Re: [R] Reading in .aux (ESRI raster files) into R
Thank you Barry -- View this message in context: http://r.789695.n4.nabble.com/Reading-in-aux-ESRI-raster-files-into-R-tp2553544p2715015.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] textplot
Hi all, Please can you help me to add text to a plot. I would like 5 graphical plots and 1 text plot. I use par(mfrow=c(2,3)) then textplot() I would like to add various pieces of information to the text plot, i.e. some individual values, a table of statistics, some sentences. Is there any easy way of doing this or am I forced to construct a very complicated dataframe? Is this even possible? Examples would be appreciated... Thanks... -- View this message in context: http://r.789695.n4.nabble.com/textplot-tp2715028p2715028.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] a very basic question: calling c function from R
Hi, Duncan Murdoch... i m facing same pro here.. n i m a very new user for R... i m just start using it now for my final year project... i wan to know how to run outside R??? i have no idea about it... can u help me??? appreciate for your helping... -- View this message in context: http://r.789695.n4.nabble.com/a-very-basic-question-calling-c-function-from-R-tp883154p2714989.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Object Browser
Hi
I noticed that nobody answered your question yet so here is my try.
If you want to see what objects are in your environment you can use ls()
but its output is only names of objects. Here is a function I use a long
time for checking what objects are there, their type, size and possibly
rows columns. You can modify it to give you some more info but usually it
is not needed.
Regards
Petr
ls.objects - function (pos = 1, pattern, order.by)
{
napply - function(names, fn) sapply(names, function(x) fn(get(x,
pos = pos)))
names - ls(pos = pos, pattern = pattern)
obj.class - napply(names, function(x) as.character(class(x))[1])
obj.mode - napply(names, mode)
obj.type - ifelse(is.na(obj.class), obj.mode, obj.class)
obj.size - napply(names, object.size)
obj.dim - t(napply(names, function(x) as.numeric(dim(x))[1:2]))
vec - is.na(obj.dim)[, 1] (obj.type != function)
obj.dim[vec, 1] - napply(names, length)[vec]
out - data.frame(obj.type, obj.size, obj.dim)
names(out) - c(Type, Size, Rows, Columns)
if (!missing(order.by))
out - out[order(out[[order.by]]), ]
out
}
r-help-boun...@r-project.org napsal dne 24.09.2010 23:04:20:
What's the best object browser?
Dear all,
I have tried all the popular R IDE or editors like Eclipse, Komodo, JGR,
Revolution...
They all have fancy fucntions like auto completion, syntax highlight
BUT, I JUST WANT A OBJECT BROWSER!
The easiest way to view objects in R console is fix(), but you have no
global view of all the objects in the workspace.
Revolution has the best object browser so far, but this thing is way too
big...
Eclipse all has automatically object browser, but you can't only see the
basic summary info. and it's really tricky to configure StatET.
Jgr is very handy, when you double click the object name, you can view
the
object in a spreadsheet like using fix(), but you have to press the
Refresh button each time you want to see the updated objects...
So, is there any thing like the combination of eclipse and Jgr?
If not, I am interested to develope something to fulfill this simple but
very important function. But right now I have no idea where to start.
Any
suggestions?
Peter
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Re: [R] textplot
Have a look at ?textplot From the gplots package: http://cran.r-project.org/web/packages/gplots/index.html Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Mon, Sep 27, 2010 at 9:48 AM, trb1 thomasrbol...@yahoo.co.uk wrote: Hi all, Please can you help me to add text to a plot. I would like 5 graphical plots and 1 text plot. I use par(mfrow=c(2,3)) then textplot() I would like to add various pieces of information to the text plot, i.e. some individual values, a table of statistics, some sentences. Is there any easy way of doing this or am I forced to construct a very complicated dataframe? Is this even possible? Examples would be appreciated... Thanks... -- View this message in context: http://r.789695.n4.nabble.com/textplot-tp2715028p2715028.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Uncertainty propagation
Thanks for the help, I start to get reasonable errors on the model... I finally turned to the simpler lm() fitting. As my data from which I fit has only 8 points in each case, I guess it does not make much sense to downweight outliers and use rlm() in this case. -- View this message in context: http://r.789695.n4.nabble.com/Uncertainty-propagation-tp2713499p2715085.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fitting with error on data
As this forum proved to be very helpful, I got another question... I'd like to fit data points on which I have an error, dx and dy, on each x and y. What would be the common procedure to fit this data by a linear model taking into account uncertainty on each point? Would weighting each point by 1/sqrt(dx2+dy2) (and taking dx and dy as relative errors) in a lm() fit do the job? I would like to propagate uncertainty of the points into the uncertainty of the fit, would that be the case? Thanks for all the help -- View this message in context: http://r.789695.n4.nabble.com/Fitting-with-error-on-data-tp2715100p2715100.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Some questions about string processing
Thanks Phil, very helpful and works as advertised.
Any thoughts on the second question?
(P.S. for anyone digging this up in the future, there's a comma missing after
the formula in lm())
On 24 Sep 2010, at 18:16, Phil Spector wrote:
Michael -
You're doing too much work half the time, and not enough
the other half :-). Try this (untested):
auto_io = function(data_name,factors){
resp_data = read.table(data_name, header = TRUE )
temp_model = lm(formula(paste('y',factors,sep='~')) data = resp_data )
. . .
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Fri, 24 Sep 2010, Michael Hopkins wrote:
Hi all
A couple of questions about string processing from someone who has only
scratched the surface so far.
1) I am wanting to send some strings into a function to allow flexibility
inside. My first idea has been e.g.
auto_io - function( var_string, factors ) {
# e.g. var_string sent as test_file.txt factors sent as x1 + x2 + x3
# input
data_name - get( var_string )
#term_list - get( factors )
resp_data - read.table( data_name,
header = TRUE )
# fit
temp_model - lm( y ~ factors, data = resp_data )
etc...
Neither the read.table() nor the lm() are working because in each case the
string is not converted or understood properly. I'm sure this is possible
(and in fact probably easy) but haven't yet seen the light. Could someone
illuminate me here?
2) I will be wanting to process strings as lists of strings. For instance,
I might instead want to send factors above as x1 x2 x3 and add the +
when I need them, or perhaps * instead, and also look at sub-models by
removing parts of the string etc. What functions should I be looking at
here and are there any examples available?
Thanks in advance. Feel free to CC me on your reply.
Michael Hopkins
Algorithm and Statistical Modelling Expert
Upstream
23 Old Bond Street
London
W1S 4PZ
Mob +44 0782 578 7220
DL +44 0207 290 1326
Fax +44 0207 290 1321
hopk...@upstreamsystems.com
www.upstreamsystems.com
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Upstream
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[R] store matrix in an arrary
Dear All I want to store matrix in an array Suppose s-array(0,4) for(i in 1:4) s[i] - read_matrix(a,2,2) But the error - number of items to replace is not a multiple of replacement length. Can you suggest me any alternative method for storing a matrix in an array. Thanks In advance. Kind Regards Wesley C Mathew [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] store matrix in an arrary
On Mon, 27 Sep 2010 11:12:26 +0100 wesley mathew wesleycmat...@gmail.com wrote: Dear All I want to store matrix in an array Suppose s-array(0,4) for(i in 1:4) s[i] - read_matrix(a,2,2) But the error - number of items to replace is not a multiple of replacement length. Can you suggest me any alternative method for storing a matrix in an array. Thanks In advance. Kind Regards Wesley C Mathew Wesley, I don't know what your read_matrix() function is doing. What you made is a 1-dimensional array of 4 elements. If your read_matrix() function returns more than one element per call, that is the source of your error message. It means you tried to store something in s[] that goes beyond its dimensions. Edwin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dr. Edwin Groot, postdoctoral associate AG Laux Institut fuer Biologie III Schaenzlestr. 1 79104 Freiburg, Deutschland +49 761-2032945 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] store matrix in an arrary
Hi Wesley, Try this (untested): s - array(0, dim = c(2, 2, 4)) for(i in 1:4) s[, ,i] - read_matrix(a, 2, 2) Another (untested) option wold be s0 - lapply(1:4, function(i) read_matrix(a, 2, 2)) s0 HTH, Jorge On Mon, Sep 27, 2010 at 6:12 AM, wesley mathew wrote: Dear All I want to store matrix in an array Suppose s-array(0,4) for(i in 1:4) s[i] - read_matrix(a,2,2) But the error - number of items to replace is not a multiple of replacement length. Can you suggest me any alternative method for storing a matrix in an array. Thanks In advance. Kind Regards Wesley C Mathew [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] a very basic question: calling c function from R
On 27/09/2010 2:45 AM, teetee wrote: Hi, Duncan Murdoch... i m facing same pro here.. n i m a very new user for R... i m just start using it now for my final year project... i wan to know how to run outside R??? i have no idea about it... can u help me??? appreciate for your helping... You should ask your instructor. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fitting problem
Hi,
I have a function that generates a set of data but I am having problems
determining the parameters using the nls fitting procedure.
MH-function(field,diameter,mu=10e-7,sig=0.1,Ms=100,chi=0){
#variables mu, sig, chi, Ms
#input: field and diameter
#all in CGS
rho - 5
kb - 1.38e-16
t - 300
length.d-length(diameter)
length.H-length(field)
M-double(length.H)
for (i in 1:length.H){
S1-0
S2-0
H - field[i]
for (j in 1:length.d){
d-diameter[j]
vol - 4/3*pi*(d/2)^3
lognorm -
1/(d*sig*sqrt(2*pi))*exp(-(log(d)-log(mu))^2/(2*sig^2))
lang - 1/tanh(Ms*rho*vol*H/(kb*t))-1/(Ms*rho*vol*H/(kb*t))
S1 - S1 + lognorm*vol*lang
S2 - S2 + lognorm*vol
}
M[i] - Ms*S1/S2 + chi*H
}
M
}
### I can calculate a set of data:
htest- (-10:10)*200
dtest- (5:15)*1e-7
mtest- MH(field=htest,diameter=dtest)
### However when I try to reverse engineer to calculate the parameters mu,
sig, chi and Ms I run into problems. Could anyone shed some light on this
problem?
fit - nls(M~MH(H,(5:15)*1e-7,mu,sig,Ms,chi),data=df,start=list(mu=10e-7,
sig=0.1, chi=0, Ms=100))
### Thanks, Paul
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Re: [R] legend
Thanks David, works fine! robert -- View this message in context: http://r.789695.n4.nabble.com/legend-tp2550747p2715250.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data frame self-join with no duplicates
Suppose I have the following data frame (df): YearPrice --- 200110 200220 200330 I would like to produce another data frame like this: a.Yeara.Priceb.Yearb.Price 200220200110 200330200110 200330200220 In SQL, this can be done as select a.*, b.* from df as a, df as b where a.Year b.Year How do I do this efficiently in R? Many thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Fitting problem
Hi
r-help-boun...@r-project.org napsal dne 27.09.2010 13:13:28:
Hi,
I have a function that generates a set of data but I am having problems
determining the parameters using the nls fitting procedure.
MH-function(field,diameter,mu=10e-7,sig=0.1,Ms=100,chi=0){
#variables mu, sig, chi, Ms
#input: field and diameter
#all in CGS
rho - 5
kb - 1.38e-16
t - 300
length.d-length(diameter)
length.H-length(field)
M-double(length.H)
for (i in 1:length.H){
S1-0
S2-0
H - field[i]
for (j in 1:length.d){
d-diameter[j]
vol - 4/3*pi*(d/2)^3
lognorm - 1/(d*sig*sqrt(2*pi))*exp(-(log(d)-log(mu))^2/(2*sig^2))
lang - 1/tanh(Ms*rho*vol*H/(kb*t))-1/(Ms*rho*vol*H/(kb*t))
S1 - S1 + lognorm*vol*lang
S2 - S2 + lognorm*vol
}
M[i] - Ms*S1/S2 + chi*H
}
M
}
### I can calculate a set of data:
htest- (-10:10)*200
dtest- (5:15)*1e-7
mtest- MH(field=htest,diameter=dtest)
### However when I try to reverse engineer to calculate the parameters
mu,
sig, chi and Ms I run into problems. Could anyone shed some light on
this
problem?
What problem? I get
fit -
nls(M~MH(H,(5:15)*1e-7,mu,sig,Ms,chi),data=df,start=list(mu=10e-7,
+ sig=0.1, chi=0, Ms=100))
Error in nls(M ~ MH(H, (5:15) * 1e-07, mu, sig, Ms, chi), data = df, start
= list(mu = 1e-06, :
'data' must be a list or an environment
evidently I have df only as function for density of F distribution.
One comment. If you looked into help page you could find this warning
Warning
Do not use nls on artificial zero-residual data.
Which I suppose you have. So add some noise to it.
Regards
Petr
fit -
nls(M~MH(H,(5:15)*1e-7,mu,sig,Ms,chi),data=df,start=list(mu=10e-7,
sig=0.1, chi=0, Ms=100))
### Thanks, Paul
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Re: [R] Fitting problem
Oops I forgot to add another line to the code (see below)!! You could add some noise if you want to. Generating the data from the function was merely a way to test if the fitting procedure works - I have experimental data that should allow me to calculate the parameters mu, sig, chi and Ms based upon my input M, H and diameter. df - data.frame(H=htest,M=mtest) -- View this message in context: http://r.789695.n4.nabble.com/Fitting-problem-tp2715234p2715271.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data frame self-join with no duplicates
Hi: One option is to read your data frame into R and then use your SQL code in conjunction with the sqldf package. It uses SQLite as its engine. HTH, Dennis On Mon, Sep 27, 2010 at 4:29 AM, Xin Zhang xin.zh...@gmail.com wrote: Suppose I have the following data frame (df): YearPrice --- 200110 200220 200330 I would like to produce another data frame like this: a.Yeara.Priceb.Yearb.Price 200220200110 200330200110 200330200220 In SQL, this can be done as select a.*, b.* from df as a, df as b where a.Year b.Year How do I do this efficiently in R? Many thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cluster analysis and supervised classification: an alternative to knn1?
Hi Ulrich, I'm studying the principles of Affinity Propagation and I'm really glad to use your package (apcluster) in order to cluster my data. I have just an issue to solve.. If I apply the funcion: apcluster(sim) where sim is the matrix of dissimilarities, sometimes I encounter the warning message: Algorithm did not converge. Turn on details and call plot() to monitor net similarity. Consider increasing maxits and convits, and, if oscillations occur also increasing damping factor lam. with too high number of clusters. I thought to solve the problem setting the argument p of the function apcluster() to mean(PreferenceRange(sim)): apcluster(sim, p=mean(preferenceRange(sim))) and actually it seems to be a good solution because I don't receive any warning message and the number of cluster is slower. Do you think it's a good solution? I submitt that I have to use apcluster() in an automatic procedure so I can't manipulate directly the arguments of the funcion. Thanks in advance. Giuseppe -- View this message in context: http://r.789695.n4.nabble.com/cluster-analysis-and-supervised-classification-an-alternative-to-knn1-tp2231656p2715278.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data frame self-join with no duplicates
Thanks Dennis. I will explore that. On Mon, Sep 27, 2010 at 7:45 AM, Dennis Murphy djmu...@gmail.com wrote: Hi: One option is to read your data frame into R and then use your SQL code in conjunction with the sqldf package. It uses SQLite as its engine. HTH, Dennis On Mon, Sep 27, 2010 at 4:29 AM, Xin Zhang xin.zh...@gmail.com wrote: Suppose I have the following data frame (df): Year Price --- 2001 10 2002 20 2003 30 I would like to produce another data frame like this: a.Year a.Price b.Year b.Price 2002 20 2001 10 2003 30 2001 10 2003 30 2002 20 In SQL, this can be done as select a.*, b.* from df as a, df as b where a.Year b.Year How do I do this efficiently in R? Many thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting problem
r-help-boun...@r-project.org napsal dne 27.09.2010 13:36:51: Oops I forgot to add another line to the code (see below)!! You could add I do not have the previous code, I do not keep mails. some noise if you want to. Generating the data from the function was merely You want to add some noise. nls ***does not work*** for artificial data without noise as stated in help page. Regards Petr a way to test if the fitting procedure works - I have experimental data that should allow me to calculate the parameters mu, sig, chi and Ms based upon my input M, H and diameter. df - data.frame(H=htest,M=mtest) -- View this message in context: http://r.789695.n4.nabble.com/Fitting-problem- tp2715234p2715271.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with unlist
Luis Felipe Parra felipe.parra at quantil.com.co writes: Hello, I am trying to unlist a list, which is attached, and I am having the problem that when I unlist it the number of elements changes from 5065 to 5084 x - lapply(SumaPluvi, FUN=[, 1); n - sapply(x, FUN=length); print(table(n)); n 1 5065 print(which(n != 1)); integer(0) length(unlist(lapply(SumaPluvi, FUN=[, 1))) [1] 5081 I dont now why, but when I unlist it the number of elements changes from 5065 to 5084 even if there is no list element with length greater than one. Do you know what can be happening? We probably won't be able to get farther without a reproducible example. One brute-force way of finding the problem is by bisection: i.e., try the first and last halves of your list separately, and see if either one individually shows a similar problem. Proceed recursively until you localize the problem ... __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 95% CI of Kaplan-Meier survival in package survival
My result using Kaplan-Meier estimate in survival package was inconsistent with that from Minitab. The survival probabilities are same, but their 95% are different. Confidence intervals for a survival curve may be calcualted on the linear (poor performance), log (good), log-log (good), or logit (good) scales. I expect that Minitab and S default to different choices. See the conf.type argument in help(survfit.formula). Terry Therneau __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Output Graphics GIF
Dear R users, How could I managed graphics in GIF format? What I have been doing is graphics in *.ps or *.eps and after I convert them using CONVERT (from ImageMagick) but the output quality is not good. Since these graphics will be use for other users they must have a better image quality. I really appreciate any help, -- Abraço, Nilza Barros [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Output Graphics GIF
Dear R users, How could I managed graphics in GIF format? What I have been doing is graphics in *.ps or *.eps and after I convert them using CONVERT (from ImageMagick) but the output quality is not good. Since these graphics will be use for other users they must have a better image quality. I really appreciate any help, -- Abraço, Nilza Barros -- Abraço, Nilza Barros [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fitting problem
Updated!
MH-function(field,diameter,mu=10e-7,sig=0.1,Ms=100,chi=0){
#variables mu, sig, chi, Ms
#input: field and diameter
#all in CGS
rho - 5
kb - 1.38e-16
t - 300
length.d-length(diameter)
length.H-length(field)
M-double(length.H)
for (i in 1:length.H){
S1-0
S2-0
H - field[i]
for (j in 1:length.d){
d-diameter[j]
vol - 4/3*pi*(d/2)^3
lognorm -
1/(d*sig*sqrt(2*pi))*exp(-(log(d)-log(mu))^2/(2*sig^2))
lang - 1/tanh(Ms*rho*vol*H/(kb*t))-1/(Ms*rho*vol*H/(kb*t))
S1 - S1 + lognorm*vol*lang
S2 - S2 + lognorm*vol
}
M[i] - Ms*S1/S2 + chi*H
}
M
}
###
htest- (-10:10)*200
dtest- (5:15)*1e-7
mtest- MH(field=htest,diameter=dtest) + rnorm(21) ##also added some
noise!!
df - data.frame(H=htest,M=mtest)
fit - nls(M~MH(H,(5:15)*1e-7,mu,sig,Ms,chi),data=df,start=list(mu=10e-7,
sig=0.1, chi=0, Ms=100))
--
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Re: [R] Data frame self-join with no duplicates
try sqldf: require(sqldf) x - read.table(textConnection(YearPrice + 200110 + 200220 + 200330), as.is=TRUE, header=TRUE) sqldf(select a.*, b.* + from x as a, x as b + where a.Year b.Year, method='raw') Year Price Year Price 1 200220 200110 2 200330 200110 3 200330 200220 On Mon, Sep 27, 2010 at 7:29 AM, Xin Zhang xin.zh...@gmail.com wrote: Suppose I have the following data frame (df): Year Price --- 2001 10 2002 20 2003 30 I would like to produce another data frame like this: a.Year a.Price b.Year b.Price 2002 20 2001 10 2003 30 2001 10 2003 30 2002 20 In SQL, this can be done as select a.*, b.* from df as a, df as b where a.Year b.Year How do I do this efficiently in R? Many thanks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Date/Time format
Hello, Is it possible to instruct (permanently) R to write on csv (and read from csv) time series, where the time stamp has a particular format: say: -mm-dd i.e., as in format(Sys.Date(), %Y-%m-%d) Many thanks in advance, Costas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] conditional assignment of colors in xyplot()
# Dear R Community,
# I have this data frame:
df1 - data.frame(
F1 = factor( c( rep(D1,12),rep(D2,12),rep(D3,12) ) ),
F2 = factor( rep( rep( paste(O,1:6,sep=), rep(2,6) ), 3) ),
F3 = factor( rep( c(V1,V2), 18 ) ),
S1 =
c(8.840955e-02,2.546822e-01,7.562658e-01,8.801181e-01,6.041766e-02,2.172731e-01,6.542187e-98,
7.067840e-04,1.430933e-01,9.764401e-01,1.380848e-05,1.192620e-01,9.107259e-01,1.235232e-01,
3.021847e-01,1.331351e-01,5.272103e-01,3.663577e-01,1.539690e-38,2.224451e-01,2.873052e-01,
5.110313e-01,7.840802e-01,8.210762e-10,1.553356e-01,4.173335e-01,5.964021e-01,4.955694e-01,
8.849240e-02,5.739598e-01,1.879075e-17,1.071003e-03,7.298928e-01,6.347287e-01,8.884034e-01,
4.460295e-11),
S2 =
c(1.32249139,1.02831831,-0.09650252,-0.05454486,2.62105492,2.00310250,8.07269064,
-1.55397883,1.77390551,0.04161954,7.14188540,-2.98033713,-0.49904251,-0.74309058,
-0.49904251,-0.74309058,1.22492623,-1.79003492,7.60003121,-0.74549596,2.53418936,
-1.60112296,0.67131380,-15.31744351,-0.18380339,0.28628435,-0.18380339,0.28628435,
2.96108998,1.18267783,5.78419118,2.70861763,0.66287857,1.10397741,0.27160971,
-15.37506924) )
# Two of the factors are used to define an array
# of panels with the third showing up as bars along the x axis.
# How do I color the bars according to their sign (red 0, blue 0)
# conditional on the value of S2 .025 - in which case color them gray?
# Initially, I tried to pass a character vector specifying colors,
# which does not achieve the desired result:
library(lattice)
library(latticeExtra)
df1$barcols - ifelse(df1$S1 .025, gray, ifelse( df1$S2 0,
blue,red))
ctp - xyplot(S2 ~ F2 | F1 + F3,
data=df1, as.table=TRUE, ylim=c(-10,10),
panel = function(x,y){
panel.barchart(x,y,horizontal=FALSE, origin=0,
col=df1$barcols ) } )
useOuterStrips(ctp)
# Any help you can provide would be greatly appreciated.
# Thank you!
# nathan.pelleg...@gmail.com
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Re: [R] Output Graphics GIF
On Mon, Sep 27, 2010 at 1:39 PM, Nilza BARROS nilzabar...@gmail.com wrote: Dear R users, How could I managed graphics in GIF format? What I have been doing is graphics in *.ps or *.eps and after I convert them using CONVERT (from ImageMagick) but the output quality is not good. Since these graphics will be use for other users they must have a better image quality. The png() graphics driver will produce bitmap graphics which should convert better to a gif. PostScript is a vector format, so there may be all sorts of things going on. Just feed png() with the width and height you want for your gif and then you wont have to resize as part of the conversion process. Also, GIFs are pretty old-tech these days, so if you could persuade your other users to switch to png files, its a big win. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting multiple animal tracks against Date/Time
On Mon, Sep 27, 2010 at 10:01 AM, Struve, Juliane
j.str...@imperial.ac.uk wrote:
Hi,
I am sorry that my question wasn not very clearly formulated. My real data
comes in 47 .csv files, one for each of 47 individual, for example:
,Fish_ID,Date,R2sqrt
1,1646,2006-08-18 08:48:59,0
2,1646,2006-08-18 09:53:20,100
I would like to read the data for all individuals in the for loop below and
then combine them in a zoo object in order to plot them together. My
question is : How do I need to set up the for loop to deal with several
individuals ? I have put question marks the bits that I am missing.
I very much appreciate your help.
Regards,
Juliane
ReleaseDates - read.csv(file=ReleaseDates,head=TRUE,sep=,)
#reads in the individuals and their release dates
for (i in 1:length(ReleaseDates)){
Fish_ID - ReleaseDates$Fish_ID[i]
??? -
read.zoo(file=paste(Results,Fish_ID,sep=_),index.column=3,header=TRUE,FUN=as.chron,sep=,)
#reads in data for each in Fish_ID
}
z - na.approx(cbind(???), na.rm = FALSE)
plot(z)
Try this (untested):
library(zoo)
filenames - paste(Results, ReleaseDates$Fish_ID, sep = _)
# list of zoo objects removing 1st two columns in each file
Lz - lapply(filenames, read.zoo, header = TRUE, FUN = as.chron,
sep = ,, colClasses = c(NULL, NULL, character, numeric))
z - do.call(merge, Lz)
colnames(z) - as.character(ReleaseDates$Fish_ID)
# plot in single panel
plot(na.approx(z), screen = 1)
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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Re: [R] Output Graphics GIF
I am guessing you are saving the plot using the menu system. If that is the case, have a look at: ?pdf ?png Generally, I like saving my graphics to pdf since it is vectorized. Cheers, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Mon, Sep 27, 2010 at 2:39 PM, Nilza BARROS nilzabar...@gmail.com wrote: Dear R users, How could I managed graphics in GIF format? What I have been doing is graphics in *.ps or *.eps and after I convert them using CONVERT (from ImageMagick) but the output quality is not good. Since these graphics will be use for other users they must have a better image quality. I really appreciate any help, -- Abraço, Nilza Barros [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting multiple animal tracks against Date/Time
Hi,
I am sorry that my question wasn not very clearly formulated. My real data
comes in 47 .csv files, one for each of 47 individual, for example:
,Fish_ID,Date,R2sqrt
1,1646,2006-08-18 08:48:59,0
2,1646,2006-08-18 09:53:20,100
I would like to read the data for all individuals in the for loop below and
then combine them in a zoo object in order to plot them together. My question
is : How do I need to set up the for loop to deal with several individuals ? I
have put question marks the bits that I am missing.
I very much appreciate your help.
Regards,
Juliane
ReleaseDates - read.csv(file=ReleaseDates,head=TRUE,sep=,)
#reads in the individuals and their release dates
for (i in 1:length(ReleaseDates)){
Fish_ID - ReleaseDates$Fish_ID[i]
??? -
read.zoo(file=paste(Results,Fish_ID,sep=_),index.column=3,header=TRUE,FUN=as.chron,sep=,)
#reads in data for each in Fish_ID
}
z - na.approx(cbind(???), na.rm = FALSE)
plot(z)
From: Gabor Grothendieck [ggrothendi...@gmail.com]
Sent: 24 September 2010 22:08
To: Struve, Juliane
Cc: r-help@r-project.org
Subject: Re: [R] plotting multiple animal tracks against Date/Time
On Fri, Sep 24, 2010 at 4:16 PM, Struve, Juliane j.str...@imperial.ac.uk
wrote:
Hi again,
when applying the code to my real data I need to deal with a large number of
individuals and massive data sets. I am using the code below to read in the
data for different individuals, and would like to create the Lines within the
loop. But Lines needs to have the variable Fish_ID somehow included in the
name, as otherwise the string will be overwritten on each execution. How can I
create a different Lines for each Fish_ID ? Or is there a better way of doing
this ? Sorry, this is surely a beginner's question.
Thank you very much for your help.
Regards,
Juliane
ReleaseDates - read.csv(file=ReleaseDates.csv,head=TRUE,sep=,)
for (i in 1:length(ReleaseDates$Fish_ID)){
Fish_ID - ReleaseDates$Fish_ID[i]
Data - read.csv(paste(Fish_ID))
Lines - paste(Data$Date,Data$Distance)
print Lines
}
That was just an example. The purpose of Lines was to make the example self
contained and reproducible. In reality you dont use Lines at all. Also your
post suggested that they were in separate files since you had headings on each
one. If that is not the case then you just read it all at once using read.zoo.
z - read.zoo(ReleaseDates.csv, split = Fish_ID, header = TRUE, sep = ,,
...whatever...)
There are three vignettes (pdf documents) that come with zoo. Read them all
and read the help page of read.zoo.
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[R] Modify the vertex label
Dear r-help, I create a graph of my baysian network. I use the package igraph. The names of vertex are within the circle, I would leave them outside the circle? E(g)$color - black tkplot(g, ,vertex.label=names,layout=layout.kamada.kawai, edge.color=E(g)$color) Best Regards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] name ONLY one column
Hi R-users I can not change the name of one column only of my matrix. my_matrix - matrix (1:12,ncol=3) colnames(my_matrix)[1] - 'myname' Error in dimnames(x) - dn : length of 'dimnames' [2] not equal to array extent thank you for your help Lorenzo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] make changes in existing vector with the apply function?
Hi, I'm trying to make some changes in a vector according to some conditions.
It takes too long time however with vector length 10 and I guess a better
way would be using the apply function. I cannot sort out how, however.
As a for/if loop:
for (i in 1:length(PrH)) {
if (is.finite(PrH[i]) == F tempHER tempSPR) {PrH[i] - 1 }
if (is.finite(PrH[i]) == F tempHER tempSPR) {PrH[i] - 0 }}
PrH, tempSPR and tempHER are equally long vectors.
Thanks in advance!Jonas
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[R] max length of a factor variable
Hi Is there a maximum length for the character string representing a level of a factor? I have a set of several million variables, each a factor of length 19. Each factor level is a character string which in some cases can be many thousands of characters long. I am trying to find out why my analysis fails - I just wanted to rule out the possibility that the internal factor conversion has a problem parsing long strings. Thanks Richard -- Richard Mott | Wellcome Trust Centre tel 01865 287588 | for Human Genetics fax 01865 287697 | Roosevelt Drive, Oxford OX3 7BN __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient equivalent to read.csv / write.csv
thank you very much for this sql package, the thing is that thoses table I read are loaded into memory once and for all, and then we work with the data.frames... Do you think then that this is going to be quicker (as I would have thougth that building the SQL DB from the flat file would already be a long process...)? For the RData file it is not possible as those files are shared to them into RData, on the write side the file writte are read by other apps so the csv format can't be changed. Looking forward to read from you Thanks -- View this message in context: http://r.789695.n4.nabble.com/efficient-equivalent-to-read-csv-write-csv-tp2714325p2715275.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Basis functions of cubic regression spline in mgcv
The functional form is given in chapter 4 of my book: Wood S.N. (2006) Generalized Additive Models: An Introduction with R. Chapman and Hall/CRC Press. (reserve your copy now for christmas) ... but note that by default mgcv reparameterises so that the identifiability constraints on the smooths are satisfied automatically (also covered in chapter 4 of the same book). You can get at the evaluated basis functions from the columns of the result of predict.gam(...,type=lpmatrix) - see help files for details. The penalty matrices are stored in the `smooth' list element of the fitted `gam' object. For example x$smooth[[1]]$S[[1]]. best, Simon Quoting Yan Li leemom.atlanta2...@gmail.com: I have a question about the basis functions of cubic regression spline in mgcv. Are there some ways I can get the exact forms of the basis functions and the penalty matrix that are used in mgcv? Thanks in advance! Yan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] bwplot superpose panel.points from another dataframe
Hi everybody,
using bwplot for producing panel boxplot with 3 dimensions
i want to add a mark on each boxplot representing one individual (on all its
dimensions)
till now, i didn't succeed getting the desired solution
I want as well to keep the median symbols as a line
Many thanks for your help
christophe
here is the tested code:
library(lattice)
ex - data.frame(v1 = log(abs(rt(180, 3)) + 1),
v2 = rep(c(2007, 2006, 2005), 60),
z = rep(c(a, b, c, d, e, f), e = 30))
# the individual to be marked
ex2 - data.frame(v1b = log(abs(rt(18, 3)) + 1),
v2 = rep(c(2007, 2006, 2005), 6),
z = rep(c(a, b, c, d, e, f), e = 3))
ex3 - merge(ex, ex2, by=c(v2,z))
bwplot(v2 ~ v1 | z, data = ex3, layout=c(3,2),
pch = |,
par.settings = list(
plot.symbol = list(alpha = 1, col = transparent,cex = 1,pch = 20)),
panel = function(x, y){
panel.bwplot(x, y)
X - tapply(ex3$v1b, ex3[,c(1,2)], max)
Y - seq(length(unique(ex3[,c(1,2)])))
panel.points(X, Y, pch = 17, col = red)
})
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Re: [R] Good documentation about Sweave
Thank you all for those great links, I will look at those. Thanks again Colin -- View this message in context: http://r.789695.n4.nabble.com/Good-documentation-about-Sweave-tp2714326p2715280.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] pairs and mfrow
Is there an alternative to par(mfrow=c(2,1)) to get stacked scatterplot matrixes generated with pairs? I am using version 2.11.1 on Windows XP. The logic I am using follows, and the second pairs plot replaces the first plot in the current graphics device, which is not what I expected (or desired). par(mfrow=c(2,1)) pairs(b2007, main=6/2000 - 12/2006) pairs(a2007, main=1/2007 - 06/2009) Thanks in advance! Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] name ONLY one column
Hi Lorenzo, The problem is that my_matrix does not have dimnames. See below. my_matrix - matrix (1:12,ncol=3) str(my_matrix) ## does not have dimnames dimnames(my_matrix) ## dimnames is NULL colnames(my_matrix) - myname # fails because you are trying to alter the value of something that does not exist ## solution: Set colnames colnames(my_matrix) - 1:dim(my_matrix)[2] str(my_matrix) # my_matrix now has colnames colnames(my_matrix)[1] - myname # and now we can alter them On Mon, Sep 27, 2010 at 8:26 AM, Lorenzo Cattarino l.cattar...@uq.edu.au wrote: Hi R-users I can not change the name of one column only of my matrix. my_matrix - matrix (1:12,ncol=3) colnames(my_matrix)[1] - 'myname' Error in dimnames(x) - dn : length of 'dimnames' [2] not equal to array extent thank you for your help Lorenzo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] name ONLY one column
Hi, It is because the column names do not exist. If you cast the matrix as a data frame your code would work. jon -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Lorenzo Cattarino Sent: 27. september 2010 10:27 To: r-help@r-project.org Subject: [R] name ONLY one column Hi R-users I can not change the name of one column only of my matrix. my_matrix - matrix (1:12,ncol=3) colnames(my_matrix)[1] - 'myname' Error in dimnames(x) - dn : length of 'dimnames' [2] not equal to array extent thank you for your help Lorenzo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] name ONLY one column
Hi, I'm not sure it's even possible (and if it is I don't know how, but I'm no expert). But I think it doesn't make much sense to have only one named column. Just give it a vector: vect_names - c(myname1, myname2, myname3) colnames(my_matrix) - vect_names HTH, Ivan Le 9/27/2010 10:26, Lorenzo Cattarino a écrit : Hi R-users I can not change the name of one column only of my matrix. my_matrix- matrix (1:12,ncol=3) colnames(my_matrix)[1]- 'myname' Error in dimnames(x)- dn : length of 'dimnames' [2] not equal to array extent thank you for your help Lorenzo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient equivalent to read.csv / write.csv
On Mon, Sep 27, 2010 at 7:49 AM, statquant2 statqu...@gmail.com wrote: thank you very much for this sql package, the thing is that thoses table I read are loaded into memory once and for all, and then we work with the data.frames... Do you think then that this is going to be quicker (as I would have thougth that building the SQL DB from the flat file would already be a long process...)? Even including that read.csv.sql is typically several times faster than unoptimized read.csv for large files. See the introductory remarks on the sqldf home page which specifically address that. http://sqldf.googlecode.com In fact, just try it and see whether its ok for you. Its just one line of code to read in a file. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Modify the vertex label
Hi, set the 'vertex.label.dist' parameter: g - graph.ring(10) tkplot(g, vertex.label.dist=1, layout=layout.circle) See ?igraph.plotting for details. Best, Gabor On Mon, Sep 27, 2010 at 11:18 AM, anderson nuel anderson@gmail.com wrote: Dear r-help, I create a graph of my baysian network. I use the package igraph. The names of vertex are within the circle, I would leave them outside the circle? E(g)$color - black tkplot(g, ,vertex.label=names,layout=layout.kamada.kawai, edge.color=E(g)$color) Best Regards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Gabor Csardi gabor.csa...@unil.ch UNIL DGM __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] smooth contour lines
Is there an easy way to control smoothness of the contour lines?
In the plot I am working on due to the undersampling the contour
lines I am getting are jugged, but it is clear by eye these should
be basically straight lines.
In maps package I found smooth.map function, but maybe there is a more
generic way
of accomplishing the same thing.
Ideally there would be an option to control smoothness of the contour
lines
in contourplot(), or levelplot(), but I cannot find a way to do it.
Best regards,
Ryszard Czerminski
AstraZeneca Pharmaceuticals LP
35 Gatehouse Drive
Waltham, MA 02451
USA
781-839-4304
ryszard.czermin...@astrazeneca.com
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[R] rimage package problems
Hi all, I tried to install the rimage in order to get to the function ?read.jpeg. However, I get this error, independent what mirror I choose: install.packages(rimage) --- Please select a CRAN mirror for use in this session --- Warning message: In getDependencies(pkgs, dependencies, available, lib) : package ‘rimage’ is not available Does anybody know what happend with the package? Is there an alternative, I simply want to draw a background picture for a plot using the standard graphics package. Thanks, Ralf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] hcluster with linkage median
Hi, I want to perform a hierarchical clustering using the median as linkage metric. As I understand it the function hcluster in package amap have this option but it does not produce the results that I expect. In the example below M is a matrix of similarities that is transformed into a matrix of dissimilarities D. D [,1] [,2] [,3] [,4] [,5] [1,] 1.0 0.9 0.2 0.2 0.1 [2,] 0.9 1.0 0.7 1.0 0.0 [3,] 0.2 0.7 1.0 0.8 0.8 [4,] 0.2 1.0 0.8 1.0 0.5 [5,] 0.1 0.0 0.8 0.5 1.0 Since [2,5]=0 the objects 2 and 5 should be grouped together in the first step as is done by the agnes function but hcluster start by clustering objects 3 and 4. Why is this? Regards Henrik library(cluster) library(amap) # Create matrix M M - matrix(nr=5,nc=5) M[,1] - c(0,1,8,8,9) M[,2] - c(1,0,3,0,10) M[,3] - c(8,3,0,2,2) M[,4] - c(8,0,2,0,5) M[,5] - c(9,10,2,5,0) # Create matrix D n - dim(M)[1] o - matrix(1,n,n) mn - (1/max(M))*M D - o-mn # Clustering using hcluster ce - hcluster(D,link=median) plot(ce) # Clustering using agnes av - agnes(D,diss=T,method=average) pltree(av) -- View this message in context: http://r.789695.n4.nabble.com/hcluster-with-linkage-median-tp2715585p2715585.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] smooth contour lines
On 27/09/2010 11:11 AM, Czerminski, Ryszard wrote:
Is there an easy way to control smoothness of the contour lines?
In the plot I am working on due to the undersampling the contour
lines I am getting are jugged, but it is clear by eye these should
be basically straight lines.
Straight lines should come out straight in our current scheme. I
suspect your data contains rounding error or other noise, so the actual
contours aren't straight.
Curves are currently an issue: the contouring scheme will generate
polygonal contours. If the grid you pass is too coarse, you'll see it.
For example,
x - 1:5
y - 1:5
z - outer(x, y, function(x,y) (x-2)^2 + (y-3)^2)
contour(x,y,z)
gives ugly contours. The solution to this problem is to use a finer grid:
x - seq(1, 5, len=20)
y - seq(1, 5, len=20)
z - outer(x, y, function(x,y) (x-2)^2 + (y-3)^2)
contour(x,y,z)
makes them look quite smooth. So for your problem, I'd fit a nice
smooth model to your data, then evaluate it on a fine grid, and you'll
get nice smooth contours.
Duncan Murdoch
In maps package I found smooth.map function, but maybe there is a more
generic way
of accomplishing the same thing.
Ideally there would be an option to control smoothness of the contour
lines
in contourplot(), or levelplot(), but I cannot find a way to do it.
Best regards,
Ryszard Czerminski
AstraZeneca Pharmaceuticals LP
35 Gatehouse Drive
Waltham, MA 02451
USA
781-839-4304
ryszard.czermin...@astrazeneca.com
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Re: [R] hcluster with linkage median
On Mon, Sep 27, 2010 at 8:22 AM, Kennedy henrik.aldb...@gmail.com wrote: Hi, I want to perform a hierarchical clustering using the median as linkage metric. As I understand it the function hcluster in package amap have this option but it does not produce the results that I expect. In the example below M is a matrix of similarities that is transformed into a matrix of dissimilarities D. D [,1] [,2] [,3] [,4] [,5] [1,] 1.0 0.9 0.2 0.2 0.1 [2,] 0.9 1.0 0.7 1.0 0.0 [3,] 0.2 0.7 1.0 0.8 0.8 [4,] 0.2 1.0 0.8 1.0 0.5 [5,] 0.1 0.0 0.8 0.5 1.0 Since [2,5]=0 the objects 2 and 5 should be grouped together in the first step as is done by the agnes function but hcluster start by clustering objects 3 and 4. Why is this? From reading the hcluster help file I get the sense that the input is _not_ the distance matrix, but a numeric matrix from which the distance is computed. I think you should simply look at hclust since that does implement the median method. Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie Correspondence Analysis Question
On Sun, 2010-09-26 at 09:41 -0700, Vik Rubenfeld wrote: I'm experienced in statistics, but I am a first-time R user. I would like to use R for correspondence analysis. I have installed R (Mac OSX). I have used the package installer to install the CA package. I have run the following line with no errors to read in the data for a table: NonLuxury - read.table(/Users/myUserName/Desktop/nonLuxury.data.txt) The R online help appears to suggest that the following line should come next: corresp(NonLuxury) However, I get the error message: Error: could not find function corresp You do need to load packages before you can use functions that are contained within those packages. For `corresp`, try require(MASS) corresp(NonLuxury) The MASS packages comes with R as a recommended package, so it should already be installed. HTH G The CA manual appears to suggest that the following line should come next: ca(NonLuxury) Again, I get the error message: Error: could not find function ca What am I missing? Thanks very much in advance to all for any info. -Vik __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] hcluster with linkage median
On Mon, Sep 27, 2010 at 8:22 AM, Kennedy henrik.aldb...@gmail.com wrote: Hi, I want to perform a hierarchical clustering using the median as linkage metric. As I understand it the function hcluster in package amap have this option but it does not produce the results that I expect. Also, if you have a large(r) data set, the package flashClust provides a much faster (n^2 vs. n^3) replacement for hclust with exactly the same results. Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Issue increasing DPI on a png output of a plot
Howdy, I have created a set of plots, but I wish to increase the dpi to 300 (instead of the default 72). From the documentation, I thought that the res parameter to png should accomplish this, but it appears to greatly alter the appearance of my plot. (plot area becomes smaller, plot lines become thicker, etc.) It is my understanding that increasing the dpi should not change the look of the plot, just the quality. Any help in recreating my plots in 300 dpi would be greatly appreciated. - Fincher __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Issue increasing DPI on a png output of a plot
Howdy, I have created a set of plots, but I wish to increase the dpi to 300 (instead of the default 72). From the documentation, I thought that the res parameter to png should accomplish this, but it appears to greatly alter the appearance of my plot. (plot area becomes smaller, plot lines become thicker, etc.) It is my understanding that increasing the dpi should not change the look of the plot, just the quality. Any help in recreating my plots in 300 dpi would be greatly appreciated. - Fincher __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Regular expressions: offsets of groups
Dear list! gregexpr(a+(b+), abcdaabbc) [[1]] [1] 1 5 attr(,match.length) [1] 2 4 What I want is the offsets of the matches for the group (b+), i.e. 2 and 7, not the offsets of the complete matches. Is there a way in R to get that? I know about gsubgn and strapply, but they only give me the strings matched by groups not their offsets. I could write something myself that first takes the above matches (ab and aabb) and then searches again using only the group (b+). For this to work, I'd have to parse the regular expression and search several times ( 2, for nested groups) instead of just once. But I'm sure there is a better way to do this. Thanks for any suggestion! Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] stacked area chart
Dear R-ers! Asking for your help with building the stacked area chart for the following simple data (several variables - with date on the X axis): ### Creating a data set my.data-data.frame(date=c(20080301,20080402,20080503,20090301,20090402,20090503,20100301,20100402,20100503), x=c(1.1,1.0,1.6,1,2,1.5,2.1,1.3,1.9),y=c(-4,-3,-6,-5,-7,-5.2,-6,-4,-4.9), z=c(-0.2,-0.3,-0.4,-0.1,-0.2,-0.05,-0.2,-0.15,-0.06),a=c(4,3,5,5,6,7,5,6,4)) my.data$date-as.character(my.data$date) my.data$date-as.Date(my.data$date,%Y%m%d) (my.data) I'd like the variables whose column values sum up to a negative number to be below zero on that chart and those that add up to a positive number to be above zero in the chart. I am calculating values for ylim and for the order of the variable entry (bottom up) like this: positives-which(colSums(my.data[2:ncol(my.data)])0) # which vars have positive column sums? negatives-which(colSums(my.data[2:ncol(my.data)])0) # which vars have negative column sums? y.max-1.1*max(rowSums(my.data[names(positives)])) # the max on the y axis of the chart y.min-1.1*min(rowSums(my.data[names(negatives)])) # the min on the y axis of the chart ylim - c(y.min, y.max) # ylim for the stacked area chart order.positives-rev(rank(positives)) order.negatives-rev(rank(negatives)) order-c(order.negatives,order.positives) order.of.vars-names(order) # the order of variables on the chart - bottom up ### so, the bottom-most area should be for z, and the second from the bottom area- for y (above z) - they'll be below zero ### and above zero we'll have a first and x second (on top of a). Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Issue increasing DPI on a png output of a plot
On Mon, Sep 27, 2010 at 8:48 AM, Justin Fincher finc...@cs.fsu.edu wrote: Howdy, I have created a set of plots, but I wish to increase the dpi to 300 (instead of the default 72). From the documentation, I thought that the res parameter to png should accomplish this, but it appears to greatly alter the appearance of my plot. (plot area becomes smaller, plot lines become thicker, etc.) It is my understanding that increasing the dpi should not change the look of the plot, just the quality. Any help in recreating my plots in 300 dpi would be greatly appreciated. In short, instead of increasing res, increase width and height. The difference is that if you increase res, font sizes, line thicknesses etc scale up in pixels so as to remain constant in terms of inches (since you are increasing the number of pixels per inch). When you increase width and height, the font sizes etc. remain constant in terms of pixels since resolution also remains constant. Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Normalizing Vector with Negative Numbers
You can use the scale function, just use the minimum instead of the mean and the range instead of the standard deviation. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Gundala Viswanath Sent: Sunday, September 26, 2010 6:23 PM To: r-h...@stat.math.ethz.ch Subject: [R] Normalizing Vector with Negative Numbers Dear expert, I have a series of number that looks like this x - c(-0.005282, 0.000314, 0.002851, -2.5059217162, -0.007545, -1.0317758496, 0.001598, -1.2981735068, 0.072411) How can I normalize it in R so that the new numbers is ranging from 0 to 1 ? - G.V. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] make changes in existing vector with the apply function?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jonas Sundberg
Sent: Monday, September 27, 2010 1:43 AM
To: r-help@r-project.org
Subject: [R] make changes in existing vector with the apply function?
Hi, I'm trying to make some changes in a vector according to
some conditions. It takes too long time however with vector
length 10 and I guess a better way would be using the
apply function. I cannot sort out how, however.
As a for/if loop:
for (i in 1:length(PrH)) {
if (is.finite(PrH[i]) == F tempHER tempSPR) {PrH[i] - 1 }
if (is.finite(PrH[i]) == F tempHER tempSPR) {PrH[i] - 0 }}
PrH, tempSPR and tempHER are equally long vectors.
Did that code ever work? I would have thought that
your tempHER and tempSPR would require a [i] after them.
(Otherwise you would get warnings and incorrect results.)
If so, a first step is to replace your for loop with
PrH[!is.finite(PrH) tempHERtempSPR] - 1
PrH[!is.finite(PrH) tempHERtempSPR] - 0
Read the '[' as 'such that' when using logical-valued
subscripts.
If time or memory is at a premium you may get slightly
better mileage from more complicated code. E.g.,
in the following we compute is.inite(PrH) only once
and use the fact that as.integer maps TRUE and FALSE
to 1 and 0, respectively:
PrHIsInfinite - !is.finite(PrH)
PrH[PrHIsInfinite] -
as.integer(tempHER[PrHIsInfinite]tempSPR[PrHIsInfinite])
I haven't tested any of this.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Thanks in advance!Jonas
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Re: [R] bwplot superpose panel.points from another dataframe
On 2010-09-27 4:54, Christophe Bouffioux wrote:
bwplot(v2 ~ v1 | z, data = ex3, layout=c(3,2),
pch = |,
par.settings = list(
plot.symbol = list(alpha = 1, col = transparent,cex = 1,pch = 20)),
panel = function(x, y){
panel.bwplot(x, y)
X- tapply(ex3$v1b, ex3[,c(1,2)], max)
Y- seq(length(unique(ex3[,c(1,2)])))
panel.points(X, Y, pch = 17, col = red)
})
Perhaps this is what you're trying to achieve:
bwplot(v2 ~ v1 | z, data = ex3, layout=c(3,2),
panel = function(x, y){
panel.bwplot(x, y, pch=|)
X - tapply(ex3$v1b, ex3[, 1:2], max)
Y - seq(nrow(unique(ex3[, 1:2])))
panel.points(X, Y, pch = 17, col = red)
})
(I didn't see any need for your par.settings.)
I'm not crazy about the way you define X,Y. I think
I would augment the data frame appropriately instead.
-Peter Ehlers
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Re: [R] rimage package problems
The rimage package appears to have been abandoned. One option is the EBImage package from Bioconductor. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Ralf B Sent: Monday, September 27, 2010 9:12 AM To: r-help Mailing List Subject: [R] rimage package problems Hi all, I tried to install the rimage in order to get to the function ?read.jpeg. However, I get this error, independent what mirror I choose: install.packages(rimage) --- Please select a CRAN mirror for use in this session --- Warning message: In getDependencies(pkgs, dependencies, available, lib) : package 'rimage' is not available Does anybody know what happend with the package? Is there an alternative, I simply want to draw a background picture for a plot using the standard graphics package. Thanks, Ralf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] pairs and mfrow
Why do you want 2 pairs plots on the same device? There may be a better approach to what you want to do. You could use splom from the lattice package along with the print.trellis function to put 2 on the same page. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Mike Harwood Sent: Monday, September 27, 2010 6:51 AM To: r-help@r-project.org Subject: [R] pairs and mfrow Is there an alternative to par(mfrow=c(2,1)) to get stacked scatterplot matrixes generated with pairs? I am using version 2.11.1 on Windows XP. The logic I am using follows, and the second pairs plot replaces the first plot in the current graphics device, which is not what I expected (or desired). par(mfrow=c(2,1)) pairs(b2007, main=6/2000 - 12/2006) pairs(a2007, main=1/2007 - 06/2009) Thanks in advance! Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] calculating mean and s.d. from a two-column table
I have a two-column table as follows where age is in the 1st column and the number of individuals is in the 2nd. age;no 1;21 2;31 3;9 4;12 5;6 Can I use mean() and sd() to calculate the mean and standard deviation from this or do I have to manually multiplicate 21*1+31*2 etc. / N? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] calculating mean and s.d. from a two-column table
On Mon, Sep 27, 2010 at 9:34 AM, Jonas Josefsson jo...@runtimerecords.net wrote: I have a two-column table as follows where age is in the 1st column and the number of individuals is in the 2nd. age;no 1;21 2;31 3;9 4;12 5;6 You can use the following trick: x = rep(age, no) This repeats age[1] no[1]-times, age[2] no[2]-times, etc... then use the usual mean and sd on x. Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
try this: x - gregexpr(a+(b+), abcdaabbcaaacaaab) justA - gregexpr(a+, abcdaabbcaaacaaab) # find matches in 'x' for 'justA' indx - which(justA[[1]] %in% x[[1]]) # now determine where 'b' starts justA[[1]][indx] + attr(justA[[1]], 'match.length')[indx] [1] 2 7 17 On Mon, Sep 27, 2010 at 11:48 AM, Titus von der Malsburg malsb...@gmail.com wrote: Dear list! gregexpr(a+(b+), abcdaabbc) [[1]] [1] 1 5 attr(,match.length) [1] 2 4 What I want is the offsets of the matches for the group (b+), i.e. 2 and 7, not the offsets of the complete matches. Is there a way in R to get that? I know about gsubgn and strapply, but they only give me the strings matched by groups not their offsets. I could write something myself that first takes the above matches (ab and aabb) and then searches again using only the group (b+). For this to work, I'd have to parse the regular expression and search several times ( 2, for nested groups) instead of just once. But I'm sure there is a better way to do this. Thanks for any suggestion! Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
Thank you Jim, but just as the solution that I discussed, your proposal involves deconstructing the pattern and searching several times. I'm looking for a general and efficient solution. Internally, the regexpr engine has all necessary information after one pass through the string. What I need is an interface that exposes this information. Titus On Mon, Sep 27, 2010 at 6:43 PM, jim holtman jholt...@gmail.com wrote: try this: x - gregexpr(a+(b+), abcdaabbcaaacaaab) justA - gregexpr(a+, abcdaabbcaaacaaab) # find matches in 'x' for 'justA' indx - which(justA[[1]] %in% x[[1]]) # now determine where 'b' starts justA[[1]][indx] + attr(justA[[1]], 'match.length')[indx] [1] 2 7 17 On Mon, Sep 27, 2010 at 11:48 AM, Titus von der Malsburg malsb...@gmail.com wrote: Dear list! gregexpr(a+(b+), abcdaabbc) [[1]] [1] 1 5 attr(,match.length) [1] 2 4 What I want is the offsets of the matches for the group (b+), i.e. 2 and 7, not the offsets of the complete matches. Is there a way in R to get that? I know about gsubgn and strapply, but they only give me the strings matched by groups not their offsets. I could write something myself that first takes the above matches (ab and aabb) and then searches again using only the group (b+). For this to work, I'd have to parse the regular expression and search several times ( 2, for nested groups) instead of just once. But I'm sure there is a better way to do this. Thanks for any suggestion! Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna www...@gmail.com wrote: You've tried: gregexpr(b+, abcdaabbc) But this would match the third occurrence of b+ in abcdaabbcbb. But in this example I'm only interested in b+ if it's preceded by a+. Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 11:48 AM, Titus von der Malsburg malsb...@gmail.com wrote: Dear list! gregexpr(a+(b+), abcdaabbc) [[1]] [1] 1 5 attr(,match.length) [1] 2 4 What I want is the offsets of the matches for the group (b+), i.e. 2 and 7, not the offsets of the complete matches. Is there a way in R to get that? I know about gsubgn and strapply, but they only give me the strings matched by groups not their offsets. I could write something myself that first takes the above matches (ab and aabb) and then searches again using only the group (b+). For this to work, I'd have to parse the regular expression and search several times ( 2, for nested groups) instead of just once. But I'm sure there is a better way to do this. Try this zero width negative look behind expression: gregexpr((?!a+)(b+), abcdaabbc, perl = TRUE) [[1]] [1] 2 7 attr(,match.length) [1] 1 2 See ?regexp for more info. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try this zero width negative look behind expression: gregexpr((?!a+)(b+), abcdaabbc, perl = TRUE) [[1]] [1] 2 7 attr(,match.length) [1] 1 2 Thanks Gabor, but this gives me the same result as gregexpr(b+, abcdaabbc, perl = TRUE) which is wrong if the string is abcdaabbcbbb. Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
You could do this: gregexpr(ab+, abcdaabbcbb)[[1]] + 1 On Mon, Sep 27, 2010 at 2:25 PM, Titus von der Malsburg malsb...@gmail.comwrote: On Mon, Sep 27, 2010 at 7:16 PM, Henrique Dallazuanna www...@gmail.com wrote: You've tried: gregexpr(b+, abcdaabbc) But this would match the third occurrence of b+ in abcdaabbcbb. But in this example I'm only interested in b+ if it's preceded by a+. Titus -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Bayesian Fractional Polynomials package bfp on CRAN
Fractional polynomials (FPs) are an automatic way of fitting non-linear, parametric effects. The R-package mfp implements a frequentist inference approach for FP models. Recently, we have proposed a Bayesian inference approach for normal FP models, which is based on the quasi-default hyper-/g/ prior for the regression coefficients [1]. This approach is implemented in the new R-package bfp. The R-package bfp (current version: 0.0-17) is now available on CRAN [2]. The current development version is (still) available on R-Forge [3]. For a quick start try: install.packages(bfp) library(bfp) example(BmaSamples) and if you have more time, to reproduce results from the paper [1]: demo(ozone) Questions, suggestions or critique are very welcome! Best regards, Daniel [1] Statistics Computing paper: http://dx.doi.org/10.1007/s11222-010-9170-7 [2] CRAN: http://cran.r-project.org/web/packages/bfp/index.html [3] R-Forge: http://r-forge.r-project.org/projects/bfp/ ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Deducer 0.4-1 and JGR 1.7-2 released
Hi All, I would like to announce the release of Deducer 0.4-1 and JGR 1.7-2 to CRAN. The updates should be propagating through the mirrors over the next few days. On the Deducer side we have a number of nice improvements: 1. A new Text Field Widget for plug-ins is included, which is better suited for entering/displaying numeric and short string fields than TextAreaWidget. 2. A new Object Chooser Widget is included, which allows the user to select an object from their workspace, possibly of a specific class. 3. Fixed combobox bug on Mac OS 64-bit. 4. Factor editor and Recode dialog handle empty strings. 5. Created menu items for common plotting templates. The new JGR includes stability fixes along with a few new features. Bug fixes: 1. Fixed MacOS JavaGD resize deadlock bug 2. Fixed 100% CPU on start-up bug reported on some UNIX systems. 3. Interrupt R button fixed on unix systems New Features: 1. Graphics device save support for png, jpeg, bmp, and tiff 2. User control over the menu system has been improved. Added ability to insert new menus, menu items, separators, and sub-menus. Menus and menu items can also be removed. Ian Fellows ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] finding big matrix size and SVD
I have 500,00 rows in my matrix and i was wondering whether there is any way to get its SVD without breaking it to parts because if R can only read about 1000 columns then to have a rectangular matrix (diagonal i think they are called) I will need to have only 1000 rows I want to know how i can do this? natasha From: Steve Lianoglou mailinglist.honey...@gmail.com Cc: R group r-help@r-project.org Sent: Mon, 27 September, 2010 3:23:58 Subject: Re: [R] finding big matrix size and SVD Hi, rote: Dear R helpers I have a big data sheet (CSV) which I use âread.csvâ to read it When im trying to get the Dim() it says 38 column which is not correct it should be something about 400. I am wondering whether there is any way I can read it right⦠I have used ncol() and itâs the same answer It seems that perhaps the input file isn't well formed, and R can't correctly identify that each row should have 400 columns? Maybe you can try to read it in manually (using readLines, for instance), and strsplit each line so that you can get finer control of reading the file. Alternatively, you can try and edit the file by hand to fix it. On the other hand I use to get the SVD of it but I have about 500,000 rows so considering that it should be rectangular matrix⦠im wondering how this can be possible I don't follow this part, sorry ... what's the question here? -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
You've tried: gregexpr(b+, abcdaabbc) On Mon, Sep 27, 2010 at 12:48 PM, Titus von der Malsburg malsb...@gmail.com wrote: Dear list! gregexpr(a+(b+), abcdaabbc) [[1]] [1] 1 5 attr(,match.length) [1] 2 4 What I want is the offsets of the matches for the group (b+), i.e. 2 and 7, not the offsets of the complete matches. Is there a way in R to get that? I know about gsubgn and strapply, but they only give me the strings matched by groups not their offsets. I could write something myself that first takes the above matches (ab and aabb) and then searches again using only the group (b+). For this to work, I'd have to parse the regular expression and search several times ( 2, for nested groups) instead of just once. But I'm sure there is a better way to do this. Thanks for any suggestion! Titus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Split-split plot design with aov function in R
*Hello, I'm new to R and trying to do Split Split Plot Design analysis with aov function in R. Sharing any worked example and suggestion will be highly appreciated. Thanks Regards! * -- * Muhammad Yaseen * [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Split-split plot design with aov function in R
http://www.amazon.com/Statistical-Design-George-Casella/dp/1441926143/ref=sr_1_1?s=gatewayie=UTF8qid=1285609902sr=8-1 Hello Mohd. Yaseen, Please check out the book by Dr. Casella and his website www.stat.ufl.edu/~casella for the relevant R codes. Chapter 5 of this book talks about Split Split Plot designs. Thanks and Best Regards, S. P.S. : Please refrain from reposting your questions. Senior members of this list are not too happy when you post the same question twice. It is just a friendly advice. http://www.amazon.com/Statistical-Design-George-Casella/dp/1441926143/ref=sr_1_1?s=gatewayie=UTF8qid=1285609902sr=8-1 On Mon, Sep 27, 2010 at 12:42 PM, Muhammad Yaseen myaseen...@gmail.comwrote: *Hello, I'm new to R and trying to do Split Split Plot Design analysis with aov function in R. Sharing any worked example and suggestion will be highly appreciated. Thanks Regards! * -- * Muhammad Yaseen * [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] compare two matrices
Hi everyone: I have a kinda easy question but i do not know how to solve that in a simple way. I want to compare the rows of two matrices. col1 - c(1,2,3,4,5,6) col2 - c(6,5,4,3,2,1) m - cbind(col1, col2) col3 - c(1,3,2,6) col4 - c(6,3,5,1) n - cbind(col3, col4) In matrix n, for example the first row is (1,6), it is also some row in matrix m, i want the code results TRUE. then for the 2nd row (3,3), it should be FALSE. So in this case it should be (TRUE, FALSE, TRUE, TRUE) I tried %in% or is.element for a row in n and matrix m, but it does not work. I think I can also write two loops to compare the rows of m and n one by one, but it takes a long time to run. Could anyone tell me how to solve this? Thank you very much!! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacked area chart
I found a solution to my original question (see code below).
But I have a question about cosmetics, which I always find very challenging.
1. How can I make all dates appear on the X axis (rotated at 90
degrees vs. horizontal)?
2. How can I create vertical grid lines so that at each date there is
a gridline?
3. How can I create a legend for colors, but not on top of the graph
but on the right side, outside of the graph (because in my real data
set I have a lot of variables - so that there'll never be enough space
for the legend in the graph itself)
Thanks a lot!
Dimitri
### Creating a data set with both positives and negatives
my.data-data.frame(date=c(20080301,20080402,20080503,20090301,20090402,20090503,20100301,20100402,20100503),
x=c(1.1,1.0,1.6,1,2,1.5,2.1,1.3,1.9),y=c(-4,-3,-6,-5,-7,-5.2,-6,-4,-4.9),
z=c(-0.2,-0.3,-0.4,-0.1,-0.2,-0.05,-0.2,-0.15,-0.06),a=c(10,13,15,15,16,17,15,16,14))
my.data$date-as.character(my.data$date)
my.data$date-as.Date(my.data$date,%Y%m%d)
(my.data)
positives-which(colSums(my.data[2:ncol(my.data)])0) # which vars
have positive column sums?
negatives-which(colSums(my.data[2:ncol(my.data)])0) # which vars
have negative column sums?
y.max-1.1*max(rowSums(my.data[names(positives)])) # the max on the y
axis of the chart
y.min-1.1*min(rowSums(my.data[names(negatives)])) # the min on the y
axis of the chart
ylim - c(y.min, y.max)
order.positives-rev(rank(positives))
order.of.pos.vars-names(order.positives)
order.negatives-rev(rank(negatives))
order.of.neg.vars-names(order.negatives)
order-c(order.negatives,order.positives)
order.of.vars-names(order) # the order of variables on the chart -
from the bottom up
### so, the bottom-most area should be for z, the second from the
bottom area- for y (above z)
all.colors-c('red','blue','green','orange','yellow','purple')
xx - c(my.data$date, rev(my.data$date))
bottom.y.coordinates-rowSums(my.data[names(negatives)])
plot(x=my.data$date, y=bottom.y.coordinates, ylim=ylim, col='white',
type='l', xaxt='n',
ylab='Title for Y', xlab='Date', main='Chart Title')
for(var in order.of.neg.vars){
top.line.coords-bottom.y.coordinates-my.data[[var]]
bottom.coords-c(bottom.y.coordinates,rev(top.line.coords))
polygon(xx,bottom.coords,col=all.colors[which(names(my.data) %in% var)])
bottom.y.coordinates-top.line.coords
}
for(var in order.of.pos.vars){
top.line.coords-bottom.y.coordinates+my.data[[var]]
bottom.coords-c(bottom.y.coordinates,rev(top.line.coords))
polygon(xx,bottom.coords,col=all.colors[which(names(my.data) %in% var)])
bottom.y.coordinates-top.line.coords
}
On Mon, Sep 27, 2010 at 11:47 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Dear R-ers!
Asking for your help with building the stacked area chart for the
following simple data (several variables - with date on the X axis):
### Creating a data set
my.data-data.frame(date=c(20080301,20080402,20080503,20090301,20090402,20090503,20100301,20100402,20100503),
x=c(1.1,1.0,1.6,1,2,1.5,2.1,1.3,1.9),y=c(-4,-3,-6,-5,-7,-5.2,-6,-4,-4.9),
z=c(-0.2,-0.3,-0.4,-0.1,-0.2,-0.05,-0.2,-0.15,-0.06),a=c(4,3,5,5,6,7,5,6,4))
my.data$date-as.character(my.data$date)
my.data$date-as.Date(my.data$date,%Y%m%d)
(my.data)
I'd like the variables whose column values sum up to a negative number
to be below zero on that chart and those that add up to a positive
number to be above zero in the chart. I am calculating values for ylim
and for the order of the variable entry (bottom up) like this:
positives-which(colSums(my.data[2:ncol(my.data)])0) # which vars
have positive column sums?
negatives-which(colSums(my.data[2:ncol(my.data)])0) # which vars
have negative column sums?
y.max-1.1*max(rowSums(my.data[names(positives)])) # the max on the y
axis of the chart
y.min-1.1*min(rowSums(my.data[names(negatives)])) # the min on the y
axis of the chart
ylim - c(y.min, y.max) # ylim for the stacked area chart
order.positives-rev(rank(positives))
order.negatives-rev(rank(negatives))
order-c(order.negatives,order.positives)
order.of.vars-names(order) # the order of variables on the chart - bottom
up
### so, the bottom-most area should be for z, and the second from the
bottom area- for y (above z) - they'll be below zero
### and above zero we'll have a first and x second (on top of a).
Thanks a lot for your advice!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] compare two matrices
one way is the following: col1 - c(1,2,3,4,5,6) col2 - c(6,5,4,3,2,1) m - cbind(col1, col2) col3 - c(1,3,2,6) col4 - c(6,3,5,1) n - cbind(col3, col4) ind.n - do.call(paste, c(as.data.frame(n), sep = \r)) ind.m - do.call(paste, c(as.data.frame(m), sep = \r)) ind.n %in% ind.m I hope it helps. Best, Dimitris On 9/27/2010 8:00 PM, xinxin xx wrote: Hi everyone: I have a kinda easy question but i do not know how to solve that in a simple way. I want to compare the rows of two matrices. col1- c(1,2,3,4,5,6) col2- c(6,5,4,3,2,1) m- cbind(col1, col2) col3- c(1,3,2,6) col4- c(6,3,5,1) n- cbind(col3, col4) In matrix n, for example the first row is (1,6), it is also some row in matrix m, i want the code results TRUE. then for the 2nd row (3,3), it should be FALSE. So in this case it should be (TRUE, FALSE, TRUE, TRUE) I tried %in% or is.element for a row in n and matrix m, but it does not work. I think I can also write two loops to compare the rows of m and n one by one, but it takes a long time to run. Could anyone tell me how to solve this? Thank you very much!! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Regular expressions: offsets of groups
On Mon, Sep 27, 2010 at 1:34 PM, Titus von der Malsburg malsb...@gmail.com wrote: On Mon, Sep 27, 2010 at 7:29 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try this zero width negative look behind expression: gregexpr((?!a+)(b+), abcdaabbc, perl = TRUE) [[1]] [1] 2 7 attr(,match.length) [1] 1 2 Thanks Gabor, but this gives me the same result as gregexpr(b+, abcdaabbc, perl = TRUE) which is wrong if the string is abcdaabbcbbb. Sorry, try this: gregexpr((?=a)b+, abcdaabbcbbb, perl = TRUE) [[1]] [1] 2 7 attr(,match.length) [1] 1 2 Note that it does not give the same answer as: gregexpr(b+, abcdaabbcbbb, perl = TRUE) [[1]] [1] 2 7 10 attr(,match.length) [1] 1 2 3 gregexpr((?=a)b+, abcdaabbcbbb, perl = TRUE) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with unlist
On Mon, Sep 27, 2010 at 5:27 AM, Ben Bolker bbol...@gmail.com wrote: Luis Felipe Parra felipe.parra at quantil.com.co writes: Hello, I am trying to unlist a list, which is attached, and I am having the problem that when I unlist it the number of elements changes from 5065 to 5084 x - lapply(SumaPluvi, FUN=[, 1); n - sapply(x, FUN=length); print(table(n)); n 1 5065 print(which(n != 1)); integer(0) length(unlist(lapply(SumaPluvi, FUN=[, 1))) [1] 5081 I dont now why, but when I unlist it the number of elements changes from 5065 to 5084 even if there is no list element with length greater than one. Do you know what can be happening? We probably won't be able to get farther without a reproducible example. One brute-force way of finding the problem is by bisection: i.e., try the first and last halves of your list separately, and see if either one individually shows a similar problem. Proceed recursively until you localize the problem ... ...and as alternative, my most recent post did contain an updated code snippet that is likely to find list elements generating more than one value. /Henrik __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Ordered logit with polr won't match SPSS output
I am learning R via a textbook that performs analysis with SPSS and SAS. In trying to reproduce the results for an ordinal logit model, I get very similar point estimates for my cut-off points, but the parameters for the covariate q60 do not match. The estimate for q51 also matches. Is this because I need to change a base case for the ordered covariate q60? Can this be done in or is it always the first case? mod-polr(as.ordered(q43j)~as.ordered(q60)+q51, method=logistic) Perhaps a book using R would be a better idea, but it's the content and price (free) of this book that I like. Thanks a lot, Ben [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] margin control in lattice package
Hi Peter, Thank you for your thoughtful reply. I am tweaking the setting print settings you suggested. It looks like this is going to solve my problem. Thanks very much for help. Jonathan On Sat, Sep 25, 2010 at 6:00 PM, Peter Ehlers ehl...@ucalgary.ca wrote: On 2010-09-25 8:59, Jonathan Flowers wrote: Hi all, I am difficulty with simple layout of plots in the lattice package I have created a series of levelplots and would like to plot them to a single device, but need to reduce the margin areas. This is easily accomplished with par(oma) and par(mar) in the base graphics package but I am having problems finding the equivalent features in the lattice package. Ideally, I would like to reduce the amount of white space among plots in the following example. Thanks in advance. library(lattice) p1- levelplot( matrix(c(1:25),nr=5,nc=5),row. values=1:5,column.values=1:5) p2- levelplot(matrix (rnorm(25),nr=5,nc=5),row.values=1:5,column.values=1:5) p3- levelplot( matrix(c(1:25),nr=5,nc=5),row.values=1:5,column.values=1:5) p4- levelplot(matrix (rnorm(25),nr=5,nc=5),row.values=1:5,column.values=1:5) print(p1,split=c(1,1,2,2),more=T) print(p2,split=c(2,1,2,2),more=T) print(p3,split=c(1,2,2,2),more=T) print(p4,split=c(2,2,2,2)) Thanks in advance, Jonathan Here are a couple of things you can play with. First, the default for the matrix method of levelplot() is to produce square plots using the argument aspect=iso. So unless you set aspect=some other value or fill, you can only reduce the outer white space at the expense of increasing the inner white space. To reduce the outer white space but keep the aspect=iso, you can set the size of the graphics device and then fiddle with the top.padding and/or bottom.padding components of the layout.heights parameter. Something like this: ## some simple data m - matrix(1:25, nr=5) ## create 4 (identical for illustration only) plots p1 - p2 - p3 - p4 - levelplot(m, aspect=iso, par.settings=list(layout.heights=list(top.padding=-2))) ## open a trellis device (I'm on Windows) trellis.device(windows, height=6, width=7) ## print the plots print(p1, split=c(1,1,2,2), more=TRUE) print(p2, split=c(2,1,2,2), more=TRUE) print(p3, split=c(1,2,2,2), more=TRUE) print(p4, split=c(2,2,2,2)) ## alternatively, use aspect=fill and adjust size in the ## print() calls p1 - p2 - p3 - p4 - levelplot(m, aspect=fill) trellis.device(windows, height=6, width=7) print(p1, split=c(1,1,2,2), panel.height=list(x=2, units=in), more=TRUE) print(p2, split=c(2,1,2,2), panel.height=list(x=2, units=in), more=TRUE) print(p3, split=c(1,2,2,2), panel.height=list(x=2, units=in), more=TRUE) print(p4, split=c(2,2,2,2), panel.height=list(x=2, units=in)) Probably the best way, if your levels are roughly the same for all plots, is to convert your data to a data frame, define a 4-level factor, and create a standard 4-panel plot instead of using the 'split' argument. -Peter Ehlers [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] scientific vs. fixed notation in xyplot()
Hi I am using xyplot() to plot on the log scale by using scale=list(log=T) argument. For example: xyplot(1:10~1:10, scales=list(log=T)) But the axis labels are printed as scientific notation (10^0.0, etc), instead of fixed notation. How can I change that to fixed notation? options(scipen=4) doesn't work on xyplot() Thanks John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sample size estimation for non-inferiority log-rank and Wilcoxon rank-sum tests
Hello Everyone, I'm trying to conduct a couple of power analyses and was hoping someone might be able to help. I want to estimate the sample size that would be necessary to adequately power a couple of non-inferiority tests. The first would be a log-rank test and the second would be a Wilcoxon rank-sum test. I want to be able to determine the sample size that would be necessary to test for a 3 day difference in median recovery time between 2 groups of cancer patients. Both of these tests are infeasible using SAS Proc Power and I haven't been able to find information about how to do them using either SAS or R. Does anyone know how to perform either of these calculations? If so, I'd greatly appreciate it if you could share a couple of examples. Thanks, Paul [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scientific vs. fixed notation in xyplot()
Try this: xyplot(1:10~1:10, scales=list(log = T, labels = round(log(1:10), 4))) On Mon, Sep 27, 2010 at 4:10 PM, array chip arrayprof...@yahoo.com wrote: Hi I am using xyplot() to plot on the log scale by using scale=list(log=T) argument. For example: xyplot(1:10~1:10, scales=list(log=T)) But the axis labels are printed as scientific notation (10^0.0, etc), instead of fixed notation. How can I change that to fixed notation? options(scipen=4) doesn't work on xyplot() Thanks John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Percentages and lattice
What I'm trying to do is to figure out how to create lattice charts of %right by region, or alternatively, by date from a dataset of observations that looks something like this: date,location,region,correct 2010-09-10,a,r1,yes 2010-09-10,a,r1,yes 2010-09-10,a,r1,no 2010-09-11,a,r1,yes 2010-09-01,b,r1,yes 2010-09-02,b,r1,no 2010-09-01,a,r2,yes 2010-09-02,a,r2,no 2010-09-02,a,r2,yes 2010-09-02,a,r2,no 2010-09-03,a,r2,yes etc. I get that I can do something like: tmp-xtabs(~correct+region+date,t) This gets me partway by providing counts. The thing that I'm missing here is how get from this to being able to plot, for example, a family of lattice-style curves that looks something like: %right (by region) on the y axis, with the x axis being the date I've tried something like barchart(xtabs(~correct+region+date,t),stack=F,auto.key=T) which gets me close with the counts of observations, but I really am after the proportions evaluated by region, not the counts. I don't see how to marginalize prop.table properly to give me my answer either (if, in region 1, on a given day, all are correct - then that number should be 100% on the chart/graph, independent of what happened in region 2)- but I'm quite sure that I'm missing something obvious here. Any suggestions would be appreciated. Thanks, mark [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Percentages and lattice
Mark - Here's one way to get the percentages you want. Suppose your data frame is called df: correct = subset(as.data.frame(with(df,table(date,region,correct))), correct=='yes') all = as.data.frame(with(df,table(date,region))) names(all)[3] = 'Total' both = merge(correct,all) both$Pct = both$Freq / both$Total * 100 - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Mon, 27 Sep 2010, Mark Noworolski wrote: What I'm trying to do is to figure out how to create lattice charts of %right by region, or alternatively, by date from a dataset of observations that looks something like this: date,location,region,correct 2010-09-10,a,r1,yes 2010-09-10,a,r1,yes 2010-09-10,a,r1,no 2010-09-11,a,r1,yes 2010-09-01,b,r1,yes 2010-09-02,b,r1,no 2010-09-01,a,r2,yes 2010-09-02,a,r2,no 2010-09-02,a,r2,yes 2010-09-02,a,r2,no 2010-09-03,a,r2,yes etc. I get that I can do something like: tmp-xtabs(~correct+region+date,t) This gets me partway by providing counts. The thing that I'm missing here is how get from this to being able to plot, for example, a family of lattice-style curves that looks something like: %right (by region) on the y axis, with the x axis being the date I've tried something like barchart(xtabs(~correct+region+date,t),stack=F,auto.key=T) which gets me close with the counts of observations, but I really am after the proportions evaluated by region, not the counts. I don't see how to marginalize prop.table properly to give me my answer either (if, in region 1, on a given day, all are correct - then that number should be 100% on the chart/graph, independent of what happened in region 2)- but I'm quite sure that I'm missing something obvious here. Any suggestions would be appreciated. Thanks, mark [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scientific vs. fixed notation in xyplot()
Thanks for the suggestion. But my example is just an example, I would prefer to have some generalized solution, like what options(scipen=4) does in general graphics, which usually gave pretty axis labels as well. Any suggestions? Thanks John From: Henrique Dallazuanna www...@gmail.com Cc: r-help@r-project.org Sent: Mon, September 27, 2010 12:16:31 PM Subject: Re: [R] scientific vs. fixed notation in xyplot() Try this: xyplot(1:10~1:10, scales=list(log = T, labels = round(log(1:10), 4))) Hi I am using xyplot() to plot on the log scale by using scale=list(log=T) argument. For example: xyplot(1:10~1:10, scales=list(log=T)) But the axis labels are printed as scientific notation (10^0.0, etc), instead of fixed notation. How can I change that to fixed notation? options(scipen=4) doesn't work on xyplot() Thanks John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sample size estimation for non-inferiority log-rank and Wilcoxon rank-sum tests
I haven't done much with the type of data you're working with, but here is a post that lists a few packages for doing sample size calculations in R. Perhaps one of them will be helpful. https://stat.ethz.ch/pipermail/r-help/2008-February/154223.html Andrew Miles On Sep 27, 2010, at 2:09 PM, Paul Miller wrote: Hello Everyone, I'm trying to conduct a couple of power analyses and was hoping someone might be able to help. I want to estimate the sample size that would be necessary to adequately power a couple of non- inferiority tests. The first would be a log-rank test and the second would be a Wilcoxon rank-sum test. I want to be able to determine the sample size that would be necessary to test for a 3 day difference in median recovery time between 2 groups of cancer patients. Both of these tests are infeasible using SAS Proc Power and I haven't been able to find information about how to do them using either SAS or R. Does anyone know how to perform either of these calculations? If so, I'd greatly appreciate it if you could share a couple of examples. Thanks, Paul [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Variation of predictor of linear model
Hi folks, I use lm to run regression and I don't know how to predict dependent variable based on the model. I used predict.lm(model, newdata=80), but it gave me warnings. Also, how can I get the variance of dependent variable based on model. Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scientific vs. fixed notation in xyplot()
This is quite elegant (thanks) and brings up a problem I could not solve awhile back, although Dr. Sarkar did his best to help. How do I do the same thing in a panel plot? e.g., toy example temp.df - data.frame(X=seq(1,100,by=1),Y=seq(1,50.5,by=.5),class=rep (c(A,B),each=50)) xyplot(Y ~ X | class,data=temp.df,scales=list(x=round(log(1:100), 4),y=round(log(1:50.5), 4),log=T)) gives me the right points on the page but still gives axis labels in scientific notation. If I try to specify labels as a list I get an error message xyplot(Y ~ X | class,data=temp.df,scales=list(log = T, labels = list(x=round(log(1:100), 4),y=round(log(1:50.5), 4 Error in construct.scales(log = TRUE, labels = list(x = c(0, 0.6931, 1.0986, : the at and labels components of scales may not be lists when relation = same Syntax problem in this last command? Thanks On 27-Sep-10, at 12:16 PM, Henrique Dallazuanna wrote: Try this: xyplot(1:10~1:10, scales=list(log = T, labels = round(log(1:10), 4))) On Mon, Sep 27, 2010 at 4:10 PM, array chip arrayprof...@yahoo.com wrote: Hi I am using xyplot() to plot on the log scale by using scale=list (log=T) argument. For example: xyplot(1:10~1:10, scales=list(log=T)) But the axis labels are printed as scientific notation (10^0.0, etc), instead of fixed notation. How can I change that to fixed notation? options(scipen=4) doesn't work on xyplot() Thanks John -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. Don McKenzie, Research Ecologist Pacific WIldland Fire Sciences Lab US Forest Service Affiliate Professor School of Forest Resources, College of the Environment CSES Climate Impacts Group University of Washington desk: 206-732-7824 cell: 206-321-5966 d...@uw.edu donaldmcken...@fs.fed.us __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Alphabetical sequence of data along the x-axis in a box plot
Hello All, I noticed when I generated some boxplots, the data is presented in alphabetical order along the x-axis (the data in this case was the four quandrants of a sample area (NE,NW, SE, SW) that was my first column of data). Is there a way to have R plot the data in a different order? I imagine you could use a dummy variable, but didn't know if there might be a simple argument that will address this? Thanks for any guidance, Eddie Hughes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot single part of the country using gadm map
Hi carolina plescia wrote: Dear all, in GADM map there are three levels (nation, province and precinct) for each country of the world but for all of them you are never able to plot only one part of a chosen country. The Spatial object you get when you read in the shapefile has convenient subsetting methods defined. For example ... library(maptools) italy - readShapeSpatial(ITA_adm1) plot(italy) liguria - italy[italy$NAME_1 == Liguria, ] plot(liguria) Is that what you are after? Paul To be sure, I am trying to plot only one region of “Italy” and colour the different precincts in it. So far I am able to colour only the part of my interest but the programme still plot the whole country. Is that a way to have only a chosen part of the country (ie. region or state) with all the specific subunits in it? I have tried also to plot and then zoom but the “zoom” function seems to work only with “plot” while gadm map uses ”spplot”. Anyone knows if the zoom command works also with spplot? Any suggestions? Thank you very much. Carolina -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 p...@stat.auckland.ac.nz http://www.stat.auckland.ac.nz/~paul/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variation of predictor of linear model
Hi, Try this: # using the iris dataset mydat - iris mymodel - lm(Sepal.Length ~ Petal.Length + Species, data = mydat) summary(mymodel) newdat - data.frame(Petal.Length = seq(1, 10, by = .1), Species = factor(rep(virginica, 91))) results - predict(object = mymodel, newdata = newdat, se.fit = TRUE) results The main lesson is that generally newdata should be a data frame with columns that have the same name as the predictors (IVs) in your model. I'm not exactly sure what you mean by variance of dependent variable based on model. Do you want its total variance, residual variance, ___ ? Cheers, Josh On Mon, Sep 27, 2010 at 12:58 PM, Yi Du abraham...@gmail.com wrote: Hi folks, I use lm to run regression and I don't know how to predict dependent variable based on the model. I used predict.lm(model, newdata=80), but it gave me warnings. Also, how can I get the variance of dependent variable based on model. Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Alphabetical sequence of data along the x-axis in a box plot
Hi Eddie, I've been on a role with the iris data, so I figure why stop. Assuming that one variable is a factor, you can easily reverse it, and if you want fine tuned control, then just reorder the levels. Here is an example: dat - iris boxplot(Sepal.Length ~ Species, data = dat) boxplot(Sepal.Length ~ rev(Species), data = dat) # They had been ordered alphabetically, now I am changing them dat$Species - factor(dat$Species, levels = c(versicolor, virginica, setosa), labels = c(versicolor, virginica, setosa)) boxplot(Sepal.Length ~ Species, data = dat) Cheers, Josh On Mon, Sep 27, 2010 at 12:59 PM, Hughes, Ed ehug...@conshelf.com wrote: Hello All, I noticed when I generated some boxplots, the data is presented in alphabetical order along the x-axis (the data in this case was the four quandrants of a sample area (NE,NW, SE, SW) that was my first column of data). Is there a way to have R plot the data in a different order? I imagine you could use a dummy variable, but didn't know if there might be a simple argument that will address this? Thanks for any guidance, Eddie Hughes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Alphabetical sequence of data along the x-axis in a box plot
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
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Hughes, Ed
Sent: Tuesday, 28 September 2010 9:00 a.m.
To: r-help@r-project.org
Subject: [R] Alphabetical sequence of data along the x-axis in a box
plot
Hello All,
I noticed when I generated some boxplots, the data is presented in
alphabetical order along the x-axis (the data in this case was the four
quandrants of a sample area (NE,NW, SE, SW) that was my first column of
data). Is there a way to have R plot the data in a different order? I
imagine you could use a dummy variable, but didn't know if there might
be a simple argument that will address this?
Thanks for any guidance,
Eddie Hughes
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[R] spplot cuts
Hello-
After looking through ?spplot, I would expect that I could specify the
values of the cuts:
...‘cuts’ number of cuts or the actual cuts to use...
So in the following command,
spplot(lzm.krige.dir[var1.pred], scales=list(draw=TRUE),
xlab=Easting,ylab=Northing,
cuts=seq(0.0,0.4,by=0.01),key.space=right,cex=1.1,col.regions=terrain.colors(41),
main=Specific Yield Layer 1,sp.layout=list(pts))
I get the following warning messages and the plot is a solid color:
Warning messages:
1: In if (length.out == 0L) integer(0L) else if (One) seq_len(length.out)
else if (missing(by)) { :
the condition has length 1 and only the first element will be used
2: In if (length.out 2L) if (from == to) rep.int(from, length.out) else
as.vector(c(from, :
the condition has length 1 and only the first element will be used
Looking at Edzer's response to a post:
http://r.789695.n4.nabble.com/using-spplot-sp-package-with-5-quantiles-td836245.html#a836246
it reinforces to me the idea that I should be able to specify where the cuts
should occur. I'm not sure what to do with the warning message and my
google searches have been unfruitful.
Some supplemental information that might be helpful:
range(data.frame(lzm.krige.dir[var1.pred])[,1])
[1] 0.1277067 0.2933876
I give the range of the plotted values because it varies between datasets
and I would like the color scale to remain fixed between images rather than
changing image to image making visual comparison's more difficult.
Any suggestions would be greatly appreciated,
Eric
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