help.
HTH ....
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Simon, Heather
Sent: Thursday, 28 April 2016 8:17 a.m.
To: r-help@r-project.org
Subject: [R] polygon angle option perpendicular to axis
I am trying to use the angle option in polygon
So, do you have MikTeK installed (assuming you are using Windows)?
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Alnazer
Elbedairy
Sent: Friday, 26 February 2016 5:13 p.m.
To: Erin Hodgess <erinm.hodg...@gmail.com>
Cc: r-help m
isn't
moderated in the 'usual' sense.
That said, I have occasionally asked a new poster to reword their question (or
simply add a subject line) and explained that this helps ensure they get a good
answer, and not a rude one. Mostly people seem to appreciate that.
Peter Alspach
(one
Tena koe Robert
Many times in R one can do things without a loop. In this case, see ?rle. You
might also need to check substring or months depending on how you dates are
stored.
HTH
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf
. If this is correct, then use this feature (i.e., nchar(ID)-4) as
you'd want "IBBS3_MSM_HN104213" (the fifth element in ID) to split to IBBS3,
MSM, HN1 and 04213. However, if it isn't always 5 digits then split at the
first number (i.e., HN and 104213).
HTH .....
Peter Alspach
---
an exclusive list (e.g., nadiv and
hglm), but it's a start.
I don't find your question clear, and thus apologise if the above does not
address your request.
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ramendra Sarma
Sent: Wednesday, 14 October
Tena koe Maria
It seems you need to multiply Flow by 0.05+ASBClass/20 (i.e., no if calls are
necessary)
HTH
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Maria Lathouri
Sent: Tuesday, 15 September 2015 10:57 p.m.
To: r-help@r
One way
seq(test1)[-which(test1==test2)]
but I imagine there are better ones .
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of baccts
Sent: Wednesday, 29 July 2015 8:26 a.m.
To: r-help@r-project.org
Subject: [R] indices
Tena koe Samantha
You probably need to set some graphics parameters such as xpd and mar (see
?par), and then give the × and y location of the legend rather than
'bottomright' (see ?legend).
HTH
Peter Alspach
PS Please don't post in html (see the posting guide) ... P
-Original
Tena koe Brian
See ?as.Date and ?strptime (and, maybe, ?locales). For example:
as.Date('2/15/15', '%m/%d/%y')
[1] 2015-02-15
as.Date('12/15/14', '%m/%d/%y') as.Date('2/15/15', '%m/%d/%y')
[1] TRUE
as.Date('12/15/16', '%m/%d/%y') as.Date('2/15/15', '%m/%d/%y')
[1] FALSE
You might have
)
will give your final table, but it will need some tidying up.
HTH
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of smart hendsome
Sent: Thursday, 26 February 2015 11:54 a.m.
To: r-help@r-project.org
Subject: [R] Replace the value with 1 and 0
Tena koe Elias
Googling 'Bayesian Networks in R' brings up several packages, so the answer
appears to be 'Yes'. Whether any of them are relevant to you is another
question.
HTH
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf
Tena koe Mike
An alternative, which is slightly fast:
diffv - diff(v)
starts - c(1, which(diffv!=1)+1)
cbind(v[starts], c(diff(starts), length(v)-starts[length(starts)]+1))
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Mike
(ttMat[1,])
str(as.matrix(ttMat[1,]))
str(t(as.matrix(ttMat[1,])))
str(matrix(ttMat[1,], ncol=2))
HTH .
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Sachinthaka
Abeywardana
Sent: Wednesday, 17 December 2014 7:55 p.m.
To: r-help@r
Tena koe
For example:
myTime - Sys.time()
myTime
[1] 2014-11-04 14:58:10 NZDT
format(myTime, '%Y%m%d:%H%M%S')
[1] 20141104:145810
My information in ?strptime.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
to allow you to get there.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kate Ignatius
Sent: Thursday, 2 October 2014 11:11 a.m.
To: r-help
Subject: [R] How to check to see if a variable is within a range
TRUE
# 3 40 45 FALSE
# 4 10 12 FALSE
# 5 70 72 TRUE
# 6 101 90 FALSE
# 7 9 12 FALSE
Which seems correct. Can't say why you get something different without more
details.
Peter Alspach
-Original Message-
From: Kate Ignatius [mailto:kate.ignat...@gmail.com]
Sent: Thursday, 2 October
the most standard
tasks (which I can share if you like)
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of JAWADI Fredj
Sent: Wednesday, 10 September 2014 8:48 a.m.
To: r-help@R-project.org
Subject: [R] Import data
Alternatively, use sapply instead of lapply
marieData - list('30008'=c(1,0,1,0), '60008'=c(0,0,1,0), '90008'=c(0,0,1,0),
'17'=1, '130001'=c(0,1))
marieData
$`30008`
[1] 1 0 1 0
$`60008`
[1] 0 0 1 0
$`90008`
[1] 0 0 1 0
$`17`
[1] 1
$`130001`
[1] 0 1
table(sapply(marieData,
Kia ora Erin
But beware - to quote from Yihui 2013 introduction to knitr
It is easy to revert to the output with prompts (set option prompt=TRUE), and
you will quickly realize the inconvenience to the readers if they want to copy
and run the code ...
Peter Alspach
-Original Message
Tena koe Steven
R is case-sensitive. FUN is missing (you have supplied fun - and × and margin)
...
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Pfeiffer, Steven
Sent: Friday, 25 July 2014 5:08 a.m
Tena koe Julien
I don't use the maps package much, but I suspect par()$usr will allow you to do
what you want.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Julien Million
Sent: Friday, 25 July 2014 9
, as Sarah says. Try:
yourData[,10] - yourData[,9]/yourData[,8]
yourData[yourData[,8]==0,10] - yourData[yourData[,8]==0,9]
This doesn't change the 0 to 1 in column 8, but it doesn't appear you actually
need to do that.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r
0.02 47.89
test - character(10^5)
system.time(for (i in 1:10^5) test[i] - as.character(i))
# user system elapsed
# 0.250.000.25
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Byron Dom
Tena koe Whitney
tapply should work. Try:
tapply(yourData$IndividualID, yourData$FamilyID, sample, size=1) # untested
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Whitney Melroy
Sent: Friday, 28 March
Tena koe Jessie
Lots of ways of doing this. Perhaps the easiest, if your data is formatted as
you suggest, is to use substring; e.g., substring(yourData$Sample, 1, 4) should
give you the sites.
Otherwise, you might need to investigate regular expressions.
HTH .
Peter Alspach
, diff))
[,1] [,2] [,3] [,4]
[1,] 75 -10 46 38
[2,] 12 -14 -62 42
and then write.csv(diffMat, 'diffMat.csv')
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of koushiki sarkar
Sent: Tuesday
Tena koe Erynn
Have you checked tapply and aggregate?
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Erynn Call
Sent: Monday, 10 March 2014 3:40 p.m.
To: r-help@r-project.org
Subject: [R] How to obtain
$id, cumsum))
df1
If you need to get back to the original order you can sort by row.names.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jim Lemon
Sent: Monday, 10 March 2014 3:04 p.m.
To: Philip A. Viton
Cc
Another suggestion - would a square-root scale work for you?
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jim Lemon
Sent: Wednesday, 26 February 2014 3:00 p.m.
To: David Parkhurst
Cc: r-help@r-project.org
Subject
Tena koe
Not really strange: (7+2):11 is 9:11, 7+2:11 is 7+c(2,3,4,5,6,7,8,9,10,11);
i.e., 9:18
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Supriya Jain
Sent: Thursday, 30 January 2014 5:32 a.m.
To: r-help@r
Tena koe
Try
apply(mydat, 1, paste, collapse='')
HTH
Peter Alspach
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
kingsly [ecoking...@yahoo.co.in]
Sent: Monday, December 09, 2013 12:22 AM
To: r-help@r-project.org
Tena koe
I think you'll find the arcsine transformation is asin(sqrt(x/100)) where × is
the percentage. However, it might be better to ask whether the data wouldn't
be better analysed using generalised models (e.g., glm).
HTH
Peter Alspach
-Original Message-
From: r-help-boun
Tena koe
Without reading your request in detail, I will suggest you look at ?merge. It
is often the answer when 'combine' is in the question.
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of bcrombie
Sent: Thursday
.
readOnly = TRUE may allow very limited changes (to insert and update rows).
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Ahmed Attia
Sent: Wednesday, 17 July 2013 6:25 a.m.
To: r-help@r-project.org
Subject
Tena koe Jie
Try
x - ' 12ab34 '
plot(1:10, xlab=x)
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jie
Sent: Wednesday, 1 May 2013 6:46 a.m.
To: r-help@r-project.org
Subject: [R] Quote as element
the scientists involved from using
Excel for 'analysis' to R.
Thanks
Peter Alspach
The contents of this e-mail are confidential and may be ...{{dropped:17}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
system elapsed
0.170.030.20
all.equal(BEN, BEN1)
[1] TRUE
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Benjamin Caldwell
Sent: Wednesday, 6 March 2013 10:18 a.m.
To: r-help
Subject: [R
))
RiskTest1[EvHint==1 MinTex==1] - 1})
all.equal(RiskTest0, RiskTest1)
Hei kona ra
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jeff Newmiller
Sent: Wednesday, 21 November 2012 8:18 a.m.
To: Virgile Capo-Chichi; r
-Original Message-
From: Patrick Connolly
Sent: Friday, 9 November 2012 11:29 a.m.
To: Peter Alspach
Subject: Interfacing R and Weka
version
_
platform x86_64-unknown-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
status
))
with(yourData, plot(x, y1, ylim=range(unlist(yourData[,-1])), type='l'))
with(yourData, lines(x, y2, col='red3'))
with(yourData, lines(x, y3, col='blue2', lty='dashed'))
I hope this is of some help ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r
these names as I find that less error prone and more
informative.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of nrm2010
Sent: Friday, 2 November 2012 10:16 a.m.
To: r-help@r-project.org
Subject: [R] Name
of × and y (repeatedly).
Hope this helps
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bazman76
Sent: Monday, 24 September 2012 8:53 a.m.
To: r-help@r-project.org
Subject: [R] Confused by code?
x-matrix(c
(johnAgg, johnData)
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of rjb
Sent: Friday, 24 August 2012 9:19 a.m.
To: r-help@r-project.org
Subject: [R] Extracting data from dataframe with tied rows
Hi R help,
I'm
Tena koe Anna
Yes: see https://stat.ethz.ch/pipermail/r-help/2006-August/111234.html which
includes an excellent description on this written by Bill Venables back in 1997.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r
$ X4: num 3 4 6 NA 6 2
$ X5: num 3 NA 6 6 5 2
lib1 - aggregate(lib[,-1], list(lib[,1]), function(x) length(x[is.na(x)])0)
lib1[apply(lib1[,-1], 1, sum)0,1]
[1] A B
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
)])
and you'll see it is 191.9161 with a name attribute. Thus you can use it 'for
other purposes'.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of PRAGYA SUR
Sent: Friday, 13 July 2012 3:56 p.m.
To: r-help@r
[,'flag1']==0, 'flag2'] -
rbinom(nrow(natalie)-sum(natalie[,'flag1']), 1, 0.5)
tapply(natalie[,'flag2'], natalie[,'flag1'], table)
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of nqf
Sent: Tuesday, 19 June 2012 5
Tena koe Jacob
Possibly your best option is to simulate.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jacob Warren (RIT Student)
Sent: Wednesday, 13 June 2012 4:40 a.m.
To: r-help@r-project.org
Subject
Tena koe Mike
Another alternative to those already given is to use the RODBC package.
HTH
Peter Alspach.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mike Smith
Sent: Wednesday, 16 May 2012 8:11 a.m.
To: r-help@r
Tena koe Rich
Probably highly skewed to the right, with discrete values (perhaps due to the
limitations in the accuracy of the assessment equipment). But note:
library(fortunes)
fortune('chicken')
HTH .
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org
in OlPrcFl - loadWorkbook(. I'm guessing it is the
which is causing problems, although I haven't used the XLConnect package.
HTH .
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mike Hilt
Sent: Thursday, 10
(x)])
C2
C1 x y
1 2 2
2 3 2
aggregate(naomi[,3], naomi[,1:2], function(x) x[length(x)])
C1 C2 x
1 1 x 2
2 2 x 3
3 1 y 2
4 2 y 2
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Naomi Sugie
Tena koe
Possibly barplot() is what you are after.
?barplot
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of gina_alessa
Sent: Wednesday, 4 April 2012 9:08 a.m.
To: r-help@r-project.org
Subject: [R
Tena koe
?read.fwf
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of barny
Sent: Friday, 10 February 2012 9:52 a.m.
To: r-help@r-project.org
Subject: [R] Getting codebook data into R
I've been trying to get
Tena koe
See the colorspace package.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jeffrey Joh
Sent: Thursday, 26 January 2012 10:36 a.m.
To: r-help@r-project.org
Subject: [R] Gray levels
The gray
Tena koe Anna
[ is for subsetting, you need c():
x - c(10.4, 5.6, 3.1, 6.4, 21.7)
y - c(12, 5.6, 7.2, 1.0, 9.3)
plot(x, y)
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Anna Olofsson
Sent: Wednesday, 11
Tena koe Frederico
Something like
yourDF[, apply(yourDF, 2, sum)!=0]
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Frederico Mestre
Sent: Friday, 23 December 2011 10:55 a.m.
To: r-help@r-project.org
Tena koe Lara
If I understand your question correctly, I use the colorspace package for that
sort of thing, but you could also use the built-in colour palettes such as
rainbow and topo.colors
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r
- sub('^[0-9a-z]* ', '', Names)
Names1
ttReg - regexpr(' [^A-Z]', Names1)
ifelse (ttReg0, substring(Names1, 1, regexpr(' [^A-Z]', Names1)-1), Names1)
Incidentally, it is not good practice to call your objects 'names' since that
is a function in R.
HTH
Peter Alspach
-Original Message
Tena koe Vikas
If I understand you correctly, you could generate a character matrix the
elements of which are the row.names and col.names of x1 pasted together
(perhaps with : as a separator). Then the upper.tri of this will give you the
associations.
HTH ...
Peter Alspach
-Original
Tena koe Erin
http://biostat.mc.vanderbilt.edu/wiki/Main/UseR-2012 has the contact person on
the front page ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Erin Hodgess
Sent: Wednesday, 19 October 2011 5
Tena koe Sharad
If I understand you correctly, you want the lower triangle of your combined
matrix to be the lower triangle of one of the correlation matrices, and the
upper triangle to be the upper triangle from the other. If so, check
lower.tri() and upper.tri().
HTH
Peter Alspach
Tena koe Don
?expand.grid
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of darkgaze
Sent: Wednesday, 5 October 2011 11:21 a.m.
To: r-help@r-project.org
Subject: [R] Create combinations of rows
I
Ladder 3 Ladder 10
1 1 2 1 1
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Markus Weisner
Sent: Wednesday, 31 August 2011 2:00 p.m.
To: r-help@r-project.org
Tena koe Henri-Paul
Does the following do what you want?
testVector - LETTERS[1:3]
temp - matrix(c(rep(c(NA, 'foo'), each=length(testVector)), testVector,
rep(c(NA, 'bar', 'blah'), each=length(testVector))), ncol=6)
c(t(temp))
HTH
Peter Alspach
-Original Message-
From: r
Tena koe
Try something along the following lines:
chrData - vector('list', 22)
names(chrData) - paste('chr', 1:22, sep='')
for (i in 1:length(chrData))
{
chrData[[i]] - read.table(file=paste('chr', i, '.out.txt', sep=''), header=F)
...
}
HTH
Peter Alspach
-Original Message
Tena koe
Assuming your matrix is called yourMatrix, then try
apply(yourMatrix, 1, function(x) which(x=5))
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of gallon li
Sent: Thursday, 21 July 2011 8:23 a.m
Tena koe Michael
The help file for strptime suggests you should be using %b (three letter month)
rather than %m (decimal number month).
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of mdkz...@aol.com
(TRUE, rep(FALSE,19)),] will give every twentieth row, starting with
the first.
There's also seq().
HTH
Peter Alspach.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of gibberish
Sent: Thursday, 23 June 2011 11:05 a.m
Tena koe Karena
Try:
table(strsplit(GGCCCAATCGCAATTCCAATT, ''))
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of karena
Sent: Thursday, 16 June 2011 8:37 a.m.
To: r-help@r-project.org
Subject: [R
Tena koe Eric
?sub and ?gsub
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Lee, Eric
Sent: Tuesday, 14 June 2011 3:49 a.m.
To: r-help@R-project.org
Subject: [R] remove commas in a number when
Tena koe Maciek
It is possible; e.g.:
i - 3
assign(paste('myVar', i, sep=''), 1:5)
myVar3
[1] 1 2 3 4 5
Personally, I find it is more convenient to use a list so that instead of
myVar1, myVar2 etc I have a list object, myVar, with each element equivalent to
myVar1, myVar2 etc.
HTH ...
Peter
Tena koe Sam
You could use cut() to convert the depths into your categories and then
tapply().
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Sam Albers
Sent: Friday, 10 June 2011 10:10 a.m.
To: r
Kia ora Colin
I don't know if there is a package that does what you want, but they are easy
enough to create using plot(). Error bars can be added with arrows().
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org
something could be achieved with
judicious use of diag() and rle().
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Santosh
Sent: Wednesday, 27 April 2011 12:14 p.m.
To: r-help
Subject: [R] sub-matrix
is used in fitting the so-called animal
model. See Lynch, M. and B. Walsh (1998) Genetics and Analysis of
Quantitative Traits. Sunderland, MA, USA, Sinauer Associates, Inc. in
particular chapters 26 and 27 (I think) for more details.
HTH
Peter Alspach
-Original Message-
From: r
Tena koe Fabiane
?par
In particular, various cex. parameters. These can be used with plot (e.g.,
plot(x, y, cex.axis=1.2))
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Fabiane Silva
Sent
of legal names.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of John Sorkin
Sent: Monday, 11 April 2011 2:29 p.m.
To: r-help@r-project.org
Subject: [R] Specifying the color of points and the plot symbol
Tena koe
Try something like:
tmp[rep(1:nrow(tmp), each=tmp[,4]),] # untested
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of jjap
Sent: Wednesday, 30 March 2011 7:16 a.m.
To: r-help@r-project.org
Tena koe
There are many ways. I tend to use the arrows() function. See
?arrows
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Bulent Arikan
Sent: Monday, 28 March 2011 10:45 a.m.
To: r-help@r
Tena koe Steven
The ... argument of the apply series of functions allows one to pass arguments
to the called function. So:
tapply(x, l.c.1, quantile, probs=0.75)
should work (although I haven't tested it).
HTH .
Peter Alspach
-Original Message-
From: r-help-boun...@r
exact syntax or
output I would not be surprised if the above needed tweaking.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of dominique
Sent: Thursday, 24 March 2011 9:14 a.m.
To: r-help@r-project.org
Tena koe D'Arcy
You might find dist() more suited to your needs.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Darcy Webber
Sent: Wednesday, 16 February 2011 1:10 p.m.
To: r-help@r-project.org
Tena koe
?par
and check the las argument.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Longe
Sent: Friday, 4 February 2011 7:16 a.m.
To: r-help@r-project.org
Subject: [R] Changing the direction
), but I don't see that it is
a bug or undocumented (although the documentation is subtle).
HTH
Peter Alspach.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of T.D. Rudolph
Sent: Wednesday, 2 February 2011 11:02 a.m.
To: r
Tena koe Lorenz
c() combines the elements to form a vector and all elements of a vector must be
of the same type. If you elements of different types you need to use a list.
Note, a data.frame is a special type of list.
HTH
Peter Alspach
-Original Message-
From: r-help-boun
Tena koe Wendy
which(test=='')
Should do it.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Wendy
Sent: Thursday, 27 January 2011 10:59 a.m.
To: r-help@r-project.org
Subject: [R] Find the empty
Tena koe David
?polygon
should help you.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of David Hervas Marin
Sent: Thursday, 27 January 2011 12:02 p.m.
To: R-Help
Subject: [R] Colour area under
=TRUE, space=c(0,2))
tTicks - barplot(dataMat, beside=TRUE, space=c(0,2))
tTicks - tapply(tTicks, rep(1:7, each=2), mean)
axis(1, tTicks, letters[1:7])
Is that what you want?
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r
Tena koe John
I got a similar message with a different package, and solved it by loading the
latest version of the package (which was built under 2.12.0). So you may need
to get around your institutional firewall!
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r
Tena koe Nasrin
Try points() instead of plot() in your second and subsequent calls to plot().
points() and lines() adds to the current plot by default. Of course you may
have difficulties with setting the x and y limits by that's another matter.
HTH
Peter Alspach
-Original
Tena koe Judit
If it is really a matrix (and not a data.frame) and you wish to change a
randomly selected zero to a one, then
judit - matrix(0, nrow=3, ncol=4)
judit[sample(1:length(judit), 1)] - 1
will do. This uses the fact that a matrix is a vector with a dim attribute.
HTH
Peter
]
}
but I'm not sure this will work in all circumstances in which people might use
merge().
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Tal Galili
Sent: Tuesday, 9 November 2010 10:10 a.m.
To: r-help
Tena koe Greg
df$c - 0
df[apply(df[,-3], 1, function(x) any(x %in% 3)), 3] - 1
df[apply(df[,-3], 1, function(x) all(is.na(x))), 3] - NA
df
a b c
1 NA NA NA
2 2 NA 0
3 2 3 1
4 3 NA 1
5 2 1 0
6 NA 3 1
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r
Tena koe Steven
cutData - rbind(summary(Acut), summary(Bcut))
barplot(cutData, beside=TRUE)
should get you started. The challenge, as you identify, is to get the data
into the appropriate form and the simple approach I have used may not work for
your real data.
HTH
Peter Alspach
, col='blue', lty='88', lwd=2))
This may be system or graphics device dependent (I'm using Windows).
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Henrik Bengtsson
Sent: Monday, 11 October 2010 10:50 a.m
Tena koe Eddie
One way:
eddie - data.frame(grp=rep(c('small','medium','large','very large'), each=20),
wgt=rnorm(80, 100, 10))
with(eddie, plot(grp, wgt))
eddie$grp - factor(eddie$grp, levels=c('small','medium','large','very large'))
with(eddie, plot(grp, wgt))
HTH ...
Peter Alspach
will make
this process easier. Let me know if you'd like more details.
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Ralf B
Sent: Thursday, 23 September 2010 8:55 a.m.
To: r-help Mailing List
Tena koe John
?aggregate
maybe?
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of array chip
Sent: Friday, 10 September 2010 11:13 a.m.
To: David Winsemius
Cc: r-help@r-project.org
Subject: Re: [R
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