Peng Yu wrote:
Hi,
Please see the command with a comment below. I don't find
'A630039F22Rik' in y. But 'A630039F22Rik' is in z. Can somebody let me
know what the problem is?
Most obvious guess is that your factor y has a level that is not
present in data. That is perfectly normal, even
Here's the code that does the job for quartiles (0,25,50,75,100). To get to
your objective of (5,10,25,75,90) is left as an exercise. There are several
well-written introductory books in R, in addition to the freely available
presentations and other online resources. I think you should spend some
hello guys, I need to do a BLUP in the simplest model
y = Xm + Zg + e
however I have missing data in the analysis which I can´t consider as
0(zero). So I need to generate the matrix X'Z, Z'X and Z'Z step by step; I
can´t use
crossprod(x) #neither
X'X - t(x)%*%x
because I should skip the
If I have a web log file as follows:
#Software: Microsoft Internet Information Services 5.0
#Version: 1.0
#Date: 2007-12-03 13:50:17
#Fields: date time c-ip cs-username s-ip s-port cs-method cs-uri-stem
cs-uri-query sc-status sc-bytes cs-bytes time-taken cs(User-Agent)
cs(Cookie) cs(Referer)
Sorry, I sent it quickly and forgot to thank in advance
Marcio
Marcio Resende wrote:
hello guys, I need to do a BLUP in the simplest model
y = Xm + Zg + e
however I have missing data in the analysis which I can´t consider as
0(zero). So I need to generate the matrix X'Z, Z'X and Z'Z step
Hi,
I am new in R and I don´t know how to sum the product of two elements at the
time in a matrix
X=[ 1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16]
I would like to do (1*5+2*6+3*7+4*8)
I need to do it step by step because I will further put a conditional in the
formula
Hi
I am not sure if I understand what you want but if your matrix is not so
big you can try
x[,1]*x[,2]
[1] 5 12 21 32
cumsum(x[,1]*x[,2])
[1] 5 17 38 70
and than check value of cumsum according to your condition.
Regards
Petr
r-help-boun...@r-project.org napsal dne 23.09.2009
Hi,
with a different (faster) algorithm, but maximum flows are implemented
in package igraph, although for some networks only calculating the
flow value is supported, giving the flow itself is not.
Best,
Gabor
On Wed, Sep 23, 2009 at 3:37 AM, shuva gupta shuvagu...@yahoo.com wrote:
Hi,
Is
Reproducible code.???
premmad wrote:
I have to remove missing data both in character and numeric datatype.I
tried using NA condition but it is not working ,please help me to solve
this.
-
Blay S
KATH
Kumasi, Ghana.
--
View this message in context:
Replace your qfu as follows:
qfu - function(x, digits=3,sci=F,...){
c(q=fivenum(x, ...)
)
}
Look up fivenum function for more information.
cheers,
-Girish
=
premmad wrote:
Thanks for the help.I got the required quantiles by altering ur code
as follows
Hi R community, I have a question. I have 5 files in a directory. Each file has
a year as a name (file 1 -2004, file 2- 2005, ...). I want to build a for
loop where I call first file, do some calculations, go to second file, do some
calculations, etc. Somethin like this:
year-2003
nfiles -
Hi,
The short answer would be ?paste (as in paste(year, .csv, sep=) ),
but I'd suggest you try this instead,
lf - list.files(pattern = .csv)
lf
# [1] 2003.csv 2004.csv 2005.csv
ln - gsub(.csv, , lf)
ln
# [1] 2003 2004 2005
length(ln)
lapply(lf, read.csv)
?list.files
?lapply
HTH,
baptiste
one implementation is in the optmatch package as far as I remember
stefano
On 23/set/09, at 03:37, shuva gupta wrote:
Hi,
Is there any R implementation of the well-known algorithm from the
Operations Research
literature, the Ford-Fulkerson algorithm of maximum flow in networks
with
Hello,
Say I have a dataset as followed:
Category Value
b1
b2
a7
a1
Then, if I:
levels(Category)
It will return:
[a], [b]
But I want to keep the original order, i.e.:
[b], [a]
Is it possible to do it in R?
Thanks in
Thanks for the help.I got the required quantiles by altering ur code
as follows
qfu-function(x,digits=3,sci=F,...)
{c(q=quantile(x,probs=c(5,90)/100))
}
and my result of the R system is different from my sas system output for the
same function .could anyone help me in this and what is the
thank a lot it works
Hans W. Borchers wrote:
But of course, it is always possible to emulate a semi-continuous variable
by introducing a binary variable and use some big-M trick. That is, with
a new binary variable b we add the following two conditions:
x3 - 3.6 * b = 0 and
x3
I tried thanks for your help and got the same result for percentile 5 95 as
in SAS.But if i need to calculate quantiles (1,5,10,99,etc.) it will not be
possible with fivenum as explained in the help page .If i need those
quantiles what is the change i need to make in the function
Hi everyone
I want help in graduating the attached rates and checking for goodness
of fit and smoothness using R please help.
Many thnk
TOo every one around the world
This message and attachments are subject to a disclaimer. Please refer
to
You might need to change the type quantile. The default is type = 7,
whereas default for SAS is type = 3 and for SPSS type = 6. Have a look
at the helpfile of quantile() for more details on the type.
HTH,
Thierry
ir.
Hello Chris,
I had the same problem, and I ended up driving R Gui through an Autoit
script, see http://www.autoitscript.com/autoit3/ Regards,
Gabriele Franzini
-Original Message-
From: cr...@binghamton.edu [mailto:cr...@binghamton.edu]
Sent: 22 September 2009 19:37
To:
Hi,
nvars - 902
data - as.data.frame(matrix(runif(100*nvars),ncol=nvars))
colnames(data)[901] - c('phenotype')
colnames(data)[902] - c('outcome')
### catch all aic values ###
res - matrix(nrow=900,ncol=2)
for (i in 1:(length(data)-2)) {
res[i,1] - names(data)[i]
res[i,2] -
Can't you just use get()? What am I missing?
f - function(fo, data, groups) {
g - xyplot(as.formula(fo), groups = get(groups), data)
print(g)
}
f(yield ~ variety | site, data = barley, groups = year)
Peter Ehlers
Erich Neuwirth wrote:
Thanks, that completely solves the
See if this works:
qfun2 - function(x, digits=3,sci=F,...){
c(q=quantile(x, probs=c(1,5,10,95,99)/100,type=6,...)
)
}
cheers,
-Girish
===
premmad wrote:
I tried thanks for your help and got the same result for percentile 5 95
as in SAS.But if i need to calculate
Dear All,
Let:
dp: depth of the river
tp: temperature with respect to depth
We can have a simple scatter plot, between depth as y-axis and temperature as
x-axis, by using a plot function as shown below.
#
dp - c(1,4,3,2,5,7,9,8,9,2)
tp - 1:10
plot(tp,dp, type= 'l')
try this:
plot(tp,dp, type= 'l',ylim=rev(range(dp)))
On Wed, Sep 23, 2009 at 7:58 AM, FMH kagba2...@yahoo.com wrote:
Dear All,
Let:
dp: depth of the river
tp: temperature with respect to depth
We can have a simple scatter plot, between depth as y-axis and temperature as
x-axis, by
Try this:
DF$Category - factor(DF$Category, levels = c(b, a))
On Wed, Sep 23, 2009 at 4:16 AM, Chris Li chri...@austwaterenv.com.au wrote:
Hello,
Say I have a dataset as followed:
Category Value
b 1
b 2
a 7
a 1
Then, if
Here is a way of doing it
x - read.table(textConnection(Category Value
+ b1
+ b2
+ a7
+ a1), header=TRUE, as.is=TRUE)
# now keep level in original order
x$Category - factor(x$Category, levels=unique(x$Category))
str(x)
Ya it works thanks for the help
--
View this message in context:
http://www.nabble.com/use-of-class-variable-in-r-as-in-Proc-means-of-sas-tp25530654p25531102.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
On Sep 23, 2009, at 7:58 AM, FMH wrote:
Dear All,
Let:
dp: depth of the river
tp: temperature with respect to depth
We can have a simple scatter plot, between depth as y-axis and
temperature as x-axis, by using a plot function as shown below.
#
dp -
Here is a way to do it. I assume that you data has each record on a
line; it came through the email as multiple lines.
x - readLines(/tempxx.txt)
# remove '#Fields: so it can be used as a header
x - sub(^#Fields: , , x)
# remove comment lines
x - x[-grep(^#, x)]
# remove quotes
x -
yet another way,
x - read.table(textConnection(Category Value
+ b1
+ b2
+ a7
+ a1), header=TRUE)
y = transform(x, Category = relevel(Category, c(b)))
str(y)
'data.frame': 4 obs. of 2 variables:
$ Category: Factor w/ 2
On Wed, Sep 23, 2009 at 1:24 AM, Peter Dalgaard
p.dalga...@biostat.ku.dk wrote:
Peng Yu wrote:
Hi,
Please see the command with a comment below. I don't find
'A630039F22Rik' in y. But 'A630039F22Rik' is in z. Can somebody let me
know what the problem is?
Most obvious guess is that your
Dear All,
Can someone please guide me how to get the certain part from a long html
language?
e.g.
tda href='2005-01.html'2005-01/a/tdtda
href='2006-01.html'2006-01/a/tdtda
href='2007-01.html'2007-01/a/tdtda
href='2008-01.html'2008-01/a/tdtda
href='2009-01.html'2009-01/a/td
How to
Sebastian,
There is rarely a completely free lunch, but fortunately for us R has
some wonderful tools
to make this possible. R supports regular expressions with commands
like grep(),
gsub(), strsplit(), and others documented on the help pages. It's
just a matter of
constructing and algorithm
On 09/23/2009 09:58 PM, FMH wrote:
Dear All,
Let:
dp: depth of the river
tp: temperature with respect to depth
We can have a simple scatter plot, between depth as y-axis and temperature as
x-axis, by using a plot function as shown below.
#
dp- c(1,4,3,2,5,7,9,8,9,2)
tp-
Hi,
The R4X package can help you. (I have wrapped your td's into one tr)
x - xml( trtda href='2005-01.html'2005-01/a/tdtda
+ href='2006-01.html'2006-01/a/tdtda
+ href='2007-01.html'2007-01/a/tdtda
+ href='2008-01.html'2008-01/a/tdtda
+ href='2009-01.html'2009-01/a/td/tr )
x[td/a/#]
td
Hi,
It's trivial with ggplot2,
library(ggplot2)
qplot(tp,dp, geom=line) + scale_y_reverse()
HTH,
baptiste
2009/9/23 David Winsemius dwinsem...@comcast.net:
On Sep 23, 2009, at 7:58 AM, FMH wrote:
Dear All,
Let:
dp: depth of the river
tp: temperature with respect to depth
We can have
Dear R-users,
I am a new user for R. I am eager to lean about it.
I wanted to read and summary of the a simple data file
I used the following,
rel - read.table(C:/Documents and Settings/ashta/My
Documents/R_data/rel.dat, quote=,header=FALSE,sep=,col.names=
c(id,orel,nrel))
Try using XML package:
Lines - tda href='2005-01.html'2005-01/a/tdtda
href='2006-01.html'2006-01/a/tdtda
href='2007-01.html'2007-01/a/tdtda
href='2008-01.html'2008-01/a/tdtda
href='2009-01.html'2009-01/a/td
library(XML)
xpathApply(htmlParse(Lines), //a, xmlAttrs)
On Wed, Sep 23, 2009 at 9:29
On Sep 23, 2009, at 4:54 AM, MKHABELA,SN wrote:
Hi everyone
I want help in graduating the attached rates and checking for
goodness of fit and smoothness using R please help.
Mortality rates for males and
females.txt__
You have provided the
On Wed, Sep 23, 2009 at 07:29:30AM -0500, Peng Yu wrote:
On Wed, Sep 23, 2009 at 1:24 AM, Peter Dalgaard
p.dalga...@biostat.ku.dk wrote:
Peng Yu wrote:
Is there an operation on a factor to get a subset and keep only the
corresponding levels (see commented line below)?
Yes, there is: call
On Sep 23, 2009, at 8:44 AM, David Winsemius wrote:
On Sep 23, 2009, at 4:54 AM, MKHABELA,SN wrote:
Hi everyone
I want help in graduating the attached rates and checking for
goodness of fit and smoothness using R please help.
Mortality rates for males and
Hello Ashta,
You need to use double blackslashes, liike: C:\\Documents and
Settings\\ashta\\MyDocuments\\R_data\\rel.dat
I usually use the following to avoid writing the path:
#select file from a popup window
f - file.choose()
#read the file. the is Rese for any other arguments e.g.
Why are you using SANN for optimizing over a smooth function of a scalar
parameter? Simulated annealing is generally quite slow, and is typically
used for nasty functions with multiple bumps and valleys.
Try `optimize' instead.
Ravi.
Chuck, thank you, but I am not sure I understood what you meant.
There are a lot of rows in index where at least 2 columns have equal
values and a lot of rows where column 1 has 2 and some other column
has 5 - same for 3 in column 1 and 6 in some other column, etc.
Thanks a lot for clarifying!
I have the following data exported as a .txt file on Windows. Everything is
working fine, except that the the data in the 10th column is treated as a
factor.
Date Week Time Completed Work_Delta Mean_Delta Balance Total Total_Delta
Work Index Open_Bugs Bug_Delta Bug_Delta2
8/17/2009 4 11.8% 64 64
Dear R gurus,
I use the above release on my MacPro under Leopard 10.5.8 and I have
no more access to the package manager and the CRAN binary list on
package installer. Error messages in the console are :
Erreur : impossible de trouver la fonction
package.manager (impossible to find
--
Tobias Erik Reiners
Justus Liebig University
IFZ - Department of Animal Ecology
Heinrich-Buff-Ring 26-32
D-35392 Giessen Germany
www.uni-giessen.de/cms/fbz/fb08/biologie/tsz/tieroekologie/mitarbeiter/diplomanden-innen/tobias-erik-reiners
What it is telling you is that it can't find the file. This could be
because the file isn't there, or you've got a typo in the file name,
that sort of thing.
In your email, you have split the filename argument between two
lines. I don't know whether this comes from what you did in R, or
Dear Helpers,
usually I try to find the answers on my own, but this one beated me.
I have to use the package Geneland which requires coordinates in
Lambert projection.
I have latitude and longitude (please copy to an .txt and read.table() )
X Y
3458231 5544356
3458263 5544301
3459143
Resolved. Thanks.
On Wed, Sep 23, 2009 at 9:52 AM, Larry White ljw1...@gmail.com wrote:
I have the following data exported as a .txt file on Windows. Everything is
working fine, except that the the data in the 10th column is treated as a
factor.
Date Week Time Completed Work_Delta
Alessandro Carletti wrote:
Hi,
I have a problem.
I have a data frame looking like:
ID val
A? .3
B? 1.2
C? 3.4
D? 2.2
E? 2.0
I need to CREATE the following TABLE:
CASE?? DIFF
A-A??? 0
A-B??? -0.9
A-C??? -3.1
A-D??? -1.9
A-E??? -1.7
B-A??? ...
B-B??? ...
B-C
B-D
Hi
r-help-boun...@r-project.org napsal dne 23.09.2009 14:49:38:
On Wed, Sep 23, 2009 at 07:29:30AM -0500, Peng Yu wrote:
On Wed, Sep 23, 2009 at 1:24 AM, Peter Dalgaard
p.dalga...@biostat.ku.dk wrote:
Peng Yu wrote:
Is there an operation on a factor to get a subset and keep only the
Hi
You can use outer. If your data are in data frame test then
DIFF - as.vector(t(outer(test$val, test$val, -)))
returns a vector, You just need to add suitable names to rows.
CASE - as.vector(t(outer(test$ID, test$ID, paste, sep=-)))
data.frame(CASE, DIFF)
will put it together.
Regards
Thanks Petr! It is good to see multiple solutions to the same problem.
Best,
Jude
-Original Message-
From: Petr PIKAL [mailto:petr.pi...@precheza.cz]
Sent: Wednesday, September 23, 2009 10:59 AM
To: Ryan, Jude
Cc: alxmil...@yahoo.it; r-help@r-project.org
Subject: Re: [R] compute
Hello R users,
I have a basic computer programing question. I am a student
currently taking a course that uses Fortran as the main programming
language, but the instructors are open to students using any language
they are familiar with. I have used R previously, and am wondering if
there is
Dear List,
is there an easy and fast way to compute the p value from a given F
value and given degrees of freedom for an effect and the dfs for the
residuals? I think of a function like this:
compute.p(F.value, numerator.dfs, denominator.dfs)
which returns the p value.
Thanks,
Sascha
Dear R'rs,
is there a function that checks if a given vector contains a certain value.
E.g., x-c(1,2,3,4).
How can I get a TRUE or FALSE for whether x contains a 2?
--
Dimitri Liakhovitski
Ninah.com
dimitri.liakhovit...@ninah.com
__
?any
any(x==2)
Stefano
-Messaggio originale-
Da: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org]per conto di Dimitri Liakhovitski
Inviato: mercoledì 23 settembre 2009 17.38
A: R-Help List
Oggetto: [R] Function to check if a vector contains a given value?
Dear R'rs,
Hi Sascha,
Take a look at ?pf
HTH,
Jorge
On Wed, Sep 23, 2009 at 11:34 AM, Sascha Wolfer wrote:
Dear List,
is there an easy and fast way to compute the p value from a given F value
and given degrees of freedom for an effect and the dfs for the residuals? I
think of a function like this:
Hi Dimitri,
See either ?any, ?%in%, or ?intersect
any(x == 2)
# [1] TRUE
HTH,
Jorge
On Wed, Sep 23, 2009 at 11:37 AM, Dimitri Liakhovitski wrote:
Dear R'rs,
is there a function that checks if a given vector contains a certain value.
E.g., x-c(1,2,3,4).
How can I get a TRUE or FALSE for
On 9/23/2009 11:13 AM, Paul Simonin wrote:
Hello R users,
I have a basic computer programing question. I am a student
currently taking a course that uses Fortran as the main programming
language, but the instructors are open to students using any language
they are familiar with. I have used
-- Forwarded message --
From: Charlie Sharpsteen ch...@sharpsteen.net
Date: Wed, Sep 23, 2009 at 8:47 AM
Subject: Re: [R] Fortran vs R
To: Paul Simonin paul.simo...@uvm.edu
Cc: r-h...@r-project.or
On Wed, Sep 23, 2009 at 8:13 AM, Paul Simonin paul.simo...@uvm.edu wrote:
Hello
Thanks a lot, Jorge!
On Wed, Sep 23, 2009 at 11:42 AM, Jorge Ivan Velez
jorgeivanve...@gmail.com wrote:
Hi Dimitri,
See either ?any, ?%in%, or ?intersect
any(x == 2)
# [1] TRUE
HTH,
Jorge
On Wed, Sep 23, 2009 at 11:37 AM, Dimitri Liakhovitski wrote:
Dear R'rs,
is there a function
WARNING! Biased opinion.
I'm an old guy who learned programming nearly 50 years ago when FORTRAN (IV)
was it, unless you wanted to write machine language which, being an
engineer, I was less interested in than in getting an answer so I could get
on with things.
I like FORTRAN, but I can't think
For power studies you need to think about what the data will look like under
the alternative hypothesis. Is the data shifted over a certain amount? (the
most common assumption), or scaled? Or both? Or a completely different shape?
Etc.
My preferred method for power studies in this case is to
Dear R-help and ROCR developers (Tobias Sing and Oliver Sander) -
I think my first question is generic and could apply to many methods,
which is why I'm directing this initially to R-help as well as Tobias and
Oliver.
Question 1. The plot function in ROCR will average your cross validation
and note that if, instead of zip files, you were using gzip files, you
could:
conn - gzfile(file.gz, rt)
theData - read.table(conn)
close(conn)
b
On Sep 22, 2009, at 11:21 PM, Gabor Grothendieck wrote:
Linux is a type of UNIX so follow the instructions I gave for UNIX.
On Tue, Sep 22,
Hi,
I am performing a repeated measures 2-way ANOVA to assess the influence of
plant and leaf on aphid fecundity. Fecundity is measured for each aphid on a
single leaf.
Here is what I typed.
wingless - reshape(Wingless,
varying =
Try this,
d - na.omit(data.frame(tp,dp))
plot(d, t=l, ylim=rev(range(d$dp)))
?na.omit
HTH,
baptiste
2009/9/23 FMH kagba2...@yahoo.com:
Thank you for the code. I found that the coding does not work if there is an
NA in dp variable. For instance;
#
dp -
Hi,
I am trying to fit a simple two-way fixed effect linear model (o ~ y + s
- 1). However, my problem is large (length(o)=79333).
I am already using slm.fit with a dense design matrix (ddm), but still:
fit - slm.fit(ddm,o)
Error: cannot allocate vector of size 9.1 Gb
Is there a way to
maybe you could modify the following to suit your situation (i use
this xPath expression to get links from google):
?htmlTreeParse
?getNodeSet
library(XML)
link -
Thanks Bryan, sorry for the late reply - I only just noticed this
post. I'm not specifically interested in that sum, but something
related to the sum so this may also be very useful.
Dan
On 18 Sep, 18:24, Bryan Keller bskel...@wisc.edu wrote:
The combn solution offered by Bill is great. It
Hi there
I'm trying to construct a model of mortality risk in 2D space that
requires numerical integration of a hazard function, for which I'm using
the integrate function. I'm occasionally encountering parameter
combinations that cause integrate to terminate with error Error in
integrate... the
You can try this also:
is.element(2, x)
On Wed, Sep 23, 2009 at 12:37 PM, Dimitri Liakhovitski ld7...@gmail.com wrote:
Dear R'rs,
is there a function that checks if a given vector contains a certain value.
E.g., x-c(1,2,3,4).
How can I get a TRUE or FALSE for whether x contains a 2?
--
Dear Rcmdr users,
I use R-2.9.2, Rcmdr 1.5-1 and the latest Rtools bundle (and the html help
workshop) under Windows XP. I wrote a RcmdrPlugin for some important
functions of Thomas Lumley's survey package.
Problem: Package compilation works without displaying any errors. That is
1) R CMD build
Try this:
plot(tp, dp, type = 'l', ylim = rev(range(dp, na.rm = TRUE)))
On Wed, Sep 23, 2009 at 1:44 PM, FMH kagba2...@yahoo.com wrote:
Thank you for the code. I found that the coding does not work if there is an
NA in dp variable. For instance;
#
dp -
Hi Marcus,
I always use a smaller error tolerance in `integrate' than the default
value. I generally use 1.e-07, whereas the default is only about 1.e-04.
Sometimes you may also need to increase the number of subdivisions from its
default value of 100.
Your problem disappears if you use a
Version 3.18 of the survey package is now percolating through CRAN.
Since the last announcement on this list, in February, the main additions are
- standard errors for survival curves (both Kaplan-Meier and Cox model)
- svyciprop() for confidence intervals on proportions, especially in small
I saw http://cran.r-project.org/doc/FAQ/R-FAQ.html#R-Web-Interfaces
and I'm still not sure yet which platform (Linux, Windows, etc.) I'll
be working on -- and no, it's not under my control to pick.
I was wondering if anyone out there had good advice, that would save
me time and stomach acid, on
Are y and s continuous or is one of them a factor/dummy variable? From the
mle specification I grasp that s is the unit of observation and is
factor-coded. If that is so, then estimating lm(o~y+s) includes a lot of
dummy/factor variables (lots of columns of 0/1 in the X matrix), and then
there
On Wed, 23 Sep 2009, Paul Simonin wrote:
Hello R users,
I have a basic computer programing question. I am a student currently taking
a course that uses Fortran as the main programming language, but the
instructors are open to students using any language they are familiar with. I
have used R
Try this:
barplot(t(as.matrix(intersect.data[,2:5])),
beside = T, horiz = T,
names.arg = intersect.data[,1], cex.axis = 0.7, cex.names = 0.7)
On Wed, Sep 23, 2009 at 4:01 PM, Nair, Murlidharan T mn...@iusb.edu wrote:
Hi,
I am trying to plot the following data so that it
Dear Group:
I want to get the loglik of the regresion associated to the estimation of the
unit root test Zivot-Andrews of the package urca. I am new using R, then I
simply tried the next sequence:
A-ur.za(var,model=intercept,lag=2)
logLik(A)
but the result is an error, and I think it
Dear all,
consider:
###
x - round(rnorm(50))
stripchart(x, pch = 21, col = black, bg = pink, method = jitter)
points(0.5, 1, pch = 21, col = black, bg = pink, cex = 2)
###
Under R 2.9.0 the points produced by stripchart are not colored,
while points() gives the desidered output (magnified
I had tried names.arg=c(intersect.data[,1]) so that was the problem. That
solves part of what need. I there a way to rotate how it is written on the
y-axis? Also, use designs instead of gray scale and making keys for it?
Thanks for chipping in.
Cheers../Murli
-Original Message-
From:
Ok, I could make it perpendicular by specifying las=2
barplot(t(as.matrix(intersect.data[,2:5])),
beside = T, horiz = T,
names.arg = intersect.data[,1], cex.axis = 0.7, cex.names = 0.7,
las=2)
Still working on the other though.
Cheers../Murli
-Original Message-
Sorry, byt what you mean by 'designs'?
You can add a legend with:
barplot(t(as.matrix(intersect.data[,2:5])),
beside = T, horiz = T, legend.text = names(intersect.data)[-1],
names.arg = intersect.data[,1], cex.axis = 0.7, cex.names = 0.7, las=2)
On Wed, Sep 23, 2009 at 4:26
What I mean by design is a black and white lines or something that is more
distinguishing than the grey levels?
Thanks for the legends :)
Cheers../Murli
-Original Message-
From: Henrique Dallazuanna [mailto:www...@gmail.com]
Sent: Wednesday, September 23, 2009 3:32 PM
To: Nair,
Hello,
I have the following data:
gene Actualgroupsreps
11213 12
22123 23
33 2 12
44 12 2 3
51 0 1 1
62 34 2 2
73
Perhaps a white border:
barplot(t(as.matrix(intersect.data[,2:5])),
border = 'white',
beside = T, horiz = T, legend.text = names(intersect.data)[-1],
names.arg = intersect.data[,1], cex.axis = 0.7, cex.names = 0.7, las=2)
Or you can use the 'col' argument to select
The journal wants black and white only :)
-Original Message-
From: Henrique Dallazuanna [mailto:www...@gmail.com]
Sent: Wednesday, September 23, 2009 3:41 PM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] dotchart to barplots
Perhaps a white border:
The current recommendation is to not put designs/hash lines/pictures/etc. into
the bars, but to use a single solid color (gray in your case). Back when a
quality graph meant using a pen plotter, hash lines made sense as a way to
distinguish between bars, but quality graphics no longer depend
Murli,
Two points:
1. I think you might want las=1;
2. have a look at the density= argument,
i.e. add density=c(10,20,30,40) to your call.
Peter Ehlers
Henrique Dallazuanna wrote:
Perhaps a white border:
barplot(t(as.matrix(intersect.data[,2:5])),
border = 'white',
beside =
Marc Schwartz-3 wrote:
Using the data that is in the online plot rather than the above, here
is a first go. Note that I am not drawing the background grid in the
barplot or the lines for table below it. These could be added if you
really need them.
Note: I snipped out the syntax
On 23/09/2009, at 11:26 PM, Christian Schulz wrote:
Hi,
nvars - 902
data - as.data.frame(matrix(runif(100*nvars),ncol=nvars))
colnames(data)[901] - c('phenotype')
colnames(data)[902] - c('outcome')
snip
Just ***WHAT*** do you think the ``c( )'' is doing for you in
the
Thanks Peter. Where did you find that option? It's really cool
Cheers../Murli
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Wednesday, September 23, 2009 3:56 PM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] dotchart to barplots
Murli,
On Sep 23, 2009, at 3:47 PM, Jim Silverton wrote:
Hello,
I have the following data:
gene Actualgroupsreps
11213 12
22123 23
33 2 12
44 12 2 3
51 0 1
Guess, I miss the argument when I ?barplot.
Cheers../Murli
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Wednesday, September 23, 2009 3:56 PM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] dotchart to barplots
Murli,
Two points:
1. I
Well, it was easy to find: ?barplot and look at all the arguments.
But I agree with Greg that this kind of look is (and should be)
pretty much history. I'm not very fond of barplots with as many
groups as you have. Since your variable X appears to be a
discretized continuous variable, why not
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