R will know they are not numeric variables but categorical
vari
On 2010-04-23 10:53, zhenjiang xu wrote:
Yes. I put the real ranges instead of '...'. But I tried the following code
and it works. This is great! Thank you. Previously I thought you said ylim
was put inside the scales().
I did say that and I was wrong. If you put it
inside scales() as I suggest
Hi, Deepayan,
This is exactly what I want. However I couldn't find "page" argument or
"default.args" when I look up the documentation. I just want to learn a bit
more how to use these arguments. Thanks again.
Jun
On Fri, Apr 23, 2010 at 10:32 AM, Deepayan Sarkar wrote:
> On Wed, Apr 21, 2010 a
Dear R-helpers,
I need two simple functions, I guess they exist in R, but I am unable
to find them.
The first function should cyclically rotate elements in a vector.
("guyrot" in the package "wavethresh" should do it, but there's some
problem with its loading)
The second function should rotate poi
On Apr 23, 2010, at 1:06 PM, Muhammad Rahiz wrote:
Does anyone know of a method that I can get the intersection where
the red and blue curves meet i.e. the value on the x-axis?
x <- 1:10
y <- 10:1
plot(x,y)
abline(lm(y~x),col="blue")
abline(h=2.5,col="red")
Two ways :
> xy <- lm(y~x)
> xy
On 23.04.2010 19:02, Juliet Hannah wrote:
This has not worked for me, meaning I can still use R, so instead I
removed the directory
rm -fR R-2.10.1
Is one method preferable to another. And what am I doing incorrectly
with "make uninstall"?
Don't know, probably not many are using it and it
Thanks everyone for the replies, that sure cleared up some things for me.
On Fri, Apr 23, 2010 at 9:11 AM, Jan van der Laan
wrote:
> When you just want to calculate the probability of belong to class A
> or B of a new observation xi and do not have to do any new model
> estimations or other analy
Does anyone know of a method that I can get the intersection where the red and
blue curves meet i.e. the value on the x-axis?
x <- 1:10
y <- 10:1
plot(x,y)
abline(lm(y~x),col="blue")
abline(h=2.5,col="red")
Muhammad
__
R-help@r-project.org mailing li
>>
>> This has not worked for me, meaning I can still use R, so instead I
>> removed the directory
>>
>> rm -fR R-2.10.1
>>
>> Is one method preferable to another. And what am I doing incorrectly
>> with "make uninstall"?
>
> Don't know, probably not many are using it and it may be fairly untested.
On Fri, Apr 23, 2010 at 12:28 PM, arnaud Gaboury
wrote:
> TY Steve, using regular expression does the job nicely. I need now to fully
> understand your code and learn more about what a regular expression is. Any
> good ref is welcome.
Skimming through the help pages for ?regex and a quick google
Yes. I put the real ranges instead of '...'. But I tried the following code
and it works. This is great! Thank you. Previously I thought you said ylim
was put inside the scales().
library(lattice)
barchart(yield ~ variety | site,data=barley, groups = year, layout =
c(1,6),auto.key = list(points =
Hi All!
I have 2 plain questions:
1.)
I know that very primitive question, but that to grant it, that the drawing
on the screen divided up onto which part draw
for example:
layout(matrix(1:4,ncol=2, byrow=T))
plot(x, y, ...) <--- 1. screen
plot(y, z, ...) <--- 2. screen
etc...
2.)
How I c
On 23.04.2010 18:10, Juliet Hannah wrote:
Hi List,
I have a question about uninstalling and installing R on linux, which
I am new to.
sessionInfo()
R version 2.10.1 (2009-12-14)
x86_64-unknown-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8
Thanks Uwe.
Indeed - it now worked.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
And even then I'm guessing the the set complement of {2,2,2, 3,5,11}
with {2,2, 11} should not be {2, 3, 3} if R's versions of sets
works the way I was taught in high school, anyway.
> x <-set(2,2,2, 3,5,11) ; y <-set(2,2, 11)
> x
{2, 3, 5, 11} # because {2,2} == {2} in set theory
Thanks a lot!
I had not heard about the fractions() function before, it is very useful!!
Cheers!
--
View this message in context:
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__
TY Steve, using regular expression does the job nicely. I need now to fully
understand your code and learn more about what a regular expression is. Any
good ref is welcome.
> -Original Message-
> From: Steve Lianoglou [mailto:mailinglist.honey...@gmail.com]
> Sent: Friday, April 23, 2010
On Fri, 23 Apr 2010, Greg Snow wrote:
The perm.test function in exactRankTests makes reference to the StatXact
manual, so presumably that would have the exact details. The details
section of the help file refers to converting real values to integers,
so I expect that it is not doing what many
Greg has provided a solution. Just to answer the question of
why set_complement() is not doing what you think it should:
You need to change your *vectors* nom and denom to *sets*
with as.set().
-Peter Ehlers
On 2010-04-23 9:42, Greg Snow wrote:
Here is a different approach that may work for y
Hi,
On Fri, Apr 23, 2010 at 11:57 AM, arnaud Gaboury
wrote:
> Dear group,
>
> Here is my df :
>
> df <-
>
> structure(list(DESCRIPTION = c("PRM HGH GD ALUMINIUM USD 09/07/10 ",
>
> "PRM HGH GD ALUMINIUM USD 09/07/10 ", "PRIMARY NICKEL USD 04/06/10 "
>
> ), CREATED.DATE = structure(c(18361, 18361,
Hi List,
I have a question about uninstalling and installing R on linux, which
I am new to.
> sessionInfo()
R version 2.10.1 (2009-12-14)
x86_64-unknown-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C
The perm.test function in exactRankTests makes reference to the StatXact
manual, so presumably that would have the exact details. The details section
of the help file refers to converting real values to integers, so I expect that
it is not doing what many of us think of as a permutation test, a
Use regular expressions:
> df$newDesc <- sub("(\\d+/\\d+/\\d +)", '',
df$DESCRIPTION)
> df
DESCRIPTION CREATED.DATE QUANITY
CLOSING.PRICEnewDesc
1 PRM HGH GD ALUMINIUM USD 09/07/102020-04-09 -1 2,415.90 PRM
HGH GD ALUMINIUM USD
2 PRM HGH G
with(df, substr(DESCRIPTION, start=1, stop=nchar(DESCRIPTION) - 10))
?nchar
On 4/23/2010 11:57 AM, arnaud Gaboury wrote:
> Dear group,
>
>
>
> Here is my df :
>
>
>
> df <-
>
> structure(list(DESCRIPTION = c("PRM HGH GD ALUMINIUM USD 09/07/10 ",
>
> "PRM HGH GD ALUMINIUM USD 09/07/10 "
Dear group,
Here is my df :
df <-
structure(list(DESCRIPTION = c("PRM HGH GD ALUMINIUM USD 09/07/10 ",
"PRM HGH GD ALUMINIUM USD 09/07/10 ", "PRIMARY NICKEL USD 04/06/10 "
), CREATED.DATE = structure(c(18361, 18361, 18325), class = "Date"),
QUANITY = c(-1L, 1L, 1L), CLOSING.PRICE
hello,
krebs (1995) states MH as prob., but yes it's rather a ratio of probs.
at each site i had 4 blocks with 2 treatments (treat vs. control) - after
treating i looked for similarity between each of those pairs.
it is of interest if changes in similarity due to treatment differ between
stages
On Apr 23, 2010, at 11:32 AM, Deepayan Sarkar wrote:
On Wed, Apr 21, 2010 at 4:02 PM, David Winsemius > wrote:
snipped earlier code
Furthermore when I try:
mtext(date(), side=3, line=4, adj=0)
I get a datetime stamp even though I am mixing graphic
paradigms. This
would seem to
Works for me. Did you replace the '' in mylist()
with appropriate c(,) code? For example:
mylist <- list(c(0,30), c(40,80), c(0,50),
c(0,50), c(0,50), c(0,50))
-Peter Ehlers
On 2010-04-23 9:22, zhenjiang xu wrote:
Peter, thanks, but that doesn't work. Did I missed something
Dear Dr. Sarkar,
When I try to run the codes, I found the following problem:
> h<- sample(1:14, 319, rep=T)
> c<- sample(1:14, 608, rep=T)
> n<- sample(1:14, 1140, rep=T)
> vt<-c(h, c, n)
> ta<-rep(c("h", "c", "n"), c(319, 608, 1140))
>
> to<-data.frame(vt,ta)
> library(lattice)
Attaching p
Hi,
I have a data.frame object:
> a.df
Methods Score
1 Northern 1.3544227
2 Northern 0.8302436
3 RT-PCR 1.0011360
4 RT-PCR 1.1149423
If I write it out with write.table,
> write.table(a.df, file = 'data.txt', quote = FALSE, sep = '\t', row.names
= FALSE)
the data.txt is looks like:
Me
Here is a different approach that may work for you, or give you a starting
place:
> library(MASS)
> fractions(60/(220*6))
[1] 1/22
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
>
On Thu, Apr 22, 2010 at 3:16 PM, Giles wrote:
> Thank you Burt Gunter and David Winsemius,
>
> I confirmed David's comment.
> One must use type=c("p","g")
> in order to get both the grid and points.
>
> So, to use densityplot well,
> we must be quite aware of the options
> for panel.xyplot!
Actua
I'm sorry, I do not think I understand exactly what your data is and
what your end goal is. Are all your meeting dates together (i.e.,
they need to be split apart into separate cells)?
On Fri, Apr 23, 2010 at 8:18 AM, Thomas Jensen
wrote:
> Thanks Josh,
>
> But I am not sure if the reshape func
On Apr 23, 2010, at 11:26 AM, serdal ozusaglam wrote:
Dear R users,
I have a simple question (probably) but i couldnt how to find a
solution for that.
i am using 2 digit industry codes and 3 digit industry codes for my
model, and i need to create dummies for the industries. The case is
probably yes. I plotted each row individually instead. Thanks
On Fri, Apr 23, 2010 at 11:14 AM, Deepayan Sarkar wrote:
> On Wed, Apr 21, 2010 at 8:55 PM, David Winsemius
> wrote:
> >
> > On Apr 21, 2010, at 9:51 PM, zhenjiang xu wrote:
> >
> >> I tried that. It seems the bar width is already ma
On Wed, Apr 21, 2010 at 4:02 PM, David Winsemius wrote:
>
> On Apr 21, 2010, at 6:58 PM, David Winsemius wrote:
>
>> Sarkar offers a worked example of taking user input regarding location for
>> locating a grid viewport outside the plot area.
>>
>> http://lmdvr.r-forge.r-project.org/figures/figure
Dear R users,
I have a simple question (probably) but i couldnt how to find a solution for
that.
i am using 2 digit industry codes and 3 digit industry codes for my model, and
i need to create dummies for the industries. The case is simple for the 2-digit
industries since there are not that m
Peter, thanks, but that doesn't work. Did I missed something?
library(lattice)
mylist <- list(c(0,30), c(40,80), )
barchart(yield ~ variety | site,data=barley, groups = year, layout =
c(1,6),auto.key = list(points = FALSE, rectangles = TRUE, space =
"right"),ylab = "Barley Yield (bushels/acre)
On Wed, Apr 21, 2010 at 8:55 PM, David Winsemius wrote:
>
> On Apr 21, 2010, at 9:51 PM, zhenjiang xu wrote:
>
>> I tried that. It seems the bar width is already maximized, although there
>> is a lot of space between groups of bars. Thank you anyway.
>
> I apologize. It was reproducible code. I mi
On Tue, Apr 20, 2010 at 9:03 AM, Rajarshi Guha wrote:
> Hi, I've been struggling with a lattice visualiation. I have a
> data.frame with 4 columns. What I'd like to have is a set of 3 panels.
> Ecah panel will have the first column plotted against serial number
> and then will superimpose the rele
OK, now it sounds like the behavior is exactly the same for both of us. I
think Gmail sometimes gobbles up newline characters, and that must have
messed up the code I sent out before.
Someone pointed out to me that a similar problem was identified before:
https://stat.ethz.ch/pipermail/r-sig-mixe
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