[R] extract column's from different dataframe
Dear R users, I want to extract column's from different data frame with different row length. How can I do this in R? Thank you very much! best regards! CR -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str. Universitatii no. 13, Suceava, 720229, Romania office phone +4 0230 52 29 78, ext. 531 mobile phone +4 0745 53 18 01 +4 0766 71 76 58 FAX:+4 0230 52 16 64 silvic.usv.ro [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot time series data irregularly hourly-spaced
Wow!! Thank you so much for your suggestions! For now, A.K's suggestion #1 is perfect for me! Thank you very much! Best, Charles On Thu, Oct 17, 2013 at 2:34 AM, William Dunlap wdun...@tibco.com wrote: You could bump up the day each time an hour was less than the previous one. E.g., testtime - c(20:00:00,22:10:00,22:20:00,23:15:00,23:43:00,00:00:00,00:51:00,01:00:00) var - seq_along(testtime) # so you know what the plot should look like # turn it ino a POSIXlt object so you can do arithmetic on it t - strptime(testtime,format=%H:%M:%S) # now add a day each time t[i]t[i-1] td - t + .difftime(cumsum(c(FALSE, diff(t)0)), units=days) # compare plots par(mfrow=c(2,1)) plot(t,var,type=b,xlab=Time,ylab=Var) plot(td,var,type=b,xlab=Time,ylab=Var) This is dicey because you may have skipped more than one day. 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 Law, Jason Sent: Wednesday, October 16, 2013 5:04 PM To: Charles Novaes de Santana; r-help@r-project.org Subject: Re: [R] Plot time series data irregularly hourly-spaced You just need the date, otherwise how would it know what time comes first? In strptime(), a date is being assumed. Try this: testtime- c(20:00:00,22:10:00,22:20:00,23:15:00,23:43:00,00:00:00,00:51:00,01:00: 00) testday - rep(Sys.Date() - c(1,0), times = c(5,3)) plot(as.POSIXct(paste(testday, testtime)), var) Jason -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Charles Novaes de Santana Sent: Wednesday, October 16, 2013 2:58 PM To: r-help@r-project.org Subject: [R] Plot time series data irregularly hourly-spaced Dear all, I have a time series of data that I would like to represent in a plot. But I am facing some problems to do it because the time is represented in hours, it can start in one day and end in another day, and it is not regularly spaced. My problem is that when I plot my data, my X-axis always starts from the lower values of my time data. For example, I would like to plot data that starts at 20:00:00 and ends at 01:00:00, but R considers that 01:00:00 is lower than 21:00:00 and my plot is kind of crossed over time. Please try this example to see it graphically: testtime- c(20:00:00,22:10:00,22:20:00,23:15:00,23:43:00,00:00:00,00:51:00,01:00: 00) var-runif(length(testtime),0,1) plot(strptime(testtime,format=%H:%M:%S),var,type=b,xlab=Time,ylab=Var) In this case, I would like to have a plot that starts at 20:00:00 and ends at 01:00:00. Does anybody know how to make R understand that 00:00:00 comes after 20:00:00 in this case? Or at least does anybody know a tip to make a plot with this kind of X-axis? Thanks for your time and thanks in advance for any help. Best regards, Charles -- Um axé! :) -- Charles Novaes de Santana, PhD http://www.imedea.uib-csic.es/~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. -- Um axé! :) -- Charles Novaes de Santana, PhD http://www.imedea.uib-csic.es/~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] saveXML() prefix argument
Le mercredi 16 octobre 2013 à 23:45 -0400, Earl Brown a écrit :
I'm using the XML package and specifically the saveXML() function but I
can't get the prefix argument of saveXML() to work:
library(XML)
concepts - c(one, two, three)
info - c(info one, info two, info three)
root - newXMLNode(root)
for (i in 1:length(concepts)) {
cur.concept - concepts[i]
cur.info - info[i]
cur.tip - newXMLNode(tip, attrs = c(id = i), parent = root)
newXMLNode(h1, cur.concept, parent = cur.tip)
newXMLNode(p, cur.info, parent = cur.tip)
}
# None of the following output a prefix on the first line of the exported
document
saveXML(root)
saveXML(root, file = test.xml)
saveXML(root, file = test.xml, prefix = '?xml version=1.0?\n')
Am I missing something obvious? Any ideas?
It looks like the function XML:::saveXML.XMLInternalNode() does not use
the 'prefix' parameter at all. So it won't be taken into account when
calling saveXML() on objects of class XMLInternalNode.
I think you should report this to Duncan Temple Lang, as this is
probably an oversight.
Regards
Thanks in advance. Earl Brown
-
Earl K. Brown, PhD
Assistant Professor of Spanish Linguistics
Advisor, TEFL MA Program
Department of Modern Languages
Kansas State University
www-personal.ksu.edu/~ekbrown
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[R] Constraint on regression parameters
Dear all, I have been trying to find a simple solution to my problem without success, though i have a feeling a simple syntaxe detail coul make the job. I am doing a polynomial linear regression with 2 independent variables such as : lm(A ~ B + I(B^2) + I(lB^3) + C, data=Dataset)) R return me a coefficient per independent variable, and I would need the coefficient of the C parameter to equal 1. I've been loonking at parameter constraints on the internet but it's always much more complicated that just removing the fit of a coefficient (or setting it to 1). I know many package allows to not fit an intercept with a -1 parameter in the syntaxe, does that exists for independent variables ? Regards, [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Constraint on regression parameters
-Original Message-
I am doing a polynomial linear regression with 2 independent variables
such as :
lm(A ~ B + I(B^2) + I(lB^3) + C, data=Dataset))
R return me a coefficient per independent variable, and I would need
the coefficient of the C parameter to equal 1.
Leaving aside the question of fitting simple polynomial coefficients instead of
orthogonal polynomials - generally frowned upon, but not always serious - the
problem you describe is one in which you are not fitting C at all; you're
assuming C adds exactly. What you're really fitting is the difference between A
and C.
Try fitting
A-C ~ B + I(B^2) + I(lB^3)
to obtain the coefficients you're looking for. But be aware that you will still
have a constant intercept, so the model you will have fitted is
A = b0 + b1.B +b2.B^2 +b3.B^3 + C + error
S Ellison
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Re: [R] Extract a predictors form constparty object (CHAID output) in R
For the record. I have found a possible sollution: nn - nodeapply(z) n.names= names(unlist(nn[[1]])) ext - unlist(sapply(n.names, function(x) grep(split.varid., x, value=T))) ext - gsub(kids.split.varid., , ext) ext - gsub(split.varid., , ext) dep.var - as.character(terms(z)[1][[2]]) plus = paste(ext, collapse= + ) mul = paste(ext, collapse= * ) shortform - as.formula(paste (dep.var, plus, sep = ~ )) satform - as.formula(paste (dep.var, mul, sep = ~ )) mosaic(shortform, data = ContraceptiveChoice) #stp - step(glm(satform, data=ContraceptiveChoice, family=binomial), direction=both) On 16 October 2013 20:18, Christiaan Pauw cjp...@gmail.com wrote: I have a large dataset (questionnaire results) of mostly categorical variables. I have tested for dependency between the variables using chi-square test. There are an incomprehensible number of dependencies. I used the chaid() function in the CHAID package to detect interactions and separate out (what I hope to be) the underlying structure of these dependencies for each variable. What typically happens is that the chi-square test will reveal a large number of dependencies (say 10-20) for a variable and the chaid function will reduce this to something much more comprehensible (say 3-5). What I want to do is to extract the names of those variable that were shown to be relevant in the chaid() results. The chaid() output is in the form of a constparty object. My question is how to extract the variable names associated with the nodes in such an object. Here is a self contained code example: library(evtree) # for the ContraceptiveChoice dataset library(CHAID) library(vcd) library(MASS) data(ContraceptiveChoice) longform - formula(contraceptive_method_used ~ wifes_education + husbands_education + wifes_religion + wife_now_working + husbands_occupation + standard_of_living_index + media_exposure) z - chaid(longform, data = ContraceptiveChoice) # plot(z) z # This is the part I want to do programatically shortform - formula(contraceptive_method_used ~ wifes_education + husbands_occupation) # The thing I want is a programatic way to extract 'shortform' from 'z' # Examples of use of 'shortfom' loglm(shortform, data = ContraceptiveChoice) Thanks in advance Christiaan -- Christiaan Pauw Nova Institute www.nova.org.za -- Christiaan Pauw Nova Institute www.nova.org.za [[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] extract column's from different dataframe
On 10/17/2013 06:17 PM, catalin roibu wrote: Dear R users, I want to extract column's from different data frame with different row length. How can I do this in R? Hi catalin, If I understand your question, which I think is: I want to extract columns from different data frames with differing numbers of rows and store them in a single object. The answer is probably to use a list: datalist-list() datalist[[1]]-dataframe1[,variable1] datalist[[2]]-dataframe2[,variable3] ... where each element of datalist may have different numbers of values. Jim __ 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 would i sum the number of NA's in multiple vectors
mattbju2013 wrote Hi guys this is my first post, i need help summing the number of NA's in a few vectors for example.. c1-c(1,2,NA,3,4) c2-c(NA,1,2,3,4) c3-c(NA,1,2,3,4) how would i get a result that only sums the number of NA's in the vector? the.result.i.want-c(2,0,1,0,0) See ?is.na . Now, if I can interpret your question correctly, you're actually looking for the number of NA per *position* in the vectors, so let's make them into a matrix first. cmat-rbind(c1,c2,c3) then use apply over columns apply(cmat,2,function(k)sum(is.na(k))) -- View this message in context: http://r.789695.n4.nabble.com/How-would-i-sum-the-number-of-NA-s-in-multiple-vectors-tp4678411p4678432.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] match values in dependence of ID and Date
hello togehter, i have a little problem, maybe you can help me. I have a data.frame like this one: IDName 1 Andy 2 John 3 Amy and a data.frame like this: ID DateValue 12013-10-0110 12013-10-0215 22013-10-017 22013-10-0310 22013-10-0415 32013-10-0110 the result should be this one: IDName First SecondThird 1 Andy10 15 2 John 7 10 15 3 Amy 10 maybe you can help me, to do this? Thank you. Mat -- View this message in context: http://r.789695.n4.nabble.com/match-values-in-dependence-of-ID-and-Date-tp4678433.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] How would i sum the number of NA's in multiple vectors
Or faster (both computational speed and amount of code): colSums(is.na(rbind(c1, c2, c3))) On Thu, Oct 17, 2013 at 4:34 AM, Carl Witthoft c...@witthoft.com wrote: mattbju2013 wrote Hi guys this is my first post, i need help summing the number of NA's in a few vectors for example.. c1-c(1,2,NA,3,4) c2-c(NA,1,2,3,4) c3-c(NA,1,2,3,4) how would i get a result that only sums the number of NA's in the vector? the.result.i.want-c(2,0,1,0,0) See ?is.na . Now, if I can interpret your question correctly, you're actually looking for the number of NA per *position* in the vectors, so let's make them into a matrix first. cmat-rbind(c1,c2,c3) then use apply over columns apply(cmat,2,function(k)sum(is.na(k))) -- View this message in context: http://r.789695.n4.nabble.com/How-would-i-sum-the-number-of-NA-s-in-multiple-vectors-tp4678411p4678432.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. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://joshuawiley.com/ Senior Analyst - Elkhart Group Ltd. http://elkhartgroup.com [[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 obtain restricted estimates from coxph()?
Consider the function f(x) = x on the open interval (0,1). It does not have a maximum. That is what your likelihood function will look like. The MLE does not exist. Chris (Although if everything is continuous and you are okay with limits there is an extension that gets you to Terry's original answer.) -Original Message- From: Y [mailto:yuhan...@gmail.com] Sent: Wednesday, October 16, 2013 7:08 PM To: Göran Broström Cc: r-help@r-project.org Subject: Re: [R] How to obtain restricted estimates from coxph()? Thanks very much for your help, Terry and G?ran! As pointed out by G?ran, the difficult part is that it's an open set. How to obtain a valid MLE in this case? Thanks, YH On Wed, Oct 16, 2013 at 9:55 AM, G?ran Brostr?m goran.brost...@umu.sewrote: On 2013-10-16 14:33, Terry Therneau wrote: On 10/16/2013 05:00 AM, r-help-requ...@r-project.org wrote: Hello, I'm trying to use coxph() function to fit a very simple Cox proportional hazards regression model (only one covariate) but the parameter space is restricted to an open set (0, 1). Can I still obtain a valid estimate by using coxph function in this scenario? If yes, how? Any suggestion would be greatly appreciated. Thanks!!! Easily: 1. Fit the unrestricted model. If the solution is in 0-1 you are done. 2. If it is outside, fix the coefficient. Say that the solution is 1.73, then the optimal solution under contraint is 1. OK, except for the small annoyance that 1 is not a member of the open set (interval) (0, 1). Maybe the answer is No in this case? Depends on what lies in the word 'valid'. If 'MLE', the answer is No. Redo the fit adding the paramters init=1, iter=0. This forces the program to give the loglik and etc for the fixed coefficient of 1.0. Terry Therneau __** 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. __** 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]] ** Electronic Mail is not secure, may not be read every day, and should not be used for urgent or sensitive issues __ 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] S4 base class
Greetings, Meanwhile I have figured out how do do it only to find out that I have more serious problems. Generally calling Base::f on the base class object is not what you want, instead you want to call Base::f on the full object for the following reasons: If the base class is virtual, then Base::f might use virtual functions (not defined in Base but defined in derived classes). If you then call Base::f on an object of class Base the call will fail. Is it possible in R to call Base::f from within Derived (when there is also Derived::f) on the full object this? I suspect not, which would be a serious drawback to the R class mechanism. Thanks, Michael Meyer __ 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] S4 base class
Sorry, if the previous message seems without context. Indeed, the first message was bounced by filtering rules (triggered by subject heading than which nothing could be more benign or less liable to suspician). It was: Greetings, I have an S4 class B (Base) which defines a function f=f(this=B,...) Dervided from B we have a derived class D which also defines a function f=f(this=D,...) In the definition of D::f we want to call the version B::f and could do this by simply calling f(baseClassObject(this),...) The question is the following: How do I refer to the base class object from the derived class? Many thanks Michael Meyer __ 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] flatten a list of lists
I have functions that generate lists objects of class foo and lists of
lists of these, of class
foolist, similar to what is shown below.
How can I flatten something like this to remove the top-level list
structure, i.e.,
return a single-level list of foo objects, of class foolist?
foo - function(n) {
result - list(x=sample(1:10,n), y=sample(1:10,n))
class(result) - foo
result
}
multifoo - function(vec, label, ...) {
result - lapply(vec, foo, ...)
names(result) - paste0(label, vec)
class(result) - foolist
result
}
foo1 - multifoo(1:2, A)
foo2 - multifoo(1:2, B)
mfoo - list(A=foo1, B=foo2)
str(mfoo, 2)
str(mfoo, 2)
List of 2
$ A:List of 2
..$ A1:List of 2
.. ..- attr(*, class)= chr foo
..$ A2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
$ B:List of 2
..$ B1:List of 2
.. ..- attr(*, class)= chr foo
..$ B2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
In this case, what is wanted is a single-level list, of 4 foo objects,
A1, A2, B1, B2,
all of class foolist
--
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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Re: [R] S4 base class
On 17/10/2013 9:01 AM, Michael Meyer wrote: Sorry, if the previous message seems without context. Indeed, the first message was bounced by filtering rules (triggered by subject heading than which nothing could be more benign or less liable to suspician). It was: Greetings, I have an S4 class B (Base) which defines a function f=f(this=B,...) Dervided from B we have a derived class D which also defines a function f=f(this=D,...) In the definition of D::f we want to call the version B::f and could do this by simply calling f(baseClassObject(this),...) The question is the following: How do I refer to the base class object from the derived class? You're asking the wrong question. You should be asking how to call the method for the inherited class . callNextMethod() is the answer to that question. By the way, your use of the syntax D::f and B::f suggests that you're thinking from a C++ point of view. That's very likely to lead to frustration: the S4 object system is very different from C++. Methods don't belong to classes, they belong to generics. There is no such thing as D::f or B::f, only f methods with different signatures. 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.
Re: [R] flatten a list of lists
unlist(mfoo, recursive = FALSE) gets you pretty close.
Best,
Ista
On Thu, Oct 17, 2013 at 9:15 AM, Michael Friendly frien...@yorku.ca wrote:
I have functions that generate lists objects of class foo and lists of
lists of these, of class
foolist, similar to what is shown below.
How can I flatten something like this to remove the top-level list
structure, i.e.,
return a single-level list of foo objects, of class foolist?
foo - function(n) {
result - list(x=sample(1:10,n), y=sample(1:10,n))
class(result) - foo
result
}
multifoo - function(vec, label, ...) {
result - lapply(vec, foo, ...)
names(result) - paste0(label, vec)
class(result) - foolist
result
}
foo1 - multifoo(1:2, A)
foo2 - multifoo(1:2, B)
mfoo - list(A=foo1, B=foo2)
str(mfoo, 2)
str(mfoo, 2)
List of 2
$ A:List of 2
..$ A1:List of 2
.. ..- attr(*, class)= chr foo
..$ A2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
$ B:List of 2
..$ B1:List of 2
.. ..- attr(*, class)= chr foo
..$ B2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
In this case, what is wanted is a single-level list, of 4 foo objects, A1,
A2, B1, B2,
all of class foolist
--
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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Re: [R] flatten a list of lists
On 17/10/2013 9:15 AM, Michael Friendly wrote:
I have functions that generate lists objects of class foo and lists of
lists of these, of class
foolist, similar to what is shown below.
You can use c() to join lists. So in the example below,
c(mfoo$A, mfoo$B)
will give you a list with the components you want, though the class
won't be set. More generally, do.call(c, unname(mfoo)) will join any
number of components. (Without unname(), the names at the top level
will be combined with the component names;
maybe you'd actually want that, but your example didn't do it.)
This won't work if your list doesn't have the regular list of lists
structure, e.g. if it mixes foo objects with foolist objects at the same
level. Then you probably need a more complicated recursive approach.
You might be able to do it with rapply().
Duncan Murdoch
How can I flatten something like this to remove the top-level list
structure, i.e.,
return a single-level list of foo objects, of class foolist?
foo - function(n) {
result - list(x=sample(1:10,n), y=sample(1:10,n))
class(result) - foo
result
}
multifoo - function(vec, label, ...) {
result - lapply(vec, foo, ...)
names(result) - paste0(label, vec)
class(result) - foolist
result
}
foo1 - multifoo(1:2, A)
foo2 - multifoo(1:2, B)
mfoo - list(A=foo1, B=foo2)
str(mfoo, 2)
str(mfoo, 2)
List of 2
$ A:List of 2
..$ A1:List of 2
.. ..- attr(*, class)= chr foo
..$ A2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
$ B:List of 2
..$ B1:List of 2
.. ..- attr(*, class)= chr foo
..$ B2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
In this case, what is wanted is a single-level list, of 4 foo objects,
A1, A2, B1, B2,
all of class foolist
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Re: [R] flatten a list of lists
Does this get you the rest of the way?
mfoo2 - unlist(mfoo, recursive = FALSE)
names(mfoo2) - unlist(lapply(mfoo, names))
class(mfoo2) - foolist
str(mfoo2)
List of 4
$ A1:List of 2
..$ x: int 3
..$ y: int 10
..- attr(*, class)= chr foo
$ A2:List of 2
..$ x: int [1:2] 6 4
..$ y: int [1:2] 8 9
..- attr(*, class)= chr foo
$ B1:List of 2
..$ x: int 2
..$ y: int 2
..- attr(*, class)= chr foo
$ B2:List of 2
..$ x: int [1:2] 3 6
..$ y: int [1:2] 4 2
..- attr(*, class)= chr foo
- attr(*, class)= chr foolist
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Ista Zahn
Sent: Thursday, October 17, 2013 8:23 AM
To: Michael Friendly
Cc: R-help
Subject: Re: [R] flatten a list of lists
unlist(mfoo, recursive = FALSE) gets you pretty close.
Best,
Ista
On Thu, Oct 17, 2013 at 9:15 AM, Michael Friendly
frien...@yorku.ca wrote:
I have functions that generate lists objects of class foo
and lists of
lists of these, of class
foolist, similar to what is shown below.
How can I flatten something like this to remove the top-level
list
structure, i.e.,
return a single-level list of foo objects, of class
foolist?
foo - function(n) {
result - list(x=sample(1:10,n), y=sample(1:10,n))
class(result) - foo
result
}
multifoo - function(vec, label, ...) {
result - lapply(vec, foo, ...)
names(result) - paste0(label, vec)
class(result) - foolist
result
}
foo1 - multifoo(1:2, A)
foo2 - multifoo(1:2, B)
mfoo - list(A=foo1, B=foo2)
str(mfoo, 2)
str(mfoo, 2)
List of 2
$ A:List of 2
..$ A1:List of 2
.. ..- attr(*, class)= chr foo
..$ A2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
$ B:List of 2
..$ B1:List of 2
.. ..- attr(*, class)= chr foo
..$ B2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
In this case, what is wanted is a single-level list, of 4 foo
objects, A1,
A2, B1, B2,
all of class foolist
--
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
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Re: [R] flatten a list of lists
Thanks to all who replied.
Here are two versions of a function (sans sanity checks) that do what I
want:
foo1 - multifoo(1:2, A)
foo2 - multifoo(1:2, B)
mfoo - list(A=foo1, B=foo2)
class(mfoo) - c(foolist, list)
#' flatten a list of lists
# from Duncan Murdoch
flatten - function(list, unname=TRUE) {
res - do.call(c, if(unname) unname(list) else list)
class(res) - class(list)
res
}
# from David Carlson
flatten2 - function(list, unname=TRUE) {
res - unlist(list, recursive = FALSE)
if(unname) names(res) - unlist(lapply(list, names))
class(res) - class(list)
res
}
mflat1 - flatten(mfoo)
mflat2 - flatten2(mfoo)
all.equal(mflat1,mflat2)
all.equal(mflat1,mflat2)
[1] TRUE
-Michael
On 10/17/2013 9:39 AM, David Carlson wrote:
Does this get you the rest of the way?
mfoo2 - unlist(mfoo, recursive = FALSE)
names(mfoo2) - unlist(lapply(mfoo, names))
class(mfoo2) - foolist
str(mfoo2)
List of 4
$ A1:List of 2
..$ x: int 3
..$ y: int 10
..- attr(*, class)= chr foo
$ A2:List of 2
..$ x: int [1:2] 6 4
..$ y: int [1:2] 8 9
..- attr(*, class)= chr foo
$ B1:List of 2
..$ x: int 2
..$ y: int 2
..- attr(*, class)= chr foo
$ B2:List of 2
..$ x: int [1:2] 3 6
..$ y: int [1:2] 4 2
..- attr(*, class)= chr foo
- attr(*, class)= chr foolist
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Ista Zahn
Sent: Thursday, October 17, 2013 8:23 AM
To: Michael Friendly
Cc: R-help
Subject: Re: [R] flatten a list of lists
unlist(mfoo, recursive = FALSE) gets you pretty close.
Best,
Ista
On Thu, Oct 17, 2013 at 9:15 AM, Michael Friendly
frien...@yorku.ca wrote:
I have functions that generate lists objects of class foo
and lists of
lists of these, of class
foolist, similar to what is shown below.
How can I flatten something like this to remove the top-level
list
structure, i.e.,
return a single-level list of foo objects, of class
foolist?
foo - function(n) {
result - list(x=sample(1:10,n), y=sample(1:10,n))
class(result) - foo
result
}
multifoo - function(vec, label, ...) {
result - lapply(vec, foo, ...)
names(result) - paste0(label, vec)
class(result) - foolist
result
}
foo1 - multifoo(1:2, A)
foo2 - multifoo(1:2, B)
mfoo - list(A=foo1, B=foo2)
str(mfoo, 2)
str(mfoo, 2)
List of 2
$ A:List of 2
..$ A1:List of 2
.. ..- attr(*, class)= chr foo
..$ A2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
$ B:List of 2
..$ B1:List of 2
.. ..- attr(*, class)= chr foo
..$ B2:List of 2
.. ..- attr(*, class)= chr foo
..- attr(*, class)= chr foolist
In this case, what is wanted is a single-level list, of 4 foo
objects, A1,
A2, B1, B2,
all of class foolist
--
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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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
code.
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PLEASE do read the posting guide
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code.
--
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] Lattice xyplot: Fill Legend Points
When I specify pch = 19 for a scatter plot the points are filled circles. Deapite reading ?points and trial-and-error experimentation I have not found how to have the legend symbols (now open circles) filled. An example command is: xyplot(pct.quant ~ sampdate, data = ffg.st, groups = func_feed_grp, type = 'p', pch = 19, key = simpleKey(text = levels(ffg.st$func_feed_grp), space = 'right', points = T, lines = F),par.settings = list(superpose.points = list(col = rainbow(7)), superpose.lines = list(col = rainbow(7))), main = 'Functional Feeding Groups (Individuals)', xlab = 'Year', ylab = 'Proportion of Individuals') Please pass me a pointer on how to fill the legend points. TIA, Rich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice xyplot: Fill Legend Points
put the pch into the par.settings On Thu, Oct 17, 2013 at 11:17 AM, Rich Shepard rshep...@appl-ecosys.com wrote: When I specify pch = 19 for a scatter plot the points are filled circles. Deapite reading ?points and trial-and-error experimentation I have not found how to have the legend symbols (now open circles) filled. An example command is: xyplot(pct.quant ~ sampdate, data = ffg.st, groups = func_feed_grp, type = 'p', pch = 19, key = simpleKey(text = levels(ffg.st$func_feed_grp), space = 'right', points = T, lines = F),par.settings = list(superpose.points = list(col = rainbow(7)), superpose.lines = list(col = rainbow(7))), main = 'Functional Feeding Groups (Individuals)', xlab = 'Year', ylab = 'Proportion of Individuals') Please pass me a pointer on how to fill the legend points. TIA, Rich __ 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] saveXML() prefix argument
Milan is correct.
The prefix is used when saving the XML content that is represented in
a different format in R.
To get the prefix
?xml version=1.0?
on the XML content that you save, use a document object
doc = newXMLDoc()
root = newXMLNode(foo, doc = doc)
saveXML(doc)
?xml version=1.0?
foo/
Sorry for the confusion.
D
On 10/17/13 2:36 AM, Milan Bouchet-Valat wrote:
Le mercredi 16 octobre 2013 à 23:45 -0400, Earl Brown a écrit :
I'm using the XML package and specifically the saveXML() function but I
can't get the prefix argument of saveXML() to work:
library(XML)
concepts - c(one, two, three)
info - c(info one, info two, info three)
root - newXMLNode(root)
for (i in 1:length(concepts)) {
cur.concept - concepts[i]
cur.info - info[i]
cur.tip - newXMLNode(tip, attrs = c(id = i), parent = root)
newXMLNode(h1, cur.concept, parent = cur.tip)
newXMLNode(p, cur.info, parent = cur.tip)
}
# None of the following output a prefix on the first line of the exported
document
saveXML(root)
saveXML(root, file = test.xml)
saveXML(root, file = test.xml, prefix = '?xml version=1.0?\n')
Am I missing something obvious? Any ideas?
It looks like the function XML:::saveXML.XMLInternalNode() does not use
the 'prefix' parameter at all. So it won't be taken into account when
calling saveXML() on objects of class XMLInternalNode.
I think you should report this to Duncan Temple Lang, as this is
probably an oversight.
Regards
Thanks in advance. Earl Brown
-
Earl K. Brown, PhD
Assistant Professor of Spanish Linguistics
Advisor, TEFL MA Program
Department of Modern Languages
Kansas State University
www-personal.ksu.edu/~ekbrown
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Re: [R] Help with combining functions
This is the world-famous fizzbuzz problem. You should be able to find
lots of implementations by Googling that word. Here's a pointless
collection I wrote once:
# a really dumb fizzbuzz alg competition
#fbfun1 is 2.5x faster than fbfun2
# fbfun3 is 10x faster than fbfun1
# fbfun1 is 2x faster than fbfun4
# fbfun5 is 20x faster than fbrun3
# Those are user times; in most cases the system time is very small indeed.
fbfun1 - function(xfoo) {
xfoo-1:xfoo
fbfoo - 1+(!as.logical(mod(xfoo,3)))*(as.logical(mod(xfoo,5))) +
2*(as.logical(mod(xfoo,3)))*(!as.logical(mod(xfoo,5)))+3*(!as.logical(mod(xfoo,3)))*(!as.logical(mod(xfoo,5)))
fbbar - unlist(lapply(fbfoo, function(x)
switch(x,0,'fizz','buzz','fizzbuzz')))
return(fbbar)
}
fbfun3 - function(xfoo) {
xfoo-1:xfoo
fbfoo - 1+(!as.logical(mod(xfoo,3)))*(as.logical(mod(xfoo,5))) +
2*(as.logical(mod(xfoo,3)))*(!as.logical(mod(xfoo,5)))+3*(!as.logical(mod(xfoo,3)))*(!as.logical(mod(xfoo,5)))
fbtab-cbind(1:4,c('','fizz','buzz','fizzbuzz'))
fbbar - fbtab[fbfoo,2]
return(fbbar)
}
# can I do it with recycled vectors, e.g. c('','','fizz') and
c('','','','','buzz') ?
fbfun4 - function(xfoo) {
fiz- rep(c('','','fizz'),length.out=xfoo)
buz-rep(c('','','','','buzz'),length.out=xfoo)
fbbar - unlist(lapply(1:xfoo, function(j)paste(fiz[j],buz[j]) ) )
return(fbbar)
}
# or completely sleazy:
fbfun5 - function(xfoo) {
fiz-
rep(c('','','fizz','','buzz','fizz','','','fizz','buzz','','fizz','','','fizzbuzz'),length.out=xfoo)
return(fiz)
}
--
View this message in context:
http://r.789695.n4.nabble.com/Help-with-combining-functions-tp4678212p4678272.html
Sent from the R help mailing list archive at Nabble.com.
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[R] S4 base class
Quote
By the way, your use of the syntax D::f and B::f suggests that youapos;re
thinking from a C++ point of view. Thatapos;s very likely to lead to
frustration: the S4 object system is very different from C++. Methods
donapos;t belong to classes, they belong to generics. There is no such thing
as D::f or B::f, only f methods with different signatures.
Duncan Murdoch
#---#
I am aware of this.
We can probably agree that we should use S4 classes and generic functions to
duplicate more usual object oriented architecture as far as possible while
remaining conscious of the regrettable differences.
For example we can pretend we are defining a virtual function in class Base by
writing:
setGeneric(F,
function(this) standardGeneric(F)
)
where the code for Base is, even though it has nothing to do with the class
Base.
We can even use it in other functions defined in class Base by writing
setGeneric(G,
function(this) standardGeneric(G)
)
setMethod(G,
signature(this=Base),
definition=function(this){
F(this)
})
which will work on all derived classes which implement F in some fashion:
setMethod(F,
signature(this=Derived),
definition=function(this){
# do something appropriate for derived.
})
With this we can reproduce some semblance of object oriented programming
However, apparently we cannot solve in this manner a common problem of object
oriented programming (from now on C++ parlance):
Suppose you have a base class Base which implements a function Base::F
which works in most contexts but not in the context of ComplicatedDerived
class
where some preparation has to happen before this very same function can be
called.
You would then define
void ComplicatedDerived::F(...){
preparation();
Base::F();
}
You can nealry duplicate this in R via
setMethod(F,
signature(this=ComplicatedDerived),
definition=function(this){
preparation(this)
F(as(this,Base))
})
but it will fail whenever F uses virtual functions (i.e. generics) which are
only defined
for derived classes of Base, whereas this is not a problem at all in normal
object oriented
languages.
This is not a contrived problem but is rather basic.
I wonder if you can do it in R in some other way.
Many thanks,
Michael
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[R] Subseting a data.frame
Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given basel_asset_class is 2, i.e. I need to obtain subset of above data.frame as (since there is only 1 record, against basel_asset_class = 2, I want to filter it) mydat_a basel_asset_class defa_frequency 1 8 0.070 2 8 0.030 3 8 0.001 Kindly guide Katherine [[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] Comparing two groups
So why not start with some statistical textbook? There are plenty of them available in CRAN. I wasn't implying, that I haven't read any textbook, or didn't do any research. I read some textbooks/Papers/etc. during the research about what to do and came across the wilcox test. I meant to imply that I could have problems understanding some of the answers, and that maybe additional explaining would be necessary. My doubts stem from the fact, that the wilcox test is a - as far as I know - ranking test, that states if two groups are different. My assumption is, due to the fact that the second group has a much higher sample size, it is clear that it differs from the first group. I performed a t-test (just to see; I am aware that I am not allowed to perform it, because my samples aren't normally distributed) and it gave me a p-value of 0.3. Actually I am not even entirely sure, if wilcox is the right test. I just want to know if the means of the two groups are significantly different. -- View this message in context: http://r.789695.n4.nabble.com/Comparing-two-groups-tp4678190p4678277.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] Lattice xyplot: Fill Legend Points
On Thu, 17 Oct 2013, Richard M. Heiberger wrote: put the pch into the par.settings Richard, Tried this again, but I'm not finding the proper location within par.settings. par.settings = list(superpose.points = list(col = rainbow(7)), superpose.lines = list(col = rainbow(7)), pch = 19) If I put it prior to the (list ... group there's an error of an extra = ; when I put it anywhere in the list (the above is one of my tries), it has no effect on the legend symbols: they remain as outlines. What have I missed? Thanks, Rich -- Richard B. Shepard, Ph.D. | Have knowledge, will travel. Applied Ecosystem Services, Inc. | http://www.appl-ecosys.com Voice: 503-667-4517 Fax: 503-667-8863 __ 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] Subseting a data.frame
Katherine, There are multiple ways to do this and I highly recommend you look into a basic R manual or search the forums. One quick example would be: mysub - subset(mydat, basel_asset_class 2) Cheers, Charles On Thu, Oct 17, 2013 at 1:55 AM, Katherine Gobin katherine_go...@yahoo.comwrote: Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given basel_asset_class is 2, i.e. I need to obtain subset of above data.frame as (since there is only 1 record, against basel_asset_class = 2, I want to filter it) mydat_a basel_asset_class defa_frequency 1 8 0.070 2 8 0.030 3 8 0.001 Kindly guide Katherine [[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. -- Charles Determan Integrated Biosciences PhD Candidate 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] map with inset
Hi David, That worked brilliantly! Many thanks. I also had trouble getting subplot() to work with either TeachingDemos or Hmisc. Best, Mark -- View this message in context: http://r.789695.n4.nabble.com/map-with-inset-tp4678341p4678426.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Weighted regression markers on scatter plots
Hi all, I'm trying to graph the results of a weighted regression analysis. Is anyone aware of a way to make my markers appear a different sizes to be consistent with their respective weights? Thanks, -Mike Sugarman Wayne State University -- View this message in context: http://r.789695.n4.nabble.com/Weighted-regression-markers-on-scatter-plots-tp4678370.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] S4 base class
Greetings, I have an S4 class B (Base) which defines a function f=f(this=B,...) Dervided from B we have a derived class D which also defines a function f=f(this=D,...) In the definition of D::f we want to call the version B::f and could do this by simply calling f(baseClassObject(this),...) The question is the following: How do I refer to the base class object from the derived class? Many thanks Michael Meyer __ 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] Singular Matrix 'a' in solve
Hi,
I have a set of matrix data named invest consists of 450 observations (75
countries, 6 years) with 7 variables (set as I, pop, inv, gov, c, life, d;
which each is numeric[450]). The procedure is modify from code provided
by B.E. Hansen at http://www.ssc.wisc.edu/~bhansen/progs/ecnmt_00.html.
*Then the variable is being transformed to*
y- lag_v(i,0)
cf - lag_v(c,0)
lpop - lag_v(pop,0)
linv- lag_v(inv,0)
lgov - lag_v(gov,0)
d1 - lag_v(d,0)
llife- lag_v(life,0)
yt - tr(y)
ct - tr(cf)
y, cf, lpop, linv, lgov, d1, llife each is in 375x1 double matrix
yt and ct each is 300x1 double matrix
(I use R Studio so these characteristics are stated).
*The lag_v() and tr() process is as below:*
max_lag - 1
tt - t-max_lag
ty - n*(t-max_lag-1)
lag_v - function(x,lagn){
yl - matrix(c(0),nrow=n,ncol=t)
for (i in 1:n) {
yl[i,]-x[(1+(i-1)*t):(t*i)]
}
yl - yl[,(1+max_lag-lagn):(t-lagn)]
out - matrix(t(yl),nrow=nrow(yl)*ncol(yl),ncol=1)
out
}
tr - function(y){
yf - matrix(c(0),nrow=n,ncol=tt)
for (i in 1:n) {
yf[i,]-y[(1+(i-1)*tt):(tt*i)]
}
yfm - yf- colMeans(t(yf))
yfm - yfm[,1:(tt-1)]
out - matrix(t(yfm),nrow=nrow(yfm)*ncol(yfm),ncol=1)
out
}
*Then before the computation, something is being setup*
x - cbind(lpop, linv, lgov, llife, cf)
(skip as I think is unrelated with the problem encounter)
*And, in the early stage of computation:*
sse_calc - function(y,x){
e - y-x%*%qr.solve(x,y)
out - t(e)%*%e
out
}
*It comes out with*
Error in qr.solve(x, y) : singular matrix 'a' in solve
I thought only square matrix would have this kind of problem. Would qr()
help in this case? Or is there any other possible solution for this problem?
Thanks.
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[R] representing points in 3D space with trajectories over time
Dear all, I have a problem where I must represent points with XYZ coordinates changing over time. I will do a number of operations on this data such as calculating the YZ-projection distance of the points to the origin over time, the frequency spectrum of the X-T data etc. I am trying to find a good way of representing this data with an appropriate data structure. It appears like higher-dimensional data frames are not allowed and I do not know if I should use a list of data frames or if there is a better solution, possibly as part of an external package. Thank you for your time Umut Toprak [[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] Incorporate Julia into R
Hi, I have some code in R with a lot of matrix multiplication and inverting. R can be very slow for larger matrices like 5000x5000. I have seen the new programming language Julia (www.julialang.org) which is quite fast in doing matrix algebra. So my idea is to set up the simulations in R and start the first calculations, then I want to give some objects to Julia and do there some matrix algebra and give the results back to R. Is this possible or does anybody know how to do this? Is there a package available? A short example with some lines of code would be also very helpful. Thanks in advance, Timo [[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] match values in dependence of ID and Date
Hi, I think based on your title, the output you provided is not clear. If it depends on Date, there should be four columns. library(reshape2) res1 - dcast(merge(dat,dat2,by=ID),ID+Name~Date,value.var=Value) colnames(res1)[3:6] - c(First, Second, Third, Fourth) rownames(res1) - 1:nrow(res1) #or res2 - reshape(merge(dat,dat2,by=ID),idvar=c(ID,Name),timevar=Date,direction=wide) dimnames(res2) - dimnames(res1) res2 # ID Name First Second Third Fourth #1 1 Andy 10 15 NA NA #2 2 John 7 NA 10 15 #3 3 Amy 10 NA NA NA A.K. On Thursday, October 17, 2013 9:31 AM, arun smartpink...@yahoo.com wrote: Hi, Try: dat - read.table(text= ID Name 1 Andy 2 John 3 Amy,sep=,header=TRUE,stringsAsFactors=FALSE) dat2 - read.table(text= ID Date Value 1 2013-10-01 10 1 2013-10-02 15 2 2013-10-01 7 2 2013-10-03 10 2 2013-10-04 15 3 2013-10-01 10,sep=,header=TRUE,colClasses=c(numeric,Date,numeric)) library(plyr) res - reshape(ddply(merge(dat,dat2,by=ID),.(ID),mutate,id=((seq_along(ID)-1)%%3+1))[,-3],idvar=c(ID,Name),timevar=id,direction=wide) rownames(res) - 1:nrow(res) colnames(res)[3:5] - c(First, Second, Third) res # ID Name First Second Third #1 1 Andy 10 15 NA #2 2 John 7 10 15 #3 3 Amy 10 NA NA A.K. On Thursday, October 17, 2013 7:42 AM, Mat matthias.we...@fnt.de wrote: hello togehter, i have a little problem, maybe you can help me. I have a data.frame like this one: ID Name 1 Andy 2 John 3 Amy and a data.frame like this: ID Date Value 1 2013-10-01 10 1 2013-10-02 15 2 2013-10-01 7 2 2013-10-03 10 2 2013-10-04 15 3 2013-10-01 10 the result should be this one: ID Name First Second Third 1 Andy 10 15 2 John 7 10 15 3 Amy 10 maybe you can help me, to do this? Thank you. Mat -- View this message in context: http://r.789695.n4.nabble.com/match-values-in-dependence-of-ID-and-Date-tp4678433.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice xyplot: Fill Legend Points
par.settings = list( superpose.points = list(col = rainbow(7), pch = 19), superpose.lines = list(col = rainbow(7)) ) On Thu, Oct 17, 2013 at 11:48 AM, Rich Shepard rshep...@appl-ecosys.com wrote: On Thu, 17 Oct 2013, Richard M. Heiberger wrote: put the pch into the par.settings Richard, Tried this again, but I'm not finding the proper location within par.settings. par.settings = list(superpose.points = list(col = rainbow(7)), superpose.lines = list(col = rainbow(7)), pch = 19) If I put it prior to the (list ... group there's an error of an extra = ; when I put it anywhere in the list (the above is one of my tries), it has no effect on the legend symbols: they remain as outlines. What have I missed? Thanks, Rich -- Richard B. Shepard, Ph.D. | Have knowledge, will travel. Applied Ecosystem Services, Inc. | http://www.appl-ecosys.com Voice: 503-667-4517 Fax: 503-667-8863 __ 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] Lattice xyplot: Fill Legend Points
On Thu, 17 Oct 2013, Richard M. Heiberger wrote: par.settings = list( superpose.points = list(col = rainbow(7), pch = 19), superpose.lines = list(col = rainbow(7)) ) I had tried that, too. Legend symbols stubbornly remain unfilled. Thanks, Richard, Rich __ 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] match values in dependence of ID and Date
Hi, Try: dat - read.table(text= ID Name 1 Andy 2 John 3 Amy,sep=,header=TRUE,stringsAsFactors=FALSE) dat2 - read.table(text= ID Date Value 1 2013-10-01 10 1 2013-10-02 15 2 2013-10-01 7 2 2013-10-03 10 2 2013-10-04 15 3 2013-10-01 10,sep=,header=TRUE,colClasses=c(numeric,Date,numeric)) library(plyr) res - reshape(ddply(merge(dat,dat2,by=ID),.(ID),mutate,id=((seq_along(ID)-1)%%3+1))[,-3],idvar=c(ID,Name),timevar=id,direction=wide) rownames(res) - 1:nrow(res) colnames(res)[3:5] - c(First, Second, Third) res # ID Name First Second Third #1 1 Andy 10 15 NA #2 2 John 7 10 15 #3 3 Amy 10 NA NA A.K. On Thursday, October 17, 2013 7:42 AM, Mat matthias.we...@fnt.de wrote: hello togehter, i have a little problem, maybe you can help me. I have a data.frame like this one: ID Name 1 Andy 2 John 3 Amy and a data.frame like this: ID Date Value 1 2013-10-01 10 1 2013-10-02 15 2 2013-10-01 7 2 2013-10-03 10 2 2013-10-04 15 3 2013-10-01 10 the result should be this one: ID Name First Second Third 1 Andy 10 15 2 John 7 10 15 3 Amy 10 maybe you can help me, to do this? Thank you. Mat -- View this message in context: http://r.789695.n4.nabble.com/match-values-in-dependence-of-ID-and-Date-tp4678433.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot - how to vary the distances of the x axis?
Hello, my dots of 0 and 2 are quite close to the marging. So I would like to move the 0 and the 2 both towards the 1. I wish to be my dots more centered. And: I dont need so much space between 0,1 and 2. How does it work? I tried: plot (data, axes=FALSE, main=i, ylab= expression (z^2)) plot.window (xlim=c (0,2), ylim=c(0,80)) box (lwd=2) axis (side=1, at = c (0,1,2)) axis (side =2) dput (data) structure(list(Genotype = c(0, 0, 0, 1, 1, 1, 1, 1, 2), z = c(0.66429502114682, 0.258444359570075, 0.0702937908415368, 0.694376498254858, 0.0967863570760579, 0.213966209301163, 0.671497050546114, 0.60318070802847, 75.6011068681301 )), .Names = c(Genotype, z), row.names = c(NA, 9L), class = data.frame) Thanks attachment: move.png__ 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] Subseting a data.frame
Kindly guide ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given basel_asset_class is 2, i.e. I need to obtain subset of above data.frame as (since there is only 1 record, against basel_asset_class = 2, I want to filter it) mydat_a basel_asset_class defa_frequency 1 8 0.070 2 8 0.030 3 8 0.001 Kindly guide Katherine [[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 (650) 467-7374 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot - how to vary the distances of the x axis?
On 17 Oct 2013, at 13:44 , Hermann Norpois wrote: Hello, my dots of 0 and 2 are quite close to the marging. So I would like to move the 0 and the 2 both towards the 1. I wish to be my dots more centered. And: I dont need so much space between 0,1 and 2. How does it work? I tried: plot (data, axes=FALSE, main=i, ylab= expression (z^2)) plot.window (xlim=c (0,2), ylim=c(0,80)) box (lwd=2) axis (side=1, at = c (0,1,2)) axis (side =2) dput (data) structure(list(Genotype = c(0, 0, 0, 1, 1, 1, 1, 1, 2), z = c(0.66429502114682, 0.258444359570075, 0.0702937908415368, 0.694376498254858, 0.0967863570760579, 0.213966209301163, 0.671497050546114, 0.60318070802847, 75.6011068681301 )), .Names = c(Genotype, z), row.names = c(NA, 9L), class = data.frame) Thanks If I understand what you want, set xlim() a bit wider, within in the plot-statement: xlim=c (-0.4,2.4), ylim=c(0,80) Hope this helps, Best wishes, Franklin - Dr. Franklin Bretschneider Dept of Biology Utrecht Unversity Padualaan 8 3584 CH Utrecht The Netherlands f.bretschnei...@uu.nl [[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] Lattice xyplot: Fill Legend Points
That should have worked. I think something else is interfering. Did you redefine either T or F? Please send the output from dput(head(ffg.st)) so we can experiment in your setting. Rich On Thu, Oct 17, 2013 at 12:12 PM, Rich Shepard rshep...@appl-ecosys.com wrote: On Thu, 17 Oct 2013, Richard M. Heiberger wrote: par.settings = list( superpose.points = list(col = rainbow(7), pch = 19), superpose.lines = list(col = rainbow(7)) ) I had tried that, too. Legend symbols stubbornly remain unfilled. Thanks, Richard, Rich __ 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] Constraint on regression parameters
You want the offset function in the formula: lm( A ~ B + I(B^2) + offset(C), data=Dataset) This will force the coefficient on C to be 1, if you wanted a coefficient of another value then just do the multiplication yourself, e.g. offset( 2 * C ) for a slope of 2. Also you can use poly(B,2) to fit a linear and quadratic terms on B. On Thu, Oct 17, 2013 at 3:45 AM, Robert U tacsun...@yahoo.fr wrote: Dear all, I have been trying to find a simple solution to my problem without success, though i have a feeling a simple syntaxe detail coul make the job. I am doing a polynomial linear regression with 2 independent variables such as : lm(A ~ B + I(B^2) + I(lB^3) + C, data=Dataset)) R return me a coefficient per independent variable, and I would need the coefficient of the C parameter to equal 1. I've been loonking at parameter constraints on the internet but it's always much more complicated that just removing the fit of a coefficient (or setting it to 1). I know many package allows to not fit an intercept with a -1 parameter in the syntaxe, does that exists for independent variables ? Regards, [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com [[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] Selecting maximums between different variables
Hi there,
another beginners question, I'm afraid. Basically i want to selct the
maximum of values, that correspond to different variables. I have a table
of oil production that looks somewhat like this:
oil - data.frame( YEAR = c(2011, 2012),
TX = c(2, 3),
CA = c(4, 25000),
AL = c(2,
21000),
ND = c(21000,6))
Now I want to find out, which state produced most oil in a given year. I
tried this:
attach(oil)
last_year = oil[ c(YEAR == 2012), ]
max(last_year)
Which works, but it doesnt't give me the corresponding values (i.e. it just
gives me the maximum output, not what state its from).
So I tried this:
oil[c(oil == max(last_year)),]
and this:
oil[c(last_year == max(last_year)),]
and this:
oil[which.max(last_year),]
and this:
last_year[max(last_year),]
None of them work, but they don't give error messages either, the output is
just NA. The problem is, in my eyes, that I'm comparing the values of
different variables with each other. Because if i change the structure of
the dataframe (which I can't do with the real data, at least not with out
doing it by hand with a huge dataset), it looks like this and works
perfectly:
oil2 - data.frame (
names = c('YEAR', 'TX', 'CA', 'AL', 'ND'),
oil_2011 = c(2011, 2, 4, 2, 21000),
oil_2012 = c(2012, 3, 25000, 21000, 6)
)
attach(oil2)
oil2[c(oil_2012 == max(oil_2012)),]
Any help is much appreciated.
Thanks, Tim Umbach
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Re: [R] Lattice xyplot: Fill Legend Points
On Thu, 17 Oct 2013, Richard M. Heiberger wrote: That should have worked. That's what I thought when I first tried it. I think something else is interfering. Did you redefine either T or F? Not intentionally. Please send the output from dput(head(ffg.st)) so we can experiment in your setting. structure(list(sampdate = structure(c(13326, 13326, 13326, 13326, 13326, 13326), class = Date), func_feed_grp = structure(c(1L, 2L, 3L, 4L, 6L, 7L), .Label = c(Filterer, Gatherer, Grazer, Omnivore, Parasite, Predator, Shredder), class = factor), quant = c(812L, 1880L, 624L, 11L, 948L, 1540L), pct.quant = c(0.14, 0.323, 0.107, 0.002, 0.163, 0.265), num.taxa = c(11L, 28L, 4L, 1L, 12L, 3L), pct.num.taxa = c(0.186, 0.475, 0.068, 0.017, 0.203, 0.051)), .Names = c(sampdate, func_feed_grp, quant, pct.quant, num.taxa, pct.num.taxa), row.names = 102:107, class = data.frame) Rich -- Richard B. Shepard, Ph.D. | Have knowledge, will travel. Applied Ecosystem Services, Inc. | http://www.appl-ecosys.com Voice: 503-667-4517 Fax: 503-667-8863 __ 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] Comparing two groups
From your question it is not clear what your question/concerns really are, and from what we can see it could very well be that you do not understand the statistics that you are computing (not just the R implementation). We ask for a reproducible example because that helps us to help you, just a couple of boxplots let us make some guesses, but we do not know the data values or even the means and standard deviations, even the actual sample sizes could help. From the graph it is not surprising that the wilcox test say that the 2 groups are different and that the t test says that they are not (but knowing data values would help even more). The 2 tests are testing very different hypotheses. The wilcox test is testing that the 2 distributions are identical and the more specific way it tests that is by looking at all possible pairs between the 2 groups and seeing what proportion of them have each group higher, if the null were true then half the time the data point from mixed would be higher than the data point from monoculture and half the time the other way. From the boxplot we can see that the median of monoculture is below the 1st quartile of mixed, so it is not surprising at all that the wilcox test rejects the null hypothesis. The t-test (which version you used you do not say) is testing if the means are equal, since monculture is clearly skewed to the right with potential outliers, it would not be surprising if the sample means were close enough to each other that the t-test does not see a significant difference. The 2 tests give different answers because they are answering very different questions. You state that I am not allowed to perform it referring to the t-test. This indicates that you don't have a full understanding or appreciation of the Central Limit Theorem (an important enough theorem that I have a cross-stitch based on it hanging on my wall (along with 2 other cross-stitches of Bayes theorem and the mean value theorem of integration)). The plot shows 18 outliers in the monoculture group which implies a sample size of at least 72, which means the other group has a sample size of at least 14 if I interpret five times as big correctly. This is a large enough sample size for the CLT to tell us the t-test will give a reasonable approximation (provided the other assumptions hold reasonably well and you are interested in the question being answered). So, I believe that the advice to read a textbook, or otherwise get some help in basic understanding of the statistical tools is reasonable. Once you have that, then if you still need help then give us a reproducible example and make it clear what your question really is and you will be much more likely to receive an answer. On Tue, Oct 15, 2013 at 6:01 AM, Andrej andrej.g.mil...@web.de wrote: So why not start with some statistical textbook? There are plenty of them available in CRAN. I wasn't implying, that I haven't read any textbook, or didn't do any research. I read some textbooks/Papers/etc. during the research about what to do and came across the wilcox test. I meant to imply that I could have problems understanding some of the answers, and that maybe additional explaining would be necessary. My doubts stem from the fact, that the wilcox test is a - as far as I know - ranking test, that states if two groups are different. My assumption is, due to the fact that the second group has a much higher sample size, it is clear that it differs from the first group. I performed a t-test (just to see; I am aware that I am not allowed to perform it, because my samples aren't normally distributed) and it gave me a p-value of 0.3. Actually I am not even entirely sure, if wilcox is the right test. I just want to know if the means of the two groups are significantly different. -- View this message in context: http://r.789695.n4.nabble.com/Comparing-two-groups-tp4678190p4678277.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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com [[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] RWeka and multicore package
I work very mutch with the packages RWeka and multicore. If you try to run
J48 or any tree of RWeka with multicore we hava some errors.
Example I:
library(RWeka);
library(multicore);
mclapply(1:100, function(i) {
J48(Species ~., iris);
});
Output: Error in .jcall(o, \Ljava/lang/Class;\, \getClass\) : \n
java.lang.ClassFormatError: Incompatible magic value 1347093252 in class
file java/lang/ProcessEnvironment$StringEnvironment\n
Example II:
library(multicore);
mclapply(1:100, function(i) {
RWeka::J48(Species ~., iris);
});
Output: Erro em .jcall(x$classifier, S, toString) :
RcallMethod: attempt to call a method of a NULL object.
Do you know some way to work with parallel processing and RWeka? I tried
MPI and SNOW without success.
R version 3.0.2 (2013-09-25) -- Frisbee Sailing
Ubuntu 12.04 x64
--
Luís Paulo Faina Garcia
Engenheiro de Computação - Universidade de São Paulo
São Carlos - SP - Brasil
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Re: [R] S4 base class
On 10/17/2013 08:54 AM, Michael Meyer wrote:
Suppose you have a base class Base which implements a function Base::F
which works in most contexts but not in the context of ComplicatedDerived
class
where some preparation has to happen before this very same function can be
called.
You would then define
void ComplicatedDerived::F(...){
preparation();
Base::F();
}
You can nealry duplicate this in R via
setMethod(F,
signature(this=ComplicatedDerived),
definition=function(this){
preparation(this)
F(as(this,Base))
})
but it will fail whenever F uses virtual functions (i.e. generics) which are
only defined
for derived classes of Base
With
.A - setClass(A, representation(a=numeric))
.B - setClass(B, representation(b=numeric), contains=A)
setGeneric(f, function(x, ...) standardGeneric(f))
setMethod(f, A, function(x, ...) {
message(f,A-method)
g(x, ...) # generic with methods only for derived classes
})
setMethod(f, B, function(x, ...) {
message(f,B-method)
callNextMethod(x, ...) # earlier response from Duncan Murdoch
})
setGeneric(g, function(x, ...) standardGeneric(g))
setMethod(g, B, function(x, ...) {
message(g,B-method)
x
})
one has
f(.B())
f,B-method
f,A-method
g,B-method
An object of class B
Slot b:
numeric(0)
Slot a:
numeric(0)
?
--
Computational Biology / Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N.
PO Box 19024 Seattle, WA 98109
Location: Arnold Building M1 B861
Phone: (206) 667-2793
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Re: [R] RWeka and multicore package
I received the following error message with the multicore package:
install.packages(multicore)
Warning in install.packages :
package multicore is not available (for R version 3.0.2)
Warning in install.packages :
package multicore is not available (for R version 3.0.2)
Warning message:
package multicore is not available (for R version 3.0.2)
With gratitude,
CEO'Riley Jr.
Charles Ellis O'Riley Jr.
Ambition is a state of permanent dissatisfaction with the present
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Luís Paulo F. Garcia
Sent: Thursday, October 17, 2013 12:22 PM
To: r-help@r-project.org
Subject: [R] RWeka and multicore package
I work very mutch with the packages RWeka and multicore. If you try to run
J48 or any tree of RWeka with multicore we hava some errors.
Example I:
library(RWeka);
library(multicore);
mclapply(1:100, function(i) {
J48(Species ~., iris);
});
Output: Error in .jcall(o, \Ljava/lang/Class;\, \getClass\) : \n
java.lang.ClassFormatError: Incompatible magic value 1347093252 in class
file java/lang/ProcessEnvironment$StringEnvironment\n
Example II:
library(multicore);
mclapply(1:100, function(i) {
RWeka::J48(Species ~., iris);
});
Output: Erro em .jcall(x$classifier, S, toString) :
RcallMethod: attempt to call a method of a NULL object.
Do you know some way to work with parallel processing and RWeka? I tried MPI
and SNOW without success.
R version 3.0.2 (2013-09-25) -- Frisbee Sailing
Ubuntu 12.04 x64
--
Lums Paulo Faina Garcia
Engenheiro de Computagco - Universidade de Sco Paulo Sco Carlos - SP -
Brasil
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Selecting maximums between different variables
Hi,
You may try:
unlist(lapply(seq_len(nrow(oil)),function(i) oil[i,-1][which.max(oil[i,-1])]))
# CA ND
#4 6
#or
library(reshape2)
datM - melt(oil,id.var=YEAR)
datM[as.logical(with(datM,ave(value,list(YEAR),FUN= function(x) x%in%
max(x,]
# YEAR variable value
#3 2011 CA 4
#8 2012 ND 6
A.K.
On Thursday, October 17, 2013 12:50 PM, Tim Umbach tim.umb...@hufw.de wrote:
Hi there,
another beginners question, I'm afraid. Basically i want to selct the
maximum of values, that correspond to different variables. I have a table
of oil production that looks somewhat like this:
oil - data.frame( YEAR = c(2011, 2012),
TX = c(2, 3),
CA = c(4, 25000),
AL = c(2,
21000),
ND = c(21000,6))
Now I want to find out, which state produced most oil in a given year. I
tried this:
attach(oil)
last_year = oil[ c(YEAR == 2012), ]
max(last_year)
Which works, but it doesnt't give me the corresponding values (i.e. it just
gives me the maximum output, not what state its from).
So I tried this:
oil[c(oil == max(last_year)),]
and this:
oil[c(last_year == max(last_year)),]
and this:
oil[which.max(last_year),]
and this:
last_year[max(last_year),]
None of them work, but they don't give error messages either, the output is
just NA. The problem is, in my eyes, that I'm comparing the values of
different variables with each other. Because if i change the structure of
the dataframe (which I can't do with the real data, at least not with out
doing it by hand with a huge dataset), it looks like this and works
perfectly:
oil2 - data.frame (
names = c('YEAR', 'TX', 'CA', 'AL', 'ND'),
oil_2011 = c(2011, 2, 4, 2, 21000),
oil_2012 = c(2012, 3, 25000, 21000, 6)
)
attach(oil2)
oil2[c(oil_2012 == max(oil_2012)),]
Any help is much appreciated.
Thanks, Tim Umbach
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Weighted regression markers on scatter plots
The simplest approach is to specify the cex parameter in the call to plot. plot(1:3, 1:3, cex=3:1) for example will plot the 1st point 3 times as big, the 2nd 2 times as big, and the 3rd at the standard size. You can get more control by using the symbols function instead of the plot function and set the diameter of circles directly. In either case you probably want to scale by the square root of the weight. The my.symbols function in the TeachingDemos package is another option if the symbols function does not include the symbol you want or if you want a little different level of control. On Wed, Oct 16, 2013 at 11:04 AM, Msugarman mike.sugar...@wayne.edu wrote: Hi all, I'm trying to graph the results of a weighted regression analysis. Is anyone aware of a way to make my markers appear a different sizes to be consistent with their respective weights? Thanks, -Mike Sugarman Wayne State University -- View this message in context: http://r.789695.n4.nabble.com/Weighted-regression-markers-on-scatter-plots-tp4678370.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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com [[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] Selecting maximums between different variables
On 17-10-2013, at 18:48, Tim Umbach tim.umb...@hufw.de wrote:
Hi there,
another beginners question, I'm afraid. Basically i want to selct the
maximum of values, that correspond to different variables. I have a table
of oil production that looks somewhat like this:
oil - data.frame( YEAR = c(2011, 2012),
TX = c(2, 3),
CA = c(4, 25000),
AL = c(2,
21000),
ND = c(21000,6))
Now I want to find out, which state produced most oil in a given year. I
tried this:
attach(oil)
last_year = oil[ c(YEAR == 2012), ]
max(last_year)
For a single year do
year - which(oil[,YEAR]==2011)
oil[year,which.max(oil[year,]),drop=FALSE]
In the help look at base::[.data.frame
Berend
Which works, but it doesnt't give me the corresponding values (i.e. it just
gives me the maximum output, not what state its from).
So I tried this:
oil[c(oil == max(last_year)),]
and this:
oil[c(last_year == max(last_year)),]
and this:
oil[which.max(last_year),]
and this:
last_year[max(last_year),]
None of them work, but they don't give error messages either, the output is
just NA. The problem is, in my eyes, that I'm comparing the values of
different variables with each other. Because if i change the structure of
the dataframe (which I can't do with the real data, at least not with out
doing it by hand with a huge dataset), it looks like this and works
perfectly:
oil2 - data.frame (
names = c('YEAR', 'TX', 'CA', 'AL', 'ND'),
oil_2011 = c(2011, 2, 4, 2, 21000),
oil_2012 = c(2012, 3, 25000, 21000, 6)
)
attach(oil2)
oil2[c(oil_2012 == max(oil_2012)),]
Any help is much appreciated.
Thanks, Tim Umbach
[[alternative HTML version deleted]]
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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-help@r-project.org mailing list
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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] representing points in 3D space with trajectories over time
If all your data is numeric then you can use an array instead of a data frame and arrays can easily be 3, 4, or higher dimensional. Or you can use a data frame with a column each for x, y, z, and time; with possible other columns representing groups or other attributes, essentially a 3 dimensional data frame with the 3rd dimension being stacked rather than projecting out. On Thu, Oct 17, 2013 at 6:59 AM, Umut Toprak umut.top...@unige.ch wrote: Dear all, I have a problem where I must represent points with XYZ coordinates changing over time. I will do a number of operations on this data such as calculating the YZ-projection distance of the points to the origin over time, the frequency spectrum of the X-T data etc. I am trying to find a good way of representing this data with an appropriate data structure. It appears like higher-dimensional data frames are not allowed and I do not know if I should use a list of data frames or if there is a better solution, possibly as part of an external package. Thank you for your time Umut Toprak [[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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com [[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] Subseting a data.frame
I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat, mydat$basel_asset_class 2). What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, similarly there could be another basel asset class having say 5 default frequncies. Thus, I need to take subset of the data.frame s.t. the no of corresponding defa_frequencies is greater than 2. The idea is we try to fit exponential curve Y = A exp( BX ) for each of the basel asset classes and to estimate values of A and B, mathematically one needs to have at least two values of X. I hope I may be able to express my requirement. Its not that I need the subset of mydat s.t. basel asset class is 2 (now 4 in revised example), but sbuset s.t. no of default frequencies is greater than or equal to 2. This 2 is not same as basel asset class 2. Kindly guide With warm regards Katherine Gobin On Thursday, 17 October 2013 9:33 PM, Bert Gunter gunter.ber...@gene.com wrote: Kindly guide ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given basel_asset_class is 2, i.e. I need to obtain subset of above data.frame as (since there is only 1 record, against basel_asset_class = 2, I want to filter it) mydat_a basel_asset_class defa_frequency 1 8 0.070 2 8 0.030 3 8 0.001 Kindly guide Katherine [[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 (650) 467-7374 [[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] Subseting a data.frame
Correction. (2nd para first three lines) Pl read following line What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, as What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are THREE default frequencies w.r.t. basel aseet class 8, I alpologize for the incovenience. Regards KAtherine On , Katherine Gobin katherine_go...@yahoo.com wrote: I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat, mydat$basel_asset_class 2). What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, similarly there could be another basel asset class having say 5 default frequncies. Thus, I need to take subset of the data.frame s.t. the no of corresponding defa_frequencies is greater than 2. The idea is we try to fit exponential curve Y = A exp( BX ) for each of the basel asset classes and to estimate values of A and B, mathematically one needs to have at least two values of X. I hope I may be able to express my requirement. Its not that I need the subset of mydat s.t. basel asset class is 2 (now 4 in revised example), but sbuset s.t. no of default frequencies is greater than or equal to 2. This 2 is not same as basel asset class 2. Kindly guide With warm regards Katherine Gobin On Thursday, 17 October 2013 9:33 PM, Bert Gunter gunter.ber...@gene.com wrote: Kindly guide ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given basel_asset_class is 2, i.e. I need to obtain subset of above data.frame as (since there is only 1 record, against basel_asset_class = 2, I want to filter it) mydat_a basel_asset_class defa_frequency 1 8 0.070 2 8 0.030 3 8 0.001 Kindly guide Katherine [[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 (650) 467-7374 [[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] Lattice xyplot: Fill Legend Points
I always get lost in simpleKey. The approach of directly modifying the trellis object usually works. tmp - xyplot(pct.quant ~ sampdate, data = ffg.st, groups = func_feed_grp, type = + 'p', pch = 19, key = simpleKey(text = levels(ffg.st$func_feed_grp), space = + 'right', points = T, lines = F),par.settings = list(superpose.points = + list(col = rainbow(7)), superpose.lines = list(col = rainbow(7))), main = + 'Functional Feeding Groups (Individuals)', xlab = 'Year', ylab = 'Proportion + of Individuals') tmp str(tmp) tmp$legend$right$args$key$points$pch [1] 1 1 1 1 1 1 1 tmp$legend$right$args$key$points$pch[] - 19 tmp$legend$right$args$key$points$pch [1] 19 19 19 19 19 19 19 tmp Rich On Thu, Oct 17, 2013 at 12:57 PM, Rich Shepard rshep...@appl-ecosys.com wrote: On Thu, 17 Oct 2013, Richard M. Heiberger wrote: That should have worked. That's what I thought when I first tried it. I think something else is interfering. Did you redefine either T or F? Not intentionally. Please send the output from dput(head(ffg.st)) so we can experiment in your setting. structure(list(sampdate = structure(c(13326, 13326, 13326, 13326, 13326, 13326), class = Date), func_feed_grp = structure(c(1L, 2L, 3L, 4L, 6L, 7L), .Label = c(Filterer, Gatherer, Grazer, Omnivore, Parasite, Predator, Shredder), class = factor), quant = c(812L, 1880L, 624L, 11L, 948L, 1540L), pct.quant = c(0.14, 0.323, 0.107, 0.002, 0.163, 0.265), num.taxa = c(11L, 28L, 4L, 1L, 12L, 3L), pct.num.taxa = c(0.186, 0.475, 0.068, 0.017, 0.203, 0.051)), .Names = c(sampdate, func_feed_grp, quant, pct.quant, num.taxa, pct.num.taxa), row.names = 102:107, class = data.frame) Rich -- Richard B. Shepard, Ph.D. | Have knowledge, will travel. Applied Ecosystem Services, Inc. | http://www.appl-ecosys.com Voice: 503-667-4517 Fax: 503-667-8863 __ 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] S4 base class
@Martin Morgan, Duncan Murdoch: OK Thanks. I did not understand the callNextMethod. I will investigate this in detail. This is great! Thanks again, Michael Meyer __ 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] Subseting a data.frame
You may try: mydat[with(mydat,ave(seq_along(basel_asset_class),basel_asset_class,FUN=length)2),] # basel_asset_class defa_frequency #2 8 0.070 #3 8 0.030 #4 8 0.001 #or library(plyr) mydat[ddply(mydat,.(basel_asset_class),mutate,L=length(defa_frequency))[,3] 2,] #assuming it is sorted. A.K. On Thursday, October 17, 2013 1:59 PM, Katherine Gobin katherine_go...@yahoo.com wrote: I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat, mydat$basel_asset_class 2). What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, similarly there could be another basel asset class having say 5 default frequncies. Thus, I need to take subset of the data.frame s.t. the no of corresponding defa_frequencies is greater than 2. The idea is we try to fit exponential curve Y = A exp( BX ) for each of the basel asset classes and to estimate values of A and B, mathematically one needs to have at least two values of X. I hope I may be able to express my requirement. Its not that I need the subset of mydat s.t. basel asset class is 2 (now 4 in revised example), but sbuset s.t. no of default frequencies is greater than or equal to 2. This 2 is not same as basel asset class 2. Kindly guide With warm regards Katherine Gobin On Thursday, 17 October 2013 9:33 PM, Bert Gunter gunter.ber...@gene.com wrote: Kindly guide ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given basel_asset_class is 2, i.e. I need to obtain subset of above data.frame as (since there is only 1 record, against basel_asset_class = 2, I want to filter it) mydat_a basel_asset_class defa_frequency 1 8 0.070 2 8 0.030 3 8 0.001 Kindly guide Katherine [[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 (650) 467-7374 [[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] Lattice xyplot: Fill Legend Points
On Thu, 17 Oct 2013, Richard M. Heiberger wrote: I always get lost in simpleKey. As this is my first use of it I take what's offered by those more experienced than I. The approach of directly modifying the trellis object usually works. tmp - xyplot(pct.quant ~ sampdate, data = ffg.st, groups = func_feed_grp, type = + 'p', pch = 19, key = simpleKey(text = levels(ffg.st$func_feed_grp), space = + 'right', points = T, lines = F),par.settings = list(superpose.points = + list(col = rainbow(7)), superpose.lines = list(col = rainbow(7))), main = + 'Functional Feeding Groups (Individuals)', xlab = 'Year', ylab = 'Proportion + of Individuals') tmp str(tmp) tmp$legend$right$args$key$points$pch [1] 1 1 1 1 1 1 1 tmp$legend$right$args$key$points$pch[] - 19 tmp$legend$right$args$key$points$pch [1] 19 19 19 19 19 19 19 OK. More steps but it will get the plots where they need to be. Many thanks, Rich -- Richard B. Shepard, Ph.D. | Have knowledge, will travel. Applied Ecosystem Services, Inc. | http://www.appl-ecosys.com Voice: 503-667-4517 Fax: 503-667-8863 __ 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] Newb: How I find random vector index?
# Suppose I have a vector:
myvec = c(1,0,3,0,77,9,0,1,2,0)
# I want to randomly pick an element from myvec
# where element == 0
# and print the value of the corresponding index.
# So, for example I might randomly pick the 3rd 0
# and I would print the corresponding index
# which is 7,
# My initial approach is to use a for-loop.
# Also I take a short-cut which assumes myvec is short:
elm = 1
while (elm != 0) {
# Pick a random index, (it might be a 0):
rndidx = round(runif(1, min=1, max=length(myvec)))
elm = myvec[rndidx]
if(elm == 0)
print(I am done)
else
print(I am not done)
}
print(rndidx)
# If myvec is large and/or contains no zeros,
# The above loop is sub-optimal/faulty.
# I suspect that skilled R-people would approach this task differently.
# Perhaps they would use features baked into R rather than use a loop?
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and provide commented, minimal, self-contained, reproducible code.
[R] Goodness of fit for a Rietveld Refinement
Hi folks, Wondering if anyone might be able to help me on this one. I have just done some geochemistry with X-ray Diffraction and Rietveld Refinement in order to quantify the data. I have a observed spectra from my sample and a calculated spectra from the Rietveld Refinement (in the attached image, along with the background). I was wondering is there a package in R that I might be able to use that would essentially show me how well (or how poorly) fitted the Rietveld calculated spectra was with regard to my observed spectra? It's essentially a goodness of fit or R-squared value but I've been having some difficulty finding the right way to assess the model fit. I'd appreciate any information or tips anyone might have. Kind regards, Rory Flood. -- Rory Flood Postgraduate Research Student Room 02 044, Elmwood Building School of Geography, Archaeology and Palaeoecology Queen's University Belfast Belfast BT7 1NN Co. Antrim Northern Ireland Tel: +44 (0) 28 9097 3929 Email: rfloo...@qub.ac.uk __ http://r.789695.n4.nabble.com/file/n4678470/Spectra.jpg -- View this message in context: http://r.789695.n4.nabble.com/Goodness-of-fit-for-a-Rietveld-Refinement-tp4678470.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] Newb: How I find random vector index?
Not only does it not require a loop, this is a one-liner:
myvec - c(1,0,3,0,77,9,0,1,2,0)
sample(which(myvec == 0), 1)
[1] 4
sample(which(myvec == 0), 1)
[1] 7
sample(which(myvec == 0), 1)
[1] 2
If there's a possibility of not having zeros then you'll need to check
that separately, otherwise sample() will throw an error. For instance:
if(any(myvec == 0)) {
sample(which(myvec == 0), 1)
}
which() will
Sarah
On Thu, Oct 17, 2013 at 2:54 PM, Stock Beaver stockbea...@ymail.com wrote:
# Suppose I have a vector:
myvec = c(1,0,3,0,77,9,0,1,2,0)
# I want to randomly pick an element from myvec
# where element == 0
# and print the value of the corresponding index.
# So, for example I might randomly pick the 3rd 0
# and I would print the corresponding index
# which is 7,
# My initial approach is to use a for-loop.
# Also I take a short-cut which assumes myvec is short:
elm = 1
while (elm != 0) {
# Pick a random index, (it might be a 0):
rndidx = round(runif(1, min=1, max=length(myvec)))
elm = myvec[rndidx]
if(elm == 0)
print(I am done)
else
print(I am not done)
}
print(rndidx)
# If myvec is large and/or contains no zeros,
# The above loop is sub-optimal/faulty.
# I suspect that skilled R-people would approach this task differently.
# Perhaps they would use features baked into R rather than use a loop?
[[alternative HTML version deleted]]
--
Sarah Goslee
http://www.functionaldiversity.org
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newb: How I find random vector index?
Typo fix below:
On Thu, Oct 17, 2013 at 3:05 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
Not only does it not require a loop, this is a one-liner:
myvec - c(1,0,3,0,77,9,0,1,2,0)
sample(which(myvec == 0), 1)
[1] 4
sample(which(myvec == 0), 1)
[1] 7
sample(which(myvec == 0), 1)
[1] 2
If there's a possibility of not having zeros then you'll need to check
that separately, otherwise sample() will throw an error. For instance:
if(any(myvec == 0)) {
sample(which(myvec == 0), 1)
}
which() will
^ just delete this.
Sarah
On Thu, Oct 17, 2013 at 2:54 PM, Stock Beaver stockbea...@ymail.com wrote:
# Suppose I have a vector:
myvec = c(1,0,3,0,77,9,0,1,2,0)
# I want to randomly pick an element from myvec
# where element == 0
# and print the value of the corresponding index.
# So, for example I might randomly pick the 3rd 0
# and I would print the corresponding index
# which is 7,
# My initial approach is to use a for-loop.
# Also I take a short-cut which assumes myvec is short:
elm = 1
while (elm != 0) {
# Pick a random index, (it might be a 0):
rndidx = round(runif(1, min=1, max=length(myvec)))
elm = myvec[rndidx]
if(elm == 0)
print(I am done)
else
print(I am not done)
}
print(rndidx)
# If myvec is large and/or contains no zeros,
# The above loop is sub-optimal/faulty.
# I suspect that skilled R-people would approach this task differently.
# Perhaps they would use features baked into R rather than use a loop?
[[alternative HTML version deleted]]
--
Sarah Goslee
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Re: [R] Weighted regression markers on scatter plots
On Oct 16, 2013, at 10:04 AM, Msugarman wrote: Hi all, I'm trying to graph the results of a weighted regression analysis. Is anyone aware of a way to make my markers appear a different sizes to be consistent with their respective weights? You have not produced any data or code. If using base graphics then `plot.default` accepta vector for cex. Thanks, -Mike Sugarman Wayne State University -- View this message in context: http://r.789695.n4.nabble.com/Weighted-regression-markers-on-scatter-plots-tp4678370.html Sent from the R help mailing list archive at Nabble.com. 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.
Re: [R] Newb: How I find random vector index?
On 10/17/2013 11:54 AM, Stock Beaver wrote:
# Suppose I have a vector:
myvec = c(1,0,3,0,77,9,0,1,2,0)
# I want to randomly pick an element from myvec
# where element == 0
# and print the value of the corresponding index.
# So, for example I might randomly pick the 3rd 0
# and I would print the corresponding index
# which is 7,
# My initial approach is to use a for-loop.
# Also I take a short-cut which assumes myvec is short:
elm = 1
while (elm != 0) {
# Pick a random index, (it might be a 0):
rndidx = round(runif(1, min=1, max=length(myvec)))
elm = myvec[rndidx]
if(elm == 0)
print(I am done)
else
print(I am not done)
}
print(rndidx)
It's a little easier if you re-arrange your problem statement. This is
equivalent: return randomly one index of myvec for which the element of
myvec equals 0. A direct implementation of this is
sample(which(myvec==0), 1)
which(myvec==0) returns a vector of indexes of myvec for which the value
of the vector is 0. sample(..., 1) randomly selects one of those.
# If myvec is large and/or contains no zeros,
# The above loop is sub-optimal/faulty.
This approach also fails if there is no 0's in the vector. What do you
want the result to be when that is the case? If we go with the simple
answer of NA, then you can special case that (and wrap it up into a
function)
OneZeroIndex - function(myvec) {
zeros - which(myvec==0)
if (length(zeros) 0) {
sample(zeros, 1)
} else {
NA
}
}
# I suspect that skilled R-people would approach this task differently.
# Perhaps they would use features baked into R rather than use a loop?
[[alternative HTML version deleted]]
Please post plain text only.
--
Brian S. Diggs, PhD
Senior Research Associate, Department of Surgery
Oregon Health Science University
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Re: [R] Selecting maximums between different variables
Hi,
You could also check ?data.table() as it could be faster.
#Speed comparison
set.seed(498)
oilT -
data.frame(YEAR=rep(rep(1800:2012,50),100),state=rep(rep(state.abb,each=213),100),value=sample(2000:8,1065000,replace=TRUE),stringsAsFactors=FALSE)
system.time(res1 -
oilT[as.logical(with(oilT,ave(value,list(YEAR),FUN=function(x) x%in%
max(x,])
# user system elapsed
# 0.532 0.008 0.540
dim(res1) #as some years have duplicated maximums
#[1] 220 3
res1[duplicated(res1[,1])|duplicated(res1[,1],fromLast=TRUE),]
library(data.table)
dt1 - data.table(oilT,key='YEAR')
system.time( res2 - dt1[dt1[,value %in% max(value),'YEAR']$V1])
# user system elapsed
# 0.060 0.000 0.062
res1 - res1[order(res1$YEAR),]
row.names(res1) - 1:nrow(res1)
identical(res1,as.data.frame(res2))
#[1] TRUE
A.K.
On Thursday, October 17, 2013 1:35 PM, arun smartpink...@yahoo.com wrote:
Hi,
You may try:
unlist(lapply(seq_len(nrow(oil)),function(i) oil[i,-1][which.max(oil[i,-1])]))
# CA ND
#4 6
#or
library(reshape2)
datM - melt(oil,id.var=YEAR)
datM[as.logical(with(datM,ave(value,list(YEAR),FUN= function(x) x%in%
max(x,]
# YEAR variable value
#3 2011 CA 4
#8 2012 ND 6
A.K.
On Thursday, October 17, 2013 12:50 PM, Tim Umbach tim.umb...@hufw.de wrote:
Hi there,
another beginners question, I'm afraid. Basically i want to selct the
maximum of values, that correspond to different variables. I have a table
of oil production that looks somewhat like this:
oil - data.frame( YEAR = c(2011, 2012),
TX = c(2, 3),
CA = c(4, 25000),
AL = c(2,
21000),
ND = c(21000,6))
Now I want to find out, which state produced most oil in a given year. I
tried this:
attach(oil)
last_year = oil[ c(YEAR == 2012), ]
max(last_year)
Which works, but it doesnt't give me the corresponding values (i.e. it just
gives me the maximum output, not what state its from).
So I tried this:
oil[c(oil == max(last_year)),]
and this:
oil[c(last_year == max(last_year)),]
and this:
oil[which.max(last_year),]
and this:
last_year[max(last_year),]
None of them work, but they don't give error messages either, the output is
just NA. The problem is, in my eyes, that I'm comparing the values of
different variables with each other. Because if i change the structure of
the dataframe (which I can't do with the real data, at least not with out
doing it by hand with a huge dataset), it looks like this and works
perfectly:
oil2 - data.frame (
names = c('YEAR', 'TX', 'CA', 'AL', 'ND'),
oil_2011 = c(2011, 2, 4, 2, 21000),
oil_2012 = c(2012, 3, 25000, 21000, 6)
)
attach(oil2)
oil2[c(oil_2012 == max(oil_2012)),]
Any help is much appreciated.
Thanks, Tim Umbach
[[alternative HTML version deleted]]
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Re: [R] Weighted regression markers on scatter plots
On 10/17/2013 04:04 AM, Msugarman wrote: Hi all, I'm trying to graph the results of a weighted regression analysis. Is anyone aware of a way to make my markers appear a different sizes to be consistent with their respective weights? Hi Mike, Have a look at the size_n_color function in the plotrix package. Jim __ 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] Subseting a data.frame
May I ask why: count_by_class - with(dat, ave(numeric(length(basel_ asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) ? -- Bert On Thu, Oct 17, 2013 at 12:36 PM, William Dunlap wdun...@tibco.com wrote: What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Here is one way. There are many others: dat - data.frame( # slightly less trivial example basel_asset_class=c(4,8,8,8,74,3,74), defa_frequency=(1:7)/8) count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) cbind(dat, count_by_class) # see what we just computed basel_asset_class defa_frequency count_by_class 1 4 0.125 1 2 8 0.250 3 3 8 0.375 3 4 8 0.500 3 574 0.625 2 6 3 0.750 1 774 0.875 2 mydat[count_by_class1, ] # I think this is what you are asking for basel_asset_class defa_frequency 2 8 0.250 3 8 0.375 4 8 0.500 574 0.625 774 0.875 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 Katherine Gobin Sent: Thursday, October 17, 2013 11:05 AM To: Bert Gunter Cc: r-help@r-project.org Subject: Re: [R] Subseting a data.frame Correction. (2nd para first three lines) Pl read following line What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, as What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are THREE default frequencies w.r.t. basel aseet class 8, I alpologize for the incovenience. Regards KAtherine On , Katherine Gobin katherine_go...@yahoo.com wrote: I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat, mydat$basel_asset_class 2). What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, similarly there could be another basel asset class having say 5 default frequncies. Thus, I need to take subset of the data.frame s.t. the no of corresponding defa_frequencies is greater than 2. The idea is we try to fit exponential curve Y = A exp( BX ) for each of the basel asset classes and to estimate values of A and B, mathematically one needs to have at least two values of X. I hope I may be able to express my requirement. Its not that I need the subset of mydat s.t. basel asset class is 2 (now 4 in revised example), but sbuset s.t. no of default frequencies is greater than or equal to 2. This 2 is not same as basel asset class 2. Kindly guide With warm regards Katherine Gobin On Thursday, 17 October 2013 9:33 PM, Bert Gunter gunter.ber...@gene.com wrote: Kindly guide ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear Forum, I have a data
Re: [R] Subseting a data.frame
What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Here is one way. There are many others: dat - data.frame( # slightly less trivial example basel_asset_class=c(4,8,8,8,74,3,74), defa_frequency=(1:7)/8) count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) cbind(dat, count_by_class) # see what we just computed basel_asset_class defa_frequency count_by_class 1 4 0.125 1 2 8 0.250 3 3 8 0.375 3 4 8 0.500 3 574 0.625 2 6 3 0.750 1 774 0.875 2 mydat[count_by_class1, ] # I think this is what you are asking for basel_asset_class defa_frequency 2 8 0.250 3 8 0.375 4 8 0.500 574 0.625 774 0.875 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 Katherine Gobin Sent: Thursday, October 17, 2013 11:05 AM To: Bert Gunter Cc: r-help@r-project.org Subject: Re: [R] Subseting a data.frame Correction. (2nd para first three lines) Pl read following line What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, as What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are THREE default frequencies w.r.t. basel aseet class 8, I alpologize for the incovenience. Regards KAtherine On , Katherine Gobin katherine_go...@yahoo.com wrote: I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat, mydat$basel_asset_class 2). What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, similarly there could be another basel asset class having say 5 default frequncies. Thus, I need to take subset of the data.frame s.t. the no of corresponding defa_frequencies is greater than 2. The idea is we try to fit exponential curve Y = A exp( BX ) for each of the basel asset classes and to estimate values of A and B, mathematically one needs to have at least two values of X. I hope I may be able to express my requirement. Its not that I need the subset of mydat s.t. basel asset class is 2 (now 4 in revised example), but sbuset s.t. no of default frequencies is greater than or equal to 2. This 2 is not same as basel asset class 2. Kindly guide With warm regards Katherine Gobin On Thursday, 17 October 2013 9:33 PM, Bert Gunter gunter.ber...@gene.com wrote: Kindly guide ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin katherine_go...@yahoo.com wrote: Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given
Re: [R] Subseting a data.frame
May I ask why: count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) The way I did it would work if basel_asset_class were non-numeric. In ave(x, group, FUN=FUN), FUN's return value should be the same type as x (or you can get some odd type conversions). E.g., num - c(2,3,2,2) ; char - c(Two,Three,Two,Two) ave(num, num, FUN=length) # good [1] 3 1 3 3 ave(char, char, FUN=length) # bad [1] 3 1 3 3 fac - factor(char, levels=c(One,Two,Three)) ave(fac, fac, FUN=length) [1] NA NA NA NA Levels: One Two Three Warning messages: 1: In `[-.factor`(`*tmp*`, i, value = 0L) : invalid factor level, NA generated 2: In `[-.factor`(`*tmp*`, i, value = 3L) : invalid factor level, NA generated 3: In `[-.factor`(`*tmp*`, i, value = 1L) : invalid factor level, NA generated but x=integer(length(group)) works in all cases: ave(integer(length(fac)), fac, FUN=length) [1] 3 1 3 3 ave(integer(length(char)), char, FUN=length) [1] 3 1 3 3 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com From: Bert Gunter [mailto:gunter.ber...@gene.com] Sent: Thursday, October 17, 2013 1:06 PM To: William Dunlap Cc: Katherine Gobin; r-help@r-project.org Subject: Re: [R] Subseting a data.frame May I ask why: count_by_class - with(dat, ave(numeric(length(basel_ asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) ? -- Bert On Thu, Oct 17, 2013 at 12:36 PM, William Dunlap wdun...@tibco.commailto:wdun...@tibco.com wrote: What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Here is one way. There are many others: dat - data.frame( # slightly less trivial example basel_asset_class=c(4,8,8,8,74,3,74), defa_frequency=(1:7)/8) count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) cbind(dat, count_by_class) # see what we just computed basel_asset_class defa_frequency count_by_class 1 4 0.125 1 2 8 0.250 3 3 8 0.375 3 4 8 0.500 3 574 0.625 2 6 3 0.750 1 774 0.875 2 mydat[count_by_class1, ] # I think this is what you are asking for basel_asset_class defa_frequency 2 8 0.250 3 8 0.375 4 8 0.500 574 0.625 774 0.875 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.comhttp://tibco.com -Original Message- From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org] On Behalf Of Katherine Gobin Sent: Thursday, October 17, 2013 11:05 AM To: Bert Gunter Cc: r-help@r-project.orgmailto:r-help@r-project.org Subject: Re: [R] Subseting a data.frame Correction. (2nd para first three lines) Pl read following line What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, as What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are THREE default frequencies w.r.t. basel aseet class 8, I alpologize for the incovenience. Regards KAtherine On , Katherine Gobin katherine_go...@yahoo.commailto:katherine_go...@yahoo.com wrote: I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat,
Re: [R] Subseting a data.frame
Thanks, Bill. But ?ave specifically says: ave(x, ..., FUN = mean) Arguments: x A numeric. So that it should not be expected to work properly if the argument is not (coercible to) numeric. Nevertheless, defensive programming is always wise. Cheers, Bert On Thu, Oct 17, 2013 at 1:34 PM, William Dunlap wdun...@tibco.com wrote: May I ask why: count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) The way I did it would work if basel_asset_class were non-numeric. In ave(x, group, FUN=FUN), FUN's return value should be the same type as x (or you can get some odd type conversions). E.g., num - c(2,3,2,2) ; char - c(Two,Three,Two,Two) ave(num, num, FUN=length) # good [1] 3 1 3 3 ave(char, char, FUN=length) # bad [1] 3 1 3 3 fac - factor(char, levels=c(One,Two,Three)) ave(fac, fac, FUN=length) [1] NA NA NA NA Levels: One Two Three Warning messages: 1: In `[-.factor`(`*tmp*`, i, value = 0L) : invalid factor level, NA generated 2: In `[-.factor`(`*tmp*`, i, value = 3L) : invalid factor level, NA generated 3: In `[-.factor`(`*tmp*`, i, value = 1L) : invalid factor level, NA generated but x=integer(length(group)) works in all cases: ave(integer(length(fac)), fac, FUN=length) [1] 3 1 3 3 ave(integer(length(char)), char, FUN=length) [1] 3 1 3 3 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com From: Bert Gunter [mailto:gunter.ber...@gene.com] Sent: Thursday, October 17, 2013 1:06 PM To: William Dunlap Cc: Katherine Gobin; r-help@r-project.org Subject: Re: [R] Subseting a data.frame May I ask why: count_by_class - with(dat, ave(numeric(length(basel_ asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) ? -- Bert On Thu, Oct 17, 2013 at 12:36 PM, William Dunlap wdun...@tibco.com wrote: What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Here is one way. There are many others: dat - data.frame( # slightly less trivial example basel_asset_class=c(4,8,8,8,74,3,74), defa_frequency=(1:7)/8) count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) cbind(dat, count_by_class) # see what we just computed basel_asset_class defa_frequency count_by_class 1 4 0.125 1 2 8 0.250 3 3 8 0.375 3 4 8 0.500 3 574 0.625 2 6 3 0.750 1 774 0.875 2 mydat[count_by_class1, ] # I think this is what you are asking for basel_asset_class defa_frequency 2 8 0.250 3 8 0.375 4 8 0.500 574 0.625 774 0.875 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 Katherine Gobin Sent: Thursday, October 17, 2013 11:05 AM To: Bert Gunter Cc: r-help@r-project.org Subject: Re: [R] Subseting a data.frame Correction. (2nd para first three lines) Pl read following line What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, as What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are THREE default frequencies w.r.t. basel aseet class 8, I alpologize for the incovenience. Regards KAtherine On , Katherine Gobin katherine_go...@yahoo.com wrote: I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. mydat basel_asset_class defa_frequency 1 4 0.150 2 8
Re: [R] Subseting a data.frame
Hi Bill, #seq_along() worked in the cases you showed. ave(seq_along(fac),fac,FUN=length) #[1] 3 1 3 3 ave(seq_along(num), num, FUN=length) #[1] 3 1 3 3 ave(seq_along(char), char, FUN=length) #[1] 3 1 3 3 I thought, there might be some advantages in speed, but they were similar in speed. set.seed(195) num1 - sample(1e3,1e7,replace=TRUE) system.time(res1 - ave(integer(length(num1)),num1,FUN=length)) # user system elapsed #4.148 0.228 4.382 system.time(res2 - ave(seq_along(num1),num1,FUN=length)) # user system elapsed # 3.944 0.228 4.181 system.time(res3 - ave(num1,num1,FUN=length)) # user system elapsed # 3.740 0.264 4.012 identical(res1,res2) #[1] TRUE identical(res2,res3) #[1] TRUE A.K. On Thursday, October 17, 2013 4:34 PM, William Dunlap wdun...@tibco.com wrote: May I ask why: count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) The way I did it would work if basel_asset_class were non-numeric. In ave(x, group, FUN=FUN), FUN's return value should be the same type as x (or you can get some odd type conversions). E.g., num - c(2,3,2,2) ; char - c(Two,Three,Two,Two) ave(num, num, FUN=length) # good [1] 3 1 3 3 ave(char, char, FUN=length) # bad [1] 3 1 3 3 fac - factor(char, levels=c(One,Two,Three)) ave(fac, fac, FUN=length) [1] NA NA NA NA Levels: One Two Three Warning messages: 1: In `[-.factor`(`*tmp*`, i, value = 0L) : invalid factor level, NA generated 2: In `[-.factor`(`*tmp*`, i, value = 3L) : invalid factor level, NA generated 3: In `[-.factor`(`*tmp*`, i, value = 1L) : invalid factor level, NA generated but x=integer(length(group)) works in all cases: ave(integer(length(fac)), fac, FUN=length) [1] 3 1 3 3 ave(integer(length(char)), char, FUN=length) [1] 3 1 3 3 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com From: Bert Gunter [mailto:gunter.ber...@gene.com] Sent: Thursday, October 17, 2013 1:06 PM To: William Dunlap Cc: Katherine Gobin; r-help@r-project.org Subject: Re: [R] Subseting a data.frame May I ask why: count_by_class - with(dat, ave(numeric(length(basel_ asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) ? -- Bert On Thu, Oct 17, 2013 at 12:36 PM, William Dunlap wdun...@tibco.commailto:wdun...@tibco.com wrote: What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Here is one way. There are many others: dat - data.frame( # slightly less trivial example basel_asset_class=c(4,8,8,8,74,3,74), defa_frequency=(1:7)/8) count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) cbind(dat, count_by_class) # see what we just computed basel_asset_class defa_frequency count_by_class 1 4 0.125 1 2 8 0.250 3 3 8 0.375 3 4 8 0.500 3 5 74 0.625 2 6 3 0.750 1 7 74 0.875 2 mydat[count_by_class1, ] # I think this is what you are asking for basel_asset_class defa_frequency 2 8 0.250 3 8 0.375 4 8 0.500 5 74 0.625 7 74 0.875 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.comhttp://tibco.com -Original Message- From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org] On Behalf Of Katherine Gobin Sent: Thursday, October 17, 2013 11:05 AM To: Bert Gunter Cc: r-help@r-project.orgmailto:r-help@r-project.org Subject: Re: [R] Subseting a data.frame Correction. (2nd para first three lines) Pl read following line What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, as What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are THREE default frequencies w.r.t. basel aseet class 8, I alpologize for the incovenience. Regards KAtherine On , Katherine Gobin katherine_go...@yahoo.commailto:katherine_go...@yahoo.com wrote:
Re: [R] Subseting a data.frame
seq_along(x), integer(length(x)), is.na(x), or anything that produces an integer (or numeric or logical) vector the length of x would work. I use integer() or numeric() to indicate I'm not using its value: it is just a vector in which to place the return values of FUN(). Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: arun [mailto:smartpink...@yahoo.com] Sent: Thursday, October 17, 2013 2:33 PM To: R help Cc: William Dunlap; Bert Gunter Subject: Re: [R] Subseting a data.frame Hi Bill, #seq_along() worked in the cases you showed. ave(seq_along(fac),fac,FUN=length) #[1] 3 1 3 3 ave(seq_along(num), num, FUN=length) #[1] 3 1 3 3 ave(seq_along(char), char, FUN=length) #[1] 3 1 3 3 I thought, there might be some advantages in speed, but they were similar in speed. set.seed(195) num1 - sample(1e3,1e7,replace=TRUE) system.time(res1 - ave(integer(length(num1)),num1,FUN=length)) # user system elapsed #4.148 0.228 4.382 system.time(res2 - ave(seq_along(num1),num1,FUN=length)) # user system elapsed # 3.944 0.228 4.181 system.time(res3 - ave(num1,num1,FUN=length)) # user system elapsed # 3.740 0.264 4.012 identical(res1,res2) #[1] TRUE identical(res2,res3) #[1] TRUE A.K. On Thursday, October 17, 2013 4:34 PM, William Dunlap wdun...@tibco.com wrote: May I ask why: count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) The way I did it would work if basel_asset_class were non-numeric. In ave(x, group, FUN=FUN), FUN's return value should be the same type as x (or you can get some odd type conversions). E.g., num - c(2,3,2,2) ; char - c(Two,Three,Two,Two) ave(num, num, FUN=length) # good [1] 3 1 3 3 ave(char, char, FUN=length) # bad [1] 3 1 3 3 fac - factor(char, levels=c(One,Two,Three)) ave(fac, fac, FUN=length) [1] NA NA NA NA Levels: One Two Three Warning messages: 1: In `[-.factor`(`*tmp*`, i, value = 0L) : invalid factor level, NA generated 2: In `[-.factor`(`*tmp*`, i, value = 3L) : invalid factor level, NA generated 3: In `[-.factor`(`*tmp*`, i, value = 1L) : invalid factor level, NA generated but x=integer(length(group)) works in all cases: ave(integer(length(fac)), fac, FUN=length) [1] 3 1 3 3 ave(integer(length(char)), char, FUN=length) [1] 3 1 3 3 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com From: Bert Gunter [mailto:gunter.ber...@gene.com] Sent: Thursday, October 17, 2013 1:06 PM To: William Dunlap Cc: Katherine Gobin; r-help@r-project.org Subject: Re: [R] Subseting a data.frame May I ask why: count_by_class - with(dat, ave(numeric(length(basel_ asset_class)), basel_asset_class, FUN=length)) should not be more simply done as: count_by_class - with(dat, ave(basel_asset_class, basel_asset_class, FUN=length)) ? -- Bert On Thu, Oct 17, 2013 at 12:36 PM, William Dunlap wdun...@tibco.commailto:wdun...@tibco.com wrote: What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Here is one way. There are many others: dat - data.frame( # slightly less trivial example basel_asset_class=c(4,8,8,8,74,3,74), defa_frequency=(1:7)/8) count_by_class - with(dat, ave(numeric(length(basel_asset_class)), basel_asset_class, FUN=length)) cbind(dat, count_by_class) # see what we just computed basel_asset_class defa_frequency count_by_class 1 4 0.125 1 2 8 0.250 3 3 8 0.375 3 4 8 0.500 3 5 74 0.625 2 6 3 0.750 1 7 74 0.875 2 mydat[count_by_class1, ] # I think this is what you are asking for basel_asset_class defa_frequency 2 8 0.250 3 8 0.375 4 8 0.500 5 74 0.625 7 74 0.875 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.comhttp://tibco.com -Original Message- From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org [mailto:r- help-boun...@r-project.orgmailto:r-help-boun...@r-project.org] On Behalf Of Katherine Gobin Sent: Thursday, October 17, 2013 11:05 AM To: Bert Gunter Cc: r-help@r-project.orgmailto:r-help@r-project.org Subject: Re: [R] Subseting a data.frame Correction. (2nd para first three lines) Pl read following line What I need is to select only those records
[R] speeding up a loop
Hey R professionals,
I have a large dataset and I want to run a loop on it basically creating a
new column which gathers information from another reference table.
When I run the code, R just freezes and even does not response after 30min
which is really unusual. I tried sapply as well but does not improve at
all.
I am running R 3.0.2 on Windows 7. I checked the system, when I run the
code, my CPU usage is about 25%-30% that is taxing my desktop.
Here is my code:
#df1 is the data set I want to add a new column#
#b is the reference tabel#
for (i in (1:nrow(df1))) {
begin=which(b$Time2==df1$start[i] b$Date==df1$Date[i])
date=unlist(strsplit(as.character(dff$end[i]), ))[1]
end=ifelse(date==2013-10-17,
which(b$Time2==df1$end[i] b$Date==df1$Date[i]),
which(b$Time2==df1$end[i]-3600*24 b$Date==as.Date(df1$Date[i])+1))
df1$new[i] - sum(b[begin:end,]$Power)
}
And here is a mimic sample of df1 b:
df1 - structure(list(Date = structure(c(1369699200, 1369699200,
1369699200,
1369699200, 1369699200), tzone = UTC, class = c(POSIXct,
POSIXt)), start = structure(c(1381991205, 1381990247, 1382010454,
1382007281, 1381992288), tzone = UTC, class = c(POSIXct,
POSIXt)), end = structure(c(1381992405, 1381993727, 1382010694,
1382007461, 1381992468), tzone = UTC, class = c(POSIXct,
POSIXt))), .Names = c(Date, start, end), row.names = c(NA,
-5L), class = data.frame)
b - structure(list(Date = structure(c(1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200), tzone = UTC,
class = c(POSIXct,
POSIXt)), Time2 = structure(c(1381989634, 1381989694, 1381989754,
1381989814, 1381989874, 1381989934, 1381989994, 1381990054, 1381990114,
1381990174, 1381990234, 1381990294, 1381990354, 1381990414, 1381990474,
1381990534, 1381990594, 1381990654, 1381990714, 1381990774, 1381990834,
1381990894, 1381990954, 1381991014, 1381991074, 1381991134, 1381991194,
1381991254, 1381991314, 1381991374, 1381991434, 1381991494, 1381991554,
1381991614, 1381991674, 1381991734, 1381991794, 1381991854, 1381991914,
1381991974, 1381992034, 1381992094, 1381992154, 1381992214, 1381992274,
1381992334, 1381992394, 1381992454, 1381992514, 1381992574), tzone = UTC,
class = c(POSIXct,
POSIXt)), Power = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44,
45, 46, 47, 48, 49, 50)), .Names = c(Date, Time2, Power
), row.names = c(NA, -50L), class = data.frame)
Thanks for your help!
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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] Incorporate Julia into R
On 17 October 2013 15:38, Timo Schmid timo_sch...@hotmail.com wrote: I have some code in R with a lot of matrix multiplication and inverting. R can be very slow for larger matrices like 5000x5000. I have seen the new programming language Julia (www.julialang.org) which is quite fast in doing matrix algebra. Its not Julia, but LAPACK. __ 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] Selecting maximums between different variables
See ?pmax for getting the max for each year.
do.call('pmax', oil[-1])
Or equivalently:
pmax(oil$TX, oil$CA, oil$AL, oil$ND)
apply and which.max will give you the index:
i - apply(oil[-1], 1, which.max)
which you can use to extract the state:
names(oil[-1])[i]
Jason
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Tim Umbach
Sent: Thursday, October 17, 2013 9:49 AM
To: r-help@r-project.org
Subject: [R] Selecting maximums between different variables
Hi there,
another beginners question, I'm afraid. Basically i want to selct the maximum
of values, that correspond to different variables. I have a table of oil
production that looks somewhat like this:
oil - data.frame( YEAR = c(2011, 2012),
TX = c(2, 3),
CA = c(4, 25000),
AL = c(2,
21000),
ND = c(21000,6))
Now I want to find out, which state produced most oil in a given year. I tried
this:
attach(oil)
last_year = oil[ c(YEAR == 2012), ]
max(last_year)
Which works, but it doesnt't give me the corresponding values (i.e. it just
gives me the maximum output, not what state its from).
So I tried this:
oil[c(oil == max(last_year)),]
and this:
oil[c(last_year == max(last_year)),]
and this:
oil[which.max(last_year),]
and this:
last_year[max(last_year),]
None of them work, but they don't give error messages either, the output is
just NA. The problem is, in my eyes, that I'm comparing the values of
different variables with each other. Because if i change the structure of the
dataframe (which I can't do with the real data, at least not with out doing it
by hand with a huge dataset), it looks like this and works
perfectly:
oil2 - data.frame (
names = c('YEAR', 'TX', 'CA', 'AL', 'ND'),
oil_2011 = c(2011, 2, 4, 2, 21000),
oil_2012 = c(2012, 3, 25000, 21000, 6)
)
attach(oil2)
oil2[c(oil_2012 == max(oil_2012)),]
Any help is much appreciated.
Thanks, Tim Umbach
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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-help@r-project.org mailing list
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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] crr question in library(cmprsk)
Hi all I do not understand why I am getting the following error message. Can anybody help me with this? Thanks in advance. install.packages(cmprsk) library(cmprsk) result1 -crr(ftime, fstatus, cov1, failcode=1, cencode=0 ) one.pout1 = predict(result1,cov1,X=cbind(1,one.z1,one.z2)) predict.crr(result1,cov1,X=cbind(1,one.z1,one.z2)) Error: could not find function predict.crr [[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] speeding up a loop
On Oct 17, 2013, at 2:56 PM, Ye Lin wrote:
Hey R professionals,
I have a large dataset and I want to run a loop on it basically creating a
new column which gathers information from another reference table.
When I run the code, R just freezes and even does not response after 30min
which is really unusual. I tried sapply as well but does not improve at
all.
I am running R 3.0.2 on Windows 7. I checked the system, when I run the
code, my CPU usage is about 25%-30% that is taxing my desktop.
A guess: It's not your CPU use ... it's your RAM use. You've probably exhausted
your RAM and your system has paged out to virutla memory
Here is my code:
#df1 is the data set I want to add a new column#
#b is the reference tabel#
for (i in (1:nrow(df1))) {
begin=which(b$Time2==df1$start[i] b$Date==df1$Date[i])
date=unlist(strsplit(as.character(dff$end[i]), ))[1]
end=ifelse(date==2013-10-17,
which(b$Time2==df1$end[i] b$Date==df1$Date[i]),
which(b$Time2==df1$end[i]-3600*24 b$Date==as.Date(df1$Date[i])+1))
df1$new[i] - sum(b[begin:end,]$Power)
}
I get:
Error in strsplit(as.character(dff$end[i]), ) : object 'dff' not found
If I change the dff to df1, I get:
Error in begin:end : argument of length 0
--
David.
And here is a mimic sample of df1 b:
df1 - structure(list(Date = structure(c(1369699200, 1369699200,
1369699200,
1369699200, 1369699200), tzone = UTC, class = c(POSIXct,
POSIXt)), start = structure(c(1381991205, 1381990247, 1382010454,
1382007281, 1381992288), tzone = UTC, class = c(POSIXct,
POSIXt)), end = structure(c(1381992405, 1381993727, 1382010694,
1382007461, 1381992468), tzone = UTC, class = c(POSIXct,
POSIXt))), .Names = c(Date, start, end), row.names = c(NA,
-5L), class = data.frame)
b - structure(list(Date = structure(c(1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200, 1369699200,
1369699200, 1369699200, 1369699200, 1369699200, 1369699200), tzone = UTC,
class = c(POSIXct,
POSIXt)), Time2 = structure(c(1381989634, 1381989694, 1381989754,
1381989814, 1381989874, 1381989934, 1381989994, 1381990054, 1381990114,
1381990174, 1381990234, 1381990294, 1381990354, 1381990414, 1381990474,
1381990534, 1381990594, 1381990654, 1381990714, 1381990774, 1381990834,
1381990894, 1381990954, 1381991014, 1381991074, 1381991134, 1381991194,
1381991254, 1381991314, 1381991374, 1381991434, 1381991494, 1381991554,
1381991614, 1381991674, 1381991734, 1381991794, 1381991854, 1381991914,
1381991974, 1381992034, 1381992094, 1381992154, 1381992214, 1381992274,
1381992334, 1381992394, 1381992454, 1381992514, 1381992574), tzone = UTC,
class = c(POSIXct,
POSIXt)), Power = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44,
45, 46, 47, 48, 49, 50)), .Names = c(Date, Time2, Power
), row.names = c(NA, -50L), class = data.frame)
Thanks for your help!
[[alternative HTML version deleted]]
__
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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.
David Winsemius
Alameda, CA, USA
__
R-help@r-project.org mailing list
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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.
