I am aware of the most basic stuff, especially vectorization and
subscripting.
I only gave a simple example with length 4. I need to do that for vectors of
length 190.
Are there any in-built commands?
Or should I write the loops myself?
Thank you.
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I need to estimate the parameters for negative binomial distribution (pdf)
using maximun likelihood, I also need to estimate the parameter for the
Poisson by ML, which can be done by hand, but later I need to conduct a
likelihood ratio test between these two distributions and I don't know how
to
On Oct 19, 2012, at 06:05 , Thomas Lumley wrote:
On Fri, Oct 19, 2012 at 12:21 PM, Sheng Liu sheng@live.ca wrote:
Thanks a lot. It's very helpful.
I've read through the c code. Just FYI and for my completion of the
question, I post some of my thought on it:
To me it looks like the
Assuming that you actually mean
a1(b2+b3+b4) + a2(b1+b3+b4) + a3(b1+b2+b4) + a4(b1+b2+b3)
^ ^ ^
this might give you what you want:
x - data.frame( a = sample( 1:10, 4 ), b = sample( 11:20, 4 ) )
x
As I found the memory problem with local machine/micro instance(amazon) for
building SVM model in R on large dataset(2,01,478 rows with 11 variables),
then I have migrated our micro instance to large instance at Amazon. Still
I have memory issue with large amazon instance while developing R model
You asked for a loop, you got one...
Vectorized is easier and faster:
x$c - x$a * ( sum( x$b ) - x$b )
Rgds,
Rainer
On Friday 19 October 2012 09:38:35 you wrote:
Hi,
I think I solved it myself by writing loops.
What I meant is: are there in-built functions in R that calculate the
On 18-10-2012, at 21:33, djbanana wrote:
I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) +
a3(b1+b2+b4) + a4(b1+b2+b3)
or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i
I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i
* b_i
On 19-10-2012, at 10:38, Berend Hasselman wrote:
On 18-10-2012, at 21:33, djbanana wrote:
I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) +
a3(b1+b2+b4) + a4(b1+b2+b3)
or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i
I realise this is the same as:
On 19-10-2012, at 04:40, stats12 wrote:
Dear R users,
I am trying to find the mle that involves integration.
I am using the following code and get an error when I use the nlm function
d-matrix(c(1,1,0,0,0,0,0,0,2,1,0,0,1,1,0,1,2,2,1,0),nrow=10,ncol=2)
h-matrix(runif(20,0,1),10)
Dear grid-expeRts,
The goal:
I would like to construct a plot (matrix) with grid and gridBase,
which consists of four sub-plots. The sub-plots should have a square plotting
region as one would force with par(pty=s) in base graphics.
The problem:
I don't get a square plotting region, not even
?tryCatch()
Michael
On Thursday, October 18, 2012, Cess wrote:
Hi
I have created a loop to obtain data from several webpages
but the loop keeps crashing with the error
Error in function (type, msg, asError = TRUE) :
Operation timed out after 5000 milliseconds with 9196 bytes received
Hi,
I was trying to figure out how to do post-hoc tests for Two Way ANOVAs and
found the following 2 approaches:
a. Do pairwise t-tests (bonferroni corrected) if one finds significance with
the ANOVA.
Link-
http://rtutorialseries.blogspot.com/2011/01/r-tutorial-series-two-way-anova-with.html
Hi all,
Has anyone used the qua.regressCOP2 function from the copBasic package???
The default copula function used in this function is plackett copula and I
wanted to use archimedean copula. Attached below is my code:
mycop-frankCopula
V=seq(0.001,0.99,by=0.000217)
Hi everyone!!
I have dataset composed of a numbers of survival analyses.
( for batch survival analyses by using for-loop) .
Here are code !!
###
dim(svsv)
Num_t-dim(svsv)
Num-Num_t[2] # These are predictors !!
names=colnames(svsv)
for (i in 1:Num )
{
name_tt=names[i]
Hi,
I think I solved it myself by writing loops.
What I meant is: are there in-built functions in R that calculate the
following:
a1(b2+...+b190) + a2(b1+b3+...+b190) + ...
I managed to solve it, quite similar to what you just emailed.
Thanks anyway!
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Dear list,
Is there any one use MICE package deal with multilevel missing values here? I
have a question about the 2lonly.pmm() and 2lonly.norm(), I get the following
error quite often. Here is the code the error, could you give me some advice
please? Am I using it in the right way?
Had the same prob and did not work out the reason why. However, a workaround
is to add:
.character {
display:none;
}
to the file R2HTML.css or your css file respectively.
Cheers
UL
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v=IBM
library(quantmod)
v
v1=getSymbols(v)
to.yearly(v1)
===
when i pass the value through a variable in to.yearly() function it shows
the error msg like
Error in try.xts(x) :
Error in UseMethod(as.xts) : no applicable method for 'as.xts' applied
to an object of
Thanks for the reply Berend,
I am aware that I do not have to define the limits to Inf or -Inf, I just did
this
to make sure all other variables (besides 'w') in 'upper' have no limits.
I can tolerate no upper limit as my peak is fitted well, but my actual data is
a time series where the peak
Take a look at ?getSymbols which does not returne te time series by
default.
You want either
to.yearly(IBM)
or
v1 = getSymbols(v, auto.assign = FALSE)
to.yearly(v1)
Michael
On Friday, October 19, 2012, sheenmaria wrote:
v=IBM
library(quantmod)
v
v1=getSymbols(v)
to.yearly(v1)
On 12-10-18 6:11 PM, dorothy borowy wrote:
I am a new user of R and am crunching through the system. I have reached an
impasse with mapping; I want to make a bubble map and lay it over a grid that
is composed of a standard x,y axis. Within this, are 16 (4x4) gridded blocks,
numbered 1-16.
sapply(seq(4,ncol(dat)), function(i)
survdiff(Surv(time,completion==2)~dat[,i], data=dat,
subset=group==3)$chisq)
[1] 0.0944 4.4854 3.4990
Chris
-Original Message-
From: Charles Determan Jr [mailto:deter...@umn.edu]
Sent: Thursday, October 18, 2012 3:04 PM
To: r-help@r-project.org
Hello,
You should give a reproducible example showing us what you have tried,
with an example dataset.
Try the following.
# A graph I've just drawn
v - c(1,2,1,3,1,4,2,5,2,4,3,5,3,6,5,7,5,8,5,10,6,8,6,10,7,8,7,9,8,9)
g - graph(v, directed = FALSE)
plot(g, layout=layout.fruchterman.reingold)
Hi folks,
Despite the pain of migrating to 64-bit R (I have to install 64-bit Office also
due to RODBC), I'm considering making the leap due to memory issues. Is there
any place that lists packages that are 64-bit incompatible? Or, will I just
have to march through all my packages and check
On 19/10/2012 8:10 AM, Alexander Shenkin wrote:
Hi folks,
Despite the pain of migrating to 64-bit R (I have to install 64-bit Office also
due to RODBC), I'm considering making the leap due to memory issues. Is there
any place that lists packages that are 64-bit incompatible? Or, will I just
swertie v_coudrain at voila.fr writes:
I am trying to model data on species abundance (count data) with a poisson
error distribution. I have a fixed and a random variables and thus needs a
mixed model. I strongly doubt that my model is overdispersed but I don't
know how to get the
On Thu, Oct 18, 2012 at 10:14 AM, Fisher Dennis fis...@plessthan.com wrote:
Jason
Are you suggesting grep in R or grep in the system? If the latter, this
won't work because I need to implement this same procedure in Windows (sorry
about not mentioning this), in which grep does not exist.
I think this might be what you want. Kate Mullen and I have been in
correspondence over some edge cases where minpack.LM may not handle bounds
appropriately. However, though nlmrt seems to do the job here, readers
should note that R benefits hugely if we maintain some friendly
competition (and
Thanks Duncan. Here's what I come up with. Anything obviously missing?
I'd think there'd be more incompatibilities than just these few...
In 32-bit: pkgs32 = available.packages(type=win.binary)
In 64-bit: pkgs64 = available.packages(type=win.binary)
pkgnames32 = pkgs32[,Package]
Hi everyone,
I am familiar with using the chron package to work with date/time
values, but what about just time values with no date info present?
What is the best tool to convert character values to time values when
no date info is present?
Thanks!
Dan
On Fri, Oct 19, 2012 at 8:55 AM, Dan Abner dan.abne...@gmail.com wrote:
Hi everyone,
I am familiar with using the chron package to work with date/time
values, but what about just time values with no date info present?
What is the best tool to convert character values to time values when
no
The number of recent questions from umn.edu makes me wonder if there's homework
involved
Simpler for your example is to use get and subset.
dat - structure(.as found below
var.to.test - names(dat)[4:6] #variables of interest
nvar - length(var.to.test)
chisq - double(nvar)
for (i
Thank you for all your responses, I assure you this is not homework. I am
a graduate student and my classes are complete. I am trying multiple
different ways to analyze data and my lab requests different types of
scripts to accomplish various tasks. I am the most computer savy in the
lab so it
Hi all,
I have a matrix with 100 variables: each variable as a value of 0 or 1.
What i want to do is convert this matrix to a data.frame but convert all the
variables to factors (0 and 1) also.
I know i can do this one variable a time but i have 100 variables...
Any easy way of doing this??
Hi Allie,
When you are working with the ff package, the counterpart of a data.frame is
called an ffdf (ff data frame). It can handle the types you are talking
about - factor, integer but characters will be stored as factors. So this
means that your data types do not have to be of 1 specific type.
Thanks for the head-up Don.
Regards,
Phil
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I create a data.frame using :
alloc - data.frame(matrix(nrow=length(unique(mid2agi$gene)), ncol=8))
colnames(alloc) - c('agi', 'hit_len', 'q_len', 'identity', 'ratio', 'e',
'ok' ,'gene')
alloc$gene - unique(mid2agi$gene)
this results in:
head(alloc)
agi hit_len q_len identity ratio
Hi
I 'm new to R and wants to implement Egarch (1,1) with Student t distribution
where I need to plot Std. dev series.
Can you please help/provide me with the syntax/commands or any useful content?
Dheeraj
[[alternative HTML version deleted]]
Hello,
Try the following.
x - matrix(sample(0:1, 12, TRUE), ncol = 4)
y - data.frame(apply(x, 2, factor))
str(y)
Hope this helps,
Rui Barradas
Em 19-10-2012 12:04, brunosm escreveu:
Hi all,
I have a matrix with 100 variables: each variable as a value of 0 or 1.
What i want to do is convert
On 19/10/2012 8:51 AM, Alexander Shenkin wrote:
Thanks Duncan. Here's what I come up with. Anything obviously missing?
I'd think there'd be more incompatibilities than just these few...
If you are on Windows, packages need to pass tests on both 32 and 64
bits to make it onto CRAN. So
Look at the package rugarch - it's the best.
On Fri, Oct 19, 2012 at 7:52 AM, Dheeraj Pandey
dheeraj.pan...@thesmartcube.com wrote:
Hi
I 'm new to R and wants to implement Egarch (1,1) with Student t
distribution where I need to plot Std. dev series.
Can you please help/provide me with the
Well, strictly speaking, this is still doing it one variable at a
time. The interpreted loop is hidden, but it's still happening.
A loop free but clumsier approach is:
y - data.frame(matrix(as.character(x),nrow = nrow(x)))
## Note also that the original column names will be lost and will have
On Oct 19, 2012, at 16:07 , Rui Barradas wrote:
Hello,
Try the following.
x - matrix(sample(0:1, 12, TRUE), ncol = 4)
y - data.frame(apply(x, 2, factor))
str(y)
Hope this helps,
Another way, possibly more easily generalized:
x - matrix(sample(0:1, 12, TRUE), ncol = 4)
y -
Thanks a lot!
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Again, thank you all for the replies (and for the free R lesson!)
You helped me a lot and I very appreciate it.
I will work my self trough the for and apply section of my R manual again..
Thanks,
Bye, inga
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Obrigado Rui, é isso mesmo ;)
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R-help -
I'm trying to create axis breaks similar to this :
http://www.r-bloggers.com/wp-content/uploads/2010/08/bar-chart-natural-axis-split1.png
.
Is there a way to do this in R? Here's my code thus far:
structure(list(condition = structure(c(2L, 1L, 3L), .Label = c(con,
exp, unedit), class
Please look at the ?mmc example for two-way ANOVA in library(HH).
If you don't already have HH you can get it with
install.packages(HH)
library(HH)
?mmc
mmc uses the glht function in the multcomp package for its calculations
and then draws the MMC graph.
Rich
On 10/19/12, Amartya
Amartya: This is probably more of a statistics question than an R question.
I would go with the Tukey HSD, were I you as that is its intention.
Is there anyone in your organization (school, company, etc) that does
statistical consulting? I have had wonderful experience working with the
Hi everyone,
I am trying to get the factor score for each individual case from a principal
component analysis, as I understand, both princomp() and prcomp() can not
produce this factor score, the principal() in psych package has this option:
scores=T, but after running the code, I could
Hi,
No problem.
You can also try this:
set.seed(1)
mat1-matrix(sample(0:1,50,replace=TRUE),nrow=10,ncol=5)
dat1-as.data.frame(mat1)
dat2-do.call(data.frame,rapply(dat1,as.factor,classes=integer,how=replace))
# str(dat2)
#'data.frame': 10 obs. of 5 variables:
# $ X1: Factor w/ 2 levels 0,1:
Thanks for pointing that out. Made some modifications and it worked.
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djbanana karl79264...@gmail.com writes:
Hi,
I think I solved it myself by writing loops.
What I meant is: are there in-built functions in R that calculate the
following:
a1(b2+...+b190) + a2(b1+b3+...+b190) + ...
Following Rainer's setup:
x - data.frame( a = sample( 1:10, 4 ), b =
Dear all:
I am trying to center labels on my plot with not much success. I have tried
text(), mtext() but it's not working. I think I am using the wrong function
for my task.
Any help will be appreciated.
My working codes.
axis(1,
Terry: Thank you, that makes quite a bit of sense. In transposing the data
to intervals (corresponding to trapping runs) it becomes quite clear that
tag loss is very high up front, and has very good survival after the
initial period. This is what I needed to know. I think I had a case of too
Dear John,
this appears to be one of the cases where your algorithm is favourable over the
nlsLM one. The fitting is now a lot better with ssquared being lower compared
to any other fit method I was comparing to.
As I am relatively new to R, it still puzzles me that there are so many
From ?princomp
Arguments:
scores: a logical value indicating whether the score on each
principal component should be calculated.
Value:
scores: if ‘scores = TRUE’, the scores of the supplied data on the
principal components. These are non-null only if ‘x’ was
Hello,
Your problem comes from the fact that as new values are inserted in
'alloc' the column alloc$agi keeps changing (obvious!) therefore R can't
know all the factor levels beforehand. Therefore the values inserted are
the numeric codes of the original factor. Since your example doesn't
I am plotting ggplot smooth line. I would like to add dots to the regression
line, the dots are shaped under different condition.
p - ggplot(data, aes(x = x, y = y,
shape = assign, linetype = factor(sex)))
p0b - p+
scale_linetype_manual(breaks=c(0,1),
Hello,
Your working code needs a plot first:
plot(1:5, xaxt = n)
And there's no 'text' argument to axis(). As for the centering of axis
labels, they're centered on my graphics device. Exactly what is
happening to yours?
sessionInfo()
R version 2.15.1 (2012-06-22)
Platform:
I may be missing something, but if you want dots you need to ask for
them with geom_point()
Best,
Ista
On Fri, Oct 19, 2012 at 11:46 AM, autumnlin jialin...@gmail.com wrote:
I am plotting ggplot smooth line. I would like to add dots to the regression
line, the dots are shaped under different
Hello,
I have a dataframe w/ 3 variables of interest: transaction,date(tdate)
time(event_tim).
How could I create a 4th variable (last_trans) that would flag the last
transaction of the day for each day?
In SAS I use:
proc sort data=all6;
by tdate event_tim;
run;
/*Create last
On 19-10-2012, at 20:00, Thomas Schu wrote:
Dear All,
Which package/function could i use to solve following linear least square
problem?
A over determined system of linear equations is given. The nnls-function may
would be a possibility BUT:
The solving is constrained with
a
Dear collegues,
given a structure of data like this:
## Data ###
set.seed(100)
a - rnorm(60,10,3)
s - c(rep(A,20),rep(B,20),rep(C,20))
p - c(rep(d,6),rep(e,6),rep(f,6),rep(g,6),rep(h,6))
df - data.frame(a,s,p)
i would like to draw a lattice bwplot in vertical
Suppose your data frame is
d - data.frame(
stringsAsFactors = FALSE,
transaction = c(T01, T02, T03, T04, T05, T06,
T07, T08, T09, T10),
date = c(2012-10-19, 2012-10-19, 2012-10-19,
2012-10-19, 2012-10-22, 2012-10-23,
2012-10-23, 2012-10-23, 2012-10-23,
http://r.789695.n4.nabble.com/file/n4646789/1.bmp
hi is it possible to draw a graph like the attached one using ggplot?
how should i code it in r?
thanks.
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Hi there everyone! So I'm a student in college, taking a very basic
Statistics course. We're using R for most of our assignments. I've hit a
pretty big wall here. I'm attempting to create a heat map of the entire
united states which corresponds to a set of percentages I have for each
state. My
Hi Uwe,
thank you very much for this great help,
jsut perfect what I needed to know :)
cheers,
Yakamu
--- On Wed, 10/3/12, Uwe Ligges lig...@statistik.tu-dortmund.de wrote:
From: Uwe Ligges lig...@statistik.tu-dortmund.de
Subject: Re: [R] how can I adjust the ranges of my y-axis in barplot?
To:
Dear all,
I would like to make 6 barplots in one page but with a legend that applies to
all the barplots and would like to put it in the central-bottom of the page.
I know only how to make legend for individual barplot, but since all my
barplots have the same type and would be better if i just
I have a list of data.frames, and i want to iterate over this list and
generate graphs with the same title of the data.frame.
I did the graphs with:
lapply(anual, function(x) plot(x[,'chuva'], type='l', xlab= 'anos', ylab =
'Precicipatação(mm)', col='red'))
where anual is list of data.frames. I
First of all, in the command par(xpd=F), what is the value of F? If
it is the default value of FALSE then that says to not plot anything
outside of the plot region, so any attempts to put a legend in the
outer margin will not plot anything. if F is TRUE (which is a recipe
for disaster) then you
Thanks for all the help guys.
This worked for me:
all6 - arrange(all6, tdate,event_tim)
lt - ddply(all6,.(tdate),tail,1)
lt$last_trans -'Y'
all6 -merge(all6,lt, by.x=c(tdate,event_tim),
by.y=c(tdate,event_tim),all.x=TRUE)
--
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I think i have a better solution
*## Example data.frame*
d - data.frame(stringsAsFactors = FALSE, transaction = c(T01, T02,
T03, T04, T05, T06, T07, T08, T09, T10),date =
c(2012-10-19, 2012-10-19, 2012-10-19, 2012-10-19, 2012-10-22,
2012-10-23, 2012-10-23, 2012-10-23, 2012-10-23, 2012-10-23),time
Hi,
May be this helps you:
dat1-read.table(text=
tdate event_tim transaction
1/10/2012 2 14
1/10/2012 4 28
1/10/2012 6 42
1/10/2012 8 14
2/10/2012 6 46
2/10/2012 9 64
2/10/2012 8 71
3/10/2012 3 85
3/10/2012 1 14
3/10/2012 4 28
9/10/2012 5 51
Hi,
In addition to merge(), you can also use join()
dat1-read.table(text=
tdate event_tim transaction
1/10/2012 2 14
1/10/2012 4 28
1/10/2012 6 42
1/10/2012 8 14
2/10/2012 6 46
2/10/2012 9 64
2/10/2012 8 71
3/10/2012 3 85
3/10/2012 1 14
3/10/2012 4 28
I would like to add specific point on to the regression line, like
(x=1),shape=value(1)
is it possible?
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hi all,
how can i saving R output to docx or Jpeg format?
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PLEASE do read the posting guide
See
?annotate
for help on adding individual points/labels/etc to a plot.
Best,
Ista
On Fri, Oct 19, 2012 at 3:48 PM, autumnlin jialin...@gmail.com wrote:
I would like to add specific point on to the regression line, like
(x=1),shape=value(1)
is it possible?
--
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On Oct 19, 2012, at 12:30 PM, Flavio Barros wrote:
I have a list of data.frames, and i want to iterate over this list and
generate graphs with the same title of the data.frame.
I did the graphs with:
lapply(anual, function(x) plot(x[,'chuva'], type='l', xlab= 'anos', ylab =
On 10/19/2012 11:42 AM, autumnlin wrote:
http://r.789695.n4.nabble.com/file/n4646789/1.bmp
hi is it possible to draw a graph like the attached one using ggplot?
Yes
how should i code it in r?
Use a geom_point() where the shape aesthetic is mapped to the grouping
variable and also add a
On Oct 19, 2012, at 12:54 PM, javad bayat wrote:
hi all,
how can i saving R output to docx or Jpeg format?
There are some packages with functions to write to Word formats, but the one I
know of only runs on Windows, so I have not bothered remembering its name. The
sos package offers nice
Hi Javad,
saving R output to jpeg depends on what you want to save. For example
saving an `lm` object to an image would be fun :)
But you could export that quite easily to e.g. docx after installing
Pandoc[1] and pander[2] package. You can find some examples in the
README[3].
Best,
Gergely
[1]
Thanks, again! Sorry for my misleading expression, I only knew the value is
inquired from a polynomial approximation, but I have no idea how it is done
in such a great detail. It's a great lesson for people like me who want a
deep understanding of the basics.
Sheng
On Thu, Oct 18, 2012 at 11:46
On Fri, Oct 19, 2012 at 8:02 PM, Sando chocosa...@daum.net wrote:
Hi everyone!!
I have dataset composed of a numbers of survival analyses.
( for batch survival analyses by using for-loop) .
Here are code !!
###
dim(svsv)
Num_t-dim(svsv)
Num-Num_t[2] # These are predictors !!
On Oct 19, 2012, at 2:48 PM, Daróczi Gergely wrote:
Hi Javad,
saving R output to jpeg depends on what you want to save. For example
saving an `lm` object to an image would be fun :)
But you could export that quite easily to e.g. docx after installing
Pandoc[1] and pander[2] package. You
New installation seems to have behavior I cannot figure out. Here is
illustrative sequence where I load a small data set (test) from Crawley's files
and try to run a simple linear model and get an error message. Oddly, R
reports that the variable 'test$ozone' is numeric while, after attaching
Hi Michael,
Try
fit - lm(ozone~garden, data = test)
summary(fit)
See ?lm for more details.
HTH,
Jorge.-
On Sat, Oct 20, 2012 at 8:26 AM, Michael Grant wrote:
New installation seems to have behavior I cannot figure out. Here is
illustrative sequence where I load a small data set (test)
On Oct 19, 2012, at 2:26 PM, Michael Grant wrote:
New installation seems to have behavior I cannot figure out. Here is
illustrative sequence where I load a small data set (test) from Crawley's
files and try to run a simple linear model and get an error message. Oddly,
R reports that the
You have an object named ozone kicking around in your workspace/global
environment. This is apparently not of mode numeric and it is masking
the ozone object (column) from the attached data frame test.
Simply doing ls() would have told you this.
Presumably you also got a warning about this
On Fri, Oct 19, 2012 at 2:50 PM, Dhaval Adjodah nuclear...@gmail.com wrote:
[...]
Your way works best! Another way I found from stackoverflow (see the post
at
http://stackoverflow.com/questions/12964332/igraph-edge-between-two-vertices/12980550#12980550)
was
to convert the graph into an
On Wed, Oct 17, 2012 at 4:08 PM, ilai ke...@math.montana.edu wrote:
On Wed, Oct 17, 2012 at 11:10 AM, Ali Tofigh alix.tof...@gmail.com wrote:
my problem is that I usually have no choice but to mix grid and base
graphics.
What does that have to do with the answer you got ? did you even try it
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