Bioconductor,
* Bioconductor software packages
<http://www.bioconductor.org/packages/bioc/>
Under the section 7.39 How do I create a plot with two y-axes?,
* http://rwiki.sciviews.org/doku.php?id=tips:graphics-base:2yaxes
I appreciate your helps!
Chel Hee Lee
_
http://windowsvistablog.com/blogs/windowsvista/archive/2007/01/23/security-features-vs-convenience.aspx
I appreciate your help!
Chel Hee Lee
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PLEASE do re
Thank you so much, Jim. Yes, I also see that the wiki
(rwiki.sciviews.org) is not found.
Chel Hee Lee
On 15-06-25 04:54 AM, Jim Lemon wrote:
Hi Chel Hee,
The last link (for FAQ 7.39) is dead and the domain
(rwiki.sciviews.org) is not there either.
Jim
On Thu, Jun 25, 2015 at 3:45 AM, Chel
Thank you so much, David! I also confirmed that the link moved is not
found.
Chel Hee Lee
On 15-06-24 04:15 PM, David Winsemius wrote:
On Jun 24, 2015, at 10:52 AM, Chel Hee Lee wrote:
Could you also kindly check the following links in the rw-FAQ manual at
<http://cran.r-project.org/
this project. Please
contact me if you voluntarily participate in or offer your help with
this project of translating R manuals. Comments and corrections is of
course most welcome.
Chel Hee Lee
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class
[1] "rowwise_df" "tbl_df" "tbl""data.frame"
$row.names
[1] 1 2 3 4 5 6 7
> attributes(new2)
$names
[1] "id" "mrjdate" "cocdate" "inhdate" "haldate" "oiddate"
$row.names
[1] 1 2 3
1.65
2 1 3.3 1.65
3 1 3.3 1.65
4 2 10.0 5.00
5 2 10.0 5.00
6 3 30.0 15.00
7 4 40.0 20.00
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/03/2014 03:14 PM, Brock Huntsman wrote:
I apologize if this is
tX = rbind(tX, as.data.frame(cbind(GId =0,Grp = 0,
S = fm, T = fm)))
Of course you have to do little more works with this change in order to
get final bar plots. I hope this helps.
Chel Hee Lee
On 12/03/2014 12:29 PM, Charles R Parker wrote:
I am trying to create groups of barplot
This question seems to be the problem specific to Ubuntu. What if you
post the message to ?? I hope you get
answers from that mailing list.
Chel Hee Lee
On 12/02/2014 11:10 AM, VG wrote:
Hi everyone,
I was having trouble with R i installed some time ago on my local ubuntu
machine. So i
0,19.0,24.0),right=FALSE)
> levels(achtergrond$spits) <-
c("between","morning","between","evening","between")
> table(achtergrond$spits)
between morning evening
1636 579 142
>
You can also use function 'findInterval()
example_02.dat" "new_example_03.dat"
[4] "new_example_04.dat" "new_example_05.dat" "new_example_06.dat"
>
> file.rename(from=file_names, to=new_names)
[1] TRUE TRUE TRUE TRUE TRUE TRUE
> list.files(pattern="*.dat")
[1] "new_ex
I see that a function 'format()' is used in your code.
> format(c(1,5,32,100), width=3, flag="0")
[1] " 1" " 5" " 32" "100"
> formatC(c(1,5,32,100), width=3, flag="0")
[1] "001" "005" "032&quo
.5)
> abline(a=0, b=-0.5)
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/05/2014 02:30 AM, Adrien Bonvin wrote:
Bonjour
Hi everybody,
Firs of all, sorry for my terrible English,
I would like to know if it’s possible to create an “empty plot” in which i
could ad
x27; as shown in the below:
> sprintf("%05d", seq(from=200, to=1200, by=500))
[1] "00200" "00700" "01200"
I like to read the documentation by typing
> help(formatC)
> help(sprintf)
You may find answers what you wish to get. Documentation has been my
if x < 0.3. x= 0.4246928
Anyway, move to next!
[1] 0.2535644
Error is printed if x < 0.3. x= 0.5071288
Anyway, move to next!
[1] "error"
Error is printed if x < 0.3. x= 0.1941202
Anyway, move to next!
(continue)
I hope this helps, and thank you so much, Martin Morgan, for your go
It seems that you would like to make a spaghetti plot (in a longitudinal
data analysis). You can use the function 'interaction.plot()'.
> with(my.df, interaction.plot(TIME, ID, X))
I hope this helps.
Chel Hee Lee
On 12/06/2014 02:24 AM, arun wrote:
Not sure whether it is
6.970592 4 6.970592
CID_10.31 CID_10 Acidobacteria1.988448 5 1.988448
CID_10.32 CID_10 Actinobacteria1.644548 6 1.644548
CID_10.33 CID_10 Others 1.582823 7 5.546493
>
I hope this helps.
Chel Hee Lee
On 12/07/2014 08:21 AM, Morway, Eric wrote:
Usin
"5001,20001" "10001,20001" "15001,20001"
[11] "1,25001" "5001,25001" "10001,25001" "15001,25001" "20001,25001"
> x.idx2 <- upper.tri(x, diag=TRUE)
> x[x.idx2]
[1] "1,1" "1,
NANA NA
RK.Records_CL V.Other
1005317NA NA
1007183NA NA
1008833NA NA
1012281NA NA
1015285 1 NA
1015315 NA NA
1015322NA NA
>
Chel H
t in R. Is this what you are looking for? I hope
this helps. I would also appreciate it if you would provide
reproducible examples next time.
Chel Hee Lee
On 12/18/2014 11:48 AM, Jeff Newmiller wrote:
Read the posting guide. Th
iol.Type'
where 'factor()' is used. I hope this helps.
Chel Hee Lee
On 12/18/2014 08:57 PM, Crombie, Burnette N wrote:
Chel, your solution is fantastic on the dataset I submitted in my question but
it is not working when I import my real dataset into R. Do I need to vectorize
dx], collapse=" ")
+ unlist(strsplit(comb, " "))
+ })
> names(tmp) <- d0$ID
> tmp
$MCZ4325
[1] "C23.2" "T43.2"
$GDR2343
[1] "M20.64" "Y32.1" "M20.64" "T44.2"
$BZD2643
[1] "B83.2" "T43.2" "B83.2&q
0 Out
x8 0 0 0 0 0 B 0
2015-04-07
x1 0
x2 0
x3 0
x4 0
x5 0
x6 A
x7 0
x8 0
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/19/2014 10:45 PM, Kristi Glover wrote:
Hi R User, Would you suggest me
You are welcome, I am glad that I was able to help.
Chel Hee Lee, PhD
Biostatistician and Manager
Clinical Research Support Unit
College of Medicine
University of Saskatchewan
On 12/20/2014 03:53 AM, Kristi Glover wrote:
Thank you Prof. Lee for your code. I am sorry that I noticed that I sent
1, labels=paste("S", 1:6, sep="_"))
> points(x=2, y=4, cex=3)
>
Please see the output. Is this what you are looking for? I hope this
helps.
Chel Hee Lee
On 12/21/2014 01:25 PM, eliza botto wrote:
Dear UseRs,
A point was plotted by the following command
plot(2,4,y
; length(x[unlist(x)])
[1] 1
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/22/2014 7:45 AM, najuzz wrote:
#Hi guys,
#I would like to count the number of individuals that receive X=0 troughout
their observational period.
#example dataset:
ID<-c(1,1,1
#x27; and 'LC_MESSAGES'
in order to change messages in a given locale? However, it is
questionable. Please see the following example (messages in Japanese):
> Sys.setenv(LANGUAGE="ja_JP.utf8")
> a
エラー: オブジェクト 'a' がありません
Thank you so much. Your explanation is very clear and helpful to
understand. Thank you, again.
Chel Hee Lee
On 1/12/2015 5:56 AM, Prof Brian Ripley wrote:
You don't want to change the locale, rather the message language. That
is not a locale, and e.g.
LANGUAGE=en
is the correct
This approach may not be fancy as what you are looking for.
> xl <- unlist(x)
> xl[grep("A", names(xl))]
f1.x1.A f1.x2.A f2.x3.A f2.x4.A
11 12 13 14
>
I hope this helps.
Chel Hee Lee
On 01/16/2015 04:40 AM, Rainer M Krug wrote:
Hi
Consider the fo
222.247934 226.107438 230.00
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/16/2015 02:16 AM, philippe massicotte wrote:
Hi all.
How we evaluate a formula in R?
Ex.:
params <- list(a = 2, b = 3)
x <- seq(1,10, length.out = 100)
func1 <- as.formu
ine(lm(y~x), col="red")
panel.text(1, -1, bquote(R^2 == .(summary(lm(y~x))$r.squared)))
},
grid = TRUE
)
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/17/2015 07:08 AM, Naresh Gurbuxani wrote:
In a conditional xyplot, I would like to add some numeri
3.15
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/19/2015 02:14 PM, Shuhua Zhan wrote:
Dear All,
I'd like to add a list name into the list contents to make a new output. The list is a
list of data.frame derived from summary command in Rqtl. I want to add th
32 33 34
> extr.1(x=x, name="A")
f1.x1.A f1.x2.A f2.x3.A f2.x4.A
11 12 13 14
> extr.1(x=x, name="a")
named numeric(0)
> extr.1(x=x, name="Aha")
f1.x1.Aha f1.x2.Aha f2.x3.Aha f2.x4.Aha
2122232
6.75
6 heart_wt c12.loc49 12 51.2 28.2062.2 3.65
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/20/2015 9:23 AM, Shuhua Zhan wrote:
Hi Chel,
Thank you very much for your help!
I'm sorry I did not post my wanted output correctly. I only want the colnames
o
is helps. I would appreciate if you DO READ the posting guide.
Chel Hee Lee
On 1/20/2015 9:30 AM, Ragia Ibrahim wrote:
Hello,
kindly ,
how to catch this Error
Error in A[[i]] : subscript out of bounds
and check that the list is empty is.null( A[[i]] ) do no twork
thanks in advance
R.I
> x <- c("hola mundo mundo");
> table(unlist(strsplit(x, " ")))
hola mundo
1 2
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/22/2015 8:25 AM, bgnumis bgnum wrote:
Hi all,
I want to cout the different words in a text.
You
7+8)
> new1
[1] 3 3 15 15
>
> grp.id <- c(1,1, 2, 3,3,3, 4,4)
> tapply(X=a, INDEX=grp.id, FUN=sum)
1 2 3 4
3 3 15 15
> aggregate(x=a, by=list(grp.id), FUN=sum)
Group.1 x
1 1 3
2 2 3
3 3 15
4 4 15
> xtabs(formula=a~grp.id)
grp.id
1 2 3 4
> do.call(rbind, lapply(split(data, data$Name), function(x)
x[order(x$CheckInDate),][nrow(x),]))
Name CheckInDate Temp
John John 2014-04-01 99.0
Mary Mary 2014-03-01 98.1
Sam Sam 2014-04-01 97.5
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/23/2015
1
>
I hope this helps.
Chel Hee Lee
On 01/28/2015 07:38 PM, William Dunlap wrote:
with(dat1, ave(integer(length(Date)), Date, FUN=seq_along))
[1] 1 1 2 1 1 2 1 2 1 2 1
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jan 28, 2015 at 4:54 PM, Morway, Eric wrote:
The two datase
(c,lapply(tapply(ID1, gl(7,2), c), paste, collapse=""))
1234567
"AA" "TG" "CT" "GC" "GT" "CG" "TA"
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/28/2015 05:5
Hi Bert! yes, you are VERY correct!!! Why am I making this simple thing
so complicated??? ;) Thank you so much for your nice lesson!
Chel Hee Lee
On 01/28/2015 09:59 PM, Bert Gunter wrote:
eek!
Chel Hee,anything that complicated should engender fear and trembling.
Much simpler and more
2 2 b
3 3 c
4 4 d
5 5 e
> eval(parse(text=DFName))[,1]
[1] 1 2 3 4 5
> length(eval(parse(text=DFName))[,1])
[1] 5
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/29/2015 12:34 AM, Jeff Newmiller wrote:
This approach is fraught with dangers.
I reco
Or, you can also do the same job using 'colSums()' as shown in the below:
> colSums(status=="I")
2010 2011 2012
344
>
I hope this helps.
Chel Hee Lee
On 1/28/2015 7:31 PM, JSHuang wrote:
Hi,
I think you need quotation around I like the following:
st
Why don't try this?
> nams<-c("P2O5 ['%']", "D50 [mu*m]")
> pdf("test.pdf")
> for(i in 1:2)
+ plot(1,1,xlab=parse(text=(bquote(.(nams[i])
> dev.off()
Is this what you are looking for? I hope this helps.
On 1/30/2015 7:54 AM, PIKAL Petr wrote:
Thanks
So the key is to put expression(s) dir
> paste(rep(v1, each=3), 1:3, sep="_")
[1] "a_1" "a_2" "a_3" "b_1" "b_2" "b_3" "c_1" "c_2" "c_3"
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/30/2015 7:34 AM,
rmat(ex, format="%A")
[1] "Thursday" "Friday"
> weekdays(ex)
[1] "Thursday" "Friday"
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/31/2015 3:06 AM, Camilla Timo wrote:
Hi there,
I have a problem, i have a datase
"Rgui" "little"
"win.binary"
path.sep r_arch
";""x64"
>
Chel Hee Lee
On 1/30/2015 9:55 PM, Jim Lemon wrote:
Hi Ahmed,
Hmmm, this seems to work for me (R-3.1.2, Linux)
legend(0,2100, legend=c("2009","
00010
I hope this helps.
Chel Hee Lee
On 1/30/2015 11:52 AM, JS Huang wrote:
Hi,
Here is my implementation. Hope this helps.
b
[1] 1 2 3 4 5
c
[1] 1 2 1 3 5 4
sapply(b,function(x)ifelse(x==c,1,0))
[,1] [,2] [,3] [,4] [,5]
[1,]10000
[2,]
> lapply(1:2, function(x) t(A[rev(1:3),,x]))
[[1]]
[,1] [,2] [,3]
[1,] "g" "d" "a"
[2,] "h" "e" "b"
[3,] "i" "f" "c"
[[2]]
[,1] [,2] [,3]
[1,] "p" "m" "j&qu
lutions). I hope this helps.
Chel Hee Lee
On 2/10/2015 2:29 AM, Rolf Turner wrote:
On 10/02/15 14:04, Ssuhanchen wrote:
Hi!
I want to use R to calculate the variable x which is in a complex
equation
in below:
2
Σ[exp(-x/2)*(x^k)/(2^k*k!)]=0.05
k=0
how to solve this equation to get
A~ha~!! Thank you, Prof. Peter Dalgaard, so much for your wonderful
lesson!!! Learning new things everyday from this R-help mailing list!
Chel Hee Lee
On 2/11/2015 10:37 AM, peter dalgaard wrote:
On 11 Feb 2015, at 17:11 , Chel Hee Lee wrote:
The functional form given in the post
Another (simpler) way that I can think is that
> y * matrix(rep(z,3), ncol=ncol(y), byrow=TRUE)
[,1] [,2]
[1,]00
[2,]6 -3
[3,] 12 -6
I hope this helps.
Chel Hee Lee
On 14-11-19 08:43 AM, Ruima E. wrote:
> Thank you Chel Hee.
>
> Isn't there a sim
> y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2)
> z = matrix(c(12, -6),ncol=2)
> t(apply(y, 1, function(x) x*z))
[,1] [,2]
[1,]00
[2,]6 -3
[3,] 12 -6
I hope this helps.
Chel Hee Lee
On 14-11-19 08:22 AM, Ruima E. wrote:
> Hi,
>
> I have this:
>
> xx <- as.factor(c("AL", "AK", "CA", "FL"))
> xx
[1] AL AK CA FL
Levels: AK AL CA FL
> as.character(xx)
[1] "AL" "AK" "CA" "FL"
I hope this helps.
Chel Hee Lee
On 14-11-22 01:12 AM, Aditya Singh wrote:
x)))
grp y
1 2 0
2 3 0
What a surprise! Is this a bug? I would appreciate if you share the
results after testing the codes. Thank you so much for your helps in
advance!
Chel Hee Lee
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I appreciate your kind guidance! I did not read the manual carefully
(it's my fault).
Thank you so much, Prof. John Fox!
Chel Hee Lee
On 01/21/2016 12:52 AM, Fox, John wrote:
Dear Chel Hee Lee,
With the formula method, the default na.action is na.omit; thus,
aggregate(y~grp, dat
I will very appreciate if anyone can provide some materials to draw a
simplex plot of a Dirichlet distribution in R as shown in the page at
http://en.wikipedia.org/wiki/File:Dirichlet_distributions.png .
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https://stat.
Thank you, David. I found some good stuffs for which I have been
looking.
On Sun, 2012-08-26 at 09:01 -0700, David Winsemius wrote:
> On Aug 25, 2012, at 9:00 PM, Chel Hee Lee wrote:
>
> > I will very appreciate if anyone can provide some materials to draw a
> > simplex p
I have some trouble to deal the value of 'NaN'. For example,
> exp(1e3)
[1] Inf
> exp(1e3)*0
[1] NaN
The correct answer should be 0 rather than NaN. I will very appreciate
if anyone can share some technique to get a correct answer.
__
R-help@r-proje
#. Live Go...
> >Live: OO#.. Dead: OO#.. Playing
> > Research Engineer (Solar/BatteriesO.O#. #.O#. with
> > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> > ------------
I have a general question how to understand the message with "non-zero exit
status" when new package is installed. Based on my experience, this message
implies a package dependence. Am I correct to understand this message? Are
there anyone who can provide some reference or explain details about the
Oh... I see!!! I very appreciate for your clear explanation! Thank you, Uwe
Ligges.
--
View this message in context:
http://r.789695.n4.nabble.com/General-question-on-the-message-with-non-zero-exit-status-tp4551438p4568839.html
Sent from the R help mailing list archive at Nabble.com.
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