Dear list,
Say I have a data frame with NA in the variable which I want to use as a
sorting variable for splitting the data frame.
df <- data.frame(A=1:10, B=c(rep(99,5), rep(100,5)), C=c(rep(NA,3),
rep(1,3), rep(2,4)))
split(df, f=df[c('C')], drop=FALSE), I got the output as follows. I was
hopi
o separate the color schemes
for lines and points, but I can't use multiple scale_color_manual statement
in one ggplot.
Jun Shen
sample data#
df1 <- structure(list(Y2 = c(0L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 0L), X = c(27L, 8L,
33L, 5
plot(y ~ x, data=tmp, main="outside text displayed",
> >>par.settings=list(clip=list(panel=FALSE)) ## permit text
> >> outside the panel
> >> ) +
> >> ## "+" must be on same line as first statement.
> >
ia adj you can control, if the text should adjust left, center or right
> > to the coords, and above, center or bottom of them.
> >
> > HTH,
> > Rainer Hurling
> >
> >
> > Am 22.09.2016 um 16:04 schrieb Jun Sh
Dear list,
Just wonder if there is a way to add annotation text outside an xyplot,
(e.g. the bottom of the plot). the panel.text seems only add text within
the plot. Thanks.
Jun
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ewmiller
wrote:
> Here are some suggestions:
>
> test.string <- c( '240.m.g.>110.kg.geo.mean'
> , '3.mg.kg.>110.kg.P05'
> , '240.m.g.>50-70.kg.geo.mean'
> )
> # based on your literal id
sted in this thread. Now I need to figure out a
way to match the patterns to the strings so each string can be parsed
correctly. I have made some progress...
Jun
On Wed, Sep 7, 2016 at 9:34 AM, Ista Zahn wrote:
> On Tue, Sep 6, 2016 at 11:59 PM, Jun Shen wrote:
> > Hi Ista,
>
ings)) print(sub(patterns[i], "\\2", strings[i]))
>
> mapply(sub, pattern = patterns, x = strings, MoreArgs=list(replacement =
> "\\2"))
>
> library(stringi)
> stri_replace_all_regex(strings, patterns, "$2")
>
> Best,
> Ista
> On Tue, Sep
can find someone locally to
> >help you if you do not receive a satisfactory reply from the list.
> >There are many people here who are pretty good at this sort of thing,
> >but I don't know if they'll reply. Regex's are certainly complex. PERL
> >people t
are numerous
> web sites and books on them if you need to acquire expertise for your
> work.
>
> Cheers,
> Bert
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed
..dd.ef.tgf.", "foo...b.bbb.tgy")
> >
> >> pat1 <- "a+\\.*a+"
> >> pat2 <-"b+\\.*b+"
> >> pat <- c(pat1,pat2)
> >
> >> pat <- do.call(paste,c(as.list(pat), sep="|"))
> >> pat
> > [
016 at 9:30 PM, Jeff Newmiller
> wrote:
> > Your opening assertion is false.
> >
> > Provide a reproducible example and someone will demonstrate.
> > --
> > Sent from my phone. Please excuse my brevity.
> >
> > On September 4, 2016 9:06:59 PM PDT, Jun Shen
Dear list,
I have a vector of strings that cannot be described by one pattern. So
let's say I construct a vector of patterns in the same length as the vector
of strings, can I do the element wise pattern recognition and string
substitution.
For example,
pattern1 <- "([^.]*)\\.([^.]*\\.[^.]*)\\.(
y business.I'm assuming this is not the case... Again, no data...
>
>
>
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom Count
Dear list,
I have the following line of code to extract the last line of the split
data and put them back together.
do.call(rbind,lapply(split(simout.s1,simout.s1[c('SID','DOSENO')]),function(x)x[nrow(x),]))
the problem is when have a huge dataset, it takes too long to run.
(actually it's > 3 h
Thanks David.This is working perfectly!
On Fri, Jul 29, 2016 at 9:00 PM, David Winsemius
wrote:
>
> > On Jul 29, 2016, at 5:52 PM, David Winsemius
> wrote:
> >
> >
> >> On Jul 29, 2016, at 5:08 PM, Jun Shen wrote:
> >>
> >> Thanks Jeff/
2','ALPHA'))
gives
meansd median min max N
CL-0.0232 0.918 -0.0786 -2.14 3.14 99
V1-0.0410 0.936 -0.1160 -2.86 2.67 99
V2-0.1760 0.978 -0.1490 -2.31 2.15 99
ALPHA -0.1380 0.960 -0.2160 -2.41 2.20 99
there is one non-missing value in V1,V2 and ALPHA is o
iller
wrote:
> Why not remove it yourself before passing it to those functions?
> --
> Sent from my phone. Please excuse my brevity.
>
> On July 28, 2016 5:51:47 PM PDT, Jun Shen wrote:
> >Dear list,
> >
> >I write a small function to calculate multiple stats on mu
Dear list,
I write a small function to calculate multiple stats on multiple variables
and export in a format exactly the way I want. Everything seems fine until
NA appears in the data.
Here is my function:
do.stats <- function(data, stats.func, summary.var)
as.data.frame(signif(sappl
Thanks Ulrik and Bob for your reply.
gsub worked for one column!
If I want to replace the whole data frame, gsub doesn't seem to work. Any
idea
On Fri, May 27, 2016 at 11:14 AM, Ulrik Stervbo
wrote:
> use gsub()
>
> On Fri, 27 May 2016 at 17:12 Jun Shen wrote:
>
>>
Dear list,
Say I have a data frame
test <- data.frame(C1=c('a,b,c,d'),C2=c('g,h,f'))
I want to replace the commas with semicolons
sub(',',';',test$C1) -> test$C1 will only replace the first comma of a string.
How do I replace them all in one run? Thanks.
Jun
[[alternative HTML versio
Dear list,
Say I have some quantile operation like this
data.frame(ID=rep(1:10,each=10),CONC=runif(100)) -> test
lapply(c('mean','sd','quantile'), function(x)
aggregate(test['CONC'],by=test['ID'],FUN=x)) -> temp
The output temp is a list of three elements. I would like to merge the
three elemen
.*)", "\\1", test)
> [1] "aaa" "aaa" "aaa" "aaa" """"
> > sub("([^.]*)(.*)", "\\2", test)
> [1] ".bb.cc" ".dd""" "." ".eee"
Dear list,
Say I have a vector that has two different types of string
test <- c('aaa.bb.cc','aaa.dd')
I want to extract the first part of the string (aaa) as a name and save the
rest of the string as another name.
I was thinking something like
sub('(.*)\\.(.*)','\\1',test) but doesn't give me
" is dynamically constructed through paste()
id <- 1:4
# I can get the element by doing
get(test)[2]
# Now I want to assign a value to the second element of this dynamical
variable.
get(test)[2] <- 5 # doesn't work.
Thanks a lot.
Jun Shen
Dear list,
I define a function to export a bunch of plots. A sample code is something
like the follow
export.plots<-function(export.type='pdf',...){
match.fun(export.type) (...)
print(plot.func1(...))
print(plot.func2(...))
print(plot.func3(...))
...
print(plot.funcn(...))
dev.off()
}
If I d
Dear list,
I have a couple of lines (superimposed) in an xyplot and just want to add
error bars to each of the data point. It's been a while since this question
was asked last time. But the segplot from latticeExtra is not straight
forward. Just wonder if there is a better way to do this. Apprecia
Dear list,
The bwplot generates box plots with a dot in median. How do I add a line of
"mean" to the boxes? Thanks a lot.
Jun
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Thanks for everyone who replied with those wonderful solutions!
Jun
On Thu, Aug 28, 2014 at 2:11 PM, Enrico Schumann
wrote:
> On Thu, 28 Aug 2014, Jun Shen writes:
>
> > Hi everyone,
> >
> > I believe I am not the first one to have this problem but couldn't find
Hi everyone,
I believe I am not the first one to have this problem but couldn't find a
relevant thread on the list.
Say I have a string (actually it is the whole column in a data frame) in a
format like this:
test<- 'AF14-485-502-89-00235'
I would like to split the test string and keep the last
ot;
$ NRT :Class 'labelled' atomic [1:1991] 0 0.75 1.5 3 4 6 9 12 24 0 ...
.. ..- attr(*, "label")= chr "Nominal Relative Time"
$ TIME:Class 'labelled' atomic [1:1991] -1.1 0.8 1.5 3 4 ...
.. ..- attr(*, "label")= chr "Actual Rel
Dear list,
I used sasxport.get to import a SAS xpt file. Although it is a data frame
but i can't view it through the "fix" command. Also when I see its
structure, it brings up attributes I am not really interested in (which
seems part of the SAS labels) and it doesn't seem to tell me the mode of
e
gt;
>
> On Friday, June 20, 2014 3:43 PM, Jun Shen wrote:
> Hi all,
>
> Here is my situation. I have a dataframe, the structure would be something
> like this,
>
>
> TestData<-data.frame(ID=rep(1:10,each=10),TIME=rep(seq(0.1,1,0.1),10),VAR1=rnorm(100),VAR2=5*rnorm(100),
13.39844 16.965365 8 2.7447862 10.21613 22.545131 9 3.490395
10.83543 25.744662 10 0.4719087 11.73021 7.226687
Thanks for any help.
Jun Shen
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;
> >>
> >> -Original Message-
> >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Jun Shen
> >> Sent: Wednesday, May 21, 2014 10:32 AM
> >> To: R-help
> >> Subject: [R] How to
ygon() for such a
feature. Is there another way to do it? Thanks.
Jun Shen
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.
The strange thing is if I use level=0.8, it is working again. Why is that?
How do I get an CI value for any level? Thanks.
Jun Shen
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Hi everyone,
I am using nls() to run a simple sigmoid curve type of regression and
would like to see how the parameter values change through iterations. How
can I see those values? Or ideally, can I even extract those values?
For example,
nls(Response~E0+Emax*Conc/(EC50+Conc), data=data, start=
Hi everyone,
I am using the axis command to draw extra axis on the right. How do I
change the text size of the labels as in
axis (4, at=c(1:9), labels='some text', las=1).
I can't find an argument to change the text size for the labels. Thanks.
Jun
[[alternative HTML version deleted]]
Dear all,
I have a print command to export some intermediate results from a
user-defined function. It takes a while to run the function and I found I
have to press a key to see the printed results on the screen. How can I ask
the function to continuously print results on the screen without pressin
Hi everyone,
This may be very simple but I couldn't figure it out. I have this function
rsm.lm<-function(data,xvar='xCmin',yvar='yCmin') {
#some calculation..
recast(df, xvar~yvar, id.var=1:2, measure.var=3)
#some other operations
}
df is a dataframe defined internally within the functi
Thanks again, Duncan. Please allow me to ask one more question. Is it
possible to generate a contour plot overlaying with the plot3d() plot?
Jun
On Thu, Dec 5, 2013 at 11:08 AM, Duncan Murdoch wrote:
> On 05/12/2013 10:33 AM, Jun Shen wrote:
>
>> Hi Federico/Duncan/David/Bert,
>
d, Dec 4, 2013 at 4:37 PM, Bert Gunter wrote:
> ... or, more simply
>
> lm(z ~ polym(x,y, degree=2) )
>
> ?polym
>
> Cheers,
> Bert
>
>
>
>
>
> On Wed, Dec 4, 2013 at 10:30 AM, David Winsemius
> wrote:
> >
> > On Dec 4, 2013, at 8:56 AM,
Hi,
I have a dataset with two independent variables (x, y) and a response
variable (z). I was hoping to generate a response surface by plotting x, y,
z on a three dimensional plot. I can plot the data with rgl.points(x, y,
z). I understand I may not have enough data to generate a surface. Is there
Dear all,
This question was asked a few years ago. Back then, the answer was NO. Just
wonder if the package has been updated to make it possible.
An example (Theoph is a dataset coming with R)
ggplot(data=Theoph, aes(x=Time, y=conc)) + geom_point() +
facet_wrap(~Subject)
gives me all 12 subject
Hi all,
I wonder if there is a way to draw two separate legends in xyplot as I
would like to separate the legend for data and the legend for reference
lines I add. I can use key argument to draw one legend with everything
together. What I really want is to put one legend at the bottom and the
othe
NA, 5L), class = "data.frame")
Here is the calculation I did originally.
>with(test,strptime(SPDTC,format='%Y-%m-%dT%H:%M')-strptime(TimeBL,format='%Y-%m-%dT%H:%M'))
The result is in seconds.
On Fri, Aug 9, 2013 at 1:55 PM, David Winsemius wrote:
>
> On Aug 9, 2
Hi all,
I used strptime() to convert character strings to time and did some
subtraction calculation.
a<-'2012-07-17T07:50'
b<-'2012-08-27T09:30'
strptime(a,format='%Y-%m-%dT%H:%M')-strptime(b,format='%Y-%m-%dT%H:%M')
The result shows
Time difference of -41.06944 days.
However when these opera
Dear list,
Just wonder if there is a limit on the number of code line in .Rhistory.
When I check the .Rhistory, some code I enter at earlier time disappeared.
If there is limit, how do I change it? Thanks.
Jun
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Dear list,
Let's say we have a data frame as follows,
>expand.grid(a=1:5,b=c(1,5,10,20),DV=c(0.1,0.2,0.3))->df1
>df1$DV<-rgamma(60,shape=10)
columns a and b are two levels. DV is the column with values we are
interested in.
Then another data frame df2 with values as follows
data.frame(a=c(2,2,
;^d'&'[[:digit:]]$') is not working.
Appreciate any comment.
Jun Shen
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et with 46 subjects.
Now I want to simulate 100 subjects. So the degrees of freedom should be
46-6=40 or 100-6=94 or some other value? Thanks a lot.
Jun Shen
Seventh Wave Labs
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Ehlers wrote:
> On 2012-04-16 08:51, David Winsemius wrote:
>
>>
>> On Apr 16, 2012, at 11:43 AM, Jun Shen wrote:
>>
>> Dear list,
>>>
>>> As the title indicates if I draw the legend outside of the plot, how
>>> do I
>>>
Dear list,
As the title indicates if I draw the legend outside of the plot, how do I
adjust the distance between the legend and the plot? The default setting is
too close. Thanks.
Here is the sample code I have
==
xyplot(1~1,key=list(space=
Dear list,
If I know the standard error for k1 and k2, is there anything I can call in
R to calculate the standard error of k1/k2? Thanks.
Jun
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Dear list,
Let's say I have data
a=c(37.961,38.214,57.68)
b=c(77.56,61.875,67.683)
wilcox.test(a,b)
the wilcoxon test only gives me a p value with one decimal place. Is this
normal? Thanks.
Jun
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R-he
lar/BatteriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> -------
> Sent from my phone. Please excuse my brevity.
>
> Jun Shen wrote:
>
> >Dea
Dear all,
In a plot command like
xyplot(Y~X|ID*PERIOD,data=...)
xyplot will generate all the possible ID*PERIOD combinations. But not all
of them have data in there. So I have a lot of empty plots. How can I
suppress those empty plots and ask xyplot only to generate plots actually
with data. Tha
Thanks for sharing the idea, Walmes.
I am also trying to add err bars to the plots. So instead of using plot(), I
use errbar() from Hmisc package. Everything seems to work fine except I
couldn't find the argument to control the color of error bars. If I specify
col='red' in the errbar() it only co
7;,'KOUT','GAMMA'),function(x)confint(lm(get(x)~RR0,dataset2)))
> names(dat) <- c('EMAX','EC50','KOUT','GAMMA')
>
> On Fri, May 27, 2011 at 10:03 AM, Jun Shen wrote:
> > Dear list,
> >
> > I am running
(Intercept) -7.75121986 15.20827567
RR0 -0.01174097 0.02220318
Thanks a lot.
Jun Shen
=
structure(list(ID = c(50, 51, 52, 53, 55, 56, 57, 59, 60, 61,
100, 101, 104, 107, 108, 110, 112, 114), KOUT = c(3.8336, 3.7519,
3.7136, 3.
Thanks a bunch, Walmes.
One more concern, the new Y axes added do not extend all the way down to
cross with x axis. Is there anyway to make them look like the very first Y
axis on the left?
Jun
On Thu, May 26, 2011 at 1:24 PM, Walmes Zeviani wrote:
> You can use mtext()
>
> par(mar=c(5.1,4.1,4.
David,
I surely tried the labels argument. But it seems for tick marks not for a
text label. Did you see a different outcome? Thanks.
Jun
On Thu, May 26, 2011 at 12:38 PM, David Winsemius wrote:
>
> On May 26, 2011, at 1:25 PM, Jun Shen wrote:
>
> Jim,
>>
>> One more
= '', col =
> 'blue')
> axis(4, col='blue', line = -3)
>
>
> You have to not plot the axises on the secondary plots.
>
> On Thu, May 26, 2011 at 12:11 PM, Jun Shen wrote:
> > Hi, Jim,
> >
> > Thanks fo
I just found out by setting bty='l' to get rid of the border line on the
top.
On Thu, May 26, 2011 at 11:41 AM, Jun Shen wrote:
> Hi, jim
>
> That's exactly what I wanted. One more trivial thing. How do I get rid the
> border line on the top? Thanks again.
>
>
#x27;', col =
> 'blue')
> axis(4, col='blue', line = -3)
>
>
> You have to not plot the axises on the secondary plots.
>
> On Thu, May 26, 2011 at 12:11 PM, Jun Shen wrote:
> > Hi, Jim,
> >
> > Thanks fo
confusing, but it can be done. What have you tried and why
> do you say "guess not"? With the use of par(new=TRUE) or by doing
> your own scaling, you can use 'axis' to put as many axises as you want
> on your graph.
>
> On Thu, May 26, 2011 at 11:35 AM, Jun
Dear list,
We have three time course profiles with very different scales, and we want
to show them in one plot. Is it possible to have three y axis? I guess not,
then what would be other options? something like two 2-y axis plots on a
three dimensional view? Appreciate any comment.
Jun Shen
Hi, Gabor,
The interim fix solved the problem! Now it's working perfectly. And thanks
for everyone's input.
Jun
On Tue, May 10, 2011 at 10:39 AM, Gabor Grothendieck <
ggrothendi...@gmail.com> wrote:
> On Tue, May 10, 2011 at 12:12 AM, Jun Shen wrote:
> > Kenneth,
>
sub
> function.
>
> HTH
>
> Kenneth
> El lun, 09-05-2011 a las 22:44 -0500, Jun Shen escribió:
> > Dear list,
> >
> > I used read.xls to read in an excel file, which has both character and
> > numeric columns. Everything seems fine except in the last column,
e first and the last white spaces with gsub
> function.
>
> HTH
>
> Kenneth
> El lun, 09-05-2011 a las 22:44 -0500, Jun Shen escribió:
> > Dear list,
> >
> > I used read.xls to read in an excel file, which has both character and
> > numeric columns. Every
Dear list,
I used read.xls to read in an excel file, which has both character and
numeric columns. Everything seems fine except in the last column, I have
this character value "ng/ml". When reading in, read.xls seems to add a space
at the end of it, became "ng/ml ". How can I prevent read.xls doin
Dear list,
This may sound silly. What is the right way to change the names of a
dataframe? Let's say I have this data frame (dose) with four columns with
names "ID", "DOSE", "TIME" "CMT". I want to change "DOSE" to "AMT". So I did
names(dose[2])<-'AMT'
But nothing happened. The name of the secon
Apr 30, 2011 at 6:12 AM, David Winsemius wrote:
>
> On Apr 29, 2011, at 7:17 PM, Jun Shen wrote:
>
> Hi, Dennis,
>>
>> Thanks for the reply. I tried to upgrade to R 2.13.0. Then when I tried to
>> load the library(nparcomp), I got an error
>>
>> Error: pack
] base
>
> other attached packages:
> [1] nparcomp_1.0-1 multcomp_1.2-5 survival_2.36-9 mvtnorm_0.9-999
> [5] sos_1.3-0 brew_1.0-6 plyr_1.5.2
>
> loaded via a namespace (and not attached):
> [1] tcltk_2.13.0 tools_2.13.0
>
> Check your version of R and the npar
Dear list,
I tried to use the nparcomp to run some post hoc non-parametric comparison
and got and error.
Error in uniroot(pfct, interval = interval) :
f() values at end points not of opposite sign
Appreciate any comments.
the command line:
>nparcomp(Ulceration~Group,data=test,type='Dunnett'
call()[[2]][[3]]])
>
> On Fri, Apr 15, 2011 at 3:14 PM, Jun Shen wrote:
> > Dear list,
> >
> > I want to loop over a list through lapply and at the same time I want to
> > extract some information from the names of the list elements. lapply does
> > not seem
Dear list,
I want to loop over a list through lapply and at the same time I want to
extract some information from the names of the list elements. lapply does
not seem to carry the names. Is there anyway to access the names of a list
within lapply? Thanks.
Jun
[[alternative HTML version d
Hi,
I have this symmetric matrix, at least I think so.
col1 col2 col3
[1,] 0.20 0.05 0.06
[2,] 0.05 0.10 0.03
[3,] 0.06 0.03 0.08
or
structure(c(0.2, 0.05, 0.06, 0.05, 0.1, 0.03, 0.06, 0.03, 0.08
), .Dim = c(3L, 3L), .Dimnames = list(NULL, c("var1", "var2",
"var3")))
But isSymmetric() d
Hi, Dennis,
Does this give what you want?
>sum(resid(FIT)^2)
Jun
Senior Pharmacokineticist
Seventh Wave Labs
On Fri, Aug 13, 2010 at 4:21 PM, Dennis Fisher wrote:
> Colleagues,
>
> I am using "nls" successfully (2.11.1, OS X) but I am having difficulties
> retrieving part of the output - resi
;t see it on the screen. Say
>summary(horton.nlme)
give the output back to console
>sink()
Hope this helps.
Jun Shen
Senior Pharmacokineticist
Seventh Wave Labs
On Thu, Aug 5, 2010 at 2:30 PM, Ronald Wendt wrote:
> I'm trying to export the results of my summary data for the o
Hi, Bill,
Thanks. This is what I am looking for.
Jun
On Thu, Jul 22, 2010 at 1:54 PM, William Dunlap wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of Jun Shen
>> Sent: Thursday, July 22, 201
Dear all,
Is there anyway I can fix slope in some value and only estimate
intercept for a linear regression? I understand lm(y~1, data) will NOT
have a slope at all. This is not what I want. Thanks.
Jun
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Hi, Juliet,
Something like this?
mymat[mymat<0]<-min(mymat[mymat>0])
Jun
On Thu, Jul 15, 2010 at 10:55 AM, Juliet Hannah wrote:
> Hi Group,
>
> I have a matrix, and I would like to replace numbers less than 0 by
> the smallest minimum number. Below is an
> small matrix, and the loop I used. I
issues.
>
> Bottom line: there may be no good way to do what you want.
>
> Note to experts: Please view this post as an invitation to correct my errors
> and provide authoritative info.
>
> Cheers to all,
>
> Bert
>
> Bert Gunter
> Genentech Nonclinical Biostatist
Dear all,
I want to do a F test, which involves calculation of the degrees of
freedom for the residuals. Now say, I have a nlme object "mod.nlme". I
have two questions
1.How do I extract the degrees of freedom?
2.How is this degrees of freedom calculated in an nlme model?
Thanks.
6,
2.99, 3.75, 3.02, 5.44, 3.08, 8.31, 10.85, 13.79, 32.06, 50.22,
63.7, 81.34, 89.59, 93.06, 92.47, 3.32, 1.14, 2.43, 2.75, 3.02,
5.4, 8.49, 7.91, 15.17, 35.01, 53.91, 68.51, 83.12, 86.85, 92.17,
95.72, 3.58, 0.02, 3.69, 4.34, 6.32, 5.15, 9.7, 11.39, 23.38,
42.9, 61.91, 71.82, 87.83)), .Names
efined function. It
seems the "mod.nlme" created in a user defined function can not be
called in xyplot(). Why is that? Appreciate any comment. (The error
message says " Error in using packet 1, object "model" not found")
Thanks.
Jun Shen
___
in a positive definite diagonal
matrix by:
random = list(pdDiag(a+b~1))
My question is how I can restart nlme within my function. If random = a+b~1
fails, it won't quit and will go back to try random = list(pdDiag(a+b~1))
automatically. Currently I just have to do this manually. Thanks a lot.
sample(1:20,4,replace=TRUE) should do it.
Jun
On Wed, Jun 16, 2010 at 9:20 AM, Somnath Somnath wrote:
> Dear all, good morning,
>
> I have a population, let say members are tagged with some simple number
> like
> 1,2,3,...20. I want to draw a sample with replacement of size 4 (say, can
> be
> mo
width=5, height=5)
>
> which shouldn't be there.
>
> HTH,
>
> baptiste
>
> On 17 May 2010 22:23, Jun Shen wrote:
>> If you do plotting in a loop, then you need to print it to the device.
>>
>> print(plot(xj,y))
>>
>> On Mon, May 17, 2010
ause it has no pages."
>
> Here is what I did in plotting and saving the file:
>
> pdf(file="C:/figure.pdf")
> for (j in 1:numColumns)
> {
> windows(width=5, height=5)
>
> plot(xj,y)
> }
> dev.off()
>
> Any ideas what might cause the problem? Th
1.Open pdf device
>pdf()
2.Do your plotting as many as you want, you won't see the plots on the
screen because they go directly to the pdf() device.
3.Turn off the pdf()
>dev.off()
Then you can review your plots in the pdf file. For more details see ?pdf
Jun
On Mon, May 17, 2010 at 2:41 PM, Shirl
, 7L, 12L, 15L,
21L, 27L, 33L, 39L, 45L, 48L, 54L, 60L, 65L, 67L, 71L, 76L, 81L,
86L, 91L, 93L, 98L, 103L, 108L, 112L), class = "data.frame")
On Fri, May 7, 2010 at 7:51 PM, David Winsemius wrote:
>
> On May 7, 2010, at 5:05 PM, David Winsemius wrote:
>
>
>> On May 7, 2010
}
conf <- if (do.conf)
stats[3L] + c(-1.58, 1.58) * iqr/sqrt(n)
list(stats = stats, n = n, conf = conf, out = if (do.out) x[out &
nna] else numeric(0L))
}
On Fri, May 7, 2010 at 3:09 PM, David Winsemius wrote:
>
> On May 7, 2010, at 3:30 PM, Jun Shen w
;
the part of the args list of 'is.na' being evaluated was:
(x)
Thanks a lot.
Jun Shen from Millipore Corporation
my.boxplot.stats<-function (x, coef = 1.5, *type=6*, do.conf = TRUE, do.out
= TRUE)
{
if (coef < 0)
ny other letter. Can anyone explain it a little bit?
Jun
On Sat, Apr 24, 2010 at 4:14 PM, David Winsemius wrote:
>
> On Apr 24, 2010, at 4:52 PM, Jun Shen wrote:
>
> Hi,
>>
>> I also want to show user's name at the lower-left corner. So I added one
>> more
Hi,
I also want to show user's name at the lower-left corner. So I added one
more line to Deepayan's code but it didn't work. I am not sure how to use
the "page" argument here. Appreciate any comment.
lattice.options(default.args = list(page = function(n) {
panel.text(lab = sprintf("%s", date(
Hi, Deepayan,
This is exactly what I want. However I couldn't find "page" argument or
"default.args" when I look up the documentation. I just want to learn a bit
more how to use these arguments. Thanks again.
Jun
On Fri, Apr 23, 2010 at 10:32 AM, Deepayan Sarkar wrote:
> On Wed, Apr 21, 2010 a
David, Duncan,
Thanks. I didn't realize the functions in "graphic" also work in "lattice".
I thought they didn't work with each other.
Jun
On Wed, Apr 21, 2010 at 6:51 PM, Duncan Murdoch wrote:
> On 21/04/2010 5:41 PM, Jun Shen wrote:
>
>> Dear all,
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