Hi,
Could anybody please help me with this? In the following function, I try to
return a good format of x, i.e. if my x[1] is 800, I don't want it to give
me 800.0.
I tried format(x, trim = TRUE), but didn't work.
If my x has only first three components, which are all integers, I don't run
into th
Dimitri Shvorob gmail.com> writes:
>
>
> ?!! Erwin, may I ask for a working code sample? (Including appropriate
> package(s))
>
> Thank you.
I used GAMS + CPlex. The models are here:
http://yetanothermathprogrammingconsultant.blogspot.com/2010/01/solving-
optimization-problem-selecting.html
On Jan 30, 2010, at 8:24 PM, Mathew, Abraham T wrote:
I'm an R newbie and had the following question. I'm trying to run a
logit model because I have a dependant variable that is a factor
with three levels. Each of my independent variables are factor
variables and have two or more levels,
Same request to Hans: I am afraid I need a little more spoon-feeding
following
> I sent a GAMS script modeling this problem to the NEOS solvers
Thanks a lot!
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?!! Erwin, may I ask for a working code sample? (Including appropriate
package(s))
Thank you.
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I'm an R newbie and had the following question. I'm trying to run a logit model
because I have a dependant variable that is a factor with three levels. Each of
my independent variables are factor variables and have two or more levels,
which is why I'm unsure about the error.
regone <- glm(da
Hi:
You could complete the entire process in one shot with the plyr package,
using
function ddply. Using the piece of data supplied,
> ddply(motifList, .(events), summarize, freq = length(events), score =
median(score))
events freq score
1 aehjm1 0.0833
2 aeijm2 -0.2500
On Jan 30, 2010, at 4:38 PM, Brad Patrick Schneid wrote:
You rock!
thank you.. you saved me hours of hunting around.
--
You should read the r-help list more often. Gabor Grothendieck gave an
example of the use of xtabs() this very morning.
David Winsemius, MD
Heritage Laboratories
Wes
On Thu, Jan 28, 2010 at 8:59 PM, Santosh wrote:
> Dear Rxperts..
>
> what settings in barplot and histogram do I use, to show bars in an
> monotonously increasing or decreasing order of the frequency of a
> categorical variable?
See ?reorder. Typical usage would be (in lattice)
barchart(y ~ reor
Hi,
When using splom() in the lattice package, I would like to be able to
access the row and column number of each individual pairs plot ,
similar to the way that current.row() and current.column() can be used
in other lattice plotting functions such as xyplot. For example, I
would like
Gabriele,
Take a look at this:
http://research.stowers-institute.org/efg/R/Math/VennDiagram.htm
Best,
Jorge
On Sat, Jan 30, 2010 at 10:13 PM, Zoppoli, Gabriele (NIH/NCI) [G] <> wrote:
> Hello!
>
> I have this problem: I want to create a Venn's diagram with three lists of
> genes'names. The fir
Hello!
I have this problem: I want to create a Venn's diagram with three lists of
genes'names. The first is all the genome, the second a subset of it comprising
all mitochondrial genes, and the third including all genes that correlate with
a given gene.
This is what I do:
> library(gplots)
>
Detach the coda library. Its HPDinterval is conflicting with the
HPDinterval in lme4. You should have gotten a warning message about
the conflict when you tried to load coda.
On Sat, Jan 30, 2010 at 9:34 PM, Doug Adams wrote:
>
> Hi,
>
> I've got a linear mixed model created using lmer:
>
> A6m
Hi,
I've got a linear mixed model created using lmer:
A6mlm <- lmer(Score ~ division + (1|school), data=Age6m)
(To those of you to whom this model looks familiar, thanks for your patience
with this & my other questions.) Anyway, I was trying this to look at the
significance of my fixed effects
I've found the functions in the psych package easier to use than the
built in functions for principal components analysis.
-Ista
On Sat, Jan 30, 2010 at 2:09 PM, ogbos okike wrote:
> Hi,
> I am learning how to do principal component analysis in R. However, since I
> am family with only a few bui
as a motivating example, consider the case where multiple/many R
packages are being developed that all use the same shared object
functions.
each time (during development) that the shared object file changes we
could go alter every single R package directory, or simply have the R
package utilize t
Hello,
I'm a programmer, but brand new to R. I want to create a citation network
using data from ISI Web of Science.
Are there any libraries out there to assist with this process?
Thanks,
Nick
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Hi, I am pretty new to R. I'm trying run a regression repeatedly, adding a
new data point each time, and then storing the predicted Y values. For
example, let's say I have 500 data points and I run the regression. I would
then like to store the Y value, run the regression again using 501 data
poin
You rock!
thank you.. you saved me hours of hunting around.
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look at plyr package
colwise (column wise) function
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for (i in c('aum', 'num','aum.core','num.core')) t[[i]][is.na(t[[i]])] <- 0
On Sat, Jan 30, 2010 at 3:34 PM, Dimitri Shvorob
wrote:
>
> How can one simplify the folowing?
>
> t$aum[is.na(t$aum)] = 0; t$aum.core[is.na(t$aum.core)] = 0
> t$num[is.na(t$num)] = 0; t$num.core[is.na(t$num.core)] = 0
>
Hi Elizabeth,
Try using sprintf (?sprintf) instead of round().
HTH,
Jorge
On Sat, Jan 30, 2010 at 5:50 PM, Elizabeth Stanny <> wrote:
> Hi,
>
> I want 3 digits printed even if the last one is zero with \Sexpr{}
>
> For example:
> > ghg.06[12]
> [1] 0.2501008
>
> I want: 0.250
>
> What I've tri
Dimitri Shvorob gmail.com> writes:
>
> > This is a "subset sum" problem and has been discussed here in December
>
> Thanks a lot! Will investigate.
>
> > Can you settle for an approximate solution?
>
> Absolutely.
You can use the script from the thread "subset sum problem" to find
approxima
On Sat, Jan 30, 2010 at 12:39 PM, ATANU wrote:
> how can i draw two regression curves in one graph??i can draw them seperately
> using scatter.smooth.but can"t adjust them in one single graph
Using the built in iris data frame:
library(lattice)
xyplot(Sepal.Length ~ Sepal.Width, iris, group = ir
Dimitri Shvorob gmail.com> writes:
>
>
> A 40-element subset proves too much :(
>
> > Error: cannot allocate vector of size 554.1 Mb
>
> Thanks, Bart!
I could solve a problem with n=200, k=40 with Cplex's MIQP solver in less than
a
minute, but that may depend heavily on the data. The mode
The primary reason is for development.
I do install via the traditional R CMD INSTALL (or variant) method,
but I want to keep the C code external from the R package.
In particular, I want to be able to modify the C code (and thus the
compiled .so library functions) without having to constantly re-
Hi,
I want 3 digits printed even if the last one is zero with \Sexpr{}
For example:
> ghg.06[12]
[1] 0.2501008
I want: 0.250
What I've tried:
\Sexpr{round(ghg.06[12],3)}
0.25
\Sexpr{format(ghg.06[12])} % set options(digits=3) in earlier rnw chunk
0.25
\Sexpr{round(ghg.06[12],4} %but I want
On Jan 30, 2010, at 4:46 PM, david hilton shanabrook wrote:
On 30 Jan 2010, at 4:20 PM, David Winsemius wrote:
On Jan 30, 2010, at 4:09 PM, david hilton shanabrook wrote:
I have a data frame with two columns, a factor and a numeric. I
want to create data frame with the factor, its frequ
Erin Hodgess wrote:
Dear R People:
I ran an ARIMA(1,0,1) on a particular series in R and got a negative
MA(1) estimate.
Then I ran an ARIMA(1,0,1) on the same series in Minitab and got a
positive MA(1) estimate.
The values are about -0.69 and 0.70.
Does R show the opposite value, please?
Uh,
Also this looks like homework, so I can not really reply with a
solution. BTW, once you have the normalized matrix, barplot will
create your output without the complications of steps 8-13. You will
have to use the data to put the text, but that again is relatively
easy with the data.
On Sat, Jan
On 30 Jan 2010, at 4:20 PM, David Winsemius wrote:
>
> On Jan 30, 2010, at 4:09 PM, david hilton shanabrook wrote:
>
>> I have a data frame with two columns, a factor and a numeric. I want to
>> create data frame with the factor, its frequency and the median of the
>> numeric column
>>> head
One quick comment about looking at the graphs you provided, why aren't
all 8 columns the same height given that each column should have the
same number of amino acids in them. FOr the cleaved case is it 114
and even after normalizing, the column sums should be the same -- 100.
Are the graphs real
Dear R People:
I ran an ARIMA(1,0,1) on a particular series in R and got a negative
MA(1) estimate.
Then I ran an ARIMA(1,0,1) on the same series in Minitab and got a
positive MA(1) estimate.
The values are about -0.69 and 0.70.
Does R show the opposite value, please?
Thanks,
Erin
--
Erin H
On Jan 30, 2010, at 12:00 PM, Brad Patrick Schneid wrote:
I have a data.frame with the following:
ID Species Count
1 A 3
1 B 2
1 E 12
2 A 13
2 C 5
2 F 4
3
On Jan 30, 2010, at 4:09 PM, david hilton shanabrook wrote:
I have a data frame with two columns, a factor and a numeric. I
want to create data frame with the factor, its frequency and the
median of the numeric column
head(motifList)
events score
1 aeijm -0.2500
2 begjm -0.2500
On 1/30/2010 3:45 PM, Jonathan Greenberg wrote:
> Rhelpers:
>
>Having a problem solving this. I have an xyplot call that looks like
> this:
>
>
> print(xyplot(temp_species_EAM_Pred_Pop$x+temp_species_NULL_Pred_Pop$x~temp_species_EAM_Pred_Pop$Action,main=current_species,
>
>
I have a data frame with two columns, a factor and a numeric. I want to create
data frame with the factor, its frequency and the median of the numeric column
> head(motifList)
events score
1 aeijm -0.2500
2 begjm -0.2500
3 afgjm -0.2500
4 afhjm -0.2500
5 aeijm -0.25
A 40-element subset proves too much :(
> Error: cannot allocate vector of size 554.1 Mb
Thanks, Bart!
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I am announcing the release of the exactci package. It calculates exact tests
and confidence intervals for binomial and Poisson tests. Here is an example to
motivate the package:
Suppose you want to see if the observed rates of 2/17877 for group A are
significantly different from the observe
I have a data.frame with the following:
ID Species Count
1 A 3
1 B 2
1 E 12
2 A 13
2 C 5
2 F 4
3 B 5
3 D 3
I need it in thi
Hi all,
I have updated the exact2x2 package which calculates exact conditional tests
and the matching confidence intervals for 2x2 tables. The big change in the
update is to allow the option paired=TRUE which gives an exact McNemar's test
together with odds ratio estimates and confidence inter
How can one simplify the folowing?
t$aum[is.na(t$aum)] = 0; t$aum.core[is.na(t$aum.core)] = 0
t$num[is.na(t$num)] = 0; t$num.core[is.na(t$num.core)] = 0
Thank you.
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how can i draw two regression curves in one graph??i can draw them seperately
using scatter.smooth.but can"t adjust them in one single graph
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On Sat, Jan 30, 2010 at 12:45 PM, Jonathan Greenberg
wrote:
> Rhelpers:
>
> Having a problem solving this. I have an xyplot call that looks like
> this:
>
>
> print(xyplot(temp_species_EAM_Pred_Pop$x+temp_species_NULL_Pred_Pop$x~temp_species_EAM_Pred_Pop$Action,main=current_species,
>
Rhelpers:
Having a problem solving this. I have an xyplot call that looks
like this:
print(xyplot(temp_species_EAM_Pred_Pop$x+temp_species_NULL_Pred_Pop$x~temp_species_EAM_Pred_Pop$Action,main=current_species,
xlab="Action",ylab="Predicted Pop",
Here is the the written instruction as i managed to get it from my professor,
the graphs and data are attached:
The graph below shows an example of the expected outcome of this course
work. You may
procude a better one. The graph for analysing the motifs of a set of
peptides is designed
this way
Robyn, I've found this page helpful in understand the details of the kinds
of plots you want to make:
http://research.stowers-institute.org/efg/R/Graphics/Basics/mar-oma/index.ht
m but in general, if your subplots are related to each other, you should
probably switch from base graphics to the latt
On Jan 29, 2010, at 9:31 AM, johannes rara wrote:
How to vectorize this for loop and how can I assign result to vector
instead of using print function?
> as.vector( sapply(mylist$a, function(x)
sapply(mylist$b, function (y)
{ c
It is not entirely clear what you are trying to do. Can you explain
what the matrix that you are creating out of 'cleaved' represents?
"Tell me what you want to do; not how you want to do it". It is hard
to follow code when you have not explained what it is doing. THere
appear to be all kinds o
Hello,
I'm fairly new to R and having trouble displaying my data graphically to a
publishable quality.
I have a multivariate data-set (columns all the same length), 8
environmental variables and 3 species diversity variables.
I'm simply trying to display bivariate plots of the environmental variab
On Fri, 29-Jan-2010 at 03:16PM -0800, Cat Morning wrote:
|> Hi all,
|> I want to write a function to create multiple lists (over 100
|> lists). For example:
Could it be that you want to create multiple vectors? That's what it
looks like to me. If your question is clear, the answer probably wil
Try his:
xtabs(corrs ~., myData)
On Sat, Jan 30, 2010 at 2:11 PM, Juliet Hannah wrote:
> Hi Group,
>
> Consider a data frame like this:
>
> mylabel1 <- rep(c("A","B","C"),each=3)
> mylabel2 <- rep(c("A","B","C"),3)
> corrs <- c(1,.8,.7,.8,1,.7,.7,.7,1)
> myData <- data.frame(mylabel1,mylabel2,
Is this what you want:
> mylist <- list(a = letters[1:3], b = LETTERS[1:3], c = c("1", "2", "3"))
>
> x <- expand.grid(seq(length(mylist$a)), seq(length(mylist$b)))
> result <- apply(x, 1, function(.row){
+ list(mylist[[1]][.row[1]], mylist[[2]][.row[2]], mylist[[3]])
+ })
>
> result
[[1]]
[[1
On Jan 30, 2010, at 2:11 PM, Juliet Hannah wrote:
Hi Group,
Consider a data frame like this:
mylabel1 <- rep(c("A","B","C"),each=3)
mylabel2 <- rep(c("A","B","C"),3)
corrs <- c(1,.8,.7,.8,1,.7,.7,.7,1)
myData <- data.frame(mylabel1,mylabel2,corrs)
myData
mylabel1 mylabel2 corrs
1A
Hi Group,
Consider a data frame like this:
mylabel1 <- rep(c("A","B","C"),each=3)
mylabel2 <- rep(c("A","B","C"),3)
corrs <- c(1,.8,.7,.8,1,.7,.7,.7,1)
myData <- data.frame(mylabel1,mylabel2,corrs)
myData
mylabel1 mylabel2 corrs
1AA 1.0
2AB 0.8
3A
I have been able to implement the Mahalanobis distance function, and I have
also been able to generate code that computes the mahalanobis distance
calculations. Both have resulted in the same results, though these results
differ from the true results.
I believe my problem is in the formulation of
Thanks a lot; this helps a lot.
On Sat, Jan 30, 2010 at 11:07 AM, Gabor Grothendieck <
ggrothendi...@gmail.com> wrote:
> See ?window.ts, e.g.
>
> > # 1
> > tt <- ts(c(0.71, 0.63, 0.85, 0.44, 0.61, 0.69, 0.92, 0.55),
> + start = 1960, frequency = 4)
> > at <- c(1961, 3)
> > window(tt, at, at)[1]
Hi,
I am learning how to do principal component analysis in R. However, since I
am family with only a few built-in functions like prcomp, sd, cor, I started
manually with examples in text books while trying to use the few functions I
know to manipulate what they have in the text. From the example i
fortune("reverse")
On Sat, Jan 30, 2010 at 1:38 PM, Yong Zhang wrote:
> Hi,
>
>
>
> Does anyone know how to call a Matlab file (M-file) in R?
>
> Thanks!
>
>
>
> Yong
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailin
On Jan 30, 2010, at 11:38 AM, Yong Zhang wrote:
Hi,
Does anyone know how to call a Matlab file (M-file) in R?
You are expecting R to read Matlab source files directly? Or do you
(more reasonably) want to read datafiles or use R as a client with
Matlab as a server?
Package R.matlab
h
On Jan 30, 2010, at 10:33 AM, Jamie Smith wrote:
I am graphing longitudinal data from three time points. I'd like to
draw a
solid line from point 1 to point 2, and then a dashed line from
point 2 to
point 3. It works if I do it in two steps:
first.vector <- c(mean(year1$variable1), mean(y
Hi Chuck,
It looks like a scoping bug in spmx to me:
f <- function() {
x <- data[data$id=="111",]
print(spm(x$value ~ f(1:nrow(x
}
f()
I'd suggest you contact the package maintainer directly.
Hadley
On Fri, Jan 29, 2010 at 1:34 PM, Chuck White wrote:
> Hello -- I posted this question
Here some kind of a brute force attack:
#brute force solution, only working with relative small subsets:
n <- 200
elem <- 3
target <- 200
x <- rnorm(n,100,10)
x.combinations <- combn(x,elem)
sums <- apply(x.combinations,2,function(x) (sum(x)-target)^2)
ans <- (x.combinations[,which.min(sums)])
kayj wrote:
Hi All,
I am working on an example where the electric utility is investigating the
effect of size of household and the type of air conditioning on electricity
consumption. I fit a multiple linear regression
Electricity consumption=size of the house hold + air conditioning type
Th
I may be able to help if you upload to winbuilder again and let me know
when you did (don't have your source package any more).
I guess there are roughly 1-5 errors:
error 1: Where is the static library from? If copied from numerical
recipes: Please do look if you have the right licenses.
er
Hi,
Does anyone know how to call a Matlab file (M-file) in R?
Thanks!
Yong
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PLEASE do read the posting guide http://
On 29.01.2010 11:44, joez wrote:
Hello,
Im trying to combine 3 affybatches (1x hgu133+2 array and 2x hgu133a array)
Im useing this script:
library(matchprobes)
library(affy)
library(AnnotationDbi)
library(hgu133plus2probe)
library(hgu133aprobe)
library(hgu133a.db)
u133p2 = ReadAffy() # readin
On Sat, Jan 30, 2010 at 10:39 AM, Jennifer Young
wrote:
> handling of variables in R to be very straightforward; sometimes I pine
> for Maple to do my algebra for me...
There are several interfaces to Computer Algebra Systems in R. Try
this (but read instructions on home page
http://ryacas.goo
See ?window.ts, e.g.
> # 1
> tt <- ts(c(0.71, 0.63, 0.85, 0.44, 0.61, 0.69, 0.92, 0.55),
+ start = 1960, frequency = 4)
> at <- c(1961, 3)
> window(tt, at, at)[1]
[1] 0.92
Note that 1961.00, 1961.25, 1961.50 and 1961.75 represent the 4
quarters of 1961 so this also works:
# 2
> window(tt, 1961
On Fri, 29 Jan 2010, Paul wrote:
P > I'm currently using r scripts in sweave to grab some data via ODBC, process
P > it then generate some tables. I'd like to be able to give someone the files
P > and let them reproduce what I've done. Is there some way to store the data
P > that is gathered b
Many thanks I will try to check it latter¡¡¡
One more think...
To add a box on a plot (and locate where you want (I think you will say
floating)) you know the function? Not a Lengend but something similar but
writting what you want inside to indicate wat you want.
2010/1/30 Jim Lemon
> On 0
Found this
http://n4.nabble.com/Subset-sum-problem-td954423.html#a954423
http://n4.nabble.com/The-subset-matching-challenge-td861840.html#a861840
and learnt/remebered about 'subset sum' and 'knapsack' problems.
My case is different (simpler) in that subset size is fixed.
--
View this message
> This is a "subset sum" problem and has been discussed here in December
Thanks a lot! Will investigate.
> Can you settle for an approximate solution?
Absolutely.
> Rcplex: This is a combinatorial problem and cannot be formulated as a
> quadratic optimization problem.
If the objective func
I am graphing longitudinal data from three time points. I'd like to draw a
solid line from point 1 to point 2, and then a dashed line from point 2 to
point 3. It works if I do it in two steps:
> first.vector <- c(mean(year1$variable1), mean(year2$variable1))
> second.vector <- c(NA, mean(year2$var
Thanks so much everyone!
Bert's shorted example does what I need, but I'm filing away Gabor's
solution for when I inevitably need it some day. I've never found the
handling of variables in R to be very straightforward; sometimes I pine
for Maple to do my algebra for me...
> If its good enough t
On Jan 30, 2010, at 9:22 AM, Dipankar Basu wrote:
Hi All,
I have a very simple question about a time series object: how to
access
values for a particular year and quarter (say)?
Suppose, following
http://www.stat.pitt.edu/stoffer/tsa2/R_time_series_quick_fix.htm
I have read in data as a
Great,
works perfect!!!
Thanks a lot
Maxim
2010/1/30 baptiste auguie
> Hi again,
>
> Below are two versions, depending on whether you want to use scan or
> read.table,
>
> ## with scan
> library(reshape)
> listOfFiles <- list.files()
> d <- llply(listOfFiles, scan)
> names(d) <- basename(l
Its a bug. Its now fixed it in the development source so try this:
library(chron)
library(zoo)
source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/pkg/zoo/R/na.approx.R?rev=653&root=zoo";)
source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/pkg/zoo/R
If the goal is to *use* the Mahalanobis distance, rather than to learn
how to write your own code, there are several existing implementations.
rseek.org is a good place to find functions.
Sarah
On Fri, Jan 29, 2010 at 9:48 PM, Robert Lonsinger
wrote:
> Hello,
> I am a new R user and trying to le
Hi All,
I have a very simple question about a time series object: how to access
values for a particular year and quarter (say)?
Suppose, following
http://www.stat.pitt.edu/stoffer/tsa2/R_time_series_quick_fix.htm
I have read in data as a time series; here is how it looks.
* Qtr1 Qtr2 Q
Hi again,
Below are two versions, depending on whether you want to use scan or read.table,
## with scan
library(reshape)
listOfFiles <- list.files()
d <- llply(listOfFiles, scan)
names(d) <- basename(listOfFiles)
melt(d)
## with read.table
listOfFiles <- list.files()
names(listOfFiles) <- base
Hi,
my data is really not spectacular, each of the 6 files (later several
hundred) contains correlation coefficients in plain text format like:
0.923960073
0.923960073
0.612571344
0.064183275
0.007733399
-0.315444372
-0.064591277
-0.268336142
...
with between 1000-13000 rows.
Scanning f
Dimitri Shvorob gmail.com> writes:
>
> > Is it a subset of a vector containing 100 elements, or 1ths?
>
> I need to pick 2-40 elements out of a 50-200-element-long vector.
>
> > A random number of elements that should be chosen, or the best 10 values
> > which sums up to a defined value?
Hi,
Hadley recently proposed a strategy using plyr for a very similar problem,
listOfFiles <- list.files()
names(listOfFiles) <- basename(listOfFiles)
library(plyr)
d <- ldply(listOfFiles, scan)
Even if you don't want to use plyr, it's always better to group things
in a list rather than clutter
> Is it a subset of a vector containing 100 elements, or 1ths?
I need to pick 2-40 elements out of a 50-200-element-long vector.
> A random number of elements that should be chosen, or the best 10 values
> which sums up to a defined value?
The best 10 values.
I still think that Rcplex
Oh, I have considered this, but just don't see how the problem can be cast as
LSAP. (In the example, sum of squared pairwise distances ('costs') is
minimized, whereas here I am squaring the distance of a sum from the target
number. (How do I define individual assignment's cost here?)
Thanks anyway
Hi,
I have many files containing one column of data. I like to use the scan
function to parse the data. Next I like to bind to a large vector.
I try this like:
count<-1
files <- list.files() # all files in the working directory
for(i in files) {
tmp <- scan(i)
assign(files[count],
Exactly how do you want to convert it? It looks like you have columns
that are not integers, but factors. Can you provide an 'str' of the
dataframe so that we can see what the structure is. Can you also
provide an idea of what you want as output; you say a multiple list of
integers, but I am not
> library(chron)
> library(zoo)
> z1<-read.zoo("~/path/to/test1.csv",header=TRUE,sep=",",FUN=times)
> z2<-read.zoo("~/path/to/test2.csv",header=TRUE,sep=",",FUN=times)
> z3<-(na.approx(merge(z1,z2),time(z1)))
Error in approx(along[!na], y[!na], along[na], ...) :
need at least two non-NA val
Could you please specifiy your problem, if possible with some data?
Is it a subset of a vector containing 100 elements, or 1ths?
A random number of elements that should be chosen, or the best 10 values
which sums up to a defined value?
Bart
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On 01/30/2010 02:22 AM, Lu Wang wrote:
...
My
question is How do I remove the outside frame. It seems that Box() can
only generate four sided boders. Or is there any way that I can draw
the inner border without using box().
Try using box only on the second and third plot:
pie(value,labels=lbls
Hello,
Here are three examples. I have a hunch you are looking for the last one.
> test <- matrix(LETTERS[1:6], nrow=2, dimnames=list(c("First", "Second")))
> #explicitly stating what the rownames are before transposing
> test
[,1] [,2] [,3]
First "A" "C" "E"
Second "B" "D" "F"
> t
On Fri, Jan 29, 2010 at 2:29 PM, Zoppoli, Gabriele (NIH/NCI) [G]
wrote:
> Hi,
>
> I've lost my mind on it... I have to scatterplot two vectors, grouped by a
> third variable, with two different dimensions according to whether each cell
> line in the plot is sensitive or resistant to a given drug
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