Yep. You are right. That is better.
-tgs
On Thu, May 29, 2014 at 5:23 PM, Ista Zahn wrote:
> 10Hi Thomas,
>
> On Thu, May 29, 2014 at 9:15 AM, Thomas Stewart
> wrote:
> > Thanks to to Ista Zahn, I was able to find a work around solution. The
> key
> > seems to
65",string1); new_string1
new_string2 <- substring(new_string1,2); new_string2
If you know of a less hackish way to accomplish this, I'm interested to
hear it. However, this work around is sufficient for now.
Thanks,
-tgs
On Wed, May 28, 2014 at 10:25 PM, Thomas Stewart
wrote:
> Can anyon
Can anyone help me understand the following behavior?
I want to replace the letter 'X' in
âthe string â
'text X' with 'â¥' (\u226
â5
). The output from gsub is not what I expect. It gives: "text ââ°Â¥".
Now, suppose I want to replace the character 'â¤' in
â the stringâ
'text â¤'
I do not think Carl's solution answers your question. Try this:
z <- textConnection(
"education gender agree disagree sexe
1 0 Male 420
2 1 Male 200
3 2 Male 400
4 3 Male 630
5 4 Male
How about this?
require(FNN)
#FOR DEMONSTRATION PURPOSES, GENERATE 2D DATA
set.seed(3242)
X <- matrix(runif(50),ncol=2)
plot(X,pch=16,cex=1.5, asp=1)
#PLOT GRID
grid <- as.matrix(expand.grid(seq(0,1,by=.1),seq(0,1,by=.1)) )
abline(v=unique(grid[,1]),col="#FF30")
abline(h=unique(grid[,2]),co
I'd recommend something along these lines:
set.seed(32438786)
plot.new()
plot.window(xlim=c(-.5,.5),ylim=c(1,50))
for(i in 1:50){
X <- rnorm(100)
CI <- confint(lm(X~1))
ifelse( sum(0 > CI) == 2 | sum(0 wrote:
> Hi,
>
> I have 100 observations X1,X2,..,X100 and the confidence interval
> lim
How about using the asp option? For example,
A <- matrix(rnorm(200),20)
image(A, asp=ncol(A)/nrow(A))
-tgs
On Tue, Nov 19, 2013 at 3:56 AM, Mercutio Florid
wrote:
>
> Today at 10:03 AM
> I use several different versions of R, including RGui on Windows and
> rstudio on Linux. In all cases, I
One solution is to format the data as if the follow-up period ended on day
2190. For example,
TTT <- Survival_days
DDD <- Outcome
DDD[ TTT>2190 ] <- 0
TTT[ TTT>2190 ] <- 2190
survfit(Surv(TTT, DDD) ~ Gender)
-tgs
On Wed, Nov 20, 2013 at 3:01 PM, Dr.Vinay Pitchika
wrote:
> Hello R users
>
>
H
ow about a custom error handling function? As in,
options(error = quote( ...CODE TO SEND EMAIL ... ))
On Wed, Jul 24, 2013 at 4:06 PM, brt wrote:
> I have a web service that uses an R script to perform the analysis. At the
> end of the R script, I use the system function to call an external
You need to be careful how you index. In the example code, y is a 'zoo
series' and y[i] is gives the ith row, as in y[i, ]. This means t(y[i])
%*% y[i] is actually a 2X2 matrix. Also, the code c(S)[c(1,4,2)] picks
off the diagonal and lower triangular elements. The example involves two
stocks,
Short answer: Yes.
Long answer: Your question does not provide specific information;
therefore, I cannot provide a specific answer.
On Mon, Jun 10, 2013 at 1:23 PM, edelance wrote:
> I have run a PCA on one data set. I need the standard deviation of the
> first
> two bands for my analysis. I
Mike-
You can use the traceback function to see where the error is:
> bob <- matrix(rnorm(100*180), nrow=180)
> yyy <- rnorm(180)
> fit1 <- cv.glmnet(bob, yyy, family="mgaussian")
Error in rep("(Intercept)", nrow(a0)) : invalid 'times' argument
> traceback()
6: predict.multnet(object, newx, s
Antonio-
What exactly do you want as output? You stated you wanted a scatter plot,
but which variable do you want on the X axis and which variable do you want
on the Y axis?
-tgs
On Sat, May 25, 2013 at 3:36 PM, António Camacho wrote:
> Hello
>
>
> I am novice to R and i was learning how to
Soyeon-
A possible solution:
get(lambda.rule,envir=list2env(cvtest))
On Tue, Dec 18, 2012 at 12:34 PM, Soyeon Kim wrote:
> Dear my R friends,
>
> I want to get a number from a list using paste function.
> In my example,
> lambda.rule <- "lambda.1se"
> cvtest is a list (result from cv.glmnet)
e approach. Either way, I
> still need to figure out how to do the true/false color coding so any
> pointers on that are welcome.
>
> ** **
>
> ** **
>
> ** **
>
> *Neal Humphrey*
>
> Tel: +1 202.662.7241 | Skype: neal.s.humphrey | nhumph...@clasponline.org*
&
Neal-
Perhaps the following code is a start for what you want.
-tgs
par(mar=c(1,1,1,1),
oma = c(0,0,0,0),
mgp=c(1.5,.2,0),
tcl=0,
cex.axis=.75,
col.axis="black",
pch=16)
Z <- textConnection("
country A1 A2 A3
A 3 4 5
B 6 9 8
C 6 9 5")
ddd <- read.table(Z,header=TRUE)
close(Z)
Count
One solution that does not require matrix multiplication:
Remember that the steady state vector is in the nullspace of I - T.
Therefore:
require(MASS)
n1 <- Null(t(diag(nrow(T)) - T))
n1 / sum(n1)
On Wed, Dec 12, 2012 at 2:19 AM, annek wrote:
> Hi,
> I have a transition matrix T for which I
Chad-
The plot you hope to create is meaningful when the variable temperature is
treated as a continuous variable. Below is some code that treats
temperature as a continuous variable and creates the plot. Note that
temperature enters the model "linearly", and you may want to explore other
possib
Why not use an indicator variable?
P1 <- ... # prediction from model 1 (Setosa) for entire dataset
P2 <- ... # prediction from model 2 for entire dataset
I <- Species=="setosa" #
Predictions <- P1 * I + P2 * ( 1 - I )
On Monday, December 10, 2012, Brian Feeny wrote:
>
> I have a dataset and I
One option is to consider a Kronecker-type expansion. See code below.
-tgs
perhaps <- function(A,B){
nA <- nrow(A)
nB <- nrow(B)
C <-
kronecker(matrix(1,nrow=nA,ncol=1),B) >=
kronecker(A,matrix(1,nrow=nB,ncol=1))
matrix(rowSums(C) == ncol(A), nA, nB, byrow=TRUE)
}
Marius <- function(A,B) apply
Kyong-
It isn't clear to me why are you using a double integral. You have the pdf
and your goal is to calculate the expected value of a univariate random
variable (the first order statistic).
Shouldn't the expected value be E[X_(1)] = int x * 5 * f(x) *
(1-F(x))^4 dx? A single integral?
Here
You can directly use the tapply function.
-tgs
tapply(dat[,3],dat[,-3],mean)
On Thu, Dec 6, 2012 at 3:03 PM, Sarah Goslee wrote:
> If I understand what you want correctly, aggregate() should do it.
>
> > aggregate(V3 ~ V1 + V2, "mean", data=dat)
>V1 V2V3
> 1 C 0 0.500
> 2
Mindy-
>
> If the sub-directories are named in a structured way --- say sd1, sd2,
> sd3, ..., sd100 --- then one option is a simple for-loop. Something like
> the code below.
>
> -tgs
>
> directory.path <- '~/blahblah/' # or 'C:/blahblah/' for windows
> sd.names <- paste(directory.path,'sd',1:10
Irucka-
I got the following, which seems to be what you want. Perhaps you could
provide reproducible code.
-tgs
> as.matrix(data)
Time.day X1 X2 X3 X4
[1,]0 0.0 0.0 0.0 0.0
[2,]1 0.16387 0.60612 0.87705 0.83798
[3,]2 0.32774 1.212
Jonathan-
First, consider starting the algorithm at this alternative solution. You
do this with the optim.start option.
HoltWinters( , optim.start = c(alpha = 0.99, beta = 0.001, gamma =
0.001))
If this solution is indeed better, the function should not converge to the
old solution. If i
This is a hack which uses the output of the density function.
rrr <- cumsum(diff(cumsum(denspoints$y))*diff(denspoints$x))
lines(denspoints$x[-512],rrr)
I don't believe this is the best solution, because it is not a direct
estimate of the cumulative density function. I think there are methods fo
For question 2,
TTT <- rt(1000,3)
mean(TTT[rank(TTT) <= 975 & rank(TTT) >25])
On Thu, Jan 27, 2011 at 8:16 AM, Ben Boyadjian wrote:
> Hello I am trying to solve these problems and I am not allowed to use loops
> or ifs.
>
> 1st Question
> My first question is that I have generated 100 random num
Along with imdb.com, boxofficemojo.com might be a potential database.
-tgs
On Sun, Jan 23, 2011 at 3:46 AM, Barry Rowlingson <
b.rowling...@lancaster.ac.uk> wrote:
> On Sat, Jan 22, 2011 at 10:07 PM, Matt Curcio
> wrote:
> > Greetings all,
> > I am wondering if anyone is aware of any studies th
Here is a hack, crude solution to get the digits:
Digits <- function(a){
out <- rep(NA,nchar(a))
for(i in 1:nchar(a)) out[i] <- substr(a,i,i)
return(as.numeric(out))
}
Digits(183429)
If all you want is the last three numbers, consider another hack solution:
Last3Digits <- function(a) a
It isn't clear to me what you want to do. Do you want the axes to show? Do
you want labels for the lines? Do you want a legend? What is your desired
output?
-tgs
On Wed, Dec 8, 2010 at 2:42 PM, casperyc wrote:
>
> Hi All,
>
> How do I add these axis labels?
>
> #
Try this modification of your code.
-tgs
ynames <-
base.dat.sel2$base.dat.Covariate[order(base.dat.sel2$base.dat.US.Num.Obs.to.Achieve.Starting.Residual,decreasing=F)]
ynames <- as.character(ynames)
ynames[12]<-expression(theta[r])
ynames[13]<-expression(EC[gw])
ttt<-barplot(base.dat.sel2$base.d
You may want to consider using eurodist in matrix form. As in,
ED <- as.matrix(eurodist)
Then you could manipulate the matrix using standard or homemade functions,
like this one:
max.matrix <- function(A) {
column <- ceiling(which.max(A)/nrow(A))
row <- which.max(A) - (column-1)*nrow(A)
A
You may want to try something like this:
x1<-rnorm(500)
plot(x1)
test<-seq(1987, 2002, by=1)
test_2<-seq(2003, 2006, by=1)
test<-format(c(test, test_2), width=5)
xxx<-seq(1,500,length=length(test))
axis(1,at=xxx,labels=test,line=1,col=0)
You'll need to specify where you want the labels (in this c
Let f be your estimated function. Suppose we have a root function, say
root(). You are looking for
b = root(f-a)
where a is some constant.
Now suppose we consider the inverse of f, call it f.inv. Then the following
holds:
a = root(f.inv-b).
In your code, you find
b = root(f-a)
and
c = ro
CZ-
How exactly are you using the regression output? Are you using the
regression parameter estimates? Or, are you using the predicted (or fitted)
response variable.
The fact that your model matrix (sometimes called the X matrix) has
dimension 60 X 2000 means there is not a unique least square
I would be helpful if you provided a more complete, reproducible example.
Consider the following code. It colors the boxes according to the first 47
colors listed in the color() vector.
-tgs
data<-as.data.frame(matrix(rnorm(47*23),ncol=47))
boxplot(data,col=colors()[1:47])
On Thu, Oct 7, 20
Paula-
Look at this example. R in this case is your data matrix.
-tgs
R<-matrix(sample(0:2,100,replace=T),nrow=10)
R[1,]<-round(runif(10),3)
f<-function(x){
show(x)
y<-x[-1]
y[y!=2]<--999
y[y==2]<-x[1]
c(x[1],y)
}
apply(R,2,f)
On Tue, Oct 5, 2010 at 11:54 AM, Paula Fergnani Salvi
> through var, it grabs col=all.colors[which(names(my.data) %in% var)].
> I am wondering if it's possible to do the same for the legend? Because
> I just want the legend to have the same colors as in the graph... (in
> real life I'll have a lot of colors and it'll be hard
Right before the plot statement add:
par(mar=c(5,4,4,6),xpd=F)
then at the end add
legend(par()$usr[2],
mean(par()$usr[3:4]),
c("Blue","Yellow","Green","Orange"),
xpd=T,
bty="n",
pch=15,
col=c("Blue","Yellow","Green","Orange"))
On Mon, Oct 4, 2010 at 5:16 PM, Dimitri Liakhovitski <
I'm not sure I understand what you want, but here is a guess.
Let y be the hold out response values. Let y.hat be the model predictions
for the corresponding ys.
The key is to remember that R^2 = cor( y , y.hat )^2.
So,
cor( cbind(y,y.hat))[1,2]^2
should give you a measure you want.
-tgs
On
To find the probability directly, you'll need to clearly state how you are
sampling. Are you sampling with replacement? (Draw then put back.) Or are
you sampling without replacement? (Draw. Draw from the remaining. Draw
from the remaining.)
Also, of the N genes, do you know how many are of the
Try
expression(paste(integral(),"f(",tau,") d",tau,sep="")))
it works for me.
-tgs
On Mon, Oct 4, 2010 at 9:25 AM, Czerminski, Ryszard <
ryszard.czermin...@astrazeneca.com> wrote:
> I would like to use greek "tau" as a symbol of variable to integrate
> over in plotmath
>
> expression(integral(
There are several ways to do this. The following is only one of the ways.
One of the advantages of this approach is that it allows including both
continuous and categorical variables.
I'll demonstrate with the iris dataset. Place your variables in a dataframe
with the y variable in the first co
you know of another solution, please let me know.
Thanks,
-tgs
On Fri, Oct 1, 2010 at 3:01 PM, Thomas Stewart wrote:
> I'm trying to run R in batch mode on an LSF managed cluster. In simple
> settings, I can do it just fine. The trouble I'm having is when I try to
> pass ar
There are several ways to perform a MANOVA analysis in R. You'll have to be
more specific about your approach. For example, which function are you
using?
-tgs
On Fri, Oct 1, 2010 at 3:31 PM, Ali S wrote:
> Can anyone tell me how to run contrasts for MANOVA?
> Thanks!
>
>
>
>[[alternati
I'm trying to run R in batch mode on an LSF managed cluster. In simple
settings, I can do it just fine. The trouble I'm having is when I try to
pass arguments to the batch job. For example,
bsub R CMD BATCH --args 1 3 commandfile.R &
LSF doesn't like the &, and it doesn't run. I'm hoping one
Ethan-
You need to be more explicit about what you mean by 'background'. Do you
mean:
(a) the entire plot including margins?, or
(b) only the plotting area?, or
(c) a different color for both margins and plotting area?
If you want (a), the solution is par(bg = '#003D79').
If you want (b), the s
HH-
I'm not familiar with the plots you mention, but the following is a quick
attempt to create the plot you describe.
data<-data.frame(
org=1:10,
q1=sample(1:10,replace=T),
q2=sample(1:10,replace=T),
q3=sample(1:10,replace=T))
# This generates a random data set like the one you describ
Josh-
The link you included with your post provides the code for the example plot.
Simply click on the icon in the "Download Source Code" section.
I think you'll be able to learn a lot by playing with that source code.
-tgs
On Thu, Sep 16, 2010 at 9:40 AM, Josh B wrote:
> Hi all,
>
> Please
Try
par(xaxs="i")
barplot(rnorm(10), space=0)
-tgs
On Wed, Sep 15, 2010 at 7:54 AM, Daniel Stepputtis <
daniel.stepput...@vti.bund.de> wrote:
> Hi all,
> I have a problem with a rather simple plot (which I have used several
> times) - barplot.
>
> I want to create a barplot, where no space is b
kage, as you mentioned.
>
> Thanks.
>
> Cliff
>
>
> On Mon, Sep 13, 2010 at 10:02 PM, Thomas Stewart wrote:
>
>> Allow me to add to Michael's and Clifford's responses.
>>
>> If you fit the same regression model for each group, then you are also
>
Allow me to add to Michael's and Clifford's responses.
If you fit the same regression model for each group, then you are also
fitting a standard deviation parameter for each model. The solution
proposed by Michael and Clifford is a good one, but the solution assumes
that the standard deviation pa
This is only a guess because I don't have your data:
sigma is must be positive in the dnorm function. My guess is that optim may
attempt an iteration with a negative sigma.
You may want to see help(optim) for dealing with this constraint.
Specifically see the lower argument.
If you specify th
IIona-
I think you may be misinterpreting the t-test.
In model 1, consider the speciesH coefficient. A test that speciesH = 0,
essentially asks: Is speciesH the same as speciesB? The test statistic for
this hypothesis is the t-value reported in the table. (t-value= -4.2,
p-value=0.0001)
In mo
Because contourplot comes from the lattice package, I think you'll want to
look at these help pages:
+ help(trellis.focus)
+ help(lpoints)
Below, I've used the example from help(contourplot) to demonstrate how one
might add points and text to a lattice plot.
-tgs
#
#
Hopefully I understand your goal.
You have several datasets of the same dimension, and you want to calculate
the pairwise correlations of each dataset's row mean.
Depending on the dataset names, it may be easy to put all the data in a 1359
X 15 X 40 dimension array. Then, the following code woul
One possibility:
n.A<-nrow(A)
n.B<-nrow(B)
abs( kronecker(A,rep(1,n.B)) - kronecker(rep(1,n.A),B) )
-tgs
On Sat, May 22, 2010 at 3:20 PM, Shi, Tao wrote:
> One way to do it:
>
> apply(B, 1, function(x) t(apply(A, 1, function(y) abs(y-x) )) )
>
>
>
>
>
> - Original Message
> > From: D
Looks like you have some numerical precision issues. Why not use the svd
function directly? (See below.)
-tgs
x <- read.table(
textConnection(
"Sample1 0.7329881 0.76912670 2.45906143 -0.06411602 1.2427801
0.3785717 2.34508664 1.1043552 -0.1883830 0.6503095
Sample2 -2.0446131 1.727832
I know this was asked and answered in 2005, I just want to make sure the
answer still holds in 2010.
Question:
If I have the mixed-model with
lmer( y ~ x1 + x2 + (1 | id) + (z1 | w ) , ... )
there is no way for me to specify a correlation structure for the random
effects? Right?
If I want to s
7 data points? What's wrong with with doing it manually?
x<-c(173,193,194,167,119,80,96)
labs<-c("Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday")
pie3D(x,labels=labs)
-tgs
On Mon, May 17, 2010 at 8:46 AM, someone wrote:
>
> I'm an R noob and have a (maybe) stupid questi
Maybe this will lead you to an acceptable solution. Note that changed how
the data set is created. (In your example, the numeric variables were being
converted to factor variables. It seems to me that you want something
different.) The key difference between my code and yours is that I use the
Maybe this?
group <- factor(c("A", "B","B","C","C","C"))
model.matrix(~0+group)
-tgs
On Sat, May 15, 2010 at 2:02 PM, Noah Silverman wrote:
> Hi,
>
> I'm looking for an easy way to discretize factors in R
>
> I've noticed that the lm function does this automatically with a nice
> result.
>
> If
r maybe
directly calculate the bin percentages and plot them directly. I think you
have a lot of options to plot what you want.
-tgs
On Fri, May 14, 2010 at 11:51 AM, Federico Calboli wrote:
> On 14 May 2010, at 16:09, Thomas Stewart wrote:
>
> > Please be more specific with your questi
Please be more specific with your question. Perhaps a simple subset of the
data you are trying to plot? Here is some non-specific advice:
Plotting histograms as percentages instead of frequency counts is already an
option of the hist function. For example,
pop1<-rnorm(100)
hist(pop1,freq=F)
I
Try
x<-rowMeans(matrix((rbinom(1000,4,.45)-4*.45)/sqrt(.45*.55/4),ncol=10))
hist(x,freq=F,ylim=c(0,.5)) ### The key is the freq=F option.
curve(dnorm(x),add=T) ### You can use curve to plot a function
lines(density(x)) ### Or density for a kernel density estimate.
-tgs
On Thu, May 6, 2010
f<-function(x) 1/dnorm(qnorm(x))
for x in (0,1)
-tgs
On Thu, May 6, 2010 at 4:40 PM, Andrew Redd wrote:
> Is there a function to compute the derivative of the probit (qnorm)
> function
> in R, or in any of the packages?
>
> Thanks,
> -Andrew
>
>[[alternative HTML version deleted]]
>
> _
This is definitely a hack, but it gets the job done.
X<-model.matrix(~0+pheno,data=data)
data2<-apply(X,2,function(X){tapply(X,data$ID,sum)})
data2
phenoAppendicitis phenoAutism phenoBreast Cancer phenoMicrocephaly
phenoPolyps
A 1 0 1 1
For binary w.r.t. continuous, how about a smoothing spline? As in,
x<-rnorm(100)
y<-rbinom(100,1,exp(.3*x-.07*x^2)/(1+exp(.3*x-.07*x^2)))
plot(x,y)
lines(smooth.spline(x,y))
OR how about a more parametric approach, logistic regression? As in,
glm1<-glm(y~x+I(x^2),family=binomial)
plot(x,y)
lin
Something like this?
plot(0,0)
legend.text<-c("VAR 1","Higher","Average","Lower","VAR
2","Higher","Average","Lower")
legend.pch=1:8
legend.col=c(0,1,1,1,0,1,1,1)
legend("bottomright",legend=legend.text,pch=legend.pch,col=legend.col,ncol=2)
-tgs
On Mon, May 3, 2010 at 3:10 PM, Nish wrote:
>
> H
See
example(gam)
particularly
plot(gam.object,se=TRUE).
On Mon, May 3, 2010 at 5:20 AM, Oscar Saenz de Miera wrote:
>
> To whoever it may correspond,
>
> My name is Oscar Saenz and I am working on my thesis in Spain.
>
> I am using GAMs in "R" and, now that I have estimated my models, I need t
See
help(lmer, package="lme4")
for a mixed-effects model approach. See
help(geeglm, package="geepack")
for a GEE approach.
-tgs
On Sat, May 1, 2010 at 1:52 PM, kende jan wrote:
> Hello
>
> I would like to perform with R, a binary logistic regression analysis
> taking account clustering
>
Or, you can modify Romain's function to account for sequential NAs.
x <- c(1,2,NA,1,1,2,NA,NA,4,5,2,3)
foo <- function( x ){
idx <- 1 + cumsum( is.na( x ) )
not.na <- ! is.na( x )
f<-factor(idx[not.na],levels=1:max(idx))
split( x[not.na], f )
}
$`1`
[1] 1 2
$`2`
[1] 1 1 2
$`3`
nu
>From the GEE article in R News, Vol. 2/3, December 2002:
"Allows different covariates in separate models
for the mean, scale, and correlation via various
link functions."
Geepack offers link functions for the scale, correlation, and mean models.
As the output suggests,
"Correlation: Structure
Henrik-
A coding solutions may be
... + (1/(2*stdev*stdev))*sum( ( y-(rev/12)- c(0,y[-n]) *exp(-lap/12) )^2
)
where n is the number of observations in y.
Personally, I would use lm. Your model can be written as a linear function.
Let x=c(0,y[-n]). Then run lm(y~x). The parameter estimat
Try the duplicated() function. As in
m[!duplicated(id), ]
-tgs
On Wed, Apr 21, 2010 at 10:17 PM, gallon li wrote:
> Dear r-helpers,
>
> I have a very simple question. Suppose my data is like
>
> id=c(rep(1,2),rep(2,2))
> b=c(2,3,4,5)
> m=cbind(id,b)
>
> > m
> id b
> [1,] 1 2
> [2,] 1 3
Two possible problems:
(a) If you're working with a normal likelihood---and it seems that you
are---the exponent should be squared. As in:
... + (1/(2*stdev*stdev))*sum( ( y-(rev/12)-lag(y)*exp(-lap/12) )^2 )
(b) lag may not be working like you think it should. Consider this silly
example:
y<
I think you'll need the offset option.
ln( Y ) = ln( K ) + b1 ln( X1 ) + b2 ln( X2 / X3 ) + 1 ln( 1 / X3 )
as in:
glm( log(Y) ~ log( X1 ) + I(log( X2 / X3 )),
offset=I(log( 1 / X3 )))
-tgs
On Tue, Apr 20, 2010 at 7:14 AM, rajiv guha wrote:
> I am trying to estimate a demand function:
>
The stretching of plot two occurs because you are allotting more space for
plot two.
You set
Plot 1: mar=c(0,4,4,2)
Plot 2: mar=c(0,4,0,2)
Plot 3: mar=c(4,4,0,2)
In plot one your are dedicating 4 lines to the top margin, in plot three you
are dedicating 4 lines to the bottom margin. In plot two,
Maybe you want something like this.
temp<-data.frame(id=rbinom(10,5,.5),grp=rbinom(10,3,.3),stdid=rnorm(10))
kk<-"stdid"
temp[,kk]
-tgs
On Tue, Apr 20, 2010 at 7:32 AM, venkata kirankumar
wrote:
> Hi all,
> I have a problem with retreaving column from a data.frame that is
>
> I have a data.fra
Try border=c(0,0,1,0).
-tgs
On Mon, Apr 19, 2010 at 4:21 AM, Steve Murray wrote:
>
> Dear all,
>
> Thanks for the response, however I'm getting the following error message
> when I execute the legend command using the 'border' argument:
>
> Error in legend(10, par("usr")[4], c("A", "B", :
> un
I'm not sure I completely understand your question, but I think the solution
to your problem is the reshape function in the reshape package. Here is a
silly example of how it would work:
> V<-matrix(rbinom(15,4,.5),nrow=3)
> X<-data.frame(A=c("A","B","C"),V=V)
> X
A V.1 V.2 V.3 V.4 V.5
1 A 1
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