Re: [R] Multiple left hand side variables in a formula
On Fri, 1 Mar 2013, Frank Harrell wrote:
Thank you Bill. A temporary re-arrangement of the formula will allow me to
do the usual subset= na.action= processing afterwards. Nice idea. I don't
need the dot notation very often for this application.
That's what the Formula package provides. It allows for multiple parts
and multiple responses on both side of the ~. Internally, the formula is
decomposed and separate terms are produced. And using an auxiliary formula
it is assured that a single model frame (with unified NA processing). And
all of this is hidden from the user by providing methods that are as
standard as possible, see:
vignette(Formula, package = Formula)
hth,
Z
Frank
William Dunlap wrote
I don't know how much of the information that model.frame supplies you
need,
but you could make a data.frame containing all the variables on both sides
of them
formula by changing lhs~rhs into ~lhs+rsh before calling model.frame.
E.g.,
f - function (formula) {
if (length(formula) == 3) { # has left hand side
envir - environment(formula)
formula - formula(call(~, call(+, formula[[2]],
formula[[3]])))
environment(formula) - envir
}
formula
}
This doesn't quite take care of the wild-card dot in the formula: straight
variables are omitted from dot's expansion but functions of variables are
not:
colnames(model.frame(f(log(mpg)+hp ~ .), data=mtcars))
[1] log(mpg) hp mpg cyl
[5] disp drat wt qsec
[9] vs am gear carb
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From:
r-help-bounces@
[mailto:
r-help-bounces@
] On Behalf
Of Frank Harrell
Sent: Friday, March 01, 2013 4:17 PM
To:
r-help@
Subject: [R] Multiple left hand side variables in a formula
The lattice package uses special logic to allow for multiple
left-hand-side
variables in a formula, e.g. y1 + y2 ~ x. Is there an elegant way to do
this outside of lattice? I'm trying to implement a data summarization
function that logically takes multiple dependent variables. The usual
invocation of model.frame( ) causes R to try to do arithmetic addition to
create a single dependent variable.
Thanks
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Multiple-left-hand-side-
variables-in-a-formula-tp4660060.html
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-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
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[R] replace zeros for NA in a column based on values of another column
Hi everyone, Imagine that I have a data frame with four columns: data- a b c d 0 1 1 0 1 1 1 1 1 0 0 1 I want to replace the zeros in columns a:b for NA only for the rows in which column d are zero. So a b c d NA 1 1 0 1 1 1 1 1 0 0 1 I am trying this: data[,1:3][data[4] == 0] - NA But get this error: Error in `[-.data.frame`(`*tmp*`, Data[4] == 0, value = NA) : only logical matrix subscripts are allowed in replacement Does anyone knows the reason of this error or is there an alternative to replace the values in one column based on the values of another? Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii http://www.soc.hawaii.edu/mora/ __ 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] using reserved words in R, and reuse variable names in different functions
On 13-03-01 8:35 PM, C W wrote: Thanks, everyone, I will definitely avoid it. Is there any tips on naming variables? I've seen the Google R style guider and Hadley R style guide. Name them in ways that are meaningful to you. R standard function names are famous for not following any naming pattern consistently. Avoid using dots in the name unless you are defining methods (e.g. print.lm is the print method for lm objects). For example, I want to use pie_t, to denote stationary distribution pie at time t. Both pi and pie are function names themselves. Actually pi is not a function, but it is a variable. If you write a package that exports a function or variable named pi, it would mask the standard one, and that could cause big problems for users. If you use it internally, it will only mask the standard one in your code, and that may not matter to you. pie is a function, but all it does is draw pie charts, so who cares if you mask it? :-). Duncan Murdoch Mike On Fri, Mar 1, 2013 at 8:12 PM, William Dunlap wdun...@tibco.com wrote: See fortune(dog). To wit: Firstly, don't call your matrix 'matrix'. Would you call your dog 'dog'? Anyway, it might clash with the function 'matrix' I once had a cat named kitty and she never had a problem with it. Clashes between non-functions and functions that cause problems are not that common. With 4000 packages, each with a number of functions, it is hard to avoid using a name that someone has used for a function. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Rolf Turner Sent: Saturday, March 02, 2013 5:01 AM To: Sarah Goslee Cc: r-help Subject: Re: [R] using reserved words in R, and reuse variable names in different functions On 03/02/2013 01:12 PM, Sarah Goslee wrote: On Fri, Mar 1, 2013 at 7:06 PM, C W tmrs...@gmail.com wrote: Thanks, that was just an example I came up with. I was just curious if using same variable names in different functions would cause problems. No. The environment of a function is independent of other functions. Especially with reserved words. Yes. Using reserved words can cause all kinds of subtle problems. Avoid it. Very sound advice. But it should be noted that t, c, and matrix to which the OP referred are *not* technically reserved words. Nonetheless their use as names of user-defined objects should be eschewed. See fortune(dog). You *can't* actually assign values to reserved words. E.g. TRUE - 42 throws an error. (Whereas matrix - 42, bad form though it may be, does not throw an error.) cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace zeros for NA in a column based on values of another column
you want to replace all rows where the 4th column is zero.. (data[ , 4 ] == 0) and you want to perform that replacement in the first column.. so try data[ data[ , 4 ] == 0 , 1 ] - NA On Sat, Mar 2, 2013 at 5:26 AM, Camilo Mora cm...@dal.ca wrote: Hi everyone, Imagine that I have a data frame with four columns: data- a b c d 0 1 1 0 1 1 1 1 1 0 0 1 I want to replace the zeros in columns a:b for NA only for the rows in which column d are zero. So a b c d NA 1 1 0 1 1 1 1 1 0 0 1 I am trying this: data[,1:3][data[4] == 0] - NA But get this error: Error in `[-.data.frame`(`*tmp*`, Data[4] == 0, value = NA) : only logical matrix subscripts are allowed in replacement Does anyone knows the reason of this error or is there an alternative to replace the values in one column based on the values of another? Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii http://www.soc.hawaii.edu/**mora/ http://www.soc.hawaii.edu/mora/ __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace zeros for NA in a column based on values of another column
Hello, Try # for columns a.b it's 1:2, not 1:3 data[data[,4] == 0, 1:3] - NA # columns a, b and c Hope this helps, Rui Barradas Em 02-03-2013 10:26, Camilo Mora escreveu: Hi everyone, Imagine that I have a data frame with four columns: data- a b c d 0 1 1 0 1 1 1 1 1 0 0 1 I want to replace the zeros in columns a:b for NA only for the rows in which column d are zero. So a b c d NA 1 1 0 1 1 1 1 1 0 0 1 I am trying this: data[,1:3][data[4] == 0] - NA But get this error: Error in `[-.data.frame`(`*tmp*`, Data[4] == 0, value = NA) : only logical matrix subscripts are allowed in replacement Does anyone knows the reason of this error or is there an alternative to replace the values in one column based on the values of another? Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii http://www.soc.hawaii.edu/mora/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multiple left hand side variables in a formula
Achim this is perfect. I had not seen Formula before. Thanks for writing
it!
Frank
Achim Zeileis-4 wrote
On Fri, 1 Mar 2013, Frank Harrell wrote:
Thank you Bill. A temporary re-arrangement of the formula will allow me
to
do the usual subset= na.action= processing afterwards. Nice idea. I
don't
need the dot notation very often for this application.
That's what the Formula package provides. It allows for multiple parts
and multiple responses on both side of the ~. Internally, the formula is
decomposed and separate terms are produced. And using an auxiliary formula
it is assured that a single model frame (with unified NA processing). And
all of this is hidden from the user by providing methods that are as
standard as possible, see:
vignette(Formula, package = Formula)
hth,
Z
Frank
William Dunlap wrote
I don't know how much of the information that model.frame supplies you
need,
but you could make a data.frame containing all the variables on both
sides
of them
formula by changing lhs~rhs into ~lhs+rsh before calling model.frame.
E.g.,
f - function (formula) {
if (length(formula) == 3) { # has left hand side
envir - environment(formula)
formula - formula(call(~, call(+, formula[[2]],
formula[[3]])))
environment(formula) - envir
}
formula
}
This doesn't quite take care of the wild-card dot in the formula:
straight
variables are omitted from dot's expansion but functions of variables
are
not:
colnames(model.frame(f(log(mpg)+hp ~ .), data=mtcars))
[1] log(mpg) hp mpg cyl
[5] disp drat wt qsec
[9] vs am gear carb
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From:
r-help-bounces@
[mailto:
r-help-bounces@
] On Behalf
Of Frank Harrell
Sent: Friday, March 01, 2013 4:17 PM
To:
r-help@
Subject: [R] Multiple left hand side variables in a formula
The lattice package uses special logic to allow for multiple
left-hand-side
variables in a formula, e.g. y1 + y2 ~ x. Is there an elegant way to
do
this outside of lattice? I'm trying to implement a data summarization
function that logically takes multiple dependent variables. The usual
invocation of model.frame( ) causes R to try to do arithmetic addition
to
create a single dependent variable.
Thanks
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Multiple-left-hand-side-
variables-in-a-formula-tp4660060.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@
mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@
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.
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Multiple-left-hand-side-variables-in-a-formula-tp4660060p4660065.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
__
R-help@
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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.
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Multiple-left-hand-side-variables-in-a-formula-tp4660060p4660080.html
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Re: [R] Multiple left hand side variables in a formula
Hi Gabor, This is not for a regression function but for a major update I'm working on for the summary.formula function in the Hmisc package. So I need to handle several data types in the formula. Thanks Frank Gabor Grothendieck wrote Gabor Grothendieck wrote On Fri, Mar 1, 2013 at 7:16 PM, Frank Harrell lt; f.harrell@ gt; wrote: The lattice package uses special logic to allow for multiple left-hand-side variables in a formula, e.g. y1 + y2 ~ x. Is there an elegant way to do this outside of lattice? I'm trying to implement a data summarization function that logically takes multiple dependent variables. The usual invocation of model.frame( ) causes R to try to do arithmetic addition to create a single dependent variable. Try: lm( cbind(Sepal.Length, Sepal.Width) ~., iris) On Fri, Mar 1, 2013 at 8:02 PM, Frank Harrell lt; f.harrell@ gt; wrote: Thanks for your reply Gabor. That doesn't handle a mixture of factor and numeric variables on the left hand side. Frank It can handle 2 level factors lm(cbind(Sepal.Length, setosa = Species == setosa) ~ ., iris) and more with some manual effort: lm(cbind(virginica = Species == virginica, setosa = Species == setosa) ~ ., iris) Typically you don't see more than that as a dependent variable. Do you actually need more? -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@ 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. - Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/Multiple-left-hand-side-variables-in-a-formula-tp4660060p4660081.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dealing with parentheses within variable names
On 13-03-01 12:57 PM, Duncan Murdoch wrote: On 01/03/2013 11:20 AM, William Dunlap wrote: A core R function that fails with odd names is reformulate(): reformulate(c(P/E, % Growth), response=+-) Error in parse(text = termtext) : text:1:16: unexpected input 1: response ~ P/E+% Growth ^ Thanks. That one looks relatively easy to fix. After taking a closer look, I realized that it is actually behaving as designed. The first input to reformulate is supposed to be a character vector, not bits of R language. For example, we want reformulate(x*w) to return ~ x*w not ~ `x*w` So you might think your example should have been entered as reformulate(c(`P/E`, `% Growth`), response=`+-`) However, this doesn't quite work: it ends up with response having double backticks. The way to get what you want is to use reformulate(c(`P/E`, `% Growth`), response=as.name(+-)) Definitely not my favourite design for a function, but I think it's not a code issue. Maybe something should be added to the help page, but this is such an obscure issue that I'm not sure I could make things clearer. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to start console output with a comment sign (as in knitr)
Dear all, knitr writes a comment sign in front of each line of console output. This extremely useful, especially for beginners who are starting to write scripts. One could just compose their script in the console (getting immediate output) and then copy the whole chunk into their .r-file. I have been able to configure the options such that the prompt and the continuing line labels are replaced by blanks. However, I do not know how to automatically comment out each line of output. Ideally, my console would behave exactly like knitr. At my work place I only have access to the R console that is included the standard download for windows. I cannot replace it by an editor that may have additional features. Would there be a way to achieve the knitr-look in the standard console? Many thanks, Marcus __ 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] replace zeros for NA in a column based on values of another column
HI,
Not sure I understand it correctly,
data1-read.table(text=
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
,sep=,header=TRUE)
data2- data1
data3- data1
If i follow this logic for the 1st and 2nd columns,
data1[ data1[ , 4 ] == 0 , 1 ] - NA
data1[ data1[ , 4 ] == 0 , 2 ] - NA
data1
# a b c d
#1 NA NA 1 0
#2 1 1 1 1
#3 1 0 0 1
Still, the column 'b' with 0 element is left as such while and `1` in b is
replaced with NA
data2[,1][data2[,4]==0 data2[,1]==0]- NA
data2[,2][data2[,4]==0 data2[,2]==0]- NA
data2
# a b c d
#1 NA 1 1 0
#2 1 1 1 1
#3 1 0 0 1
#or you can try
data3[,1:2]- lapply(letters[1:2],function(x)
{x1-cbind(data3[,x],data3[,4]);colnames(x1)- c(x,d);x1;
x1[rowSums(x1)==0,1]-NA;x1[,1]})
data3
# a b c d
#1 NA 1 1 0
#2 1 1 1 1
#3 1 0 0 1
A.K.
- Original Message -
From: Anthony Damico ajdam...@gmail.com
To: Camilo Mora cm...@dal.ca
Cc: R help r-help@r-project.org
Sent: Saturday, March 2, 2013 6:10 AM
Subject: Re: [R] replace zeros for NA in a column based on values of another
column
you want to replace all rows where the 4th column is zero.. (data[ , 4 ]
== 0)
and you want to perform that replacement in the first column..
so try
data[ data[ , 4 ] == 0 , 1 ] - NA
On Sat, Mar 2, 2013 at 5:26 AM, Camilo Mora cm...@dal.ca wrote:
Hi everyone,
Imagine that I have a data frame with four columns:
data-
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
I want to replace the zeros in columns a:b for NA only for the rows in
which column d are zero. So
a b c d
NA 1 1 0
1 1 1 1
1 0 0 1
I am trying this:
data[,1:3][data[4] == 0] - NA
But get this error:
Error in `[-.data.frame`(`*tmp*`, Data[4] == 0, value = NA) :
only logical matrix subscripts are allowed in replacement
Does anyone knows the reason of this error or is there an alternative to
replace the values in one column based on the values of another?
Thanks,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
http://www.soc.hawaii.edu/**mora/ http://www.soc.hawaii.edu/mora/
__**
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posting-guide.html http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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[R] Expressions in lattice conditional variables
I would like to have a lattice conditioning ( | var ) variable have
expression() as values because I want panel labels to be able to use
plotmath notation for subscripts, etc. lattice barks at this. Does anyone
know of a trick workaround? An attempted example program is below. Thanks
-Frank
require(lattice)
set.seed(1)
var - c(rep('A', 100), rep('B', 100))
trt - sample(c('T1','T2'), 200, TRUE)
x - c(runif(100), 10*runif(100))
y - x + c(runif(100)/10, runif(100))
N - tapply(x, llist(var, trt), function(x) sum(!is.na(x)))
print(N)
vn - vector('expression', length(var))
for(v in unique(var)) {
i - var == v
n - tapply(!is.na(x[i]), trt[i], sum)
nam - names(n)
w - sprintf('paste(%s, (, n[%s]==%g,~~n[%s]==%g,))',
v, nam[1], n[1], nam[2], n[2])
cat(w, '\n')
vn[var == v] - parse(text=w)
n - sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
vn[var == v] - n
}
trt - factor(trt)
xyplot(as.integer(trt) ~ x | vn, panel=panel.bpplot, ylim=c(0,3),
scale=list(y=list(at=1:2, labels=levels(trt)),
x=list(relation='free', limits=list(c(0,1),c(0,13,
ylab='Treatment', layout=c(1,2))
Error in unique.default(x) :
unimplemented type 'expression' in 'HashTableSetup'
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Expressions-in-lattice-conditional-variables-tp4660089.html
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Expressions in lattice conditional variables
Whoops - these 2 lines should have been omitted from the program:
n - sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
vn[var == v] - n
Frank Harrell wrote
I would like to have a lattice conditioning ( | var ) variable have
expression() as values because I want panel labels to be able to use
plotmath notation for subscripts, etc. lattice barks at this. Does
anyone know of a trick workaround? An attempted example program is below.
Thanks -Frank
require(lattice)
set.seed(1)
var - c(rep('A', 100), rep('B', 100))
trt - sample(c('T1','T2'), 200, TRUE)
x - c(runif(100), 10*runif(100))
y - x + c(runif(100)/10, runif(100))
N - tapply(x, llist(var, trt), function(x) sum(!is.na(x)))
print(N)
vn - vector('expression', length(var))
for(v in unique(var)) {
i - var == v
n - tapply(!is.na(x[i]), trt[i], sum)
nam - names(n)
w - sprintf('paste(%s, (, n[%s]==%g,~~n[%s]==%g,))',
v, nam[1], n[1], nam[2], n[2])
cat(w, '\n')
vn[var == v] - parse(text=w)
n - sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
vn[var == v] - n
}
trt - factor(trt)
xyplot(as.integer(trt) ~ x | vn, panel=panel.bpplot, ylim=c(0,3),
scale=list(y=list(at=1:2, labels=levels(trt)),
x=list(relation='free', limits=list(c(0,1),c(0,13,
ylab='Treatment', layout=c(1,2))
Error in unique.default(x) :
unimplemented type 'expression' in 'HashTableSetup'
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
http://r.789695.n4.nabble.com/Expressions-in-lattice-conditional-variables-tp4660089p4660090.html
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Re: [R] replace zeros for NA in a column based on values of another column
Hello,
Arun has a point, I forgot to check the element to be replaced for a
zero. The following code will do that.
dat - read.table(text =
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
, header = TRUE)
dat - data.matrix(dat)
str(dat)
dat[, 1:2] - sapply(1:2, function(i) ifelse(dat[, i] == 0 dat[, 4] ==
0, NA, dat[, i]))
Hope this helps,
Rui Barradas
Em 02-03-2013 16:11, arun escreveu:
HI,
Not sure I understand it correctly,
data1-read.table(text=
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
,sep=,header=TRUE)
data2- data1
data3- data1
If i follow this logic for the 1st and 2nd columns,
data1[ data1[ , 4 ] == 0 , 1 ] - NA
data1[ data1[ , 4 ] == 0 , 2 ] - NA
data1
# a b c d
#1 NA NA 1 0
#2 1 1 1 1
#3 1 0 0 1
Still, the column 'b' with 0 element is left as such while and `1` in b is
replaced with NA
data2[,1][data2[,4]==0 data2[,1]==0]- NA
data2[,2][data2[,4]==0 data2[,2]==0]- NA
data2
# a b c d
#1 NA 1 1 0
#2 1 1 1 1
#3 1 0 0 1
#or you can try
data3[,1:2]- lapply(letters[1:2],function(x) {x1-cbind(data3[,x],data3[,4]);colnames(x1)-
c(x,d);x1; x1[rowSums(x1)==0,1]-NA;x1[,1]})
data3
# a b c d
#1 NA 1 1 0
#2 1 1 1 1
#3 1 0 0 1
A.K.
- Original Message -
From: Anthony Damico ajdam...@gmail.com
To: Camilo Mora cm...@dal.ca
Cc: R help r-help@r-project.org
Sent: Saturday, March 2, 2013 6:10 AM
Subject: Re: [R] replace zeros for NA in a column based on values of another
column
you want to replace all rows where the 4th column is zero.. (data[ , 4 ]
== 0)
and you want to perform that replacement in the first column..
so try
data[ data[ , 4 ] == 0 , 1 ] - NA
On Sat, Mar 2, 2013 at 5:26 AM, Camilo Mora cm...@dal.ca wrote:
Hi everyone,
Imagine that I have a data frame with four columns:
data-
a b c d
0 1 1 0
1 1 1 1
1 0 0 1
I want to replace the zeros in columns a:b for NA only for the rows in
which column d are zero. So
a b c d
NA 1 1 0
1 1 1 1
1 0 0 1
I am trying this:
data[,1:3][data[4] == 0] - NA
But get this error:
Error in `[-.data.frame`(`*tmp*`, Data[4] == 0, value = NA) :
only logical matrix subscripts are allowed in replacement
Does anyone knows the reason of this error or is there an alternative to
replace the values in one column based on the values of another?
Thanks,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
http://www.soc.hawaii.edu/**mora/ http://www.soc.hawaii.edu/mora/
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Re: [R] How to start console output with a comment sign (as in knitr)
?options Change the prompt option: options(prompt = # ) ## is I think what you want. -- Bert On Sat, Mar 2, 2013 at 5:35 AM, Marcus Kriele mkri...@me.com wrote: Dear all, knitr writes a comment sign in front of each line of console output. This extremely useful, especially for beginners who are starting to write scripts. One could just compose their script in the console (getting immediate output) and then copy the whole chunk into their .r-file. I have been able to configure the options such that the prompt and the continuing line labels are replaced by blanks. However, I do not know how to automatically comment out each line of output. Ideally, my console would behave exactly like knitr. At my work place I only have access to the R console that is included the standard download for windows. I cannot replace it by an editor that may have additional features. Would there be a way to achieve the knitr-look in the standard console? Many thanks, Marcus __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] print method like print.anova()
I have a print method for a set of statistical tests, vcdExtra::CMHtest,
for which I'd like to
have more sensible printing of pvalues, as in print.anova().
[Testing this requires the latest version of vcdExtra, from R-Forge
**|install.packages(vcdExtra, repos=http://R-Forge.R-project.org;)|**
]
With my current print method, I get results like this, but all Prob
values should better
be reported as something like '0.0001' .
CMHtest(MSPatients[,,1])
Cochran-Mantel-Haenszel Statistics for New Orleans Neurologist by
Winnipeg Neurologist
AltHypothesis Chisq Df Prob
corNonzero correlation 51.424 1 7.4426e-13
cmeans Col mean scores differ 55.393 3 5.6601e-12
rmeans Row mean scores differ 53.631 3 1.3450e-11
generalGeneral association 64.318 9 1.9580e-10
In the print.CMHtest() function below the lines before # # TODO give the
output shown above.
The lines below try to use print.anova(), but this gives something even
worse:
Cochran-Mantel-Haenszel Statistics for New Orleans Neurologist by
Winnipeg Neurologist
AltHypothesis Chisq Df Prob
cor 3 51.424 1 7.440e-13
cmeans 1 55.393 3 5.660e-12
rmeans 4 53.631 3 1.345e-11
general 2 64.318 9 1.958e-10
Here is the print method, showing the attempt to use print.anova() as well:
print.CMHtest - function(x, digits = max(getOption(digits) - 2, 3),
...) {
heading - Cochran-Mantel-Haenszel Statistics
if (!is.null(x$names)) heading - paste(heading, for,
paste(x$names, collapse= by ))
# TODO: determine score types (integer, midrank) for heading
df - x$table
types - rownames(df)
labels - list(cor=Nonzero correlation, rmeans=Row mean scores
differ,
cmeans=Col mean scores differ, general=General association)
df -
data.frame(AltHypothesis=as.character(unlist(labels[types])), df)
cat(heading,\n\n)
print(df, digits=digits, ...)
cat(\n)
## TODO: use print.anova() method, but this screws up the AltHyp
attr(df, heading) - paste(heading,\n)
class(df) - c(anova, data.frame)
print(df)
invisible(x)
}
--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept. Chair, Quantitative Methods
York University Voice: 416 736-2100 x66249 Fax: 416 736-5814
4700 Keele StreetWeb: http://www.datavis.ca
Toronto, ONT M3J 1P3 CANADA
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to start console output with a comment sign (as in knitr)
Try the following to see if it does what you want:
## init
con - textConnection(output, w)
options(echo=FALSE)
sink(con)
addTaskCallback( function(expr, out, err, vis) {
sink()
close(con)
if(vis) {
cat(paste(#, output, collapse=\n), \n)
}
con - textConnection(output, w)
sink(con)
TRUE
})
3+4
5*6
tmp - 7-8
ls()
### clean up when finished
removeTaskCallback(1)
sink()
options(echo=TRUE)
This misses errors and warnings, but should do what you want for everything
else.
On Sat, Mar 2, 2013 at 6:35 AM, Marcus Kriele mkri...@me.com wrote:
Dear all,
knitr writes a comment sign in front of each line of console output. This
extremely useful, especially for beginners who are starting to write
scripts. One could just compose their script in the console (getting
immediate output) and then copy the whole chunk into their .r-file.
I have been able to configure the options such that the prompt and the
continuing line labels are replaced by blanks. However, I do not know how
to automatically comment out each line of output. Ideally, my console
would behave exactly like knitr.
At my work place I only have access to the R console that is included the
standard download for windows. I cannot replace it by an editor that may
have additional features.
Would there be a way to achieve the knitr-look in the standard console?
Many thanks, Marcus
__
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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.
--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] print method like print.anova()
Le samedi 02 mars 2013 à 12:37 -0500, Michael Friendly a écrit :
I have a print method for a set of statistical tests, vcdExtra::CMHtest,
for which I'd like to
have more sensible printing of pvalues, as in print.anova().
[Testing this requires the latest version of vcdExtra, from R-Forge
**|install.packages(vcdExtra, repos=http://R-Forge.R-project.org;)|**
]
With my current print method, I get results like this, but all Prob
values should better
be reported as something like '0.0001' .
I think you want to use format.pval(), which is intended exactly for
this use case:
format.pval(c(7.4426e-13, 5.6601e-12, 1.3450e-11, 1.9580e-10), eps=1e-3)
# [1] 0.001 0.001 0.001 0.001
My two cents
CMHtest(MSPatients[,,1])
Cochran-Mantel-Haenszel Statistics for New Orleans Neurologist by
Winnipeg Neurologist
AltHypothesis Chisq Df Prob
corNonzero correlation 51.424 1 7.4426e-13
cmeans Col mean scores differ 55.393 3 5.6601e-12
rmeans Row mean scores differ 53.631 3 1.3450e-11
generalGeneral association 64.318 9 1.9580e-10
In the print.CMHtest() function below the lines before # # TODO give the
output shown above.
The lines below try to use print.anova(), but this gives something even
worse:
Cochran-Mantel-Haenszel Statistics for New Orleans Neurologist by
Winnipeg Neurologist
AltHypothesis Chisq Df Prob
cor 3 51.424 1 7.440e-13
cmeans 1 55.393 3 5.660e-12
rmeans 4 53.631 3 1.345e-11
general 2 64.318 9 1.958e-10
Here is the print method, showing the attempt to use print.anova() as well:
print.CMHtest - function(x, digits = max(getOption(digits) - 2, 3),
...) {
heading - Cochran-Mantel-Haenszel Statistics
if (!is.null(x$names)) heading - paste(heading, for,
paste(x$names, collapse= by ))
# TODO: determine score types (integer, midrank) for heading
df - x$table
types - rownames(df)
labels - list(cor=Nonzero correlation, rmeans=Row mean scores
differ,
cmeans=Col mean scores differ, general=General association)
df -
data.frame(AltHypothesis=as.character(unlist(labels[types])), df)
cat(heading,\n\n)
print(df, digits=digits, ...)
cat(\n)
## TODO: use print.anova() method, but this screws up the AltHyp
attr(df, heading) - paste(heading,\n)
class(df) - c(anova, data.frame)
print(df)
invisible(x)
}
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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] print method like print.anova()
Hi Michael,
Take a look at ?printCoefmat.
I hope this helps,
John
On Sat, 02 Mar 2013 12:37:36 -0500
Michael Friendly frien...@yorku.ca wrote:
I have a print method for a set of statistical tests, vcdExtra::CMHtest,
for which I'd like to
have more sensible printing of pvalues, as in print.anova().
[Testing this requires the latest version of vcdExtra, from R-Forge
**|install.packages(vcdExtra, repos=http://R-Forge.R-project.org;)|**
]
With my current print method, I get results like this, but all Prob
values should better
be reported as something like '0.0001' .
CMHtest(MSPatients[,,1])
Cochran-Mantel-Haenszel Statistics for New Orleans Neurologist by
Winnipeg Neurologist
AltHypothesis Chisq Df Prob
corNonzero correlation 51.424 1 7.4426e-13
cmeans Col mean scores differ 55.393 3 5.6601e-12
rmeans Row mean scores differ 53.631 3 1.3450e-11
generalGeneral association 64.318 9 1.9580e-10
In the print.CMHtest() function below the lines before # # TODO give the
output shown above.
The lines below try to use print.anova(), but this gives something even
worse:
Cochran-Mantel-Haenszel Statistics for New Orleans Neurologist by
Winnipeg Neurologist
AltHypothesis Chisq Df Prob
cor 3 51.424 1 7.440e-13
cmeans 1 55.393 3 5.660e-12
rmeans 4 53.631 3 1.345e-11
general 2 64.318 9 1.958e-10
Here is the print method, showing the attempt to use print.anova() as well:
print.CMHtest - function(x, digits = max(getOption(digits) - 2, 3),
...) {
heading - Cochran-Mantel-Haenszel Statistics
if (!is.null(x$names)) heading - paste(heading, for,
paste(x$names, collapse= by ))
# TODO: determine score types (integer, midrank) for heading
df - x$table
types - rownames(df)
labels - list(cor=Nonzero correlation, rmeans=Row mean scores
differ,
cmeans=Col mean scores differ, general=General association)
df -
data.frame(AltHypothesis=as.character(unlist(labels[types])), df)
cat(heading,\n\n)
print(df, digits=digits, ...)
cat(\n)
## TODO: use print.anova() method, but this screws up the AltHyp
attr(df, heading) - paste(heading,\n)
class(df) - c(anova, data.frame)
print(df)
invisible(x)
}
--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept. Chair, Quantitative Methods
York University Voice: 416 736-2100 x66249 Fax: 416 736-5814
4700 Keele StreetWeb: http://www.datavis.ca
Toronto, ONT M3J 1P3 CANADA
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[R] Errors-In-Variables in R
In reference to [1], how would you solve the following regression problem: Given observations (X_i,Y_i) with known respective error distributions (e_X_i,e_Y_i) (say, 0-mean Gaussian with known STD), find the parameters a and b which maximize the Likelihood of Y = a*X + b Taking the example further, how many of the very simplified assumptions from the above example can be lifted or eased and R still has a method for finding an errors-in-variables fit? __ 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] how to log out from the conference
Hello, please log me out from the conference. I tried to contact the webmaster but the only e-mail I reached was this one. Probably fault on my side in the registration. Thanks, Petr Suvorov [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to start console output with a comment sign (as in knitr)
Thanks Greg, this is great! Regards, Marcus
On 2013-03-02, at 19:06 , Greg Snow wrote:
Try the following to see if it does what you want:
## init
con - textConnection(output, w)
options(echo=FALSE)
sink(con)
addTaskCallback( function(expr, out, err, vis) {
sink()
close(con)
if(vis) {
cat(paste(#, output, collapse=\n), \n)
}
con - textConnection(output, w)
sink(con)
TRUE
})
3+4
5*6
tmp - 7-8
ls()
### clean up when finished
removeTaskCallback(1)
sink()
options(echo=TRUE)
This misses errors and warnings, but should do what you want for everything
else.
On Sat, Mar 2, 2013 at 6:35 AM, Marcus Kriele mkri...@me.com wrote:
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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] easy way to install new R version??
Some packages have to be installed on new R version, 2.15.3 but I have 2.15.2 If I don't want to go through all download, and uninstall and install procedures, is there any simple R command that could handle this? Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] explanation of the problem..
HI Utpal,
Alight, I will look into it. I was under the impression that this is what you
wanted:
dat1- structure(list(V1 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L), V2 = c(1L,
1L, 1L, 1L, 1L, 1L, 0L), V3 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L),
V4 = c(1L, 1L, 0L, 1L, 1L, 1L, 1L), V5 = c(1L, 1L, 1L, 1L,
1L, 0L, 1L), V6 = c(1L, 1L, 1L, 0L, 1L, 1L, 1L), V7 = c(1L,
1L, 1L, 1L, 1L, 1L, 1L), V8 = c(1L, 1L, 1L, 0L, 1L, 0L, 1L
), V9 = c(1L, 0L, 1L, 1L, 1L, 1L, 1L), V10 = c(1L, 1L, 1L,
1L, 1L, 1L, 1L), V11 = c(0L, 1L, 1L, 1L, 1L, 1L, 1L), V12 = c(1L,
1L, 1L, 1L, 1L, 1L, 1L), V13 = c(1L, 1L, 1L, 1L, 1L, 1L,
1L)), .Names = c(V1, V2, V3, V4, V5, V6, V7,
V8, V9, V10, V11, V12, V13), class = data.frame, row.names = c(NA,
-7L))
dat2- dat1[,1:4]
dat2[4,]- c(3,2,3,1)
dat2[6,]- c(3,3,3,3)
dat2
# V1 V2 V3 V4
#1 1 1 1 1
#2 1 1 1 1
#3 1 1 1 0
#4 3 2 3 1
#5 1 1 1 1
#6 3 3 3 3
#7 1 0 1 1
res-
lapply(2:ncol(dat2),function(i) {x1-
combn(names(dat2),i);unlist(lapply(1:ncol(x1), function(i) {x2-
dat2[,x1[,i]];dim(x2[!as.logical(rowSums(x2==0)),])[1]}))})
res
#[[1]]
#[1] 6 7 6 6 5 6
#[[2]]
#[1] 6 5 6 5
#
#[[3]]
#[1] 5
A.K.
From: Utpal Bakshi utpalm...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Saturday, March 2, 2013 2:14 PM
Subject: Re: explanation of the problem..
may be you could use the specaccum function of vegan package for permutations..
the function is shown in pan genome script file..
On Sun, Mar 3, 2013 at 12:39 AM, arun smartpink...@yahoo.com wrote:
Another problem, is that with this combinations of many columns will be
computationally intensive.
From: Utpal Bakshi utpalm...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Saturday, March 2, 2013 1:54 PM
Subject: Re: explanation of the problem..
Hi Arun.. yes... even if the number is greater than 1 (what ever it may be),
it will still count as 1..
for example
dat1
# V1 V2 V3
#1 3 1 2
#2 1 16 0 #3 1 0 1
will considered as same as
dat1
# V1 V2 V3
#1 1 1 1
#2 1 1 0 #3 1 0 1
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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] easy way to install new R version??
On Sat, Mar 2, 2013 at 1:49 PM, capricy gao capri...@yahoo.com wrote: Some packages have to be installed on new R version, 2.15.3 but I have 2.15.2 If I don't want to go through all download, and uninstall and install procedures, is there any simple R command that could handle this? Thanks. After installing 2.15.3 from within R do this: update.packages() -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] xYplot help
xYplot has many options that are passed to panel.xYplot. Did you read the
documentation? You can suppress labels, use an automatically generated
Key() function to control where you want keys, and use several other
options. Start with label.curve=FALSE and go from there.
Frank
bwr87 wrote
I'm trying to work the lattice xYplot function to plot confidence
intervals
for two groups on the same plot. The problem is that it automatically
labels the groups (1 and 2) on the plot itself, and the labels get
obscured
by the CI lines and make it look bad.
This is my code:
xYplot(Cbind(Mean,L95,U95) ~ jitter(Time), type='b', data=A,main=Bouts
Vs.
Time,col=c('black', 'gray'), groups=Group,title=Bouts,xlab=Time,
ylab=Bouts, method=bars, lwd=3, ylim=c(0,30))
A is the data, with columns Time, Group, Mean, L95, U95. L95 and U95 are
the lower and upper confidence interval estimates. Group is either 1 or 2,
each measured at the same number of time points.
The goal is simply to take the labels off of the plot itself. Then I can
use an auto.key to put the labels on the side.
Please help!
-Ben
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-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Re: [R] Expressions in lattice conditional variables
Frank,
As you probably realize, the problem is not with lattice but with
unique (actually: .Internal(unique, ...)):
unique(expression('a','b','b'))
generates your error.
Lattice needs a factor variable for the panels, and it tries to
use unique() on your 'vn' to do that.
For example, if you replace '|vn' with '|as.character(vn)', you'll
get the plot, but of course the strip labels are a mess.
I'm not sure that I understand your need correctly, but if it's
a matter of creating the strip labels, then I would explore using
the factor.levels argument to strip.default().
I hope that this isn't totally out to lunch.
Peter Ehlers
On 2013-03-02 09:20, Frank Harrell wrote:
Whoops - these 2 lines should have been omitted from the program:
n - sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
vn[var == v] - n
Frank Harrell wrote
I would like to have a lattice conditioning ( | var ) variable have
expression() as values because I want panel labels to be able to use
plotmath notation for subscripts, etc. lattice barks at this. Does
anyone know of a trick workaround? An attempted example program is below.
Thanks -Frank
require(lattice)
set.seed(1)
var - c(rep('A', 100), rep('B', 100))
trt - sample(c('T1','T2'), 200, TRUE)
x - c(runif(100), 10*runif(100))
y - x + c(runif(100)/10, runif(100))
N - tapply(x, llist(var, trt), function(x) sum(!is.na(x)))
print(N)
vn - vector('expression', length(var))
for(v in unique(var)) {
i - var == v
n - tapply(!is.na(x[i]), trt[i], sum)
nam - names(n)
w - sprintf('paste(%s, (, n[%s]==%g,~~n[%s]==%g,))',
v, nam[1], n[1], nam[2], n[2])
cat(w, '\n')
vn[var == v] - parse(text=w)
n - sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
vn[var == v] - n
}
trt - factor(trt)
xyplot(as.integer(trt) ~ x | vn, panel=panel.bpplot, ylim=c(0,3),
scale=list(y=list(at=1:2, labels=levels(trt)),
x=list(relation='free', limits=list(c(0,1),c(0,13,
ylab='Treatment', layout=c(1,2))
Error in unique.default(x) :
unimplemented type 'expression' in 'HashTableSetup'
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Re: [R] using reserved words in R, and reuse variable names in different functions
Duncan's comment may not qualify as a fortune, but it did make me chuckle. Peter Ehlers On 2013-03-02 03:01, Duncan Murdoch wrote: On 13-03-01 8:35 PM, C W wrote: [...snip...] pie is a function, but all it does is draw pie charts, so who cares if you mask it? :-). Duncan Murdoch [...snip...] __ 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] Errors-In-Variables in R
There's a no homework policy in R-help. Rui Barradas Em 02-03-2013 18:28, Cedric Sodhi escreveu: In reference to [1], how would you solve the following regression problem: Given observations (X_i,Y_i) with known respective error distributions (e_X_i,e_Y_i) (say, 0-mean Gaussian with known STD), find the parameters a and b which maximize the Likelihood of Y = a*X + b Taking the example further, how many of the very simplified assumptions from the above example can be lifted or eased and R still has a method for finding an errors-in-variables fit? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] caret pls model statistics
Greetings, I have been exploring the use of the caret package to conduct some plsda modeling. Previously, I have come across methods that result in a R2 and Q2 for the model. Using the 'iris' data set, I wanted to see if I could accomplish this with the caret package. I use the following code: library(caret) data(iris) #needed to convert to numeric in order to do regression #I don't fully understand this but if I left as a factor I would get an error following the summary function iris$Species=as.numeric(iris$Species) inTrain1=createDataPartition(y=iris$Species, p=.75, list=FALSE) training1=iris[inTrain1,] testing1=iris[-inTrain1,] ctrl1=trainControl(method=cv, number=10) plsFit2=train(Species~., data=training1, method=pls, trControl=ctrl1, metric=Rsquared, preProc=c(scale)) data(iris) training1=iris[inTrain1,] datvars=training1[,1:4] dat.sc=scale(datvars) n=nrow(dat.sc) dat.indices=seq(1,n) timematrix=with(training1, classvec2classmat(Species[dat.indices])) pls.dat=plsr(timematrix ~ dat.sc, ncomp=3, method=oscorespls, data=training1) x=crossval(pls.dat, segments=10) summary(x) summary(plsFit2) I see two different R2 values and I cannot figure out how to get the Q2 value. Any insight as to what my errors may be would be appreciated. Regards, -- Charles [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Errors-In-Variables in R
Perhaps it would have been clearer that this is no homework if I hadn't forgotten to say what [1] is. Sorry for that. [1] https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=15225 (This is no homework but genuinely adresses the problem that R to my knowledge does not have models for error in variables) On Sat, Mar 02, 2013 at 09:34:21PM +, Rui Barradas wrote: There's a no homework policy in R-help. Rui Barradas Em 02-03-2013 18:28, Cedric Sodhi escreveu: In reference to [1], how would you solve the following regression problem: Given observations (X_i,Y_i) with known respective error distributions (e_X_i,e_Y_i) (say, 0-mean Gaussian with known STD), find the parameters a and b which maximize the Likelihood of Y = a*X + b Taking the example further, how many of the very simplified assumptions from the above example can be lifted or eased and R still has a method for finding an errors-in-variables fit? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Raster images and saving with original pixel dimensions in tiff, jpeg, or png perferablly.
Hello R-Help, I want to be able to read in a raster image, plot it with grid.raster or rasterImage and save the image with one pixel per a pixel element from my array. Saved preferably in a common image format. The real goal of my question is to eventually read in images with text on them, manipulate them with my controlled functions, save them without changing the image dimensions, and perform OCR outside R. I have read The R Journal article on raster images (http://journal.r-project.org/archive/2011-1/RJournal_2011-1_Murrell.pdf) and have experimented with some of the par() variables with no success. There are some arguments in gird.raster that elude me currently on how to use them. These include, vjust, hjust, default.units, gp and vp. Perhaps these are what I need to use but I am getting nowhere without more documentation. Here is the basic code I've used to experiment with making this work. The end result I'm hoping for would be an image exactly like the one I've read in (Rlogo.jpg for this small example). I turned off dev.copy lines because I'm not sure on the guidelines for file writing functions. Thanks, Hans Thompson Forgive me If I made any mistakes following posting guidelines. START# library(jpeg) library(grid) img - readJPEG(system.file(img, Rlogo.jpg, package=jpeg)) grid.raster(img, interpolate =F) #turnon next line for output. Don't want to accidentally write any unwanted files. #dev.copy(tiff, outputimage.tiff) dev.off() #or using package = graphics but not perfered. library(graphics) plot( c(0, dim(img)[2]), c(0, dim(img)[1]), type = n, xlab = , ylab = ) rasterImage(img, 0, 0, dim(img)[2], dim(img)[1]) #dev.copy(tiff, outputfile2.tiff) dev.off() ##STOP## __ 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] caret pls model statistics
I have discovered on of my errors. The timematrix was unnecessary and an unfortunate habit I brought from another package. The following provides the same R2 values as it should, however, I still don't know how to retrieve Q2 values. Any insight would again be appreciated: library(caret) library(pls) data(iris) #needed to convert to numeric in order to do regression #I don't fully understand this but if I left as a factor I would get an error following the summary function iris$Species=as.numeric(iris$Species) inTrain1=createDataPartition(y=iris$Species, p=.75, list=FALSE) training1=iris[inTrain1,] testing1=iris[-inTrain1,] ctrl1=trainControl(method=cv, number=10) plsFit2=train(Species~., data=training1, method=pls, trControl=ctrl1, metric=Rsquared, preProc=c(scale)) data(iris) training1=iris[inTrain1,] datvars=training1[,1:4] dat.sc=scale(datvars) pls.dat=plsr(as.numeric(training1$Species)~dat.sc, ncomp=3, method=oscorespls, data=training1) x=crossval(pls.dat, segments=10) summary(x) summary(plsFit2) Regards, Charles On Sat, Mar 2, 2013 at 3:55 PM, Charles Determan Jr deter...@umn.eduwrote: Greetings, I have been exploring the use of the caret package to conduct some plsda modeling. Previously, I have come across methods that result in a R2 and Q2 for the model. Using the 'iris' data set, I wanted to see if I could accomplish this with the caret package. I use the following code: library(caret) data(iris) #needed to convert to numeric in order to do regression #I don't fully understand this but if I left as a factor I would get an error following the summary function iris$Species=as.numeric(iris$Species) inTrain1=createDataPartition(y=iris$Species, p=.75, list=FALSE) training1=iris[inTrain1,] testing1=iris[-inTrain1,] ctrl1=trainControl(method=cv, number=10) plsFit2=train(Species~., data=training1, method=pls, trControl=ctrl1, metric=Rsquared, preProc=c(scale)) data(iris) training1=iris[inTrain1,] datvars=training1[,1:4] dat.sc=scale(datvars) n=nrow(dat.sc) dat.indices=seq(1,n) timematrix=with(training1, classvec2classmat(Species[dat.indices])) pls.dat=plsr(timematrix ~ dat.sc, ncomp=3, method=oscorespls, data=training1) x=crossval(pls.dat, segments=10) summary(x) summary(plsFit2) I see two different R2 values and I cannot figure out how to get the Q2 value. Any insight as to what my errors may be would be appreciated. Regards, -- Charles -- Charles Determan Integrated Biosciences PhD Student University of Minnesota [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Errors-In-Variables in R
Based on your comments in the (not-a-)bug report, I *think* this might help: quanttrader.info/public/betterHedgeRatios.pdf or more generally, the idea of total least squares regression. Cheers, MW On Sat, Mar 2, 2013 at 9:55 PM, Cedric Sodhi man...@gmx.net wrote: Perhaps it would have been clearer that this is no homework if I hadn't forgotten to say what [1] is. Sorry for that. [1] https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=15225 (This is no homework but genuinely adresses the problem that R to my knowledge does not have models for error in variables) On Sat, Mar 02, 2013 at 09:34:21PM +, Rui Barradas wrote: There's a no homework policy in R-help. Rui Barradas Em 02-03-2013 18:28, Cedric Sodhi escreveu: In reference to [1], how would you solve the following regression problem: Given observations (X_i,Y_i) with known respective error distributions (e_X_i,e_Y_i) (say, 0-mean Gaussian with known STD), find the parameters a and b which maximize the Likelihood of Y = a*X + b Taking the example further, how many of the very simplified assumptions from the above example can be lifted or eased and R still has a method for finding an errors-in-variables fit? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] Metafor SMCR Pre-Post Effect sizes
Dear all, I am very grateful that Wolfgang Viechtbauer implemented the standardised mean change for dependent groups. I was playing around a bit today, and I am not sure if I understand the SMCR procedure correctly. The documentation states that sd1i and sd2i are needed, but it seems to me that SMCR is ignoring sd2i (so Variances are not pooled). Instead, it uses sd1i (pre-test sd), as suggested by Becker 1988. Is that correct? Thank you all very much for your time and help, Markus [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using reserved words in R, and reuse variable names in different functions
Well *I* think it should be a fortune! cheers, Rolf On 03/03/2013 10:30 AM, Peter Ehlers wrote: Duncan's comment may not qualify as a fortune, but it did make me chuckle. Peter Ehlers On 2013-03-02 03:01, Duncan Murdoch wrote: On 13-03-01 8:35 PM, C W wrote: [...snip...] pie is a function, but all it does is draw pie charts, so who cares if you mask it? :-). Duncan Murdoch [...snip...] __ 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] Errors-In-Variables in R
Hello,
Like you say, apparently R doesn't have models for error in variables.
But R packages might have.
library(sos)
findFn('errors-in-variables')
Some look promising. Hope you find something.
Rui Barradas
Em 02-03-2013 21:55, Cedric Sodhi escreveu:
Perhaps it would have been clearer that this is no homework if I
hadn't forgotten to say what [1] is. Sorry for that.
[1] https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=15225
(This is no homework but genuinely adresses the problem that R to my
knowledge does not have models for error in variables)
On Sat, Mar 02, 2013 at 09:34:21PM +, Rui Barradas wrote:
There's a no homework policy in R-help.
Rui Barradas
Em 02-03-2013 18:28, Cedric Sodhi escreveu:
In reference to [1], how would you solve the following regression
problem:
Given observations (X_i,Y_i) with known respective error distributions
(e_X_i,e_Y_i) (say, 0-mean Gaussian with known STD), find the parameters
a and b which maximize the Likelihood of
Y = a*X + b
Taking the example further, how many of the very simplified assumptions
from the above example can be lifted or eased and R still has a method
for finding an errors-in-variables fit?
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Re: [R] Errors-In-Variables in R
On Mar 2, 2013, at 1:55 PM, Cedric Sodhi wrote: Perhaps it would have been clearer that this is no homework if I hadn't forgotten to say what [1] is. Sorry for that. [1] https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=15225 (This is no homework but genuinely adresses the problem that R to my knowledge does not have models for error in variables) In addition to searching for errors in variables you should also be searching for deming regression, 'orthogonal regression, total least squares regression, and measurement error models Here are a few links to get you started: http://markmail.org/message/4mo62jqfyudrchzi?q=list:org%2Er-project%2Er-help+deming+orthogonal http://markmail.org/message/htlptlcccunsd5mm?q=list:org%2Er-project%2Er-help+deming+orthogonal http://markmail.org/message/zhogz6337m3ofl7d?q=list:org%2Er-project%2Er-help+deming+orthogonal -- David. On Sat, Mar 02, 2013 at 09:34:21PM +, Rui Barradas wrote: There's a no homework policy in R-help. Rui Barradas Em 02-03-2013 18:28, Cedric Sodhi escreveu: In reference to [1], how would you solve the following regression problem: Given observations (X_i,Y_i) with known respective error distributions (e_X_i,e_Y_i) (say, 0-mean Gaussian with known STD), find the parameters a and b which maximize the Likelihood of Y = a*X + b Taking the example further, how many of the very simplified assumptions from the above example can be lifted or eased and R still has a method for finding an errors-in-variables fit? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ 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] if value is in vector, perform this function
Hi,
I'm trying to set up R to run a simulation of two populations in which every
3.5 days, the initial value of one of the populations is reset to 1.5. I'm
simulation an experiment we did in which we fed Daphnia populations twice a
week with algae, so I want the initial value of the algal population to reset
to 1.5 twice a week to simulate that feeding. I've use for loops and if/else
loops before but I can't figure out how to syntax if t is in this vector of
possible t values, do this command, else, do this command if that makes sense.
Here's what I have (and it doesn't work):
params = c(1, 0.15, 0.164, 1)
init = c(1.5, 0.05)
t=seq(1,60, by=0.5) #all time values, experiment ran for 60 days
#feeding sequence - every 3.5 days
feed_days = seq(1,60,by=3.5)
Daphnia - function(t,x,params){
C_D = x[2];
C_A = 0;
for(t %in% feed_days){
if t == TRUE {
C_A = 1.5
}
else{
C_A = 0
}}
gamma = params[1]; m_D = params[2]; K_q = params[3]; q_max = params[4];
M_D = m_D * C_D
I_A = (C_D * q_max * C_A) / (K_q + C_A)
r_D = gamma * I_A
return(
list(c(
- I_A,
r_D - M_D
)))
}
library(deSolve)
results - ode(init, t, Daphnia, params, method = lsoda)
Let me know if there's any other info that would be helpful and thanks very
much for your help!
[[alternative HTML version deleted]]
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Re: [R] Empirical Bayes Estimator for Poisson-Gamma Parameters
just seeing this. when you say y1 and y2, are you asking about 2 differnt models in an effort to model select? i need more info, i might be able to help. please supply a bit more info. ~n On Feb 25, 2013, at 4:39 AM, Ali A. Bromideh wrote: Dear Sir/Madam, I apologize for any cross-posting. I got a simple question, which I thought the R list may help me to find an answer. Suppose we have Y_1, Y_2, ., Y_n ~ Poisson (Lambda_i) and Lambda_i ~Gamma(alpha_i, beta_i). Empirical Bayes Estimator for hyper-parameters of the gamma distr, i.e. (alpha_t, beta_t) are needed. y=c(12,5,17,14) n=4 What about a Hierarchal B ayes estimators? Any relevant work and codes in R (or S+) is highly appreciated. Kind regards, Ali [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Empirical Bayes Estimator for Poisson-Gamma Parameters
also, kruschke at indiana has some info on this, both online and youtube. (if homework.) if not, more infor will be helpful. ~n On Feb 25, 2013, at 9:41 AM, Bert Gunter wrote: Homework? We don't do homework here. If not, search (e.g. via google -- R hierarchical Bayes -- or some such). -- Bert On Mon, Feb 25, 2013 at 1:39 AM, Ali A. Bromideh a.bromi...@ikco.com wrote: Dear Sir/Madam, I apologize for any cross-posting. I got a simple question, which I thought the R list may help me to find an answer. Suppose we have Y_1, Y_2, ., Y_n ~ Poisson (Lambda_i) and Lambda_i ~Gamma(alpha_i, beta_i). Empirical Bayes Estimator for hyper-parameters of the gamma distr, i.e. (alpha_t, beta_t) are needed. y=c(12,5,17,14) n=4 What about a Hierarchal B ayes estimators? Any relevant work and codes in R (or S+) is highly appreciated. Kind regards, Ali [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ordering Table Columns
cdouglass wrote Hello all, Totally new to this and I'm just doing a frequency distribution analysis on T-shirt sales by size. I have a .csv with 60 orders. I read in the data using read.csv. If I look at the summary() or table() of the data it looks fine, except that the shirt sizes are alphabetical rather than from S-XXL--so the bar graph loses the shape of the data based on size. All I want to do is get the table to arrange the data: S M L XL XXL Here's the code that I've run that got me closer to what I want. It seems like it should be simple, but going through the R in a Nutshell and asking Google as many different ways as I can think to phrase it are turning up nothing. shirt - read.csv(http://localhost/examples/tshirt_purchases.csv;, header=TRUE, sep = ,, nrows=60) shirt.table-summary(shirt) shirt.table Shirt.Size L :20 M :20 S :11 XL : 7 XXL: 2 What I want is: Shirt.Size S :11 M :20 L :20 XL : 7 XXL: 2 Does anyone know how to do this, or am I coming at it from the wrong direction? If this has been answered previously and I've just failed to find it in my searches, please accept my apologies. Many Thanks, Chris Think you want to have a look at factors Typing ?factor will throw up the relevant help pages # Your Data tbl - read.table(header = TRUE, text = ShirtSize Number L 20 M 20 S 11 XL 7 XXL 2 ) # ShirtSize Alphabetical tbl[tbl$ShirtSize,] # Reorder Factor tbl$ShirtSize = factor(tbl$ShirtSize, levels=c(S,M,L,XL,XXL)) # ShirtSize Order by Size tbl[tbl$ShirtSize,] Pete -- View this message in context: http://r.789695.n4.nabble.com/Ordering-Table-Columns-tp4660110p4660129.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] caret pls model statistics
Charles, You should not be treating the classes as numeric (is virginica really three times setosa?). Q^2 and/or R^2 are not appropriate for classification. Max On Sat, Mar 2, 2013 at 5:21 PM, Charles Determan Jr deter...@umn.eduwrote: I have discovered on of my errors. The timematrix was unnecessary and an unfortunate habit I brought from another package. The following provides the same R2 values as it should, however, I still don't know how to retrieve Q2 values. Any insight would again be appreciated: library(caret) library(pls) data(iris) #needed to convert to numeric in order to do regression #I don't fully understand this but if I left as a factor I would get an error following the summary function iris$Species=as.numeric(iris$Species) inTrain1=createDataPartition(y=iris$Species, p=.75, list=FALSE) training1=iris[inTrain1,] testing1=iris[-inTrain1,] ctrl1=trainControl(method=cv, number=10) plsFit2=train(Species~., data=training1, method=pls, trControl=ctrl1, metric=Rsquared, preProc=c(scale)) data(iris) training1=iris[inTrain1,] datvars=training1[,1:4] dat.sc=scale(datvars) pls.dat=plsr(as.numeric(training1$Species)~dat.sc, ncomp=3, method=oscorespls, data=training1) x=crossval(pls.dat, segments=10) summary(x) summary(plsFit2) Regards, Charles On Sat, Mar 2, 2013 at 3:55 PM, Charles Determan Jr deter...@umn.edu wrote: Greetings, I have been exploring the use of the caret package to conduct some plsda modeling. Previously, I have come across methods that result in a R2 and Q2 for the model. Using the 'iris' data set, I wanted to see if I could accomplish this with the caret package. I use the following code: library(caret) data(iris) #needed to convert to numeric in order to do regression #I don't fully understand this but if I left as a factor I would get an error following the summary function iris$Species=as.numeric(iris$Species) inTrain1=createDataPartition(y=iris$Species, p=.75, list=FALSE) training1=iris[inTrain1,] testing1=iris[-inTrain1,] ctrl1=trainControl(method=cv, number=10) plsFit2=train(Species~., data=training1, method=pls, trControl=ctrl1, metric=Rsquared, preProc=c(scale)) data(iris) training1=iris[inTrain1,] datvars=training1[,1:4] dat.sc=scale(datvars) n=nrow(dat.sc) dat.indices=seq(1,n) timematrix=with(training1, classvec2classmat(Species[dat.indices])) pls.dat=plsr(timematrix ~ dat.sc, ncomp=3, method=oscorespls, data=training1) x=crossval(pls.dat, segments=10) summary(x) summary(plsFit2) I see two different R2 values and I cannot figure out how to get the Q2 value. Any insight as to what my errors may be would be appreciated. Regards, -- Charles -- Charles Determan Integrated Biosciences PhD Student University of Minnesota [[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. -- Max [[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.
