Thank you so much for the suggestion.
I want to estimate interactions between the attributes of the choice set
and subject-specific data (e.g., gender or income).
However, y ~ x | z notation is to see the interaction between the
alternatives of choice set and subject-specific data (e.g., gender or
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
Just to be sure: I couldn't find a way of creating an interactive 3d graph
which can be embedded
in a pdf - is it correct that this can not be done with R?
Cheers,
Rainer
- --
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation
On Fri, 22 Jun 2012, Kazu Nada wrote:
Thank you so much for the suggestion.
I want to estimate interactions between the attributes of the choice set
and subject-specific data (e.g., gender or income).
However, y ~ x | z notation is to see the interaction between the
alternatives of choice set
Hi,
Try this post:
http://r.789695.n4.nabble.com/Exporting-an-rgl-graph-tp1872712p1905113.html
HTH,
b.
On 22 June 2012 19:26, Rainer M Krug r.m.k...@gmail.com wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
Just to be sure: I couldn't find a way of creating an interactive 3d
On Thu, Jun 21, 2012 at 10:56 PM, york8866 yu_y...@hotmail.com wrote:
I have a dataset like the following:
ID DV
1 0.868576818
2 0.337120116
3 0.029233775
4 0.719783525
5 0.976631182
6 0.672941605
7 0.13239462
8 0.99936475
9
Hi Simon,
Thanks for taking the time to reply. Please let me explain a few more details.
The problem that I am working on is essentially the same as the
Bristol Channel Sole Egg distribution example in your book and in the
soap paper but instead it is Herring Larvae in the English Channel -
same
On Wed, Jun 20, 2012 at 7:18 PM, John G. Bullock john.bull...@yale.edu wrote:
In your actual use case, are there any annotations outside the panel
(tick marks, labels, etc.)? Things would be much simpler if not (as in
your example). In that case, just call the suitable panel function
after
On Tue, Jun 19, 2012 at 10:55 PM, Stephen Eglen
s.j.eg...@damtp.cam.ac.uk wrote:
Justification is hard-coded, and that's not easy to change. This is
not only for the colorkey; e.g.,
xyplot(y~1, scales = list(alternating = 3))
will also give you left-aligned axes on the right.
My only
Removing rows with NAs, using na.omit(), doesn't seem to be working for me.
Dataset:
str ( ex10s )
'data.frame': 2189576 obs. of 5 variables:
$ LOPNR : int 58 58 58 58 64 64 64 64 64 64 ...
$ DIAGNOS: Factor w/ 173 levels F20,F200,F2000,..: 128 128 128 128 105
105 105 160 105 105 ...
$
The build system has rolled up R-2.15.1.tar.gz (codename Roasted
Marshmallows) at 9:00 this morning. This is a maintenance release; see the
list below for details.
You can get it from
http://cran.r-project.org/src/base/R-2/R-2.15.1.tar.gz
or wait for it to be mirrored at a CRAN site nearer to
Hi
Dear Petr,
nbsp;
Thank you very much for reply. You cannot read the Chinese characters
may
Yes, and I cannot install it either.
be because you don't have install this language. Do you have any idea
how
solve this problem? Or who can help me? May I install Linux?
What problem?
Hi
both na.omit and complete cases works for me smoothly when NA is not a
valid level in factor.
If this is the case, as it seems to be, you need reset your factor levels
so that NA is not a valid level.
ex10s$dg - factor( ex10s$dg )
both commands shall work than.
Regards
Petr
Removing
Hi,
Stackoverflow had an example. Forgot the contributor's name but it shows
percentages at the top. I use it as it is as I am not the expert.
png(Histogram.png)
pdata-read.csv(histogram.csv,header=T)
histPercent - function(x, ...) {
H - hist(pdata$y, plot = FALSE)
H$density -
Hi,
Is there a way to calculate variance directly by specifying
confidence interval using R ? I am specifically asking because I wanted
to investigate how this could be useful for project schedule variance
calculation.
Moreover I am interested in using R for monte carlo simulation as
On 22/06/2012 09:41, Stuart Leask wrote:
Removing rows with NAs, using na.omit(), doesn't seem to be working for me.
It won't if NA is a level of the factor, which is what you seems to have
here. For
table(as.factor(c(1,2,NA)))
1 2
1 1
omits NAs by default.
Dataset:
str ( ex10s )
On 2012-06-22 01:41, Stuart Leask wrote:
Removing rows with NAs, using na.omit(), doesn't seem to be working for me.
Dataset:
str ( ex10s )
'data.frame': 2189576 obs. of 5 variables:
$ LOPNR : int 58 58 58 58 64 64 64 64 64 64 ...
$ DIAGNOS: Factor w/ 173 levels F20,F200,F2000,..: 128
On 21.06.2012 23:34, Yakir Gagnon wrote:
Hi all,
I'm trying to install the mixdist package with:
install.packages(mixdist)
But I'm getting the following errors:
* installing *source* package ‘mixdist’ ...
** R
** data
** preparing package for lazy loading
** help
*** installing help
On 06/22/2012 03:26 AM, york8866 wrote:
I have a dataset like the following:
ID DV
1 0.868576818
2 0.337120116
3 0.029233775
4 0.719783525
5 0.976631182
6 0.672941605
7 0.13239462
8 0.99936475
9 0.91540604
10 0.545686514
to get a
Hi,
I really like the scatter.smooth() function - its extremeley useful.
However, as far as I understand it, there is no way to change the
properties of the smoothing line e.g. col, lty, lwd. The
scatter.smooth function accepts a ... argument, but these are not
passed to the plotting function
On 12-06-22 3:32 AM, baptiste auguie wrote:
Hi,
Try this post:
http://r.789695.n4.nabble.com/Exporting-an-rgl-graph-tp1872712p1905113.html
There's no more progress on U3D from me since that thread, but rgl can
now export WebGL format, which is viewable in most browsers (though not
IE).
Dear list members,
I get the following error when using the glht function to perform a post hoc
analysis for an ANOVA with repeated measures:
require(nlme)
lme_H2H_musicians = lme(H2H ~ Emotion*Material, data=musicians, random =
~1|Subject)
require(multcomp)
summary(glht(lme_H2H_musicians,
Hello,
I am comparing each words of a paragraph with the dictionary words (I have
a text file/csv file of dictionary words -dictionary.csv/dictionary.txt)
and if a word is not present in the dictionary, I want to store that word
in another variable .
and want output showing the list of these
HI,
I have one input file
Alpha 0 0.625
Beta1 0.107143
Gama2 0.910714
Delta 0 0
zeta0 0
eta 0 0.089286
I want to to plot bar plot like below
http://r.789695.n4.nabble.com/file/n4634182/Screenshot.png
I can plot if i have input file in
On 06/22/2012 07:35 PM, Manish Gupta wrote:
HI,
I have one input file
Alpha 0 0.625
Beta1 0.107143
Gama2 0.910714
Delta 0 0
zeta0 0
eta 0 0.089286
I want to to plot bar plot like below
Hello,
Try something along the lines of
dic - letters[1:4]
para - sample(letters, 100, TRUE)
ix - para %in% dic
para[ !ix ]# all words
unique(para[ !ix ]) # without duplicates
Hope this helps,
Rui Barradas
Em 22-06-2012 07:45, raishilpa escreveu:
Hello,
I am comparing each words of
I am trying to plot the linear fit by id of the following data (test.l):
id year incidents
1001 0
1011 1
102121
103127
1041 3
105112
1002 5
1012 5
102219
103210
1042
Given my reproducible example
myexample-structure(list(site = structure(c(4L, 2L, 2L, 4L, 2L, 4L, 4L,
3L, 1L, 3L, 1L, 1L, 3L, 4L, 5L, 2L), .Label = c(A, B, C,
D, E), class = factor), obs = c(0.302, 0.956, 0.72, 1.21,
0.887, 0.728, 1.294, 20.493, 0.902, 0.031, 0.468, 2.318, 4.795,
89.581,
There's almost certainly a more elegant way to do it, but this works:
r - raster(nrow=30, ncol=30, xmn=0)
r[]-NA
r[393:409]-99
r[423:439]-99
r[453:455]-99
r[456:460]-30
r[461:469]-99
r[483:499]-99
r[513:529]-99
plot(r,col=terrain.colors(100))
plot(edge(r, type=outer), col=c(transparent, black),
I have a graph plotted in r.The x axis tickmarks are at 0,5,10. I need a
graph with resolution of tickmarks at 0.1 interval.When i place tickmarks of
0.1 intervals using the following command axis(1, at=seq(0,5,0.1),
cex.axis=0.7, las=2) I only get the tickmarks of 0.1 interval that are very
Thank you all for the help,
York
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-make-a-histogram-with-percentage-on-top-of-each-bar-tp4634131p4634203.html
Sent from the R help mailing list archive at Nabble.com.
__
On Fri, Jun 22, 2012 at 7:51 AM, sathya7priya sathya7pr...@gmail.com wrote:
I have a graph plotted in r.The x axis tickmarks are at 0,5,10. I need a
graph with resolution of tickmarks at 0.1 interval.When i place tickmarks of
0.1 intervals using the following command axis(1, at=seq(0,5,0.1),
Hello,
Try
test - read.table(text=
id year incidents
1001 0
1011 1
102121
103127
1041 3
105112
1002 5
1012 5
102219
103210
1042 2
105212
1003
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 22/06/12 13:14, Duncan Murdoch wrote:
On 12-06-22 3:32 AM, baptiste auguie wrote:
Hi,
Try this post:
http://r.789695.n4.nabble.com/Exporting-an-rgl-graph-tp1872712p1905113.html
There's no more progress on U3D from me since that thread,
Hello,
Try the following
myex - myexample
myex$sample - as.character(myex$sample)
myex$sample - as.Date(myex$sample, format=%d/%m/%Y)
myex
Then, run the first attempt without bothering to order, lattice is
clever and will do it for you.
Hope this helps,
Rui Barradas
Em 22-06-2012 13:20,
This will get it assuming the data.frame is arranged like this
DatFile - read.table(text=Name x1 x2
Alpha 00.625
Beta1 0.107143
Gama2 0.910714
Delta 0 0
zeta0 0
eta 0 0.089286,
sep=, header=TRUE)
rownames(DatFile) - DatFile$Name
You are correct about scatter.smooth, but loess.smooth
(listed along with scatter.smooth on the same
help page) gives you the way to get what you want:
x - rnorm(25)
y - rnorm(25)
plot(x, y)
lines(loess.smooth(x,y), col=red, lty=2, lwd=2)
--
David L
Dear all,
I would like to (i) produce boxplot graphs with axis in logarithm in base 10
and (ii) showing the values on the axis in 10^exponent format rather than
10E+exponent.
To illustrate with an example, I have some widely spread data that I chart
plot using boxplot() [figure on the left];
Hello everybody,
problem solved, there was a typo.
I wrote Type instead of Material
Best
Messaggio originale
Da: angelo.arc...@virgilio.it
Data: 22-giu-2012 11.05
A: r-help@r-project.org
Ogg: Error with glht function: Error in mcp2matrix(model, linfct = linfct) :
Variable(s)
Hi,
I am trying to overlay to raster but I always got this warning message:
test-overlay(raster1,raster2,fun=function(x,y){x*y})
Warning message:
In x * y : longer object length is not a multiple of shorter object length
Here are the general information about my raster layer
raster1
class
Perhaps you meant ... at=seq(0,5,1) ?
S
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
sathya7priya [sathya7pr...@gmail.com]
Sent: 22 June 2012 12:51
To: r-help@r-project.org
Subject: [R] axis in r plot
I have a graph
Hi everybody,
I have been looking for exactly the same 3D equilateral pyramid
representation in R and couldn't find an answer in all the forum discussions
I've red.
It's been 2 years since the last post on the subject here, did anybody find
the way to make it ?
Thank you,
Julie
--
View this
Hi David,
Thanks for the reply. I agree - it just seemed that it was something
fairly obvious that was missing from an otherwise handy little
function Would it be appropriate to file it as a request for
improvement in the bug tracking system?
Mark
On 22 June 2012 16:30, David L Carlson -
thanks for your reply but I don't think your solution can accomplish to what
I need...
(I can't see any ordering of the variable sample with reference to the
variable obs)
please keep in mind that for a numer of reasons that I do not mention here
for the sake of conciseness in my case the
The key is to supply an expression, not text, to the labels argument to axis.
See help(plotmath) for details. Here is an example:
x - list(One=10^(sin(1:10)+5), Two=10^(cos(1:30)*2))
boxplot(x, log=y, yaxt=n)
ylim - par(usr)[3:4]
log10AtY - seq(ceiling(ylim[1]), floor(ylim[2]))
You might prefer placing the tick marks with
log10AtY - log10(axTicks(side=2))
log10AtY - unique(round(log10AtY))
instead of the
ylim - par(usr)[3:4]
log10AtY - seq(ceiling(ylim[1]), floor(ylim[2]))
in my original example.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Hello,
Ok, I hadn't understood. I still don't, not completely. Why do you want
to order 'sample' with reference to 'obs'? It would make sense only in
the case of 'obs' ties. Doesn't it make more sense to order 'sample'
within 'site'?
As for keeping 'sample' character it's very simple, I was
Sorry, missed the plot instruction.
dotplot(sample[ord] ~ obs[ord] | site, data=myex,
[... etc ...]
Rui Barradas
Em 22-06-2012 16:43, Rui Barradas escreveu:
Hello,
Ok, I hadn't understood. I still don't, not completely. Why do you
want to order 'sample' with reference to
Regarding regression models, there's a bit of discussion on whether or not
it is necessary to take the sample design into account (for instance, SPSS
doesn't), so you can run them just normally without much remorse. Or get
your life complicated (see below).
Your xtabs call seems OK to me.
Hi,
I have a list of mutations, called mutList, of the form:
head(mutList)
Alu
1 AluJ
2 AluJ/F(R)AM
3AluJ/FLAM
4AluJ/FRAM
5 AluJ/monomer
6AluJb
It contains about 500 elements and not all of them contain the
sequence Alu. I tried using this code:
You should use grepl() instead:
Alu-mutList[grepl(Alu,mutList)]
or some variant of the above, depending on the actual structure of
mutList (which might be a data frame, but doesn't appear to be a
generic list). str() and dput() are both very useful for asking
well-formed questions.
Sarah
On
It's really not that difficult to fix with some simple changes to the
code:
scatter.smooth.col -
function (x, y = NULL, span = 2/3, degree = 1, family = c(symmetric,
gaussian), xlab = NULL, ylab = NULL, ylim = range(y, pred$y,
na.rm = TRUE), evaluation = 50, ..., lcol, llty, llwd)
{
Yes! I'm guessing what he wants is
mutList$Alu[grepl(Alu,mutList$Alu)]
which can be simplified using with() to
with(mutList,Alu[grepl(Alu,Alu)])
But, in the absence of dput'ted data, we can't be sure. I would also
recommend that he reads up on list or data frame syntax, depending on what
this
On Fri, Jun 22, 2012 at 12:30 PM, Joshua Budman josh.bud...@gmail.com wrote:
Hi,
Thank you very much. You are correct, it is a data frame. However, the code
below is still returning the full data frame. Do you have any suggestions?
Thanks in advance,
Josh
On 22-Jun-12, at 12:22 PM, Sarah
Look at quadplot in package klaR.
--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On
The key is to supply an expression, not text, to the labels argument to axis.
See help(plotmath) for details. Here is an example:
x - list(One=10^(sin(1:10)+5), Two=10^(cos(1:30)*2))
boxplot(x, log=y, yaxt=n)
ylim - par(usr)[3:4]
log10AtY - seq(ceiling(ylim[1]), floor(ylim[2]))
Man, R has a steep learning curve (but I suppose you all know this). I have
very little programming knowledge, so when I search for answers to my
questions, I struggle with making sense of a lot of the pages.
I have a spreadsheet that I've read into R using read.csv. I've also
attached it. It
Hi,
You could try this;
dat2-read.table(text=
Alu
1 AluJ
2 AluJ/F(R)AM
3 monomer
4 AluJ/FLAM
5 AluJ/FRAM
6 AluJ/monomer
7 AluJb
8 JBF
9 FRAM
,sep=,header=TRUE)
subset(dat2,grepl(Alu,dat2$Alu))
Alu
1 AluJ
2 AluJ/F(R)AM
4 AluJ/FLAM
It's really not that difficult to fix with some simple changes to the
code:
scatter.smooth.col -
function (x, y = NULL, span = 2/3, degree = 1, family = c(symmetric,
gaussian), xlab = NULL, ylab = NULL, ylim = range(y, pred$y,
na.rm = TRUE), evaluation = 50, ..., lcol, llty,
On Jun 22, 2012, at 1:50 PM, Martin Maechler wrote:
It's really not that difficult to fix with some simple changes to the
code:
scatter.smooth.col -
function (x, y = NULL, span = 2/3, degree = 1, family =
c(symmetric,
gaussian), xlab = NULL, ylab = NULL, ylim = range(y, pred$y,
Hi and welcome to the R-help list.
It would be much better for readers to get your data in a more easily used
format.
There is a function called dput() that will output your data in a way that R
can read easily.
We don't need to see all the data but perhaps hundred lines of it would be
nice.
On Fri, Jun 22, 2012 at 2:18 PM, John Kane jrkrid...@inbox.com wrote:
Hi and welcome to the R-help list.
It would be much better for readers to get your data in a more easily used
format.
There is a function called dput() that will output your data in a way that R
can read easily.
We
I have built a R package including the C code. My question is:
Can the shared object get loaded automatically as we load the R package?
In my case, I have to do dyn.load(dirc/filename.so) in the R script to
load the shared object, which is very inconvenient.
Is there anyway to make R to load this
I believe that useDynLib() in the NAMESPACE file should do it for you.
Best,
Michael
On Fri, Jun 22, 2012 at 1:36 PM, xuan zhao xuan.z...@sentrana.com wrote:
I have built a R package including the C code. My question is:
Can the shared object get loaded automatically as we load the R package?
Why doesn't this work?
#Drop a variable name from a data frame
DropLikeSAS - function(x,df) {
df[[x]] - NULL
0
}
DropLikeSAS(VarName, DataFrameName)
Try it. The column VarName
Here is a part of the data for the first two questions:
head(Sunday,100)
SunDate SunTime SunScore
1 5/9/2010 0:000:00 127
26/12/2011 0:000:00 125
36/15/2008 0:040:04 98
4 8/3/2008 0:070:07 118
57/24/2011 0:070:07 122
6
Hi,
You could also use grep to get the same result.
Try this:
dat2-read.table(text=
Alu
1 AluJ
2 AluJ/F(R)AM
3 monomer
4 AluJ/FLAM
5 AluJ/FRAM
6 AluJ/monomer
7 AluJb
8 JBF
9 FRAM
,sep=,header=TRUE)
#grepl
dat3-subset(dat2,grepl(Alu,dat2$Alu))
On Jun 22, 2012, at 5:09 PM, Joseph Boyer wrote:
Why doesn't this work?
#Drop a variable name from a data frame
DropLikeSAS - function(x,df) {
df[[x]] - NULL
0
}
Dear all
I am trying to calculate the value of n using uniroot. Here is the message
I am having:
Error in uniroot(integ, lower = 0, upper = 1000, n) :
'interval' must be a vector of length 2
Please would you be able to give me an indication on why I am having this
error message.
Many
On Jun 22, 2012, at 5:09 PM, Joseph Boyer wrote:
DropLikeSAS - function(x,df) {
df[[x]] - NULL
0
}
DropLikeSAS(VarName, DataFrameName)
Sorry for the blank message
DropLikeSAS - function(x,df) {
Dear All,
I am using the zelig package on a Mac with OS X 10.6.8. Specifically, I am
using the ologit to do a ordinal logistic regression for a ordered categorical
dependent. The results make sense and are the same as I am getting when using
ordinal regression in SPSS. So far so good. However,
What is the difference of using 2L versus 2 as the margin argument in the
apply() function? Where can I find detailed information on all of the
possible margin arguments?
x
[,1] [,2]
[1,]12
[2,]34
[3,]56
[4,]78
[5,]9 10
sqrt(apply(x, *2L*,
On 06/22/2012 09:51 PM, sathya7priya wrote:
I have a graph plotted in r.The x axis tickmarks are at 0,5,10. I need a
graph with resolution of tickmarks at 0.1 interval.When i place tickmarks of
0.1 intervals using the following command axis(1, at=seq(0,5,0.1),
cex.axis=0.7, las=2) I only get
On 06/23/2012 12:27 AM, althalis wrote:
Hi everybody,
I have been looking for exactly the same 3D equilateral pyramid
representation in R and couldn't find an answer in all the forum discussions
I've red.
It's been 2 years since the last post on the subject here, did anybody find
the way to
Hello,
2L is an integer, 2 might be or not. In the case of apply(), there is no
difference.
All possible values for margin? Any possible(*) combination of the
dimensions of 'x' in
apply(x, margin, function)
(*) non-null. If, say, x - array(0, dim=c(2,3,4)) (3dim) then you can
call any of
I think the OP might also be tripped up on the fact that R is
pass-by-value so effects on df inside DropLikeSAS won't have impact
outside the function's scope. The df inside of DropLikeSAS() is
changed as expected, but that has no effect on the df outside that
function.
To the OP: There are ways
On 12-06-22 5:09 PM, Joseph Boyer wrote:
Why doesn't this work?
David gave you a solution that works. The answer to this question is
that unlike SAS, R is a computing language that has an idea of variable
scoping: when you modify the argument df in DropLikeSAS, you are making
local
On Fri, Jun 22, 2012 at 5:53 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
2L is an integer, 2 might be or not. In the case of apply(), there is no
difference.
All possible values for margin? Any possible(*) combination of the
dimensions of 'x' in
apply(x, margin, function)
(*)
On Fri, Jun 22, 2012 at 4:00 PM, jeka12386 jeka12...@hotmail.co.uk wrote:
Dear all
I am trying to calculate the value of n using uniroot. Here is the message
I am having:
Error in uniroot(integ, lower = 0, upper = 1000, n) :
'interval' must be a vector of length 2
Please would you
On Fri, Jun 22, 2012 at 5:13 AM, Mohan Radhakrishnan moh...@fss.co.in wrote:
Hi,
Is there a way to calculate variance directly by specifying
confidence interval using R ? I am specifically asking because I wanted
to investigate how this could be useful for project schedule variance
Right, I had missed that one.
To the op: what Michael is saying can be shown with the following example.
x - array(1:24, c(2,3,4))
r1 - apply(x, -1, sum) # remove dim 1
r2 - apply(x, 2:3, sum) # include dims 2 and 3
all.equal(r1, r2) # FALSE, different attributes
all(r1 == r2) # TRUE,
I have defined t at the beginning of my query.
I have added n on uniroot below and still getting the same error message
uniroot(integ,lower=0,upper=1000,n)
--
View this message in context:
http://r.789695.n4.nabble.com/Uniroot-error-message-with-in-intergration-tp4634247p4634264.html
Sent
Thanks Rui!
I was not aware that *number*L is an alternate to as.integer(*number*).
Luba
On Fri, Jun 22, 2012 at 3:53 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
2L is an integer, 2 might be or not. In the case of apply(), there is no
difference.
All possible values for margin? Any
Error in uniroot(integ, lower = 0, upper = 1000, n) :
'interval' must be a vector of length 2
Please would you be able to give me an indication on why I am having this
error message.
Because uniroot has a second parameter called 'interval' which overrides lower
and upper and you have
Michael,
Try
?pt
Best
Ozgur
--
View this message in context:
http://r.789695.n4.nabble.com/significance-level-p-for-t-value-in-package-zelig-tp4634252p4634271.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
This advice is almost certainly false!
A t-statistic can be calculated, but the distribution will not
necessarily be student's t nor will the df be those of the rse. See, for
example, rlm() in MASS, where values of the t-statistic are given without p
values. If Brian Ripley says that p values
FYI below.
On Fri, Jun 22, 2012 at 4:00 PM, Luba G lubaglouk...@gmail.com wrote:
Thanks Rui!
I was not aware that *number*L is an alternate to as.integer(*number*).
Luba
identical(2,as.integer(2))
[1] FALSE
identical(2L,as.integer(2))
[1] TRUE
Cheers,
Bert
On Fri, Jun 22, 2012 at
86 matches
Mail list logo
