[R] melt error that I don't understand.
I'm stumped. I have a dataset I want to melt to create a temporal sequence of
events for each subject, but in each row, I would like to retain the baseline
characteristics.
D - structure(list(ID = c(A, B, C, D, E, F, G, H, I, J),
AGE = structure(c(68L, 63L, 55L, 64L, 60L, 78L, 60L, 62L,
60L, 75L),
label = Age, class = labelled),
BMI = structure(c(25L, 27L, 27L, 28L, 32L, NA, 36L, 27L,
31L, 25L),
label = BMI (kg/m2), class = labelled),
EventDays = structure(c(722L, 738L, 707L, 751L, 735L, 728L,
731L, 717L, 728L, 735L),
label = Time to first ACM/censor
(days), class = labelled),
ImplantDays = c(NA, NA, 575, NA, NA, NA, 490, 643, NA,
NA)),
.Names = c(ID, AGE, BMI, EventDays, InterventionDays),
row.names = c(NA, 10L),
class = data.frame)
melt(D, c(ID, AGE, BMI)) # produces the following error
Error in data.frame(ids, variable, value, stringsAsFactors = FALSE) :
arguments imply differing number of rows: 10, 20
Now, I know AGE and BMI aren't exactly identifying variables, but my hope would
be that, since ID uniquely identifies the subjects, I could use this as a short
cut to getting the data set I want. I can get the data I want if I go about it
a little differently.
#* What I would like it to look like.
Timeline - melt(D[, c(ID, EventDays, InterventionDays)], ID,
na.rm=TRUE)
Timeline - arrange(Timeline, ID, value)
Timeline - merge(D[, c(ID, AGE, BMI)],
Timeline,
by=ID, all.x=TRUE)
At first I thought it might be the mixture of character and numeric variables
as IDs, but the following example works
A - data.frame(id = LETTERS[1:10],
age = c(50, NA, 51, 52, 53, 54, 55, 56, 57, 58),
meas1 = rnorm(10),
meas2 = rnorm(10, 5),
stringsAsFactors=FALSE)
melt(A, c(id, age))
I'm sure I'm missing something really obvious (kind of like how I can stare at
the dry goods aisle for 10 minutes and still not find the chocolate chips). If
anyone could help me understand why this error is occurring, I'd greatly
appreciate it.
sessionInfo()
R version 2.15.2 (2012-10-26)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] C
attached base packages:
[1] splines stats graphics grDevices utils datasets methods base
other attached packages:
[1] lazyWeave_2.2.3 Hmisc_3.10-1 survival_2.36-14 plyr_1.7.1
reshape2_1.2.2
loaded via a namespace (and not attached):
[1] cluster_1.14.3 grid_2.15.2 lattice_0.20-10 stringr_0.6.1 tools_2.15.2
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
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Re: [R] lmer, p-values and all that
My apologies for coming late into this conversation, but I'm curious about
something in your response
You use the following code to peform a likelihood ratio test between an lm
object and a mer object
fm0 - lm(distance~age,data=Orthodont)
fm2 - lmer(distance~age+(1|Subject),data=Orthodont,REML=FALSE)
ddiff - -2*c(logLik(fm0)-logLik(fm2))
pchisq(ddiff,1,lower.tail=FALSE)
It seems like this would be simple to roll up into a function, such as
lrtestGeneric - function(fit1, fit2){
chisq - -2 * c(logLik(fit1) - logLik(fit2))
df - abs(attributes(logLik(fit1))$df - attributes(logLik(fit2))$df)
p - pchisq(chisq, df, lower.tail=FALSE)
lrtest - data.frame(L.R.Chisq=chisq, d.f.=df, P=p)
return(lrtest)
}
lrtestGeneric(fm0, fm2)
My question is about whether it is appropriate to use the degrees of freedom
returned by logLik or if I should just use 1 degree of freedom when comparing a
model without the random effect to one with the random effect. For instance,
logLik returns a difference of 3 between degrees of freedom in the models.
Should I be using the 3 degrees of freedom in the likelihood ratio test, or is
it better to go with 1?
fit0 - lm(Reaction ~ Days, sleepstudy)
fit1 - lmer(Reaction ~ Days + (Days|Subject), sleepstudy)
lrtestGeneric(fit0, fit2)
Any education you can provide would be great appreciated.
Thanks
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Ben Bolker
Sent: Wednesday, March 27, 2013 10:34 PM
To: David Winsemius
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] lmer, p-values and all that
On 13-03-27 10:10 PM, David Winsemius wrote:
On Mar 27, 2013, at 7:00 PM, Ben Bolker wrote:
Michael Grant michael.grant at colorado.edu writes:
Dear Help:
I am trying to follow Professor Bates' recommendation, quoted by
Professor Crawley in The R Book, p629, to determine whether I should
model data using the 'plain old' lm function or the mixed model
function lmer by using the syntax anova(lmModel,lmerModel).
Apparently I've not understood the recommendation or the proper
likelihood ratio test in question (or both) for I get this error
message: Error: $ operator not defined for this S4 class.
I don't have the R Book handy (some more context would be extremely
useful! I would think it would count as fair use to quote the
passage you're referring to ...)
This is the quoted Rhelp entry:
http://tolstoy.newcastle.edu.au/R/help/05/01/10006.html
(I'm unable to determine whether it applies to the question at hand.)
OK, I misspoke -- sorry. I think the lmer()/lme() likelihoods are actually
comparable; it's GLMMs (glmer(), with no analogue in lme()-land except for
MASS::glmmPQL, which doesn't give you log-likelihoods at all) where the problem
arises.
You can (1) use lme(), (2) look at http://glmm.wikidot.com/faq for
suggestions about testing random-effects terms (including perhaps don't do it
at all), or (3) construct the likelihood ratio test yourself as follows:
library(nlme)
data(Orthodont)
fm1 - lme(distance~age,random=~1|Subject,data=Orthodont)
fm0 - lm(distance~age,data=Orthodont)
anova(fm1,fm0)[[p-value]]
detach(package:nlme,unload=TRUE)
library(lme4.0) ## stable version of lme4
fm2 - lmer(distance~age+(1|Subject),data=Orthodont,REML=FALSE)
anova(fm2,fm0) ## fails
ddiff - -2*c(logLik(fm0)-logLik(fm2))
pchisq(ddiff,1,lower.tail=FALSE)
## not identical to above but close enough
Would someone be kind enough to point out my blunder?
You should probably repost this to the
r-sig-mixed-mod...@r-project.org mailing list.
My short answer would be: (1) I don't think you can actually use
anova() to compare likelihoods between lm() and lme()/lmer() fits in
the way that you want: *maybe* for lme() [don't recall], but almost
certainly not for lmer(). See http://glmm.wikidot.com/faq for
methods for testing significance/inclusion of random factors (short
answer: you should *generally* try to make the decision whether to
include random factors or not on _a priori_ grounds, not on the basis
of statistical tests ...)
Ben Bolker
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[R] Comparing dcast and reshape
I'm in the middle of my own little intellectual exercise comparing
dcast() and reshape() and have successfully stumped myself. I want to
melt() a data frame, then dcast() it into a new form. After doing so, I
want to duplicate the process using reshape().
So far, I can do the melt and cast
require(reshape2)
Raw - data.frame(site = c(1, 1, 1, 1, 2, 2, 2, 2),
id = c(1, 1, 2, 2, 1, 1, 2, 2),
instrument = rep(c(beck, phq), 4),
base.score = c(27, 13, 31, 11, 22, 10, 41, 17),
score.90d = c(20, 11, 27, 12, 24, 8, 34, 15))
Full.Melt - melt(Raw, id.vars=c(site, id, instrument),
measure.vars=c(base.score, score.90d))
FullCast - dcast(Full.Melt, site + id ~ instrument + variable,
value.var=value)
FullCast
site id beck_base.score beck_score.90d phq_base.score phq_score.90d
11 1 27 20 1311
21 2 31 27 1112
32 1 22 24 10 8
42 2 41 34 1715
I can also replicate the melt using reshape, but I can't reshape it into
the same wide format.
FullLong - reshape(Raw,
varying=list(score=c(base.score, score.90d)),
idvar=c(site, id, instrument),
direction=long)
Any pointers on how to get FullLong into the same wide format as
FullCast?
Thanks for taking the time to educate me.
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic| 9500 Euclid Ave. | Cleveland, OH 44195 |
(216) 445-1365
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Re: [R] Comparing dcast and reshape
Thanks for the prompt reply, Dr. Winsemius.
reshape(FullLong)
Will return the original data frame, but that wasn't quite what I wanted. I
wanted to take it to a wider form than the original.
But I just answered my own question (typically, _after_ asking the world at
large).
FullWide - reshape(Raw,
v.names=c(base.score, score.90d),
timevar=instrument,
idvar=c(site, id),
direction=wide)
gives me what I was looking for. I suspect there isn't a way to go directly
from FullLong to FullWide, but if someone can prove me wrong, I'd love to see
how.
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Wednesday, October 17, 2012 5:18 PM
To: Nutter, Benjamin
Cc: r-help@r-project.org
Subject: Re: [R] Comparing dcast and reshape
On Oct 17, 2012, at 2:09 PM, Nutter, Benjamin wrote:
I'm in the middle of my own little intellectual exercise comparing
dcast() and reshape() and have successfully stumped myself. I want to
melt() a data frame, then dcast() it into a new form. After doing so,
I want to duplicate the process using reshape().
So far, I can do the melt and cast
require(reshape2)
Raw - data.frame(site = c(1, 1, 1, 1, 2, 2, 2, 2),
id = c(1, 1, 2, 2, 1, 1, 2, 2),
instrument = rep(c(beck, phq), 4),
base.score = c(27, 13, 31, 11, 22, 10, 41, 17),
score.90d = c(20, 11, 27, 12, 24, 8, 34, 15))
Full.Melt - melt(Raw, id.vars=c(site, id, instrument),
measure.vars=c(base.score, score.90d))
FullCast - dcast(Full.Melt, site + id ~ instrument + variable,
value.var=value)
FullCast
site id beck_base.score beck_score.90d phq_base.score phq_score.90d
11 1 27 20 1311
21 2 31 27 1112
32 1 22 24 10 8
42 2 41 34 1715
I can also replicate the melt using reshape, but I can't reshape it
into the same wide format.
FullLong - reshape(Raw,
varying=list(score=c(base.score, score.90d)),
idvar=c(site, id, instrument),
direction=long)
Any pointers on how to get FullLong into the same wide format as
FullCast?
The reshape function will recognize that the object was created as a
wide-long reshaping and if you just use this code, you will get back
the original:
reshape(FullLong)
--
David Winsemius, MD
Alameda, CA, USA
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Re: [R] Opinion: Why I find factors convenient to use
Whether I use stringsAsFactors=FALSE or stringsAsFactors=TRUE tends to rely on
where my data are coming from. If the data are coming from our Oracle
databases (well controlled data), I import the with stringsAsFactors=TRUE and
everything is great. If the data are given to me by a fellow in the form of an
Excel spreadsheet, I have a good cry and then set stringsAsFactors=FALSE.
Regardless, before I get to analyzing the data, I convert them all to factors.
I imagine people's preferences for the default setting are strongly tied to the
quality of the data with which they tend to work.
I would prefer the default argument be left as it is, however. Mostly because
1) I feel like it assumes you are importing data for analysis and not for data
management; and more importantly
2) Changing the default would mean I have to change the way I approach data
import--and I don't like to change.
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Rui Barradas
Sent: Monday, August 20, 2012 8:03 AM
To: S Ellison
Cc: r-help
Subject: Re: [R] Opinion: Why I find factors convenient to use
Hello,
Em 20-08-2012 12:30, S Ellison escreveu:
-Original Message-
Over the years, many people -- including some who I would consider
real expeRts -- have criticized factors and advocated the use
(sometimes exclusively) of character vectors instead.
Exclusive use of character vectors is not going to do the job.
The concept of a factor is fundamental to a lot of statistics; a programming
environment that does not implement factors and their associated special
behaviour is probably not a statistical programming language.
Special behaviours I have in mind include:
- Level order can be arbitrarily specified for display purposes
- A control level can be intentionally chosen for contrasts
- the option of ordered factors (for example, for polr and the like)
So I think the language does and will require a 'factor' type in one form or
another.
_When_ you decide to convert a character input to a factor is, of course,
up to the user,and for cleanup it's very often better to stick with character
early and convert to factor a bit later. But personally, I think that there
is sufficient control over the coding of data to allow user discretion. and
on the whole, it seems to me that character input gets used as factor data so
much of the time when it is used at all that the default
stringsAsFactors=TRUE setting seems the more sensible default.
I disagree with this last point. Just think of the number of questions to this
list about, say, dates. When read from file using one of the forms of
read.table, they usually cause problems. Unless the user is an experienced one,
in which case he/she might not have a question to ask.
Besides, the default TRUE is contradictory with stick with character early and
convert to factor a bit later. With both early and later.
A different thing is to have a very used function's default behavior change
from one version of R to the next one. What about all the code in use? Maybe
it's better to leave it be.
Rui Barradas
S Ellison
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its
Re: [R] Variable labels
I have my own function for doing this that is similar to the one presented
below. Others may have other ideas that work better. As a general rule, I
would caution against writing out just the label without the variable name.
The only reason I see to separate the labels and names is if you are generating
a report (and using write.table wouldn't be my first choice for doing that).
However, the function below at least allows you to decide when you call it. If
you want to exclude the names, just set names=FALSE.
#*** possibly buggy. I didn't do much testing on this function.
write.table.with.labels - function(data, file, names=TRUE, labels=TRUE,
append=FALSE, ...){
if (names) write.table(t(names(data)), file, col.names=FALSE, ...)
if (labels) write.table(t(label(data)), file, col.names=FALSE,
append=any(c(names, append)), ...)
write.table(data, file, col.names=FALSE, append=TRUE, any(c(names, labels,
append)), ...)
}
label(mtcars$mpg) - Miles per gallon
write.table.with.labels(mtcars, test.csv, sep=,)
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Francois Maurice
Sent: Tuesday, July 17, 2012 4:36 PM
To: r-help@r-project.org
Subject: [R] Variable labels
Hi,
I'm using write.table() to export dataframe to Excel. All works fine except
that I want to export the variable labels instead of variable names.
I can see the labels in the R consol using attr(), but I just don't know how
to use the labels instead of the names.
Thanks,
François Maurice
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Reading a bunch of csv files into R
For example:
myDir - some file path
filenames - list.files(myDir)
filenames - filenames[grep([.]csv, filenames)]
data_names - gsub([.]csv, , filenames)
for(i in 1:length(filenames)) assign(data_names[i], read.csv(file.path(myDir,
filenames[i])))
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kevin Wright
Sent: Friday, May 25, 2012 2:55 PM
To: HJ YAN
Cc: r-help@r-project.org
Subject: Re: [R] Reading a bunch of csv files into R
See ?dir
Assign the value to a vector and loop over the elements of the vector.
Kevin
On Fri, May 25, 2012 at 12:16 PM, HJ YAN yhj...@googlemail.com wrote:
Dear R users
I am struggling from a data importing issue:
I have some hundreds of csv files needed to be read into R for futher
analysis. All those csv files are named in one of the three formats:
(1) strings: e.g. London_Oxford street
(2) Integer: e.g. 1234_5678
(3) combined: e.g. London_1234
I intend to use read.csv(_xxx.csv) but I only dealt with sigle
documents before and if there are only no more than 20 files, I do not
bother to search a more efficient way.
Is there any claver way that I do not have to type in all these
hundreds names by hand, maybe using a R package or write some code in
some other languages if it is not too difficult to learn.
Any thoughts/hints please??
Many thanks in advance!
HJ
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--
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Re: [R] R help!
So long as x is a character vector, I tend to use the following for this
problem.
x - c(12/31/11 23:45, 1/1/12 2:15)
x.split - strsplit(x, )
x.date - sapply(x.split, function(y) return(y[1]))
x.time - sapply(x.split, function(y) if (length(y) 1) return(y[2]) else NA)
x.date
[1] 12/31/11 1/1/12
x.time
[1] 23:45 2:15
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Alex Roth
Sent: Wednesday, May 02, 2012 4:01 PM
To: r-help@r-project.org
Subject: [R] R help!
Hello there, I was wondering if you could help me with a quick R issue.
I have a data set where one of the columns has both date and time in it, e.g.
12/31/11 23:45 in one cell. I want to use R to split this column into two new
columns: date and time.
One of the problems with splitting here is that when the dates go into single
digits there are no 0's in front of months January-September (e.g., January is
represented by 1 as opposed to 01), so every entry is a different length.
Therefore, splitting by the space is the only option, I think.
Here's the coding I've developed thus far:
z$dt - z$Date#time and date is all under z$Date
foo - strsplit( , z$dt) #attempted split based on the space
And then if that were to work, I would proceed use the coding:
foo2 - matrix(unlist(foo), ncol = 2, byrow=TRUE) z$Date - foo[ ,1] z$Time -
foo[ ,2]
However, foo - strsplit( , z$dt) isn't working. Do you know what the problem
is? If you could respond soon, that would be greatly appreciated!
Thanks so much!
Alex
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Re: [R] VarCorr procedure from lme4
I've run into this situation and have been able to prevent problems by using
lme4::VarCor(...)
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Stevens
Sent: Tuesday, May 01, 2012 9:58 PM
To: r-help@r-project.org
Subject: Re: [R] VarCorr procedure from lme4
That was it - detaching 'nlme' was the trick. Thanks Walmes and the rest.
David
On 5/1/2012 7:47 PM, David Stevens wrote:
Yes - I also have nlme. Bad juju?
David
On 5/1/2012 1:32 PM, Walmes Zeviani wrote:
It could be a bad coexistence between packages in the same R session.
Are you using nlme and/or doBy packages too?
Bests.
Walmes.
=
=
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S,
49.231759 W) Departamento de Estatística - Universidade Federal do
Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: wal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/~walmes linux user number: 531218
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--
David K Stevens, P.E., Ph.D., Professor
Civil and Environmental Engineering
Utah Water Research Laboratory
8200 Old Main Hill
Logan, UT 84322-8200
435 797 3229 - voice
435 797 1363 - fax
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Re: [R] multiple values in one column
This is a function I use for these kinds of situations. Assuming the delimiter
within the column is consistent and the spelling is consistent, it is pretty
useful.
The function returns a vector of 0/1 values, 1 if the text in level is found, 0
otherwise.
var=the variable
level=The value of interest in var
'split_levels' - function(var, level, sep=,){
#*** identify level in var.
f - function(v){
v - unlist(strsplit(v,sep))
ifelse(level %in% v, return(1), return(0))
}
#*** split the variable
new.var - unlist(sapply(var,f))
names(new.var) - NULL
#*** assign NA's where they were in original variable
new.var[is.na(var)] - NA
return(new.var)
}
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mark Grimes
Sent: Friday, April 06, 2012 11:16 AM
To: John D. Muccigrosso
Cc: r-help@r-project.org
Subject: Re: [R] multiple values in one column
John
I have to deal with this kind of thing too for my class.
# Some functions
# for ad$Full.name = Mark Grimes
get.first.name - function(cell){
x-unlist(strsplit(as.character(cell), ))
return(x[1])
}
get.last.name - function(cell){
x-unlist(strsplit(as.character(cell), ))
return(x[2])
}
# For roster$Name = Grimes, Mark L
get.first.namec - function(cell){
x-unlist(strsplit(as.character(cell), , ))
y - get.first.name(x[2])
return(y)
}
get.last.namec - function(cell){
x-unlist(strsplit(as.character(cell), , ))
return(x[1])
}
Use these functions with the apply family for processing class files.
Hope this helps,
Mark
On Apr 6, 2012, at 9:09 AM, John D. Muccigrosso wrote:
I have some data files in which some fields have multiple values. For example
first last sex major
John Smith M ANTH
Jane DoeF HIST,BIOL
What's the best R-like way to handle these data (Jane's major in my example),
so that I can do things like summarize the other fields by them (e.g., sex by
major)?
Right now I'm processing the files (in excel since they're spreadsheets) by
duplicating lines with two values in the major field, eliminating one value
per row. I suspect there's a nifty R way to do this.
Thanks in advance!
John Muccigrosso
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[R] Reading Text Files with RODBC
I'm thoroughly stumped. I've been playing with RODBC and wanted to see if I could retrieve data from text files using this package as well (for the most part, this is an intellectual exercise, but occasionally I do get data files large enough in CSV format RODBC could be helpful) . I set up a DNS called Text Files and then ran the following code in R library(RODBC) mtg - odbcConnect(Text Files) sqlTables(mtg) TABLE_CAT TABLE_SCHEMTABLE_NAME TABLE_TYPE REMARKS 1 C:\\USERS\\NUTTERB NACore2012.txt TABLE NA 2 C:\\USERS\\NUTTERB NA MTGCards.csv TABLE NA sqlFetch(mtg, MTGCards.csv) Error in odbcTableExists(channel, sqtable) : 'MTGCards.csv': table not found on channel MTGCards.csv is an export from an MS Access database, and I'm able to get it out of Access, and I'm also able to connect to our Oracle databases. So I'm not sure what it is I'm not getting about reading the text files. If anyone has done this successfully and has any pointers, I'd appreciate it. So far I've not been able to solve it with documentation from RODBC, RStudio (I get the same error messages when I use the RGui), or Microsoft ODBC drivers. Windows 7, 32 bit R 2.14.1 RODBC 1.3-4 RStudio Version 3 Benjamin Nutter | Biostatistician | Quantitative Health Sciences Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216) 445-1365 === Please consider the environment before printing this e-mail Cleveland Clinic is ranked one of the top hospitals in America by U.S.News World Report (2010). Visit us online at http://www.clevelandclinic.org for a complete listing of our services, staff and locations. Confidentiality Note: This message is intended for use only by the individual or entity to which it is addressed and may contain information that is privileged, confidential, and exempt from disclosure under applicable law. If the reader of this message is not the intended recipient or the employee or agent responsible for delivering the message to the intended recipient, you are hereby notified that any dissemination, distribution or copying of this communication is strictly prohibited. If you have received this communication in error, please contact the sender immediately and destroy the material in its entirety, whether electronic or hard copy. Thank you. __ 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] Reading Text Files with RODBC
Ah, yes. If you can't find the answer to your question, ask a different
question!
sqldf does, indeed, do what I want. Thank you
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Thursday, February 16, 2012 1:15 PM
To: Nutter, Benjamin
Cc: r-help@r-project.org
Subject: Re: [R] Reading Text Files with RODBC
On Thu, Feb 16, 2012 at 10:12 AM, Nutter, Benjamin nutt...@ccf.org wrote:
I'm thoroughly stumped. I've been playing with RODBC and wanted to see if I
could retrieve data from text files using this package as well (for the most
part, this is an intellectual exercise, but occasionally I do get data files
large enough in CSV format RODBC could be helpful) .
I set up a DNS called Text Files and then ran the following code in
R
library(RODBC)
mtg - odbcConnect(Text Files)
sqlTables(mtg)
TABLE_CAT TABLE_SCHEM TABLE_NAME
TABLE_TYPE REMARKS
1 C:\\USERS\\NUTTERB NA Core2012.txt
TABLE NA
2 C:\\USERS\\NUTTERB NA MTGCards.csv
TABLE NA
sqlFetch(mtg, MTGCards.csv)
Error in odbcTableExists(channel, sqtable) :
'MTGCards.csv': table not found on channel
MTGCards.csv is an export from an MS Access database, and I'm able to get it
out of Access, and I'm also able to connect to our Oracle databases. So I'm
not sure what it is I'm not getting about reading the text files. If anyone
has done this successfully and has any pointers, I'd appreciate it. So far
I've not been able to solve it with documentation from RODBC, RStudio (I get
the same error messages when I use the RGui), or Microsoft ODBC drivers.
This isn't precisely what you are asking for but if the idea is to apply an sql
statement to a csv file then read.csv.sql in the sqldf package can apply an sql
statement to a csv file reading the result into R. If you omit the sql
statement then it reads it all in.
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] R not recognizing words
I've developed a preference for
x$y %in% Low
when subsetting.
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of arabarkev
Sent: Friday, January 06, 2012 12:20 AM
To: r-help@r-project.org
Subject: [R] R not recognizing words
Hello all
I'm new to R and am experiencing a problem with a categorical variable. All
the data of this variable are Low, High, or NA. When I put summary(x$y),
it gives me the number of High, Low, and NA entries. However,
when I try to subset by writing x$y==Low or x$y==High, R does not
recognize the word and it writes FALSE for all the entries (but not the NA
entries).
Can anybody help me out?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/R-not-recognizing-words-tp4268283p4268283.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] nice report generator?
With any sort of reproducible report, you'll have to 'manually place' all of
the tables and figures at least once. If done well, however, you'll only have
to ever do it once.
I'm not an Sweave expert (yet, regrettably), but using lazyWeave, you could
generate a customized ANOVA table using the code below. If you're familiar
with Sweave, I'd recommend using it.
library(lazyWeave)
#*** Function to produce the latex code for an ANOVA table
anova.table - function(aov.object, caption=, ...){
out - data.frame(Source = rownames(anova(aov.object)),
SS = anova(aov.object)[, 2],
D.F. = anova(aov.object)[, 1],
stringsAsFactors=FALSE)
out$MS - out$SS / out$D.F.
out$F - NA
out$F[-nrow(out)] - out$MS[-nrow(out)] / out$MS[nrow(out)]
out$p - 1 - pf(out$F, out$D.F., out$D.F.[nrow(out)])
out - rbind(out, data.frame(Source = Total, SS=sum(out$SS),
D.F.=sum(out$D.F.), MS=NA, F=NA, p=NA))
out$SS - round(out$SS, 3)
out$MS - round(out$MS, 3)
out$F - round(out$F, 3)
rownames(out) - NULL
out[is.na(out)] -
paste(lazy.table(colnames(out), rborder=c(0, 0, 1), open=TRUE, close=FALSE,
caption=caption),
lazy.table(out, rborder=2, open=FALSE, close=TRUE), sep=\n)
}
#***
#*** Analysis of Variance
m1 - aov(mpg ~ cyl, data=mtcars)
#***
#*** Write the result to a file
lazy.write(
lazy.file.start(),
anova.table(m1, caption=An example ANOVA table from the dataset \\tt
mtcars\\rm.),
lazy.file.end(),
OutFile=TestFile.tex)
lazy.build(TestFile.tex)
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 | (216)
445-1365
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Sarah Goslee
Sent: Wednesday, December 07, 2011 2:34 PM
Cc: r-help; Janko Thyson
Subject: Re: [R] nice report generator?
On Wed, Dec 7, 2011 at 1:50 PM, Michael comtech@gmail.com wrote:
Thanks a lot!
Typically the console output will be something like this:
Analysis of Variance Table
Response: as.double(my.delta)
Df Sum Sq Mean Sq F value Pr(F) my.factor 1 0
0.005 2e-04 0.988 Residuals 1484 32630 21.988
I am looking for a report-generator to make this look nicer and tabularize
it...
It doesn't matter if the outputs are in HTML, PDF, etc.
As long as it looks fantastic, that's what we are looking for ... and
also the convenience is important... we dont' want to create Latex
document with manual insertaion of tables and plots... These need
automation... thx a lot!
That's what sweave is *for* - automation and reproducible research. It takes
care of the heavy lifting of generating and inserting tables and figures.
You will almost certainly need some tweaking to match your own definition of
looks fantastic though.
Sarah
On 07.12.2011 19:19, Sarah Goslee wrote:
Sweave.
Or ODFWeave, if Sweave/LaTeX are too much overhead.
But really, it depends on what kind of report output you need to
deliver. Printed? HTML? PDF?
Sarah
On Wed, Dec 7, 2011 at 1:14 PM, Michaelcomtech@gmail.com wrote:
Hi all,
I am looking for recommendations/pointers about best report
generator you think that are currently available?
i.e. the package that can help turn console output into
nice-looking neat report to send to bosses?
thanks a lot!
.
--
Sarah Goslee
http://www.functionaldiversity.org
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Please consider the environment before printing this e-mail
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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.
[R] Loading an S object into R
I hope you'll all forgive me for displaying my severe lack of knowledge on this
topic, and I can't really provide much in the way of reproducible code.
A colleague of mine has asked if I know how to import an S object into R. The
object is stored in a file named 'pre3.f' When I open the file as a text
document I get what's printed below. Being one of those young bucks with no
experience with S, and not having access to a license of S in which to play
with this, does anyone have any suggestions on how to get this (what I assume
is) data from the .f file into R?
Benjamin
S data
.Data class 8 €- ß coefficients var loglik score iter
linear.predictors residuals means terms n call Design assign na.action fail
non.slopes stats method maxtime time.inc units fitFunction center scale.pred x
y time surv std.err surv.summary ,
À À ƒ € K [ c g d` a· Á·
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T.STG=T2A T.STG=T2B T.STG=T2C T.STG=T3 G1= 3 G2= 3 BX.POS BX.NEG RP.YEAR G1=
3 * G2= 3 ÔwÎ
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Re: [R] Loading an S object into R
Shoot! I was really hoping this would be something not in an obvious
documentation. Feel free to stamp my forehead with One of Them
And you were right, it turned out to be a cph model.
Thanks, Josh.
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Tuesday, May 24, 2011 12:20 PM
To: Nutter, Benjamin
Cc: r-help@r-project.org
Subject: Re: [R] Loading an S object into R
Hi Benjamin,
I also have no experience with S plus, but the suggestions on the R
Import/Export manual seem worth trying:
http://cran.r-project.org/doc/manuals/R-data.html#EpiInfo-Minitab-SAS-S_002dPLUS-SPSS-Stata-Systat
I also question whether this is just data, words like loglik, iter, etc.
seem more like a model object of some kind (which I suppose is a type of data,
but...)
Good luck,
Josh
On Tue, May 24, 2011 at 8:24 AM, Nutter, Benjamin nutt...@ccf.org wrote:
I hope you'll all forgive me for displaying my severe lack of knowledge on
this topic, and I can't really provide much in the way of reproducible code.
A colleague of mine has asked if I know how to import an S object into R.
The object is stored in a file named 'pre3.f' When I open the file as a
text document I get what's printed below. Being one of those young bucks
with no experience with S, and not having access to a license of S in which
to play with this, does anyone have any suggestions on how to get this (what
I assume is) data from the .f file into R?
Benjamin
S data
.Data class 8 €- ß coefficients var loglik
score iter linear.predictors residuals means terms n call Design
assign na.action fail non.slopes stats method maxtime time.inc units
fitFunction center scale.pred x y time surv std.err surv.summary
,
À À ƒ € K [ c g d`
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Re: [R] help with \ in strings
Depending on what else you're writing around the %, you might consider
using the latexTranslate() function in Hmisc.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of viostorm
Sent: Sunday, April 24, 2011 8:48 AM
To: r-help@r-project.org
Subject: Re: [R] help with \ in strings
Josh,
Thank you so much!!! Works perfectly!
-Rob
Robert Schutt III, MD, MCS
Resident - Department of Internal Medicine University of Virginia,
Charlottesville, Virginia
--
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l
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Re: [R] Fibonacci
Fibonacci - c(1, 1)
while (max (Fibonacci) 500){
Fibonacci - c(Fibonacci, sum(tail(Fibonacci, 2))) }
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Georgina Imberger
Sent: Wednesday, April 20, 2011 5:43 AM
To: r-help@r-project.org
Subject: [R] Fibonacci
Hi!
I am trying to work out the code to get a Fibonacci sequence, using the
while() loop and only one variable. And I can't figure it out.
Fibonacci-c(1,1)
while (max(Fibonacci)500){
Fibonacci-c(Fibonacci, (max(Fibonacci) + ?(Fibanacci))) }
How can I tell R to take the value one before the max value? (Without
defining another variable)
(Probably super easy... I am a beginner...)
Thanks,
Georgie
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Re: [R] the first. from SAS in R
My apologies for coming to the party so late. I'm sure this question has been answered a couple of times. The attached function is one I pulled from the help archives, but I can't seem to duplicate the search that led me to it. In any case, I've attached the function I found, and an .Rd file I use as part of a local package. I've also attached a pair of accompanying records to retrieve the last record and the nth record. These have the advantage of not requiring data frames to be sorted prior to extraction--the function will sort them for you. Benjamin -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of David Katz Sent: Wednesday, November 24, 2010 10:17 AM To: r-help@r-project.org Subject: Re: [R] the first. from SAS in R Often the purpose of first/last in sas is to facilitate grouping of observations in a sequential algorithm. This purpose is better served in R by using vectorized methods like those in package plyr. Also, note that first/last has different meanings in the context of by x; versus by x notsorted;. R duplicated does not address the latter, which splits noncontiguous records with equal x. Regards, David -- View this message in context: http://r.789695.n4.nabble.com/the-first-from-SAS-in-R-tp3055417p3057476. 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. === P Please consider the environment before printing this e-mail Cleveland Clinic is ranked one of the top hospitals in America by U.S.News World Report (2009). Visit us online at http://www.clevelandclinic.org for a complete listing of our services, staff and locations. Confidentiality Note: This message is intended for use only by the individual or entity to which it is addressed and may contain information that is privileged, confidential, and exempt from disclosure under applicable law. If the reader of this message is not the intended recipient or the employee or agent responsible for delivering the message to the intended recipient, you are hereby notified that any dissemination, distribution or copying of this communication is strictly prohibited. If you have received this communication in error, please contact the sender immediately and destroy the material in its entirety, whether electronic or hard copy. Thank you. __ 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] Tinn-R looses connection to R (Windows Vista)
I've noticed it, but I haven't looked into it much since I rarely work
on Vista. I have found that opening R before I open Tinn-R tends to
work better than using Tinn-R to open the preferred GUI.
Benjamin
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Agi Richard
Sent: Wednesday, November 17, 2010 4:43 AM
To: R-help@r-project.org
Subject: [R] Tinn-R looses connection to R (Windows Vista)
Hi everyone,
I wonder, if anyone can help me with this annoying issue, although it is
related to Tinn-R and Windows Vista...
I am running R 2.11.1 and Tinn-R 2.3.5.2 together. I saved 2 R-hotkeys:
ALT+A for sending the whole content of one file and ALT+S for a
ALT+selection
from one file from Tinn-R to R. Everything works fine with other windows
versions, also with Vista at the beginning of every session, but after a
certain time (not always the same, neither the number of times, that I
used the hotkeys is the same, so it occurs randomly, but quite fast,
lets say on average within half an hour - 2hs) ALT+S does not work
anymore, but instead this hotkey opens the start menue of Vista!! (The
same function the windows key has!) I would be very happy, if anyone has
an idea why this could be!! Is it related to R, to Tinn-R or to Vista?
Thank you very very much in advance!
Carol
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Re: [R] Opening a .R file with R (Windows)
Here's a thread that has some good discussion on the topic
http://r.789695.n4.nabble.com/Running-script-with-double-click-td1579014.html
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bert Gunter
Sent: Tuesday, September 28, 2010 1:48 PM
To: Joshua Wiley
Cc: r-help@r-project.org
Subject: Re: [R] Opening a .R file with R (Windows)
Well, try following the correct conventions ...
1. Double click on an .Rdata file, which is produced by saving from R, and it
will open.
2. Drag and drop a .R or any text file into an open R window and it will source
the contents.
This is probably documented somewhere .. maybe in the RW FAQ.
-- Bert
On Tue, Sep 28, 2010 at 10:22 AM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi Kye,
I have never gotten .R files to work quite like other types (e.g.,
double-clicking a .PDF) in Windows. AFAIK there is no simple way to
do it, because you do not edit scripts directly in R (I am happy to be
corrected if someone knows better). For general use, I would just
open R first and then open the file, or if you just want to run the
file, you can use R's batch mode from the Windows command prompt.
Best regards,
Josh
On Tue, Sep 28, 2010 at 10:11 AM, Kye Gilder kye.gil...@gmail.com wrote:
I am new to using R. I installed R on my computer (Windows) and
everything things appears to be just fine. However, I have a simple
script RTest.R that does a few simple calculations. When I
double-click the RTest.R icon, I get an Information dialong box which
says, ARGUMENT 'C:\Documents and Settings\kgilder\Desktop\RTest.R'
__ignored__ . When I choose OK, R then opens, but it does not open or
display the script. Any help or suggestions?
When I open R and use File Open New Script and path to the file,
it opens just fine.
Regards,
Kye
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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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--
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Genentech Nonclinical Biostatistics
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===
P Please consider the environment before printing this e-mail
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Re: [R] Converting date format
If x is your vector of character date variables:
orig.date - as.Date(x, format=c(%m/%d/%Y))
new.date - format(x, format=c(%m/%d/%y))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Hosack, Michael
Sent: Tuesday, March 23, 2010 10:11 AM
To: R-help@r-project.org
Subject: [R] Converting date format
R community:
Hello, I would to like to convert a character date variable from
%m/%d/%Y to %m/%d/%y. Any advice would be greatly appreciated. I have
tried functions for changing the formatting and removing the unnecessary
digits without success.
Mike
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[R] Retrieving latitude and longitude via Google Maps API
Does anyone have any experience retrieving latitutde and longitude for
an address from the Google Maps API?
I'd like to have an R script that submits a street address, city, state,
and zip code and returns the coordinates. So far, I've been submitting
the coordinates from another program, then loading the coordinates in R
and merging them back into the data frame I want to use. It'd be nice
to be able to do it all in one script, but I'm not comprehending the API
thing very well.
I'm using R 2.9.1 on Windows XP. Any suggestions or pointers?
Benjamin
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 |
(216) 445-1365
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Re: [R] Help on getting help from manuals
As has been pointed out, there are tools in R to help find the commands
you are looking for.
As a practical note, I recommend starting with '?' if you think you know
what command you need.
If you're unsure of what command you need, my next step would be
help.search(). Often, the results from here will point you to a command
that you are looking for.
If a handful of tries with help.search() doesn't get you what you need,
then try RSiteSearch(). This will search the mailing list for entries
relevant to your search criteria. Often times, you'll find that someone
has already asked your question. In fact, I can only recall a couple of
times where a search of the mailing list has failed to provide a
solution for me.
Aside from reading lots of documentation, the one thing that has helped
me the most in picking up R has been reading Peter Dalgaard's
Introductory Statistics with R. It's a very small, basic introduction
to R, but laid a solid enough foundation that I soon found myself able
to produce and refine my own solutions.
In short, the best way to improve our knowledge of R is lots of reading
and a bit of trial and error.
Benjamin
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of ManInMoon
Sent: Friday, March 12, 2010 4:41 AM
To: r-help@r-project.org
Subject: [R] Help on getting help from manuals
Hi,
A number of people have suggested I read the manuals...
Could someone help me by telling me where the primary start point is
please?
For example, I am interested in writing functions with variable number
of arguments - where should I start to look?
An introduction to R only show a brief example - with no pointer to
where to find further data.
I can't do ?xxx from R console in most cases - as I don't know what the
function name is that I am looking for!!!
People have helped me find substitute to get some metadata out - BUT
how could I have found that without guidance from nice people in Nabble?
Any help on this very much appreciated.
--
View this message in context:
http://n4.nabble.com/Help-on-getting-help-from-manuals-tp1590323p1590323
.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Row-wisely converting a data frame into a list
as.data.frame(t(df))
For example
x - as.data.frame(t(mtcars))
typeof(x)
[1] list
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Sebastian Bauer
Sent: Tuesday, March 02, 2010 8:12 AM
To: r-help@r-project.org
Subject: [R] Row-wisely converting a data frame into a list
Hello,
is there an elegant way, how I can convert each row of a data frame into
distinct elements of a list?
In essence, what I'm looking for is something like
rows.to.lists - function( df ) {
ll - NULL
for( i in 1:nrow(df) )
ll - append( ll, list(df[i,]) )
return (ll)
}
but more done more efficiently (the data frame may contain ten-thousands
of rows). I thought about using apply() but this function always returns
a matrix.
Thanks in advance!
Bye,
Sebastian
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Re: [R] counting the number of ones in a vector
What you did works well. You could also try the following.
table(x)[1]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Randall Wrong
Sent: Friday, February 26, 2010 9:41 AM
To: r-help@r-project.org
Subject: [R] counting the number of ones in a vector
Dear R users,
I want to count the number of ones in a vector x.
That's what I did : length( x[x==1] )
Is that a good solution ?
Thank you very much,
Randall
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Re: [R] counting the number of ones in a vector
But if x has any missing values:
x - c(1, 1, 1, NA, NA, 2, 1, NA)
sum( x == 1)
[1] NA
sum(x==1, na.rm=TRUE)
[1] 4
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Henrique Dallazuanna
Sent: Friday, February 26, 2010 9:47 AM
To: Randall Wrong
Cc: r-help@r-project.org
Subject: Re: [R] counting the number of ones in a vector
Try:
sum(x == 1)
On Fri, Feb 26, 2010 at 11:40 AM, Randall Wrong randall.wr...@gmail.com wrote:
Dear R users,
I want to count the number of ones in a vector x.
That's what I did : length( x[x==1] )
Is that a good solution ?
Thank you very much,
Randall
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Re: [R] First. Last. Data row selection
I've attached some functions I've written based on previous questions that have been posted here. Unfortunately, I was too lazy to give credit to previous commenters in my Rd file, and for that I hope they'll forgive me. In any case, please be assured that the functions I've attached are in no way my original work. Benjamin -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of wookie1976 Sent: Tuesday, February 23, 2010 12:40 PM To: r-help@r-project.org Subject: [R] First. Last. Data row selection I am in the process of switching from SAS over to R. I am working on very large CSV datasets that contain vehicle information. As I am processing the data, I need to select the first (or sometimes the second) record (by date) for any records that have the same license plate number. In SAS, there is a function called 'first.' that can be used on sorted datasets to pull out those first entries for each occurrence of a particular variable (in this case the variable is 'license plate') found in the data. I have spent some time looking around and cannot seem to find an equivalent function in R. Can anyone recommend an efficient technique that would pull this off? I assume the database must first be sorted by vehicle plate and date, and then apply the filter or function. Any help would be greatly appreciated. Thanks, Joe -- View this message in context: http://n4.nabble.com/First-Last-Data-row-selection-tp1566260p1566260.htm l 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. === P Please consider the environment before printing this e-mail Cleveland Clinic is ranked one of the top hospitals in America by U.S.News World Report (2009). Visit us online at http://www.clevelandclinic.org for a complete listing of our services, staff and locations. Confidentiality Note: This message is intended for use only by the individual or entity to which it is addressed and may contain information that is privileged, confidential, and exempt from disclosure under applicable law. If the reader of this message is not the intended recipient or the employee or agent responsible for delivering the message to the intended recipient, you are hereby notified that any dissemination, distribution or copying of this communication is strictly prohibited. If you have received this communication in error, please contact the sender immediately and destroy the material in its entirety, whether electronic or hard copy. Thank you. __ 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] Circles around letters or numbers in plot title
Has anyone ever tried putting a circle around a letter or a number in a
plot title?
For instance, if I have a plot title Scatterplot for Subject 24, I
want to put a circle around 24 to distinguish that plot from the other
30 I've generated. Any tips or ideas beyond plotting a circle in the
margin?
Benjamin
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 |
(216) 445-1365
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Re: [R] all possible subsets, with AIC
I've dabbled in this a little bit, and the result of my dabbling is attached. I'll give you fair warning, however. The attached function can take a long time to run, and if your model has 10 or more predictors, you may be retired before it finishes running. In any case, it will models for all possible subsets of predictors in lm, glm, or coxph. If requested, it will also plot the R-squared, Adjusted R-squared, AIC, or BIC of those models (when the values are applicable to the model). It might give you a good starting point. Benjamin -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of kcleary2 Sent: Friday, February 12, 2010 3:19 PM To: r-help@r-project.org Subject: [R] all possible subsets, with AIC Hello, I have a question about doing ALL possible subsets regression with a general linear model. My goal is to produce cumulative Akaike weights for each of 7 predictor variables-to obtain this I need R to: 1. Show me ALL possible subsets, not just the best possible subsets 2. Give me an AIC value for each model (instead of a BIC value). I have tried to do this in library(RcmdrPlugin.HH), and using the leaps code below. With the leaps code my problem is that my response is not a vector, it's a single value (density of a species) ANy help would be greatly appreciated. Thanks a lot, Kate ALL-SUBSETS REGRESSIOM DESCRIPTION leaps() performs an exhaustive search for the best subsets of the variables in x for predicting y in linear regression, using an efficient branch-and-bound algorithm. It is a compatibility wrapper for regsubsets [1] does the same thing better. Since the algorithm returns a best model of each size, the results do not depend on a penalty model for model size: it doesn't make any difference whether you want to use AIC, BIC, CIC, DIC, ... USAGE leaps(x=, y=, wt=rep(1, NROW(x)), int=TRUE, method=c(Cp, adjr2, r2), nbest=10, names=NULL, df=NROW(x), strictly.compatible=TRUE) ARGUMENTS x A matrix of predictors y A response vector wt Optional weight vector int Add an intercept to the model method Calculate Cp, adjusted R-squared or R-squared nbest Number of subsets of each size to report names vector of names for columns of x df Total degrees of freedom to use instead of nrow(x) in calculating Cp and adjusted R-squared strictly.compatible Implement misfeatures of leaps() in S -- Kate Cleary MS Candidate Department of Fish, Wildlife, and Conservation Biology Colorado State University Fort Collins, CO 970-491-3535 Links: -- [1] https://webmail.warnercnr.colostate.edu/leaps/help/regsubsets [[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. === P Please consider the environment before printing this e-mail Cleveland Clinic is ranked one of the top hospitals in America by U.S.News World Report (2009). Visit us online at http://www.clevelandclinic.org for a complete listing of our services, staff and locations. Confidentiality Note: This message is intended for use only by the individual or entity to which it is addressed and may contain information that is privileged, confidential, and exempt from disclosure under applicable law. If the reader of this message is not the intended recipient or the employee or agent responsible for delivering the message to the intended recipient, you are hereby notified that any dissemination, distribution or copying of this communication is strictly prohibited. If you have received this communication in error, please contact the sender immediately and destroy the material in its entirety, whether electronic or hard copy. Thank you. __ 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] Drop last numeral
Data-c(1131, 1132, 1731 ,1732 ,1821 ,1822, 2221 ,,
2241 ,2242,414342 ,414371 ,414372)
Bldgid-substring(as.character(Data),1,nchar(Data)-1)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of LCOG1
Sent: Tuesday, January 12, 2010 1:37 PM
To: r-help@r-project.org
Subject: [R] Drop last numeral
Hello all,
Frustrated and i know you can help
I need to drop the last numeral of each of my values in my data set. So
for the following i have tried the ?substring but since i have to
specify the length, but because my data are of varying lengths it doenst
work so well
Data-c(1131, 1132, 1731 ,1732 ,1821 ,1822, 2221 ,,
2241 ,2242,414342 ,414371 ,414372)
Bldgid-substring(as.character(Data),1,3)
returns:
113 113 173 173 182 182 222 222 224 224 414 414
414
but i want
113, 113, 173 ,173 ,182 ,182, 222 ,222, 224
,224,41434
,41437 ,41437)
The values thats have more than 4 numerals are whats messing things up.
Tried ?formatC as well but couldn't get it to coerce things correctly.
Thanks for the help
JR
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Re: [R] %d/%m/%Y can not be displayed in a .rd file
Try \%d/\%m/\%Y
Escaping the % should do the trick. If I remember correctly, Latex uses
the % as the comment delimiter. I don't know if that's the cause of the
error, but escaping it has always solve the problem for me.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of rusers.sh
Sent: Thursday, January 07, 2010 2:09 PM
To: r-help@r-project.org
Subject: [R] %d/%m/%Y can not be displayed in a .rd file
Hi all,
I found the date format (e.g.%d/%m/%Y) in the .rd file cannot be
displayed after building the package. See below, ###.rd file
\examples{ a-10/20/1999
DateConversion(a,DateIn=%m/%d/%Y,DateOut=%d/%m/%Y)
}
The result is
Examples:
a-10-20-1999
DateConversion(a,DateIn=
??%m/%d/%Y seems cannot be recognized.
Is there some method to solve this and make it visible?
Thanks a lot.
--
-
Jane Chang
Queen's
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Re: [R] Newbie needs to count elements in a row
For a single row where mat is your matrix and r is the row
sum(!is.na(mat[r,]))
For every row in the matrix
rowSums(!is.na(mat))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Verena Weber
Sent: Tuesday, December 29, 2009 8:50 AM
To: r-help@r-project.org
Subject: [R] Newbie needs to count elements in a row
Hi,
I have a n*m matrix and would like to count the number of elements not
equal to NA in a ROW.
e.g.
x 1 2 3 NA 10
y 2 NA 8 9 NA
Which function can I use to obtain
4 for row x and
3 for row y?
Could you help me? I found some functions for columns but not for
rows...
Thank you very much!
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Re: [R] Remove double quotation marks
It seems from your example that you're assuming all of the vectors have
the same length. If this is the case, then a data.frame might be your
friend.
df - data.frame(
v1 = c(0, 1, 0),
v2 = c(1, 1, 0),
v3 = c(2, 1, 2),
v4 = c(2, 2, 1),
v5 = c(0, 1, 1) )
x - 5
get.var - c(v1, v2, paste(v, x, sep=))
df[, get.var]
Or, if you know your variables in the data.frame will be named
sequentially
df[, c(1, 2, x)]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Lisa
Sent: Tuesday, December 29, 2009 3:55 PM
To: r-help@r-project.org
Subject: Re: [R] Remove double quotation marks
Thank you for your help. But here I just want to combine some vectors by
column based on the numbers determined by other R script.
For example, I have five vectors:
v1 - c(0, 1, 0)
v2 - c(1, 1, 0)
v3 - c(2, 1, 2)
v4 - c(2, 2, 1)
v5 - c(0, 1, 1)
If I am going to combine the first two vectors, and one more other
vector determined by my other R script, say, vector 5. Then my R script
is
x - 5
cbind(v1, v2, paste(v, x, sep = ))
The output is
v1 v2
[1,] 0 1 v5
[2,] 1 1 v5
[3,] 0 0 v5
This is not what I want. I want to get this:
v1 v2 v5
[1,] 0 1 0
[2,] 1 1 1
[3,] 0 0 1
Can you give me further suggestions or comments? Thanks a lot.
Lisa
Barry Rowlingson wrote:
On Tue, Dec 29, 2009 at 6:31 PM, Lisa lisa...@gmail.com wrote:
Thank you for your reply. But in the following case, cat() or
print()
doesn't work.
data.frame(cbind(variable 1, variable 2, cat(paste(variable, x),
\n))), where x is a random number generated by other R script.
Lisa
Yes, because you are Doing It Wrong. If you have data that is indexed
by an integer, don't store it in variables called variable1, variable2
etc, because very soon you will be posting a message to R-help that is
covered in the R FAQ...
Store it in a list:
v = list()
v[[1]] = c(1,2,3,4,5)
v[[2]] = c(4,5,6,7,8,9,9,9)
then you can do v[[i]] for integer values of i.
If you really really must get values of variable by name, perhaps
because someone has given you a data file with variables called
variable1 to variable99, then use the paste() construction together
with the 'get' function:
[ not tested, but should work ]
v1=99
v2=102
i=2
get(paste(v,i,sep=))
[1] 102
Barry
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Re: [R] extracting the last row of each group in a data frame
I usually use the following function:
last.record - function(data, id, ..., na.last=TRUE, decreasing=FALSE){
#*** Make vector of variables to sort by
v - c(id, unlist(list(...)))
#*** Sort Data Frame
data - data[do.call(order,
c(data[,v, drop=FALSE], na.last=na.last,
decreasing=decreasing)),]
#*** Extract last record for each id
data[!duplicated(data[,id], fromLast=TRUE),]
}
DataData Frame from which the record is to be extracted
Id ID variable from which the record is to be extracted. The
data frame is automatically
sorted by this variable. May be either a character string
or an integer.
... Names of variables (or indices) in additon to id by which
data should be sorted.
na.last Argument passed to order(). Determines if missing values
are placed at the end
of the sorting.
Decreasing Argument passed to order(). Determines if data frame is
sorted in descending
order.
So, in your example
df - data.frame(Name = c(A, A, A, B, B, C, D),
Value = c(1, 2, 3, 4, 8, 2, 3))
last.record(df, Name, Value)
last.record(df, 1, 2)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Hao Cen
Sent: Monday, November 16, 2009 2:43 PM
To: r-help@r-project.org
Subject: [R] extracting the last row of each group in a data frame
Hi,
I would like to extract the last row of each group in a data frame.
The data frame is as follows
Name Value
A 1
A 2
A 3
B 4
B 8
C 2
D 3
I would like to get a data frame as
Name Value
A 3
B 8
C 2
D 3
Thank you for your suggestions in advance
Jeff
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Re: [R] row selection
sub3 - x[-seq(1, nrow(x), by=5), ]
Notice the '-' in front of the seq() command. This will select
everything but what is in the sequence.
From: Ashta [mailto:sewa...@gmail.com]
Sent: Friday, October 09, 2009 12:42 PM
To: Nutter, Benjamin
Cc: r-help@r-project.org
Subject: Re: [R] row selection
Hi all,
Thank you for your help. Now I am able to select every 5th row of the
data from the main data set (x)
using
sub1- x[seq(1, nrow(x), by=5), ]
So sub1 contains one fith of the data set X. I want also create
another data set that will contain the remaining data set from X (ie.,
four fifth of the data set).
Any help is highly appreciated.
I have a matrix named x with N by C
I want to select every 5 th rrow from matrix x I used the following
code
n- nrow(x)
for(i in 1: n){
+ b - a[i+5,]
b
}
sc x[seq(1, nrow(x), by=5), ]
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org]
On Behalf Of David Winsemius
Sent: Thursday, October 08, 2009 4:19 PM
To: Ashta
Cc: R help
Subject: Re: [R] row selection
On Oct 8, 2009, at 4:14 PM, Ashta wrote:
Hi all,
I have a matrix named x with N by C
I want to select every 5 th rrow from matrix x I used the
following
code
n- nrow(x)
for(i in 1: n){
+ b - a[i+5,]
b
}
Error: subscript out of bounds
What did you expect when i in your loop counter became one
greater
than the number of rows?
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Drawing lines in margins
Look at the xpd option in ?par. If you set par(xpd=TRUE) you should be
able to add a segment for what you want. But please let me know if
someone gives you a better way to do this.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Alan Cohen
Sent: Wednesday, July 29, 2009 10:22 AM
To: r-help@r-project.org
Subject: [R] Drawing lines in margins
Hi all,
Quick question: What function can I use to draw a line in the margin of
a plot? segments() and lines() both stop at the margin.
In case the answer depends on exactly what I'm trying to do, see below.
I'm using R v. 2.8.1 on Windows XP.
Cheers,
Alan
I'm trying to make a horizontal barplot with a column of numbers on the
right side. I'd like to put a line between the column header and the
numbers. The following reconstructs the idea - just copy and paste it
in:
aa - 1:10
plot.mtx2-cbind(aa,aa+1)
colnames(plot.mtx2)-c(Male,Female)
lci2- cbind(aa-1,aa)
uci2- cbind(aa+1,aa+2)
par(mar=c(5,6,4,5))
cols - c(grey79,grey41)
bplot2-barplot(t(plot.mtx2),beside=TRUE,xlab=Malaria death rates per
100,000,
names.arg=paste(state,aa,sep=),legend.text=F,las=1,xlim=c(0,13),
horiz=T, col=cols,
main=Malaria death rates by state and sex)
legend(8,6,legend=c(Female,Male),fill=cols[order(2:1)])
segments(y0=bplot2, y1=bplot2, x0=t(lci2), x1=t(uci2))
mtext(10*(aa+1),side=4,line=4,at=seq(3,3*length(aa),by=3)-0.35,padj=0.5,
adj=1,las=1,cex=0.85)
mtext(10*aa,side=4,line=4,at=seq(2,3*length(aa)-1,by=3)-0.65,padj=0.5,ad
j=1,las=1,cex=0.85)
mtext(Estimated,side=4,line=3,at=3*length(aa)+2.75,padj=0.5,adj=0.5,la
s=1,cex=0.85)
mtext(Deaths,side=4,line=3,at=3*length(aa)+1.25,padj=0.5,adj=0.5,las=1
,cex=0.85)
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Re: [R] numbers on barplot
The only thing you're missing is the midpoints of the bars. Since you
specified
Graph - barplot(dat$Average)
You can get the midpoints from the Graph object. So to put the number
on top of each bar you might use something like:
text(Graph, dat$Average, dat$Average)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mohsen Jafarikia
Sent: Monday, July 27, 2009 10:02 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] numbers on barplot
Hello all,
I have this simple barplot code:
ifn - id.dat
dat - read.table(ifn)
ofn - id.png
bitmap(ofn, type = png256, width = 30, height = 30, pointsize = 30, bg
=
white,res=50)
par(mar=c(5, 5, 3, 2),lwd=5)
par(cex.main=1.6,cex.lab=1.6,cex.axis=1.6)
names(dat)-c(NumberOfPeople,Average)
Graph-barplot(dat$Average)
dev.off()
and here is the data (id.dat):
150.08
60.09
70.37
I want to write down the NumberOfPeople on top of each of the bars.
Can
anybody help me on this?
Thanks,
Mohsen
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Re: [R] help with as.numeric
as.numeric() doesn't convert factors to the explicit value, nor should
it. Under what you're expecting, ff you have a factor where the levels
are Female and Male, using as.numeric() wouldn't produce anything
meaningful.
However, as.numeric() does something much smarter. It converts Female
to 1, and Male to 2. More generally, if you have n levels, it will
produce a vector of values between 1 and n. This is referred to as the
'internal coding.'
If you want to convert your height and bmi variables to their numeric
values, you need to do
as.numeric(as.character(height))
This will get you around the internal coding.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of deanj2k
Sent: Friday, May 15, 2009 7:58 AM
To: r-help@r-project.org
Subject: [R] help with as.numeric
hi everyone, wondering if you could help me with a novice problem. I
have a
data frame called subjects with a height and weight variable and want to
calculate a bmi variable from the two. i have tried:
attach(subjects)
bmi - (weight)/((height/100)^2)
but it comes up with the error:
Warning messages:
1: In Ops.factor(height, 100) : / not meaningful for factors
2: In Ops.factor((weight), ((height/100)^2)) :
/ not meaningful for factors
I presume that this means the vectors height and weight are not in
numeric
form (confirmed by is.numeric) so i changed the code to:
bmi - (as.numeric(weight))/((as.numeric(height)/100)^2)
but this just comes up with a result which doesnt make sense i.e.
numbers
such as 4 within bmi vector. Ive looked at
as.numeric(height)/as.numeric(weight) and these numbers just arnt the
same
as height/weight which is the reason for the incorrect bmi. Cant anyone
tell me where I am going wrong? Its quiet frustrating because I cant
understand why a function claiming to convert to numeric would come up
with
such a bizarre result.
--
View this message in context:
http://www.nabble.com/help-with-as.numeric-tp23558326p23558326.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Help with a cumullative Hazrd Ratio plot
?mtext
You may need to adjust the margins. For this I recommend adjusting that
mar option in par (see ?par).
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Bernardo Rangel Tura
Sent: Wednesday, May 13, 2009 6:31 AM
To: r-help
Subject: Re: [R] Help with a cumullative Hazrd Ratio plot
On Wed, 2009-05-13 at 07:19 -0300, Bernardo Rangel Tura wrote:
Hi R-masters
I need help to make modified cumulative hazard ratio plot.
I need create a common plot but with the number of subjects in risk
each
ticks times for two different groups in bottom of plot (I put one
example in attach).
Do you know a routine for this?
Is possible create a routine for this?
In this case with how commands?
Thanks in advance!
Sorry I put attach in jpeg format
In this mail a attach in PDF format
--
Bernardo Rangel Tura, M.D,MPH,Ph.D
National Institute of Cardiology
Brazil
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Re: [R] Selecting all rows of factors which have at least one positive value?
x -
data.frame(matrix(c(rep(11,4),rep(12,3),rep(13,3),rep(0,3),1,rep(0,4),re
p(1,2)),ncol=2))
id.keep - unique(subset(x,X20)$X1)
x2 - subset(x,X1 %in% id.keep)
x2
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Stephan Lindner
Sent: Thursday, April 02, 2009 11:26 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] Selecting all rows of factors which have at least one
positive value?
Dear all,
I'm trying to select from a dataframe all rows which correspond to a
factor (the id variable) for which there exists at least one positive
value of a certain variable. As an example:
x -
data.frame(matrix(c(rep(11,4),rep(12,3),rep(13,3),rep(0,3),1,rep(0,4),re
p(1,2)),ncol=2))
x
X1 X2
1 11 0
2 11 0
3 11 0
4 11 1
5 12 0
6 12 0
7 12 0
8 13 0
9 13 1
10 13 1
and I want to select all rows pertaining to factor levels of X1 for
which exists at least one 1 for X2. To be clear, I want rows 1:4
(since there exists at least one observation for X1==11 for which
X2==1) and rows 8:10 (likewise).
It is easy to obtain the corresponding factor levels (i.e.,
unique(x$X1[x$X2==1])), but I got stalled selecting the corresponding
rows. I tried grep, but then I have to loop and concatenate the
resulting vector. Any ideas?
Thanks a lot!
Stephan
--
---
Stephan Lindner
University of Michigan
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Re: [R] adding matrices with common column names
Shucks, Dimitris beat me to it. And his code is a bit more elegant than
mine. But since I did the work I may as well post it, right?
This version incorporates a couple of error checks to make sure all your
arguments are matrices with the same number of rows.
add.by.name - function(...){
args - list(...)
mat.test - sapply(args,is.matrix)
if(FALSE %in% mat.test) stop(All arguments must be matrices)
mat.row - unique(sapply(args,nrow))
if(length(mat.row)1) stop(All matrices must have the same number of
rows)
all.names - unique(as.vector(sapply(args,colnames)))
sum.mat - matrix(0,nrow=mat.row,ncol=length(all.names))
colnames(sum.mat) - all.names
for(i in 1:length(args)){
tmp - args[[i]]
sum.mat[,colnames(tmp)] - sum.mat[,colnames(tmp)] + tmp
}
return(sum.mat)
}
m1 - matrix(1:20,ncol=4); colnames(m1) - c(a,b,c,d)
m2 - matrix(1:20,ncol=4); colnames(m2) - c(b,c,d,e)
m3 - matrix(1:20,ncol=4); colnames(m3) - c(a,b,d,e)
add.by.name(m1,m2,m3)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of murali.me...@fortisinvestments.com
Sent: Friday, March 27, 2009 9:25 AM
To: r-help@r-project.org
Subject: [R] adding matrices with common column names
folks,
if i have three matrices, a, b, cc with some colnames in common, and i
want to create a matrix which consists of the common columns added up,
and the other columns tacked on, what's a good way to do it? i've got
the following roundabout code for two matrices, but if the number of
matrices increases, then i'm a bit stymied.
a - matrix(1:20,ncol=4); colnames(a) - c(a,b,c,d) b -
matrix(1:20,ncol=4); colnames(b) - c(b,c,d, e)
cbind(a[,!(colnames(a) %in% colnames(b)), drop = FALSE],
a[,intersect(colnames(a),colnames(b))] +
b[,intersect(colnames(a),colnames(b)), drop = FALSE],
b[,!(colnames(b) %in% colnames(a)), drop = FALSE])
a b c d e
[1,] 1 7 17 27 16
[2,] 2 9 19 29 17
[3,] 3 11 21 31 18
[4,] 4 13 23 33 19
[5,] 5 15 25 35 20
now, what if i had a matrix cc? i want to perform the above operation on
all three matrices a, b, cc.
cc - matrix(1:10,ncol=2); colnames(cc) - c(e,f)
i need to end up with:
a b c d e f
[1,] 1 7 17 27 17 6
[2,] 2 9 19 29 19 7
[3,] 3 11 21 31 21 8
[4,] 4 13 23 33 23 9
[5,] 5 15 25 35 25 10
and, in general, with multiple matrices with intersecting colnames?
thanks,
murali
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Re: [R] Plot inside For loop
I understood this to mean you want to open a new plotting window on each
iteration of the loop. If this is correct, I usually go about it by
using x11()
If you're looking to add additional lines or points, then you may want
to look at the aptly named functions lines() and points().
If neither of these are your goal, then I'm afraid I'll need more
clarification on what you're trying to do.
Benjamin
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mohan Singh
Sent: Wednesday, March 25, 2009 1:54 PM
To: r-help@r-project.org
Subject: [R] Plot inside For loop
Hi
I am plotting a set of data inside a for loop.
Is it possible to use plot in for loop without redrawing the whole
plot? Am using par(new=TRUE) but that draws on top of the previous plot.
Couldn't find any threads about the topic.
Thanks
Mohan
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Re: [R] Plot inside For loop
This might not be particularly elegant, but if you put your initial
plot() outside of the loop and then use the loop to place your points,
you might get what you want. If others have better solutions, I'd be
interested as well. Note that if you take this approach, you might want
to specify xaxt=n and yaxt=n and draw your own axes.
plot(NA,NA,xlim=c(-30,100), ylim=c(-30,90), ...);
for(i in 1:query) {
points(..);
#count increments
}
axis(...)
From: Mohan Singh [mailto:mohan.si...@ucd.ie]
Sent: Wednesday, March 25, 2009 3:00 PM
To: Nutter, Benjamin
Cc: r-help@r-project.org
Subject: Re: [R] Plot inside For loop
basically the for loop goes something like this
setCounters(135);
for(i in 1:query) {
plot(c2data[start1:count,],c3data[start1:count,],xlim=c(-30,100),
ylim=c(-30,90), sub=i);
points(..);
#count increments
}
so if i use windows() or x11(), i get different plot windows
if I use par(), i get one plot, but it redraws on top of previous plots,
so if my query goes from 1 to 5, it plots 5 plots on top of each other,
which in some ways, is ok, but doesn't go well for putting it in a
publication.
i want to draw a single plot with additional data from each for loop..
as the loop proceeds, it adds another set of data onto the previous plot
, i suppose, it doesn't redraw the axis and labels and just plot the
data
Mohan
On 25 Mar 2009, at 18:43, Nutter, Benjamin wrote:
I understood this to mean you want to open a new plotting window on each
iteration of the loop. If this is correct, I usually go about it by
using x11()
If you're looking to add additional lines or points, then you may want
to look at the aptly named functions lines() and points().
If neither of these are your goal, then I'm afraid I'll need more
clarification on what you're trying to do.
Benjamin
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mohan Singh
Sent: Wednesday, March 25, 2009 1:54 PM
To: r-help@r-project.org
Subject: [R] Plot inside For loop
Hi
I am plotting a set of data inside a for loop.
Is it possible to use plot in for loop without redrawing the whole
plot? Am using par(new=TRUE) but that draws on top of the previous plot.
Couldn't find any threads about the topic.
Thanks
Mohan
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Confidentiality Note: This message is intended for use
only by the individual or entity to which it is addressed
and may contain information that is privileged,
confidential, and exempt from disclosure under applicable
law. If the reader of this message is not the intended
recipient or the employee or agent responsible for
delivering the message to the intended recipient, you are
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you have received this communication in error, please
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+353-85-7279622
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you have received this communication in error, please
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Re: [R] help with loop
Why use a loop? Try using diff()
x - c(4, 19, 21, 45, 50, 73, 78, 83, 87, 94)
sum(diff(x))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Rafael Moral
Sent: Thursday, March 12, 2009 9:04 AM
To: r-help@r-project.org
Subject: [R] help with loop
Dear useRs,
I'm trying to write a loop to sum my data in the following way:
(the second - the first) + (the third - the second) + (the fourth - the third)
+ ...
for each column.
So, I wrote something like this:
c - list()
for(i in 1:ncol(mydata)) {
for(j in 2:nrow(mydata)) {
c[[i]] - sum(yc[j,i] - yc[(j-1),i])
}}}
As for the columns it works pretty fine, but it only returns the last
subtraction, however, I need the sum of all subtractions.
Any ideas?
Regards,
Rafael.
Veja quais são os assuntos do momento no Yahoo! +Buscados
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Re: [R] The Origins of R
snip
Those of us on this list (with the possible exception of one or two
nutters)
would take it that it goes without saying that R was developed on the
basis
of S --- we all ***know*** that.
snip
Just want to clarify that the nutters referred to here are not the same
as the Nutters that bear my name :-)
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Re: [R] parsing problem
-Original Message-
From: Nutter, Benjamin
Sent: Monday, February 02, 2009 7:37 AM
To: 'venkata kirankumar'
Subject: RE: [R] parsing problem
This is how I would approach it. I'd be happy to know if there's a
better way.
#*** Initial Data Frame
x -
c(Kontrolle,Placebo,125mg/kg,250mg/kg,500mg/kg,1000mg/kg))
x
#*** Convert alphabetical and punctuation characters to (see ?regex)
y - gsub([[:alpha:],[:punct:]],,x)
y
#*** Convert character strings to numeric
z - as.numeric(y)
z
#*** Calculate minimum
min(df$z,na.rm=TRUE)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of venkata kirankumar
Sent: Monday, February 02, 2009 7:18 AM
To: r-help@r-project.org
Subject: [R] parsing problem
Hi all,
I am trying to parse a vector for caliculating minimum in that vector
the
vector having values like
1Kontrolle
2 Placebo
3 125mg/kg
4 250mg/kg
5 500mg/kg
61000mg/kg
hear i tries for comverting it into numeric with using as.numaric()
function
but i got values like
5
6
2
3
4
1
it gives 1000mg/kg is the least one
but i have toget 125mg/kg as the minimum value
for that i have to remove all the strings and spetial charecters from
that
for that i used parse() but i am not able to get the out put
can anyone suggest how I will solve it
thanks in advance
regards;
kiran
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Re: [R] Text data
Jim's solution is more elegant than the following (and probably more
efficient) but you could also try the following (This let's you sort by
AN/HN, and then by the number at the start of the filename):
text - c( 26M_AN_C.bmp, 22M_AN_C.bmp, 20M_HA_O.bmp,
20M_AN_C.bmp, 26M_HA_O.bmp, 22M_HA_O.bmp,
31M_AN_C.bmp, 38M_HA_O.bmp)
split - do.call(rbind,strsplit(text,_))
o - order(split[,2],split[,1],split[,3])
text[o]
[1] 20M_AN_C.bmp 22M_AN_C.bmp 26M_AN_C.bmp 31M_AN_C.bmp
20M_HA_O.bmp
[6] 22M_HA_O.bmp 26M_HA_O.bmp 38M_HA_O.bmp
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Alice Lin
Sent: Wednesday, January 28, 2009 3:38 PM
To: r-help@r-project.org
Subject: [R] Text data
i have a data column of text entries:
26M_AN_C.bmp
22M_AN_C.bmp
20M_HA_O.bmp
20M_AN_C.bmp
26M_HA_O.bmp
22M_HA_O.bmp
31M_AN_C.bmp
38M_HA_O.bmp
.
.
.
.
And I would like to sort by the middle tag: AN, HA, etc.
Is there a way to parse text data in R?
In excel, I would have used the left and right function to cut out
just
the middle two letters out and put into another column to sort by.
Thanks!
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Re: [R] help
?floor
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of zahid khan
Sent: Wednesday, January 14, 2009 8:34 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] help
Dear ALL
suppose x=7.5,and i need of only integer part of variable x that is
7 only then what command i can use in R.
THANKS
Zahid Khan
Lecturer in Statistics
Department of Mathematics
Hazara University Mansehra.
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Re: [R] Colors in barplot
I believe the defaults in barplot are found using
grey.colors{GrDevices}.
?grey.colors
(For some reason my machine won't pull up the help files for grey.colors
from the command line, but I can still access it through the html help).
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Antje
Sent: Thursday, December 04, 2008 6:32 AM
To: [EMAIL PROTECTED]
Subject: [R] Colors in barplot
Hello,
Can anybody help me to find out which colors are used automatically when
calling barplot (e.g. 3 series beside each other will get different gray
values).
I want to apply a legend but I don't know the colors used...
Antje
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Re: [R] Changing the position of the origin
If I understand what you're seeking to do, you might also consider the
rplot() function written by Rolf Turner. It can be found at
http://tolstoy.newcastle.edu.au/R/help/02a/1174.html
Benjamin
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Plantky
Sent: Monday, November 17, 2008 11:46 PM
To: r-help@r-project.org
Subject: [R] Changing the position of the origin
Hi all,
Can anyone tell me how I can make 0,0 start at the top left hand
corner of a graph, instead of the typical lower left hand corner? I've
tried to plot with axes=F and then putting on the axes later, but I
want the points to correspond to the axes.
Thanks,
Kang Min
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Re: [R] Is there a way to vectorize this? [with correction]
** Sorry to repost. I forgot to include a function necessary to make
the example work **
I apologize up front for this being a little long. I hope it's
understandable. Please let me know if I need to clarify anything.
Several months ago I wrote a series of functions to help me take my R
analyses and build custom reports in html files. Each function either
builds or modifies a string of html code that can then be written to a
file to produce the desired output.
To make modifications in the html code, I've placed 'markers' around
certain characteristics that I might want to change. For instance, the
alignment characteristics have an 'algnmark' on either side of them.
When I wish to change the alignment, I can find where these markers are,
determine their location, and replace the contents between them.
I've been using the functions for a few months now, and am pleased with
the utility. Unfortunately, as I was writing these, I wasn't very
strong with my vectorization skills and relied on for loops (lots of for
loops) to get through the work. So while I'm pleased with the utility,
I've been trying to optimize the functions by vectorizing the for loops.
At this point, I've hit a small snag. I have a situation where I can't
seem to figure out how to vectorize the loop. Part of me wonders if it
is even possible.
The scenario is this: I run a string of code through the loop, on each
pass, the section of code in need of modification is identified and the
changes are made. When this is done, however, the length of the string
changes. The change in length needs to be recognized in the next pass
through the loop.
Okay, some code to illustrate what I mean. This first function formats
the html file. I only include it because it will be necessary to create
illustrate what the function is doing. I am eliminating all comments
and spacing from the code for brevity.
#*** Start of html.file.start
'html.file.start' - function(title, size=11, font=Times New Roman){
size - format(floor(size),nsmall=1)
code - paste(
html xmlns:o='urn:schemas-microsoft-com:office:office\'
xmlns:w=\'urn:schemas-microsoft-com:office:word\'
xmlns=\'http://www.w3.org/TR/REC-html40\'
head
meta http-equiv=Content-Type content=\'text/html;
charset=windows-1252\'
meta name=ProgId content=Word.Document
meta name=Generator content=\'Microsoft Word 11\'
meta name=Originator content=\'Microsoft Word 11\'
style
!--
/* Style Definitions */
p.MsoNormal, li.MsoNormal, div.MsoNormal
p.MsoEndnoteText, li.MsoEndnoteText, div.MsoEndnoteText
{margin-top:2.0pt;
margin-right:0in;
margin-bottom:0in;
margin-left:.15in;
margin-bottom:.0001pt;
text-indent:-.15in;
mso-pagination:none;
font-size:9.0pt;
mso-bidi-font-size:10.0pt;
font-family:'Times New Roman';
mso-fareast-font-family:'Times New Roman';}
p.Textbody, li.Textbody, div.Textbody--
/style
,
title,title,/title
/head
body lang=EN-US style=\'tab-interval:.5in;,
textmark; font-size:,size,pt; textmark;,
fontmark; font-family:,font,; fontmark;\', sep=)
return(code)
} # End of html.file.start
# Start of html.text
'html.text' - function(text, size=11, font=Times New Roman,
align=left, title){
size - format(floor(size),nsmall=1)
if(missing(title)) title - else title - paste(br/,title)
title - paste(b,title,/bbr/\n,sep=)
code - paste(
p class=MsoNormal ,
algnmark align=,align, algnmark
span class=GramE style=\',
textmark; font-size:,size,pt; textmark;,
fontmark; font-family:,font,; fontmark;,
stylemark; font-weight:normal; font-style:normal;,
text-decoration:none; stylemark;\',
title,text,
/span
/p,sep=)
return(code)
} #** End of html.text
So here is the function I'm trying to vectorize.
#*** Start of html.align
html.align - function(code,new.align=left){
#* Create a string to replace the current alignment setting.
align - paste( align=,new.align, ,sep=)
#* Function to pass to sapply. This is handy when 'code'
#* is a vector.
f1 - function(code,align=align){
mark - unlist(gregexpr(algnmark,code)) #* Get positions of
markers
if(mark[1]0){
odd - seq(1,length(mark),by=2) #* odd elements are starting
marker
evn - seq(2,length(mark),by=2) #* even elements are ending marker
mark[odd] - mark[odd]+9 #* These two lines determine the
starting
mark[evn] - mark[evn]-1 #* and ending elements of the substring
to
#* be replaced
for(i in 1:length(odd)){
l.old - nchar(code) #* store the length of the code segment.
old.align - substr(code,mark[odd[i]],mark[evn[i]])
Re: [R] Standard deviation for rows
?apply
e.g. apply(matrix,1,sd)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Alex99
Sent: Wednesday, October 15, 2008 1:17 PM
To: r-help@r-project.org
Subject: [R] Standard deviation for rows
Hi everyone,
I have just started using R, and I have a simple question.
How can I get the Standard deviation for rows. basically I am looking
for
something like rowMeans()
but for Standard deviation (I tried rowSds() didn't exist)
Thanks,
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ml
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P Please consider the environment before printing this e-mail
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Re: [R] Reading Data
Perhaps it would be easier to try something like this:
df1 - read.table(filename, ... ) # All columns read as characters
df1 - t(df1) # Transpose df1
write.table - (df1, newfile, ...) # Write the transposed data
# to a new file
df2 - read.table(newfile, header=TRUE, ...) # Read in transposed
data
I doubt it's the most computation-time-efficient, but conceptually it's
pretty quick.
Benjamin
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of [EMAIL PROTECTED]
Sent: Tuesday, October 07, 2008 8:28 AM
To: [EMAIL PROTECTED]; r-help@r-project.org
Subject: [R] FW: Reading Data
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
hi let me explain you the problem
we have a database which is in this format
Stocks 30-Jan-08 28-Feb-08 31-Mar-08 30-Apr-08
a1.003.007.003.00
b2.004.004.007.00
c3.008.00655.00 3.00
d4.0023.00 4.005.00
e5.0078.00 6.005.00
and we have a query which is in this format
Identifier weight Start_Date End_Date
a6.7631-Jan-06 31-Jan-07
e2.8628-Feb-06 28-Feb-07
f22.94 31-Mar-06 30-Mar-07
y30.05 28-Apr-06 30-Apr-07
h20.55 31-May-06 31-May-07
d6.76
f2.86
r22.94
okay now my task is to calculate returns for all the indentifiers for
a respective start and end date from table 1.
now length of start date and end date column is same and that of weight
and identifier is same.
i hope everything is clear now.
let me also send you the code that i have written but in my code i have
problem with the date format and also with the stocks name
data=read.table(H:/Rahul/london/david/rexcel/price.txt)
query=read.table(H:/Rahul/london/david/rexcel/prac.txt,header=TRUE)
data=as.matrix(data)
instrument=data[,1]
date=data[1,]
query=as.matrix(query)
q_ins=query[,1]
wt=query[,2]
q_sd=query[,3]
q_ed=query[,4]
returns=function(I,SD,ED){
p=rep(0,2)
for(i in 2:length(instrument))
{
if(instrument[i]==I)
{
for(j in 2:length(date))
{
if(date[j]==SD)
p[1]=data[i,j]
}
for(j in 2:length(date))
{
if(date[j]==ED)
p[2]=data[i,j]
}
}
returns=(p[2]/p[1])-1
}
#print(p)
#print(I)
return(returns)
}
##The original Funda##
matrix_ret=matrix(0,length(q_ins),length(q_sd))
for(i in 1:length(q_sd))
{
for(j in 1:length(q_ins))
{
matrix_ret[j,i]=returns(q_ins[j],q_sd[i],q_ed[i])
}
}
#Removing NA from the matrix
matrix_ret1=sapply(X=matrix_ret, FUN=function(x)
ifelse(is.na(x),0.00,x))
matrix_ret=matrix(matrix_ret1,length(q_ins),length(q_sd))
wt_ret=matrix(0,length(q_sd),1)
for(i in 1:length(q_sd))
{
for(j in 1:length(q_ins))
{
wt_ret[i]=wt_ret[i]+(wt[j]*matrix_ret[j,i])
}
}
result=cbind(q_ed,wt_ret)
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
-Original Message-
From: Gustaf Rydevik [mailto:[EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] ]
Sent: Tuesday, October 07, 2008 2:19 PM
To: Agarwal, Rahul-A
Cc: [EMAIL PROTECTED]; r-help@r-project.org
Subject: Re: [R] Reading Data
On Tue, Oct 7, 2008 at 10:36 AM, [EMAIL PROTECTED] wrote:
Hi,
I have a data in which the first row is in date format and the first
column is in text format and rest all the entries are numeric.
Whenever I am trying to read the data using read.table, the whole of
my data is converted in to the text format.
Please suggest what shall I do because using the numeric data which
are prices I need to calculate the return but if these prices are not
numeric then calculating return will be a problem
regards
Rahul Agarwal
Analyst
Equities Quantitative Research
UBS_ISC, Hyderabad
On Net: 19 533 6363
Hi,
A single column in a data frame can't contain mixed formats.
In the absence of example data, would guess one of the following could
work :
1)
read.table(data.txt,skip=1, header=T) ## If you have headers
2)
read.table(data.txt, header=T) ## If the date row is supposed to be
variable names.
3)
read.table(data.txt,skip=1) ## If there are no headers, and you want
to ignore the date
regards,
Gustaf
--
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tel: +46(0)703 051 451
address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik
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Re: [R] lsmeans
I hope you'll forgive me for resurrecting this thread. My question
refers to John Fox's comments in the discussion of lsmeans from
https://stat.ethz.ch/pipermail/r-help/2008-June/164106.html
John you said, It wouldn't be hard, however, to do the computations
yourself, using the coefficient vector for the fixed effects and a
suitably constructed model-matrix to compute the effects; you could also
get standard errors by using the covariance matrix for the fixed
effects.
I've been able to make use of all of that except for the 'suitably
constructed model-matrix' part. I've looked through some other threads
on this topic, but am still a little in the dark as to what I'd need to
do to construct a suitable matrix.
I would like to use the least squares means to develop parameter
estimates for a parametric ROC analysis, as described by Mithat Gonen's
book (Analyzing Receiver Operating Characteristic Curves with SAS,
2007).
Any suggestions on references that would explain how to go about
constructing the suitable model matrix?
Many Thanks
Benjamin
P Please consider the environment before printing this e-mail
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Re: [R] Coefficients, OR and 95% CL
You might also consider
?confint
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Jorge Ivan Velez
Sent: Monday, September 22, 2008 5:36 PM
To: Luciano La Sala
Cc: R mailing list
Subject: Re: [R] Coefficients, OR and 95% CL
Dear Luciano,
See ?logistic.display in the epicalc package. If glm1 is your model,
something like
logistic.display(glm1)
should do the job.
HTH,
Jorge
On Mon, Sep 22, 2008 at 5:28 PM, Luciano La Sala
[EMAIL PROTECTED]wrote:
Dear R-users,
After running a logistic regression, I need to calculate OR by
exponentiating the coefficient, and then I need the 95% CL for the OR
as
well. For the following example (taken from P. Dalaagard's book), what
would
be the most straightforward method of getting what I need? Could
anyone
enlight me please?
Thank you!
Lucho
summary(glm(menarche~age,binomial))
Call:
glm(formula = menarche ~ age, family = binomial)
Deviance Residuals:
Min1QMedian3Q Max
-4.68654 -0.13049 -0.01067 0.09608 2.35254
Coefficients:
Estimate Std. Error z value Pr(|z|)
(Intercept) -17.9175 1.7074 -10.49 2e-16 ***
age 1.3549 0.1296 10.45 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 974.31 on 703 degrees of freedom
Residual deviance: 223.95 on 702 degrees of freedom
(635 observations deleted due to missingness)
AIC: 227.95
Number of Fisher Scoring iterations: 9
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Re: [R] HI
Perhaps a simpler way might be to use the na argument in read.table. for
instance:
read.table( filename, na=0, ...)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Amit Patel
Sent: Tuesday, September 16, 2008 9:32 AM
To: r-help@r-project.org
Subject: [R] HI
Does anyone know an easy way to convert all the zero values in a
imported csv table into NA's
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Re: [R] Color of box frame in Legend (Was: Matrix barplot)
Try sourcing in the 'new.legend' function below. It's the legend
function with a new argument called 'box.col'. The argument will change
the color of the box surrounding the legend. If I understand what it is
you are looking for, this should work. Also, I didn't see a way to
change the axis bar in your code, so I suppressed the axis in the call
to barplot, and manually replaced the axis using the axis function. I
hope this works for you.
Benjamin
data - data.frame(Year=c(2000,2001,2002),
A=c(2,2,1),
B=c(3,1,2),
C=c(0,3,5))
data.mat - as.matrix(data)[,2:4]
rownames(data.mat) - data[['Year']]
data.mat - t(data.mat)
textcolor=yellow
par(col.axis=textcolor,col.main=textcolor)
barplot(data.mat,beside=TRUE,col=rainbow(3),main=Test,yaxt=n)
axis(2,at=0:5,col=textcolor)
new.legend(x=topleft, colnames(data[,2:4]),fill=rainbow(3),
inset=0.05,
text.col=textcolor,box.col=textcolor
)
###
###
new.legend - function (x, y = NULL, legend, fill = NULL, col =
par(col),
lty, lwd, pch, angle = 45, density = NULL, bty = o, bg =
par(bg),
box.lwd = par(lwd), box.lty = par(lty), box.col=black,
pt.bg = NA, cex = 1,
pt.cex = cex, pt.lwd = lwd, xjust = 0, yjust = 1, x.intersp = 1,
y.intersp = 1, adj = c(0, 0.5), text.width = NULL, text.col =
par(col),
merge = do.lines has.pch, trace = FALSE, plot = TRUE,
ncol = 1, horiz = FALSE, title = NULL, inset = 0)
{
if (missing(legend) !missing(y) (is.character(y) ||
is.expression(y))) {
legend - y
y - NULL
}
mfill - !missing(fill) || !missing(density)
title - as.graphicsAnnot(title)
if (length(title) 1)
stop(invalid title)
legend - as.graphicsAnnot(legend)
n.leg - if (is.call(legend))
1
else length(legend)
if (n.leg == 0)
stop('legend' is of length 0)
auto - if (is.character(x))
match.arg(x, c(bottomright, bottom, bottomleft,
left, topleft, top, topright, right, center))
else NA
if (is.na(auto)) {
xy - xy.coords(x, y)
x - xy$x
y - xy$y
nx - length(x)
if (nx 1 || nx 2)
stop(invalid coordinate lengths)
}
else nx - 0
xlog - par(xlog)
ylog - par(ylog)
rect2 - function(left, top, dx, dy, density = NULL, angle,
...) {
r - left + dx
if (xlog) {
left - 10^left
r - 10^r
}
b - top - dy
if (ylog) {
top - 10^top
b - 10^b
}
rect(left, top, r, b, angle = angle, density = density,
...)
}
segments2 - function(x1, y1, dx, dy, ...) {
x2 - x1 + dx
if (xlog) {
x1 - 10^x1
x2 - 10^x2
}
y2 - y1 + dy
if (ylog) {
y1 - 10^y1
y2 - 10^y2
}
segments(x1, y1, x2, y2, ...)
}
points2 - function(x, y, ...) {
if (xlog)
x - 10^x
if (ylog)
y - 10^y
points(x, y, ...)
}
text2 - function(x, y, ...) {
if (xlog)
x - 10^x
if (ylog)
y - 10^y
text(x, y, ...)
}
if (trace)
catn - function(...) do.call(cat, c(lapply(list(...),
formatC), list(\n)))
cin - par(cin)
Cex - cex * par(cex)
if (is.null(text.width))
text.width - max(abs(strwidth(legend, units = user,
cex = cex)))
else if (!is.numeric(text.width) || text.width 0)
stop('text.width' must be numeric, = 0)
xc - Cex * xinch(cin[1], warn.log = FALSE)
yc - Cex * yinch(cin[2], warn.log = FALSE)
if (xc 0)
text.width - -text.width
xchar - xc
xextra - 0
yextra - yc * (y.intersp - 1)
ymax - yc * max(1, strheight(legend, units = user, cex = cex)/yc)
ychar - yextra + ymax
if (trace)
catn( xchar=, xchar, ; (yextra,ychar)=, c(yextra,
ychar))
if (mfill) {
xbox - xc * 0.8
ybox - yc * 0.5
dx.fill - xbox
}
do.lines - (!missing(lty) (is.character(lty) || any(lty
0))) || !missing(lwd)
n.legpercol - if (horiz) {
if (ncol != 1)
warning(horizontal specification overrides: Number of
columns := ,
n.leg)
ncol - n.leg
1
}
else ceiling(n.leg/ncol)
if (has.pch - !missing(pch) length(pch) 0) {
if (is.character(pch) !is.na(pch[1]) nchar(pch[1],
type = c) 1) {
if (length(pch) 1)
warning(not using pch[2..] since pch[1] has multiple
chars)
np - nchar(pch[1], type = c)
pch - substr(rep.int(pch[1], np), 1:np, 1:np)
}
if (!merge)
dx.pch -
Re: [R] Simple... but...
x - c(1,3,5)
y - c(2,4,6)
xy - sort(c(x,y))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Shubha Vishwanath Karanth
Sent: Wednesday, July 23, 2008 8:55 AM
To: [EMAIL PROTECTED]
Subject: [R] Simple... but...
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do it? I mean the best way
Thanks, Shubha
This e-mail may contain confidential and/or privileged
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and provide commented, minimal, self-contained, reproducible code.
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Cleveland Clinic is ranked one of the top hospitals
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Visit us online at http://www.clevelandclinic.org for
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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] Bland-Altman method to measure agreement with repeated measures
The function given below is one I've written to handle repeated
measures. I've also included the Help File. If you happen to see any
potential improvements, I would be open to suggestions.
###
### Function Code
###
'Bland.Altman' - function(x,y,alpha=.05,rep.meas=FALSE,subject,...){
#**
#* Construct a Bland Altman Plot
#* 1. Set a few constants
#* 2. Calculate mean difference
#* 3. Calculate difference standard deviation
#* 4. Calculate upper and lower confidence limits
#* 5. Make Plot
#**
#*** 1. Set a few constants
z - qnorm(1-alpha/2) ## value of z corresponding to alpha
d - x-y ## pair-wise differences
m - (x+y)/2 ## pair-wise means
#*** 2. Calculate mean difference
d.mn - mean(d,na.rm=TRUE)
#*** 3. Calculate difference standard deviation
if(rep.meas==FALSE){ d.sd=sqrt(var(d,na.rm=TRUE)) }
else{
#*** 3a. Ensure subject is a factor variable
if(!is.factor(subject)) subject - as.factor(subject)
#*** 3b. Extract model information
n - length(levels(subject)) # Number of subjects
model - aov(d~subject) # One way analysis of variance
MSB - anova(model)[[3]][1] # Degrees of Freedom
MSW - anova(model)[[3]][2] # Sums of Squares
#*** 3c. Calculate number of complete pairs for each subject
pairs - NULL
for(i in 1:length(levels(as.factor(subject{
pairs[i] - sum(is.na(d[subject==levels(subject)[i]])==FALSE)
}
Sig.dl - (MSB-MSW)/((sum(pairs)^2-sum(pairs^2))/((n-1)*sum(pairs)))
d.sd - sqrt(Sig.dl+MSW)
}
#*** 4. Calculate lower and upper confidence limits
ucl - d.mn+z*d.sd
lcl - d.mn-z*d.sd
#*** 5. Make Plot
plot(m, d,abline(h=c(d.mn,ucl,lcl)), ...)
values - round(cbind(lcl,d.mn,ucl),4)
colnames(values) - c(LCL,Mean,UCL)
if(rep.meas==FALSE) Output - list(limits=values,Var=d.sd^2)
else Output - list(limits=values,Var=Sig.dl)
return(Output)
}
###
### Help File
###
Bland Altman Plots
Description:
Constructs a Bland-Altman Plot.
Usage:
Bland.Altman(x,y,alpha=.05,rep.meas=FALSE,subject,...)
Arguments:
x,y: vectors of values to be compared.
alpha: Significance level for determining confidence limits.
Defaults to 0.05
rep.meas: Toggles if data provided should be considered as repeated
measures. Defaults to 'FALSE'
subject: Required if 'rep.meas=TRUE'. A vector of the same length of
'x' and 'y' that denotes which subject/group the measurement
belongs to.
...: Other arguments to be passed to the 'plot' method.
Details:
When 'rep.meas=TRUE', the confidence limits are calculated using a
method proposed by Bland and Altman. These limits are slightly
wider, allowing for the correlation within subjects/factors. The
standard deviation used to compute these limits is:
sigma^2[d] = sigma^2[dI] + sigma^2[dw]
where sigma^2[d] is the variance of the differences, sigma^2[dI]
is the variance of the subjects and methods interaction, and
sigma^2[dw] is the within subject variation. Estimates of these
values can be found with
s^2[dw] = MSw
s^2[dI] = (MSb - MSw) / ((sum(m[i])^2 - sum(m[i]^2)) /
((n-1)*sum(m[i]))
)
Where MSb and MSw are the between and within subject variance of
the one way analysis of variance and m[i] is the number of pairs
for the ith subject. The sum of these two estimates provides the
estimate for s^2[d] .
Value:
limits: A vector containing the Mean Bias and confidence limits.
Var.dl: The Variance of the Bias. If 'rep.meas=TRUE', this is
s^2[dI]
.
Author(s):
Benjamin Nutter [EMAIL PROTECTED]
Created: December 2007
References:
J Martin Bland and Douglass G Altman, Measuring Agreement in
Method Comparison Studies, _Statistical Methods in Medical
Research_, 1999; 8: 135 - 160.
J Martin Bland and Doublas G. Altman, Agreement Between Methods
of Measurement with Multiple Observations per Individual _Journal
of Biopharmaceutical Statistics_ 17:571-582, 2007. (Corrects the
formula given in the 1999 paper).
Burdick RK, Graybill FA. _Confidence Intervals on Variance
Components_. New York: Dekker, 1992.
Examples:
observer1=rnorm(500,5,2)
observer2=rnorm(500,10,4)
ID=rep(1:50,10)
Bland.Altman(observer1,observer2)
Bland.Altman(observer1,observer2,rep.meas=TRUE,subject=ID)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of kende jan
Sent: Saturday, July 05, 2008 4:18 AM
To: R-help@r-project.org
Subject: [Possible SPAM] [R] Bland-Altman method to measure agreement
with repeated measures
Dear all,
I want to
Re: [R] [Possible SPAM] Reading selected lines in an .html file
I've tried to tackle a similar question at the request of a coworker. Unfortunately, it is difficult to read in HTML code because it lacks character that can consistently be used as a delimiter. The only guideline I can offer is that any text you're interested in is going to be between a and a . So the goal is to eliminate anything between and . What's more, if you really want to read in HTML code, you'll need a good grasp on HTML itself, and some familiarity with how the code you're reading in is structured. For instance, I'm attaching code that I wrote to read in HTML tables that were generated by other functions commonly used in my work place. But my code assumes that the tables are written by row (using the tr tag. Essentially, after studying the code I was going to read in, I hand picked the markers that I could use to isolate the text I wanted. I then proceeded to play a game of Simon Says to break down the code to smaller and smaller pieces until I got what I wanted. Unless you're going to be doing this a lot, I wouldn't recommend taking the time to try and write a function like this. In most cases it's probably faster just to copy the data by hand. But if you are determined to make it work, I hope the ideas help. Benjamin -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of vittorio Sent: Wednesday, June 04, 2008 3:50 PM To: [EMAIL PROTECTED] Subject: [Possible SPAM] [R] Reading selected lines in an .html file Dear friend, In an R program running permanently on a server I would like to read hour by hour the temperature in *C and the humidity from a site like this (actually, from many of such sites): http://www.wunderground.com/global/stations/16239.html How can I read the content of the site and select the info I need? Ciao Vittorio __ 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. === P Please consider the environment before printing this e-mail Cleveland Clinic is ranked one of the top hospitals in America by U.S. News World Report (2007). Visit us online at http://www.clevelandclinic.org for a complete listing of our services, staff and locations. Confidentiality Note: This message is intended for use only by the individual or entity to which it is addressed and may contain information that is privileged, confidential, and exempt from disclosure under applicable law. If the reader of this message is not the intended recipient or the employee or agent responsible for delivering the message to the intended recipient, you are hereby notified that any dissemination, distribution or copying of this communication is strictly prohibited. If you have received this communication in error, please contact the sender immediately and destroy the material in its entirety, whether electronic or hard copy. Thank you. __ 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] How do I import packages with the package I've built?
So if I understand you correctly, if I include in my namespace file the
commandimportFrom(MASS, stepAIC), whenever stepAIC is called from
my package, it will still run even if I haven't explicitly imported the
MASS package?
Or if I use the command import(survival) then the functions in
the survival package will work within my functions even without
explicitly loading the package?
Thanks for the help.
Benjamin
-Original Message-
From: Uwe Ligges [mailto:[EMAIL PROTECTED]
Sent: Saturday, November 17, 2007 12:04 PM
To: Nutter, Benjamin
Cc: [EMAIL PROTECTED]
Subject: Re: [R] How do I import packages with the package I've built?
There is a difference between importing a namespace and loading a
package:
If you import functions from a namespace into your package, these
functions are known within that namespace, i.e. functions from your
package's namespace know the other functions and can use them. The user
still won't see those imported functions while working on the top level.
If you want the latter, you have to load the other package explicitly,
e.g. in your .onLoad directives.
Uwe Ligges
Nutter, Benjamin wrote:
I have successfully completed building a package to contain the
functions I commonly use. However, I need to have other packages
installed in order for some of my functions to work. I've been
studying
the instructions on installing packages for about a month now, but
still
haven't figured this one out. From what I do understand, to import
additional packages I need some combination of commands in my
DESCRIPTION and NAMESPACE files. Currently, these are what mine look
like:
DESCRIPTION
Package: myPackage
Type: Package
Title: Commonly used functions.
Version: 1.0
Date: 2007-11-08
Author: My Name
Maintainer: My Name [EMAIL PROTECTED]
Description: Blah Blah Blah
License: R 2.3.0
Depends: MASS, survival
NAMESPACE
export(func1, func2, ... , funcx)
import(MASS, survival)
These pass all the checks, and the package installs just fine, but it
doesn't load MASS and survival when I load myPackage. What am I doing
wrong?
Benjamin
P Please consider the environment before printing this e-mail
Cleveland Clinic is ranked one of the top hospitals
in America by U.S. News World Report (2007).
Visit us online at http://www.clevelandclinic.org for
a complete listing of our services, staff and
locations.
Confidentiality Note: This message is intended for
use\...{{dropped:13}}
__
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.
===
P Please consider the environment before printing this e-mail
Cleveland Clinic is ranked one of the top hospitals
in America by U.S. News World Report (2007).
Visit us online at http://www.clevelandclinic.org for
a complete listing of our services, staff and
locations.
Confidentiality Note: This message is intended for use\...{{dropped:13}}
__
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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.
