[R] Re : Re: Re : Privacy rights of an old user of this list
As I said previously, I'm only concerned about what was published in the archive of stat.ethz.ch. I visited https://stat.ethz.ch/mailman/listinfo/r-help to find out where I had to send my request to remove those messages and I read: R-help list run by maechler at stat.math.ethz.ch, bates at r-project.org So I emailed these addresses and r-help-owner at r-project.org. Only David got back to me, denying my request, so this is why I'm discussing it here. If anybody could provide me a contact of who is controlling this archive, I would definitely appreciate it. Regards, John - Message d'origine - De : Uwe Ligges Envoyés : 21.09.13 10:26 à : Adan Leobardo Martinez Cruz Objet : Re: [R] Re : Privacy rights of an old user of this list The locations where the mailing list are archived are not even controlled by any member of R-core or the R foundation. Everybody, including John, could collect mails from the list and publish them. This is a well known property of a public mailing list. Uwe Ligges On 17.09.2013 13:29, Adan Leobardo Martinez Cruz wrote: Dear all, I will express my opinion without knowing the details of the posts John would like to be removed. In the current state, people posting on this and other servers have no clear way to go when trying to remove their posts. It is a likely event that the number of people attempting the removal of their past posts will increase. Their reasons will vary and may or not may be reasonable to us. It seems that a discussion on how the R-server will handle this likely situation is needed (including the possibility of keeping the current policy, of course) Once the decision has been taken, a warning note would be helpful for newcomers (something in! big, black letters saying that whatever we post will not be removed or something like that). Best regards to all, adan From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] on behalf of John Gonzalez [john.gonza...@gmx.fr] Sent: Monday, September 16, 2013 5:14 PM To: David Winsemius; Albin Blaschka; Duncan Murdoch; Jeff Newmiller; S Ellison; Jim Lemon Cc: r-help@r-project.org Subject: [R] Re : Privacy rights of an old user of this list I would like to thank David for letting me publish this and discuss it openly. I must acknowledge from the answers that I received to my post, that the administrators of this list are doing what seems to be fair to me: what most people demand or understand that is right. However I don't share your views and I honestly think you are making a mistake which may hurt you just as much as it is hurting me now) in the long term. Let me develop my point. First of ! all let me clarify for those who accuse me of being desinformed or inn ocent about my request, I'm not asking for your collaboration to remove what I published from the internet, its google records or any of the infinite copies that may be lying around. I'm asking for a very simple thing: There are 7 messages (sorry it wasn't 3...) written by me and hosted at the server stat.ethz.ch https://stat.ethz.ch/pipermail/r-help/2009-March/190367.html which I would like to have removed. Now, Steve E makes a good point: I am also of the opinion that the list owner was not showing disrespect by describing the state of affairs you agreed to on signing up, or by declining to act beyond the requirements of the conditions applicable to the list. Steve Fair enough. But that doesn't mean that those conditions are right and should never be modified. I'm probably something similar to an unhappy customer who has bought a product with no money-back policy but with an important distinction: I'm going to be wearing this product for the rest of my life. So th! at makes me, if anything, a very unhappy customer. Now let me explain why in this world I'm spending time on requesting the removal of these 7 messages in that server. I have a MS in Computer Science and a 5 years long Telecommunications degree, I know quite well how the internet works. This is not the first time that I request this. I already requested it in another mailing list, where they were kind enough to aprove it after I verified my identity and they checked that they weren't removing anything critical. The result was that that piece information was obviously not erased from the entire internet but was not showing up in the first 12 pages of google when you would look up my name (when it was on the first page previously). It took me 5 requests to different servers but I managed. There is nothing impossible about it and it made a difference in my life. So why is this important for me (something like not showing up on the first pages of google?). Well please u! nderstand that there is a difference between publishing an article and writing an email to a list. An article goes through several personal revisions and is examined by a professional
[R] Setting a new method for generic function to satisfy R CMD check
Dear All,
Can somebody explain me how to correctly set a new class-specific method
for generic functions (e.g. for plot, summary and print). I simply
programmed these functions with the class names and it works perfectly.
E.g.:
x-2
y-3
class(x)-newclass
class(y)-newclass
print.newclass-function(x){ cat(x*10) }
print(x)
print(y)
However, I need the new methods for a new package, and the R CMD check
gives me warnings that my approach is not correct. I have to define
methods and do not use full names as print.newclass.
I read the Writing R Extensions and it is not explained there (at least
to such level that I could get it) and I didn't get anything understandable
by googling.
Maybe someone can help me?
Many thanks in advance!
Mikhail
[[alternative HTML version deleted]]
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[R] problem with showMethods()
This message could be useful for people fairly new to S4 classes.
Therefore r-help and not r-devel.
One use for showMethods(), important to me,
is to find all methods that use a particular class in their signatures.
This works:
showMethods(class=Action, where=package:CTDesignExplorer)
It finds the methods in the package.
This works:
showMethods(class=Action, where=topenv(parent.frame()))
in the sense that it finds nothing (looking in .GlobalEnv).
no applicable functions
The default for where is where=topenv(parent.frame()),
so you'd think this would also give nothing.
showMethods(class=Action)
Instead, it prints names of hundreds of S4 generics, reporting that they are
not.
For example,
Function xvcopy:
not an S4 generic function
that come from packages installed even though they are not loaded.
Since this was the first thing I tried,
getting to the bottom of it derailed me for hours.
The problem seems to be from this, in the definition of showMethod:
if (length(f) == 0L) {
f - if (missing(.where)) ### This is the problem. It's missing,
even though the default value is set.
getGenerics()
else getGenerics(.where)
}
I think better would be just
f - getGenerics(.where)
Then the default value would apply.
This would forestall a vast downpour of reports about generics claiming that
they are not generics.
Or maybe someone can explain the thinking behind the if clause.
Thanks.
-Roger Day
sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: x86_64-apple-darwin10.8.0 (64-bit)
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] CTDesignExplorer_1.3.9
loaded via a namespace (and not attached):
[1] BiocGenerics_0.6.0 Biostrings_2.28.0 grid_3.0.1 IRanges_1.18.1
[5] lattice_0.20-15parallel_3.0.1 stats4_3.0.1 tools_3.0.1
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Re: [R] Filtering out data from list
Hi, list1-as.list(c(hi I am here!,my name is bob!, I am mary!!, bob likes mary!!)) list2- list1[grepl(bob,unlist(list1))] list2 #[[1]] #[1] my name is bob! # #[[2]] #[1] bob likes mary!! A.K. Hm thanks :P but i was actually trying to say is like... if my list were more random like below and i'm trying to filter out this list which doesnt have the specific word such as bob. list1 [[1]] [1] hi I am here! [[2]] [1] my name is bob! [[3]] [1] I am mary!! [[4]] [1] bob likes mary!! will become list1 [[1]] [1] my name is bob! [[2]] [1] bob likes mary!! - Original Message - From: arun smartpink...@yahoo.com To: R help r-help@r-project.org Cc: Sent: Saturday, September 21, 2013 12:08 PM Subject: Re: Filtering out data from list Hi, list1-lapply(c(,2:3),function(x) paste0(hi I am here!,x)) #or list1- as.list(paste0(hi I am here!,c(,2:3))) str(list1) #List of 3 # $ : chr hi I am here! # $ : chr hi I am here!2 # $ : chr hi I am here!3 It is not a list within a list list2-list1[grepl(2,unlist(list1))] list2 #[[1]] #[1] hi I am here!2 #list within a list listNew- list( as.list(paste0(hi I am here!,c(,2:3 str(listNew) #List of 1 # $ :List of 3 # ..$ : chr hi I am here! #..$ : chr hi I am here!2 #..$ : chr hi I am here!3 listNew2- lapply(1:3,function(x) as.list(paste0(hi I am here!,c(,2:3 str(listNew2) #List of 3 # $ :List of 3 # ..$ : chr hi I am here! #..$ : chr hi I am here!2 #..$ : chr hi I am here!3 #$ :List of 3 # ..$ : chr hi I am here! # ..$ : chr hi I am here!2 # ..$ : chr hi I am here!3 # $ :List of 3 # ..$ : chr hi I am here! # ..$ : chr hi I am here!2 # ..$ : chr hi I am here!3 A.K. Okay im pretty new in r... and im having trouble figuring how to filtering out data just say for example i have a list which prints out list1 [[1]] [1] hi I am here! [[2]] [1] hi I am here!2 [[3]] [1] hi I am here!3 1) okay first I need to confirm , is that a list within a list ? since its displaying [[1]] [1] -- in this fashion 2) and now just say i want to have list1 to only contain the list with the number 2 in it how do i do it? or any methods to suggest ? __ 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] Translating recoding syntax from SPSS to R
Colleagues, I am in the process of learning R. I've been able to import my dataset (from Stata) and do some simple coding. I have now come to coding situation that requires some assistance. This is some code in SPSS that I would like to be able to execute in R: if (race eq 1 and usborn=0) confused=1 . if (race eq 2 and usborn=0) confused=1 . if (race eq 1 and usborn=1) confused=0 . if (race eq 2 and usborn=1) confused=0 . if (race eq 3 and usborn=1) confused=0 . if (race eq 3 and cohort=1) confused=0 . if (race eq 3 and cohort=2) confused=0 . variable labels confused R claims to be both an African American and foriegn born . value labels confused 1 Both AfAm and Foreign 2 Not . select if (confused eq 0) . Any assistance would be greatly appreciated. -- Mosi __ 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] Obtaining data from a different row of data frame
HI,
A modified code to avoid the ?sapply()
df1- structure(list(Dates = structure(c(13151, 13152, 13153, 13154,
13157, 13158, 13159, 13160, 13161, 13164), class = Date), P1 = c(10,
13, 16, 19, 22, 25, 28, 31, 34, 37), P2 = c(100, 102, 104, 106,
108, 110, 112, 114, 116, 118), P3 = c(90, 94, 98, 102, 106, 110,
114, 118, 122, 126), P4 = c(70, 75, 80, 85, 90, 95, 100, 105,
110, 115), OF1 = c(3, 3, 4, 5, 2, 2, 2, 1, 1, 5), OF2 = c(5,
3, 4, 2, 1, 2, 2, 1, 1, 0), OF3 = c(4, 3, 4, 1, 3, 2, 2, 1, 1,
0), OF4 = c(3, 5, 4, 2, 3, 1, 2, 1, 1, 0)), .Names = c(Dates,
P1, P2, P3, P4, OF1, OF2, OF3, OF4), row.names = c(NA,
-10L), class = data.frame)
df1$OF2[9]-4
df2- df1
df2[,10:13]- NA
colnames(df2)[10:13]- paste0(newPrice,1:4)
##your code
for(j in 2:5) {
df2[j+8] = df2[df2[,j+4] + row(df2)[,j], j]
}
indx1- unlist(df1[,grep(OF,colnames(df1))],use.names=FALSE)
indx1[rep(seq(nrow(df1)),4)%in% 6:10][indx1[rep(seq(nrow(df1)),4)%in% 6:10]-
rep(5:1,4)=0]- NA
val1- unlist(df1[,grep(P,colnames(df1))],use.names=FALSE)
df1[,10:13]- val1[indx1+seq_along(indx1)]
colnames(df1)[10:13]- colnames(df2)[10:13]
identical(df1[,10:13],df2[,10:13])
#[1] TRUE
###On a bigger dataset:
set.seed(29)
df2- data.frame(Dates=seq(as.Date(2006-01-03),length.out=2000,by=1
day),cbind(matrix(sample(10:120,2000*300,replace=TRUE),ncol=300),matrix(sample(0:6,2000*300,replace=TRUE),ncol=300)))
colnames(df2)[2:301]- paste0(P,1:300)
colnames(df2)[302:601]- paste0(OF,1:300)
df3- df2
df2[,602:901]-NA
colnames(df2)[602:901]- paste0(newPrice,1:300)
system.time({
for(j in grep(^P,colnames(df2))) {
df2[j+600] = df2[df2[,j+300] + row(df2)[,j], j]
}
})
# user system elapsed
# 8.508 0.000 8.523
colN_OF- ncol(df3[,grep(OF,colnames(df3))])
system.time({
indx1- unlist(df3[,grep(OF,colnames(df3))],use.names=FALSE)
indx1[rep(seq(nrow(df3)),colN_OF) %in%
1995:2000][indx1[rep(seq(nrow(df3)),colN_OF) %in% 1995:2000] -
rep(6:1,colN_OF)=0] -NA
val1- unlist(df3[,grep(P,colnames(df3))],use.names=FALSE)
df3[,602:901]- val1[indx1+seq_along(indx1)]
colnames(df3)[602:901]- colnames(df2)[602:901]
})
# user system elapsed
# 0.568 0.000 0.569
identical(df2,df3)
#[1] TRUE
A.K.
- Original Message -
From: arun smartpink...@yahoo.com
To: Ira Sharenow irasharenow...@yahoo.com
Cc:
Sent: Sunday, September 22, 2013 1:28 AM
Subject: Re: [R] Obtaining data from a different row of data frame
Ira,
I tried with a bigger dataset to look for any errors in the code:
set.seed(29)
df2- data.frame(Dates=seq(as.Date(2006-01-03),length.out=2000,by=1
day),cbind(matrix(sample(10:120,2000*300,replace=TRUE),ncol=300),matrix(sample(0:6,2000*300,replace=TRUE),ncol=300)))
colnames(df2)[2:301]- paste0(P,1:300)
colnames(df2)[302:601]- paste0(OF,1:300)
df3- df2
df2[,602:901]-NA
colnames(df2)[602:901]- paste0(newPrice,1:300)
system.time({
for(j in grep(^P,colnames(df2))) {
df2[j+600] = df2[df2[,j+300] + row(df2)[,j], j]
}
})
# user system elapsed
# 9.584 0.000 9.601
vec1- 6:1 ##change values according to the range of actual values in your rows.
vec2- 1995:2000 ##change accordingly. If the maximum value is say 100, take
100 rows from the tail end. Change the vec1 also so that both are of the same
length
system.time({
df3[vec2,grep(OF,colnames(df3))]- t(sapply(seq_along(vec1),function(i)
{x1-as.matrix(df3[vec2[i],grep(OF,colnames(df3))]); x1[x1=vec1[i]]-NA;
x1}))
indx1- unlist(df3[,grep(OF,colnames(df3))],use.names=FALSE)
val1- unlist(df3[,grep(P,colnames(df3))],use.names=FALSE)
df3[,602:901]- val1[indx1+seq_along(indx1)]
colnames(df3)[602:901]- colnames(df2)[602:901]
})
# user system elapsed
# 0.552 0.000 0.553
identical(df2[,602:901],df3[,602:901])
#[1] TRUE
A.K.
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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] Translating recoding syntax from SPSS to R
Hi, Try: set.seed(429) dat1- data.frame(race=sample(1:3,20,replace=TRUE),usborn=sample(0:2,20,replace=TRUE)) dat1$confused- 1*((dat1$race==1|dat1$race==2) dat1$usborn==0) head(dat1) # race usborn confused #1 3 2 0 #2 1 0 1 #3 2 1 0 #4 3 2 0 #5 1 2 0 #6 1 1 0 A.K. - Original Message - From: Mosi Ifatunji ifatu...@gmail.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Saturday, September 21, 2013 6:03 PM Subject: [R] Translating recoding syntax from SPSS to R Colleagues, I am in the process of learning R. I've been able to import my dataset (from Stata) and do some simple coding. I have now come to coding situation that requires some assistance. This is some code in SPSS that I would like to be able to execute in R: if (race eq 1 and usborn=0) confused=1 . if (race eq 2 and usborn=0) confused=1 . if (race eq 1 and usborn=1) confused=0 . if (race eq 2 and usborn=1) confused=0 . if (race eq 3 and usborn=1) confused=0 . if (race eq 3 and cohort=1) confused=0 . if (race eq 3 and cohort=2) confused=0 . variable labels confused R claims to be both an African American and foriegn born . value labels confused 1 Both AfAm and Foreign 2 Not . select if (confused eq 0) . Any assistance would be greatly appreciated. -- Mosi __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Translating recoding syntax from SPSS to R
Just a slight addition to Arun's answer: within() saves typing and
makes life a bit easier
#
set.seed(429)
dat1 - data.frame(
race=sample(1:3,20,replace=TRUE),
usborn=sample(0:2,20,replace=TRUE))
# within is a nice way to add variables
# and make modifications within a dataset
# saves typing lots of dat1$
dat1 - within(dat1, {
confused - 1 * ((race==1 | race==2) usborn==0)
})
head(dat1)
# only confused 0
dat1 - subset(dat1, confused == 0)
#
On Sat, Sep 21, 2013 at 11:47 PM, arun smartpink...@yahoo.com wrote:
Hi,
Try:
set.seed(429)
dat1-
data.frame(race=sample(1:3,20,replace=TRUE),usborn=sample(0:2,20,replace=TRUE))
dat1$confused- 1*((dat1$race==1|dat1$race==2) dat1$usborn==0)
head(dat1)
# race usborn confused
#13 20
#21 01
#32 10
#43 20
#51 20
#61 10
A.K.
- Original Message -
From: Mosi Ifatunji ifatu...@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Saturday, September 21, 2013 6:03 PM
Subject: [R] Translating recoding syntax from SPSS to R
Colleagues,
I am in the process of learning R. I've been able to import my dataset (from
Stata) and do some simple coding. I have now come to coding situation that
requires some assistance. This is some code in SPSS that I would like to be
able to execute in R:
if (race eq 1 and usborn=0) confused=1 .
if (race eq 2 and usborn=0) confused=1 .
if (race eq 1 and usborn=1) confused=0 .
if (race eq 2 and usborn=1) confused=0 .
if (race eq 3 and usborn=1) confused=0 .
if (race eq 3 and cohort=1) confused=0 .
if (race eq 3 and cohort=2) confused=0 .
variable labels confused R claims to be both an African American and foriegn
born .
value labels confused
1 Both AfAm and Foreign
2 Not .
select if (confused eq 0) .
Any assistance would be greatly appreciated.
-- Mosi
__
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
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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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://joshuawiley.com/
Senior Analyst - Elkhart Group Ltd.
http://elkhartgroup.com
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re : Re: Re : Privacy rights of an old user of this list
On 21.09.2013 22:40, John Gonzalez wrote: As I said previously, I'm only concerned about what was published in the archive of stat.ethz.ch. And it is OK to habe your post on Nabble (and many others) after you removed it from stat.ethz.ch? Are the swiss that bad? I visited https://stat.ethz.ch/mailman/listinfo/r-help to find out where I had to send my request to remove those messages and I read: R-help list run by maechler at stat.math.ethz.ch, bates at r-project.org So I emailed these addresses and r-help-owner at r-project.org. Right, Martin Maechler may be on vacations, and he will probably answer that he does not remove posts in general. Only David got back to me, denying my request, so this is why I'm discussing it here. If anybody could provide me a contact of who is controlling this archive, I would definitely appreciate it. Regards, John - Message d'origine - De : Uwe Ligges Envoyés : 21.09.13 10:26 À : Adan Leobardo Martinez Cruz Objet : Re: [R] Re : Privacy rights of an old user of this list The locations where the mailing list are archived are not even controlled by any member of R-core or the R foundation. Everybody, including John, could collect mails from the list and publish them. This is a well known property of a public mailing list. Uwe Ligges On 17.09.2013 13:29, Adan Leobardo Martinez Cruz wrote: Dear all, I will express my opinion without knowing the details of the posts John would like to be removed. In the current state, people posting on this and other servers have no clear way to go when trying to remove their posts. It is a likely event that the number of people attempting the removal of their past posts will increase. Their reasons will vary and may or not may be reasonable to us. It seems that a discussion on how the R-server will handle this likely situation is needed (including the possibility of keeping the current policy, of course) Once the decision has been taken, a warning note would be helpful for newcomers (something in big, black letters saying that whatever we post will not be removed or something like that). Best regards to all, adan From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] on behalf of John Gonzalez [john.gonza...@gmx.fr] Sent: Monday, September 16, 2013 5:14 PM To: David Winsemius; Albin Blaschka; Duncan Murdoch; Jeff Newmiller; S Ellison; Jim Lemon Cc: r-help@r-project.org Subject: [R] Re : Privacy rights of an old user of this list I would like to thank David for letting me publish this and discuss it openly. I must acknowledge from the answers that I received to my post, that the administrators of this list are doing what seems to be fair to me: what most people demand or understand that is right. However I don't share your views and I honestly think you are making a mistake which may hurt you just as much as it is hurting me now) in the long term. Let me develop my point. First of all let me clarify for those who accuse me of being desinformed or innocent about my request, I'm not asking for your collaboration to remove what I published from the internet, its google records or any of the infinite copies that may be lying around. I'm asking for a very simple thing: There are 7 messages (sorry it wasn't 3...) written by me and hosted at the server stat.ethz.ch https://stat.ethz.ch/pipermail/r-help/2009-March/190367.html which I would like to have removed. Now, Steve E makes a good point: I am also of the opinion that the list owner was not showing disrespect by describing the state of affairs you agreed to on signing up, or by declining to act beyond the requirements of the conditions applicable to the list. Steve Fair enough. But that doesn't mean that those conditions are right and should never be modified. I'm probably something similar to an unhappy customer who has bought a product with no money-back policy but with an important distinction: I'm going to be wearing this product for the rest of my life. So that makes me, if anything, a very unhappy customer. Now let me explain why in this world I'm spending time on requesting the removal of these 7 messages in that server. I have a MS in Computer Science and a 5 years long Telecommunications degree, I know quite well how the internet works. This is not the first time that I request this. I already requested it in another mailing list, where they were kind enough to aprove it after I verified my identity and they checked that they weren't removing anything critical. The result was that that piece information was obviously not erased from the entire internet but was not showing up in the first 12 pages of google when you would look up my name (when it was on the first page previously). It took me 5 requests to different servers but I managed. There is nothing impossible about it and it made a difference in my life. So why is this
Re: [R] Setting a new method for generic function to satisfy R CMD check
On 13-09-21 3:30 PM, Mikhail Beketov wrote:
Dear All,
Can somebody explain me how to correctly set a new class-specific method
for generic functions (e.g. for plot, summary and print). I simply
programmed these functions with the class names and it works perfectly.
E.g.:
x-2
y-3
class(x)-newclass
class(y)-newclass
print.newclass-function(x){ cat(x*10) }
print(x)
print(y)
However, I need the new methods for a new package, and the R CMD check
gives me warnings that my approach is not correct. I have to define
methods and do not use full names as print.newclass.
I read the Writing R Extensions and it is not explained there (at least
to such level that I could get it) and I didn't get anything understandable
by googling.
Maybe someone can help me?
You need to declare methods in the NAMESPACE file, and there is special
markup for them in Rd files.
Duncan Murdoch
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Re: [R] time zones from longitude, latitude, and date
On Sat, Sep 21, 2013 at 10:25 AM, David Winsemius dwinsem...@comcast.net wrote: On Sep 21, 2013, at 3:13 AM, Prof Brian Ripley wrote: On 21/09/2013 08:17, Gabor Grothendieck wrote: On Fri, Sep 20, 2013 at 4:31 PM, carlisle thacker carlisle.thac...@gmail.com wrote: I was looking for something like shown on the map: http://upload.wikimedia.org/wikipedia/commons/8/88/World_Time_Zones_Map.png Information about local daylight savings times would also help. The data are from ships, supposedly in local time, but no time-zone info is given. A function that would return time zone and whether or not daylight savings time applies at given date would would help. I'm trying to track down more information about the data and whether they can be referenced to UTC. The zone.tab file has this information. See the Examples section at the end of ?Sys.timezone for info on its whereabouts. On some OSes only. This is a couple of snippets of that file from a Mac: /usr/share/zoneinfo/zone.tab #- # This file contains a table with the following columns: # 1. ISO 3166 2-character country code. See the file `iso3166.tab'. # 2. Latitude and longitude of the zone's principal location # in ISO 6709 sign-degrees-minutes-seconds format, # either +-DDMM+-DDDMM or +-DDMMSS+-DDDMMSS, # first latitude (+ is north), then longitude (+ is east). # 3. Zone name used in value of TZ environment variable. # 4. Comments; present if and only if the country has multiple rows. #- US +340308-1181434 America/Los_Angeles Pacific Time US +611305-1495401 America/Anchorage Alaska Time US +581807-1342511 America/Juneau Alaska Time - Alaska panhandle US +593249-1394338 America/Yakutat Alaska Time - Alaska panhandle neck US +643004-1652423 America/NomeAlaska Time - west Alaska US +515248-1763929 America/AdakAleutian Islands US +211825-1575130 Pacific/HonoluluHawaii UY -3453-05611 America/Montevideo UZ +3940+06648 Asia/Samarkand west Uzbekistan UZ +4120+06918 Asia/Tashkent east Uzbekistan VA +415408+0122711 Europe/Vatican After looking at it I'm not sure it will be of much additional value for the purposes outlined. It does not provide boundaries of time zones. The idea is to take the nearest lat/long found. If the input data is lat/longs of major cities then it would likely work exactly. If not then it would only be an approximation but that might be sufficient depending on one's purpose. The nice thing is that the time zones you get out is precisely the ones that R can use. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Translating recoding syntax from SPSS to R
Thanks!
This worked wonderfully!
Very exciting!
-- Mosi
On Sep 22, 2013, at 3:20 AM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Just a slight addition to Arun's answer: within() saves typing and
makes life a bit easier
#
set.seed(429)
dat1 - data.frame(
race=sample(1:3,20,replace=TRUE),
usborn=sample(0:2,20,replace=TRUE))
# within is a nice way to add variables
# and make modifications within a dataset
# saves typing lots of dat1$
dat1 - within(dat1, {
confused - 1 * ((race==1 | race==2) usborn==0)
})
head(dat1)
# only confused 0
dat1 - subset(dat1, confused == 0)
#
On Sat, Sep 21, 2013 at 11:47 PM, arun smartpink...@yahoo.com wrote:
Hi,
Try:
set.seed(429)
dat1-
data.frame(race=sample(1:3,20,replace=TRUE),usborn=sample(0:2,20,replace=TRUE))
dat1$confused- 1*((dat1$race==1|dat1$race==2) dat1$usborn==0)
head(dat1)
# race usborn confused
#13 20
#21 01
#32 10
#43 20
#51 20
#61 10
A.K.
- Original Message -
From: Mosi Ifatunji ifatu...@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Saturday, September 21, 2013 6:03 PM
Subject: [R] Translating recoding syntax from SPSS to R
Colleagues,
I am in the process of learning R. I've been able to import my dataset (from
Stata) and do some simple coding. I have now come to coding situation that
requires some assistance. This is some code in SPSS that I would like to be
able to execute in R:
if (race eq 1 and usborn=0) confused=1 .
if (race eq 2 and usborn=0) confused=1 .
if (race eq 1 and usborn=1) confused=0 .
if (race eq 2 and usborn=1) confused=0 .
if (race eq 3 and usborn=1) confused=0 .
if (race eq 3 and cohort=1) confused=0 .
if (race eq 3 and cohort=2) confused=0 .
variable labels confused R claims to be both an African American and foriegn
born .
value labels confused
1 Both AfAm and Foreign
2 Not .
select if (confused eq 0) .
Any assistance would be greatly appreciated.
-- Mosi
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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://joshuawiley.com/
Senior Analyst - Elkhart Group Ltd.
http://elkhartgroup.com
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Re: [R] Creating rectangular plots with x and y coordinates and treatments from a matrix for a randomized block design
Is this what you were after:
## A function to generate a RCB design
rcbd-function(b,g,rb,cb,r,c)
+ {
+# b =number of blocks
+# g = a vector of treatments
+# rb = number of rows per blocks
+# cb =number of columns per block
+# r = total rows
+# c = total columns
+ library(foreach)
+ genotypes-times(b) %do% sample(g,length(g))
+ block-rep(1:b,each=length(g))
+ genotypes-factor(genotypes)
+ block-factor(block)
+ ### generate the base design
+ k-c/cb # number of blocks on the x-axis
+ x-rep(rep(1:r,each=cb),k) # X-coordinate
+ l-cb
+ p-r/rb
+ m-l+1
+ d-l*b/p
+ y-c(rep(1:l,r),rep(m:d,r)) # Y-coordinate
+ data.frame(x,y,block,genotypes)
+ }
set.seed(100)
ans - rcbd(b=4,g=1:4,rb=2,cb=2,r=4,c=4)
result - matrix(nrow = max(ans$x), ncol = max(ans$y))
result[cbind(ans$y, ans$x)] - ans$genotypes
result
[,1] [,2] [,3] [,4]
[1,]2423
[2,]1341
[3,]3223
[4,]1441
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Sun, Sep 22, 2013 at 1:07 AM, Laz lmra...@ufl.edu wrote:
Dear R users,
I have a function I created to generate randomized block designs given
below. Once I generate the design, I would like to plot it in a
rectangular form to mimic the excel style where the blocks are
represented with the treatment values placed exactly where their
appropriate X and Y coordinates are.
## A function to generate a RCB design
rcbd-function(b,g,rb,cb,r,c)
{
# b =number of blocks
# g = a vector of treatments
# rb = number of rows per blocks
# cb =number of columns per block
# r = total rows
# c = total columns
library(foreach)
genotypes-times(b) %do% sample(g,length(g))
block-rep(1:b,each=length(g))
genotypes-factor(genotypes)
block-factor(block)
### generate the base design
k-c/cb # number of blocks on the x-axis
x-rep(rep(1:r,each=cb),k) # X-coordinate
l-cb
p-r/rb
m-l+1
d-l*b/p
y-c(rep(1:l,r),rep(m:d,r)) # Y-coordinate
data.frame(x,y,block,genotypes)
}
set.seed(100)
rcbd(b=4,g=1:4,rb=2,cb=2,r=4,c=4)
x y block genotypes
1 1 1 1 2
2 1 2 1 1
3 2 1 1 4
4 2 2 1 3
5 3 1 2 2
6 3 2 2 4
7 4 1 2 3
8 4 2 2 1
9 1 3 3 3
10 1 4 3 1
11 2 3 3 2
12 2 4 3 4
13 3 3 4 2
14 3 4 4 4
15 4 3 4 3
16 4 4 4 1
How can I produce a diagram like this one below or a better one for any run
of my function?
2 4 2 3
1 3 4 1
3 2 2 3
1 4 4 1
Regards,
Laz
[[alternative HTML version deleted]]
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Re: [R] Obtaining data from a different row of data frame
Ira,
No problem.
If you change the ?for() loop to ?lapply, there should be increase in speed
(Usually, it is not the case. Here it is does, but still not as good in terms
of speed as the method I showed).
df1- structure(list(Dates = structure(c(13151, 13152, 13153, 13154,
13157, 13158, 13159, 13160, 13161, 13164), class = Date), P1 = c(10,
13, 16, 19, 22, 25, 28, 31, 34, 37), P2 = c(100, 102, 104, 106,
108, 110, 112, 114, 116, 118), P3 = c(90, 94, 98, 102, 106, 110,
114, 118, 122, 126), P4 = c(70, 75, 80, 85, 90, 95, 100, 105,
110, 115), OF1 = c(3, 3, 4, 5, 2, 2, 2, 1, 1, 5), OF2 = c(5,
3, 4, 2, 1, 2, 2, 1, 1, 0), OF3 = c(4, 3, 4, 1, 3, 2, 2, 1, 1,
0), OF4 = c(3, 5, 4, 2, 3, 1, 2, 1, 1, 0)), .Names = c(Dates,
P1, P2, P3, P4, OF1, OF2, OF3, OF4), row.names = c(NA,
-10L), class = data.frame)
df1$OF2[9]-4
df2- df1
df2[,10:13]- NA
colnames(df2)[10:13]- paste0(newPrice,1:4)
##your code
for(j in 2:5) {
df2[j+8] = df2[df2[,j+4] + row(df2)[,j], j]
}
#using ?lapply()
df1[,10:13]-lapply(2:5,function(j) {df1[df1[,j+4]+row(df2)[,j],j]})
colnames(df1)[10:13]- colnames(df2)[10:13]
identical(df1,df2)
#[1] TRUE
###Speed check:
set.seed(29)
df2- data.frame(Dates=seq(as.Date(2006-01-03),length.out=2000,by=1
day),cbind(matrix(sample(10:120,2000*300,replace=TRUE),ncol=300),matrix(sample(0:6,2000*300,replace=TRUE),ncol=300)))
colnames(df2)[2:301]- paste0(P,1:300)
colnames(df2)[302:601]- paste0(OF,1:300)
df3- df2
df2[,602:901]-NA
colnames(df2)[602:901]- paste0(newPrice,1:300)
system.time({
for(j in grep(^P,colnames(df2))) {
df2[j+600] = df2[df2[,j+300] + row(df2)[,j], j]
}
})
# user system elapsed
#11.652 0.148 11.822
system.time({df3[,602:901]-lapply(2:301,function(j)
{df3[df3[,j+300]+row(df3)[,j],j]}) })
# user system elapsed
# 2.960 0.000 2.962
colnames(df3)[602:901]- colnames(df2)[602:901]
identical(df2,df3)
#[1] TRUE
A.K.
From: Ira Sharenow irasharenow...@yahoo.com
To: arun smartpink...@yahoo.com
Sent: Sunday, September 22, 2013 10:49 AM
Subject: Re: [R] Obtaining data from a different row of data frame
Arun,
Thanks for the time you spent helping me.
I always learned to use the apply family (but maybe your strategies are
faster), and now I think I am going to learn Hadley Wickham’s methods. Right
now I need to do other parts of the project. In a few days I will take another
look at your code to see if I can get more out of my code.
For my current project once I am finished my boss will use my code and possibly
modify it, so speed is just one factor. Transparency and his future coding time
is another consideration. I need to balance things off.
I need tolerable speed and relatively easy to understand code. It is an
interesting trade off.
Thank again for your help. I’ll get back to you when I take another look at the
details of what you wrote.
Ira
On 9/21/2013 11:27 PM, arun wrote:
HI, A modified code to avoid the ?sapply()
df1- structure(list(Dates = structure(c(13151, 13152, 13153, 13154,
13157, 13158, 13159, 13160, 13161, 13164), class = Date), P1 = c(10,
13, 16, 19, 22, 25, 28, 31, 34, 37), P2 = c(100, 102, 104, 106,
108, 110, 112, 114, 116, 118), P3 = c(90, 94, 98, 102, 106, 110,
114, 118, 122, 126), P4 = c(70, 75, 80, 85, 90, 95, 100, 105,
110, 115), OF1 = c(3, 3, 4, 5, 2, 2, 2, 1, 1, 5), OF2 = c(5,
3, 4, 2, 1, 2, 2, 1, 1, 0), OF3 = c(4, 3, 4, 1, 3, 2, 2, 1, 1,
0), OF4 = c(3, 5, 4, 2, 3, 1, 2, 1, 1, 0)), .Names = c(Dates,
P1, P2, P3, P4, OF1, OF2, OF3, OF4), row.names = c(NA,
-10L), class = data.frame)
df1$OF2[9]-4 df2- df1
df2[,10:13]- NA
colnames(df2)[10:13]- paste0(newPrice,1:4) ##your code for(j in 2:5) {
df2[j+8] = df2[df2[,j+4] + row(df2)[,j], j]
}
indx1- unlist(df1[,grep(OF,colnames(df1))],use.names=FALSE)
indx1[rep(seq(nrow(df1)),4)%in% 6:10][indx1[rep(seq(nrow(df1)),4)%in% 6:10]-
rep(5:1,4)=0]- NA val1- unlist(df1[,grep(P,colnames(df1))],use.names=FALSE)
df1[,10:13]- val1[indx1+seq_along(indx1)]
colnames(df1)[10:13]- colnames(df2)[10:13]
identical(df1[,10:13],df2[,10:13])
#[1] TRUE ###On a bigger dataset:
set.seed(29)
df2- data.frame(Dates=seq(as.Date(2006-01-03),length.out=2000,by=1
day),cbind(matrix(sample(10:120,2000*300,replace=TRUE),ncol=300),matrix(sample(0:6,2000*300,replace=TRUE),ncol=300)))
colnames(df2)[2:301]- paste0(P,1:300)
colnames(df2)[302:601]- paste0(OF,1:300)
df3- df2 df2[,602:901]-NA
colnames(df2)[602:901]- paste0(newPrice,1:300)
system.time({
for(j in grep(^P,colnames(df2))) {
df2[j+600] = df2[df2[,j+300] + row(df2)[,j], j]
}
})
# user system elapsed
# 8.508 0.000 8.523 colN_OF- ncol(df3[,grep(OF,colnames(df3))])
system.time({
indx1- unlist(df3[,grep(OF,colnames(df3))],use.names=FALSE)
indx1[rep(seq(nrow(df3)),colN_OF) %in%
1995:2000][indx1[rep(seq(nrow(df3)),colN_OF) %in% 1995:2000] -
rep(6:1,colN_OF)=0] -NA
val1- unlist(df3[,grep(P,colnames(df3))],use.names=FALSE)
df3[,602:901]- val1[indx1+seq_along(indx1)]
Re: [R] PCA
Hi, see tutorial for Adegenet, http://adegenet.r-forge.r-project.org/ | Documents | adegenet-basics.pdf | section 6. It should help You. Vojtěch - Vojtěch Zeisek Department of Botany, Faculty of Science, Charles Uni., Prague, CZ Institute of Botany, Academy of Science, Czech Republic Community of the openSUSE GNU/Linux https://www.natur.cuni.cz/faculty-en?set_language=en http://www.ibot.cas.cz/?p=indexamp;site=en http://www.opensuse.org/ http://trapa.cz/ -- View this message in context: http://r.789695.n4.nabble.com/PCA-tp4676674p4676691.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] Coding several dummy variables into a single categorical variable
Colleagues, I have generated several dummy variables: n$native0 - 1 * (n$re==white n$usborn==yes) n$native1 - 1 * (n$re==afam n$usborn==yes) n$native2 - 1 * (n$re==carib n$usborn==yes) n$native3 - 1 * (n$re==carib n$usborn==no) I would now like to combine these into a single categorical variable where the new variable would be n$native. And values of native would be 0 through 3, where n$native0 would be a 0 value on n$native, n$native1 would be a 1 value on n$native etc. Any help would be greatly appreciated. -- Mosi __ 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] Coding several dummy variables into a single categorical variable
Hi,
Try:
set.seed(385)
df- data.frame(re= sample(c(white,afam,carib),20,replace=TRUE), usborn=
sample(c(yes,no),20,replace=TRUE),stringsAsFactors=FALSE)
df1-within(df,{native3- 1*(re==carib usborn==no); native2-
1*(re==carib usborn==yes); native1- 1*(re==afam usborn==yes);
native0- 1*(re==white usborn==yes)})
library(reshape2)
df2- melt(df1,id.vars=c(re,usborn))[,-3]
colnames(df2)[3]- native
df2
A.K.
- Original Message -
From: Mosi Ifatunji ifatu...@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Sunday, September 22, 2013 2:40 PM
Subject: [R] Coding several dummy variables into a single categorical variable
Colleagues,
I have generated several dummy variables:
n$native0 - 1 * (n$re==white n$usborn==yes)
n$native1 - 1 * (n$re==afam n$usborn==yes)
n$native2 - 1 * (n$re==carib n$usborn==yes)
n$native3 - 1 * (n$re==carib n$usborn==no)
I would now like to combine these into a single categorical variable where the
new variable would be n$native.
And values of native would be 0 through 3, where n$native0 would be a 0 value
on n$native, n$native1 would be a 1 value on n$native etc.
Any help would be greatly appreciated.
-- Mosi
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coding several dummy variables into a single categorical variable
Hi,
Not sure this is what you wanted.
df2- melt(df1,id.vars=c(re,usborn))
df2New-df2[df2$value==1,-4]
df2New$variable-as.numeric(gsub([[:alpha:]],,df2New$variable))
colnames(df2New)[3]- native
row.names(df2New)- 1:nrow(df2New)
df2New
# re usborn native
#1 white yes 0
#2 white yes 0
#3 white yes 0
#4 white yes 0
#5 white yes 0
#6 afam yes 1
#7 afam yes 1
#8 afam yes 1
#9 carib yes 2
#10 carib yes 2
#11 carib yes 2
#12 carib yes 2
#13 carib yes 2
#14 carib no 3
#15 carib no 3
#16 carib no 3
#17 carib no 3
#18 carib no 3
A.K.
- Original Message -
From: arun smartpink...@yahoo.com
To: Mosi Ifatunji ifatu...@gmail.com
Cc: R help r-help@r-project.org
Sent: Sunday, September 22, 2013 3:25 PM
Subject: Re: [R] Coding several dummy variables into a single categorical
variable
Hi,
Try:
set.seed(385)
df- data.frame(re= sample(c(white,afam,carib),20,replace=TRUE), usborn=
sample(c(yes,no),20,replace=TRUE),stringsAsFactors=FALSE)
df1-within(df,{native3- 1*(re==carib usborn==no); native2-
1*(re==carib usborn==yes); native1- 1*(re==afam usborn==yes);
native0- 1*(re==white usborn==yes)})
library(reshape2)
df2- melt(df1,id.vars=c(re,usborn))[,-3]
colnames(df2)[3]- native
A.K.
- Original Message -
From: Mosi Ifatunji ifatu...@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Sunday, September 22, 2013 2:40 PM
Subject: [R] Coding several dummy variables into a single categorical
variable
Colleagues,
I have generated several dummy variables:
n$native0 - 1 * (n$re==white n$usborn==yes)
n$native1 - 1 * (n$re==afam n$usborn==yes)
n$native2 - 1 * (n$re==carib n$usborn==yes)
n$native3 - 1 * (n$re==carib n$usborn==no)
I would now like to combine these into a single categorical variable where the
new variable would be n$native.
And values of native would be 0 through 3, where n$native0 would be a 0 value
on n$native, n$native1 would be a 1 value on n$native etc.
Any help would be greatly appreciated.
-- Mosi
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coding several dummy variables into a single categorical variable
Hi,
If you don't want to install any package:
colnames(df1)[grep(native,colnames(df1))]-
gsub(([[:alpha:]])(\\d+),\\1_\\2,colnames(df1)[grep(native,colnames(df1))])
df2-reshape(df1,direction=long,varying=3:ncol(df1),sep=_)[,-5]
df3-df2[as.logical(df2$native),-4]
colnames(df3)[3]- native
row.names(df3)- 1:nrow(df3)
identical(df2New,df3)
#[1] TRUE
A.K.
- Original Message -
From: arun smartpink...@yahoo.com
To: Mosi Ifatunji ifatu...@gmail.com
Cc: R help r-help@r-project.org
Sent: Sunday, September 22, 2013 3:34 PM
Subject: Re: [R] Coding several dummy variables into a single categorical
variable
Hi,
Not sure this is what you wanted.
df2- melt(df1,id.vars=c(re,usborn))
df2New-df2[df2$value==1,-4]
df2New$variable-as.numeric(gsub([[:alpha:]],,df2New$variable))
colnames(df2New)[3]- native
row.names(df2New)- 1:nrow(df2New)
df2New
# re usborn native
#1 white yes 0
#2 white yes 0
#3 white yes 0
#4 white yes 0
#5 white yes 0
#6 afam yes 1
#7 afam yes 1
#8 afam yes 1
#9 carib yes 2
#10 carib yes 2
#11 carib yes 2
#12 carib yes 2
#13 carib yes 2
#14 carib no 3
#15 carib no 3
#16 carib no 3
#17 carib no 3
#18 carib no 3
A.K.
- Original Message -
From: arun smartpink...@yahoo.com
To: Mosi Ifatunji ifatu...@gmail.com
Cc: R help r-help@r-project.org
Sent: Sunday, September 22, 2013 3:25 PM
Subject: Re: [R] Coding several dummy variables into a single categorical
variable
Hi,
Try:
set.seed(385)
df- data.frame(re= sample(c(white,afam,carib),20,replace=TRUE), usborn=
sample(c(yes,no),20,replace=TRUE),stringsAsFactors=FALSE)
df1-within(df,{native3- 1*(re==carib usborn==no); native2-
1*(re==carib usborn==yes); native1- 1*(re==afam usborn==yes);
native0- 1*(re==white usborn==yes)})
library(reshape2)
df2- melt(df1,id.vars=c(re,usborn))[,-3]
colnames(df2)[3]- native
A.K.
- Original Message -
From: Mosi Ifatunji ifatu...@gmail.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Sunday, September 22, 2013 2:40 PM
Subject: [R] Coding several dummy variables into a single categorical
variable
Colleagues,
I have generated several dummy variables:
n$native0 - 1 * (n$re==white n$usborn==yes)
n$native1 - 1 * (n$re==afam n$usborn==yes)
n$native2 - 1 * (n$re==carib n$usborn==yes)
n$native3 - 1 * (n$re==carib n$usborn==no)
I would now like to combine these into a single categorical variable where the
new variable would be n$native.
And values of native would be 0 through 3, where n$native0 would be a 0 value
on n$native, n$native1 would be a 1 value on n$native etc.
Any help would be greatly appreciated.
-- Mosi
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Re: [R] Creating rectangular plots with x and y coordinates and treatments from a matrix for a randomized block design
Thanks Jim,
What you have done is very good.
I was wondering if we could draw some horizontal and vertical lines to
separate the blocks. Is it possible using R?
Regards,
Laz
On 9/22/2013 11:10 AM, jim holtman wrote:
Is this what you were after:
## A function to generate a RCB design
rcbd-function(b,g,rb,cb,r,c)
+ {
+# b =number of blocks
+# g = a vector of treatments
+# rb = number of rows per blocks
+# cb =number of columns per block
+# r = total rows
+# c = total columns
+ library(foreach)
+ genotypes-times(b) %do% sample(g,length(g))
+ block-rep(1:b,each=length(g))
+ genotypes-factor(genotypes)
+ block-factor(block)
+ ### generate the base design
+ k-c/cb # number of blocks on the x-axis
+ x-rep(rep(1:r,each=cb),k) # X-coordinate
+ l-cb
+ p-r/rb
+ m-l+1
+ d-l*b/p
+ y-c(rep(1:l,r),rep(m:d,r)) # Y-coordinate
+ data.frame(x,y,block,genotypes)
+ }
set.seed(100)
ans - rcbd(b=4,g=1:4,rb=2,cb=2,r=4,c=4)
result - matrix(nrow = max(ans$x), ncol = max(ans$y))
result[cbind(ans$y, ans$x)] - ans$genotypes
result
[,1] [,2] [,3] [,4]
[1,]2423
[2,]1341
[3,]3223
[4,]1441
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Sun, Sep 22, 2013 at 1:07 AM, Laz lmra...@ufl.edu wrote:
Dear R users,
I have a function I created to generate randomized block designs given
below. Once I generate the design, I would like to plot it in a
rectangular form to mimic the excel style where the blocks are
represented with the treatment values placed exactly where their
appropriate X and Y coordinates are.
## A function to generate a RCB design
rcbd-function(b,g,rb,cb,r,c)
{
# b =number of blocks
# g = a vector of treatments
# rb = number of rows per blocks
# cb =number of columns per block
# r = total rows
# c = total columns
library(foreach)
genotypes-times(b) %do% sample(g,length(g))
block-rep(1:b,each=length(g))
genotypes-factor(genotypes)
block-factor(block)
### generate the base design
k-c/cb # number of blocks on the x-axis
x-rep(rep(1:r,each=cb),k) # X-coordinate
l-cb
p-r/rb
m-l+1
d-l*b/p
y-c(rep(1:l,r),rep(m:d,r)) # Y-coordinate
data.frame(x,y,block,genotypes)
}
set.seed(100)
rcbd(b=4,g=1:4,rb=2,cb=2,r=4,c=4)
x y block genotypes
1 1 1 1 2
2 1 2 1 1
3 2 1 1 4
4 2 2 1 3
5 3 1 2 2
6 3 2 2 4
7 4 1 2 3
8 4 2 2 1
9 1 3 3 3
10 1 4 3 1
11 2 3 3 2
12 2 4 3 4
13 3 3 4 2
14 3 4 4 4
15 4 3 4 3
16 4 4 4 1
How can I produce a diagram like this one below or a better one for any run of
my function?
2 4 2 3
1 3 4 1
3 2 2 3
1 4 4 1
Regards,
Laz
[[alternative HTML version deleted]]
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Re: [R] gam and optim
just to clarify how I see the error, it was the mis-definition of the
penalty term in the function dv. The following code corrects this error.
What is actually being minimised at this step is the penalised deviance
conditional on the smoothing parameter. A second issue is that the optim
default (Nelder-Mead) would have to be recycled several times to
approximate the minimum obtained by gam, however, in this example, BFGS
gets it straight away.
sorry for all the confusion!
greg
set.seed(1)
library(mgcv)
x-runif(100)
lp-exp(-2*x)*sin(8*x)
y-rpois(100,exp(lp))
m1-gam(y~s(x),poisson)
W-diag(fitted(m1))
X-predict(m1,type=lpmatrix)
S-m1$smooth[[1]]$S[[1]]
S-rbind(0,cbind(0,S))
A-X%*%solve(t(X)%*%W%*%X+m1$sp*S)%*%t(X)%*%W
sum(diag(A))
sum(m1$edf)
fit-fitted(m1)
b-m1$coef
range(exp(X%*%b)-fit)
z-y/fit-1+X%*%b
range(A%*%z-X%*%b)
s-m1$sp
dv-function(t)
{
f-exp(X%*%t)
-2*sum(y*log(f)-f-ifelse(y==0,0,y*log(y))+y)+s*t%*%S%*%t
}
dv(b)
m1$dev+m1$sp*b%*%S%*%b
t1-optim(rep(0,10),dv,method=BFGS)
t1$p
b
dv(t1$p)
dv(b)
fit1-exp(X%*%t1$p)
plot(x,y)
points(x,exp(lp),pch=16,col=green3)
points(x,fitted(m1),pch=16,cex=0.5,col=blue)
points(x,fit1,cex=1.5,col=red)
please ignore this, I see the error.
greg
hi
probably a silly mistake, but I expected gam to minimise the penalised
deviance.
thanks
greg
set.seed(1)
library(mgcv)
x-runif(100)
lp-exp(-2*x)*sin(8*x)
y-rpois(100,exp(lp))
plot(x,y)
m1-gam(y~s(x),poisson)
points(x,exp(lp),pch=16,col=green3)
points(x,fitted(m1),pch=16,cex=0.5,col=blue)
W-diag(fitted(m1))
X-predict(m1,type=lpmatrix)
S-m1$smooth[[1]]$S[[1]]
S-rbind(0,cbind(0,S))
A-X%*%solve(t(X)%*%W%*%X+m1$sp*S)%*%t(X)%*%W
sum(diag(A))
sum(m1$edf)
fit-fitted(m1)
b-m1$coef
range(exp(X%*%b)-fit)
z-y/fit-1+X%*%b
range(A%*%z-X%*%b)
dv-function(t)
{
f-exp(X%*%t)
-2*sum(y*log(f)-f-ifelse(y==0,0,y*log(y))+y)+t%*%S%*%t
}
dv(b)
m1$dev+b%*%S%*%b
#so far, so good
t1-optim(rep(0,10),dv)
t1$p
b
#different
dv(t1$p)
dv(b)
#different, and dv(t1$p) is lower!
fit1-exp(X%*%t1$p)
points(x,fit1,pch=16,cex=0.5,col=red)
# different
# gam found b which does approximate the true curve, but does not
minimise
the penalised deviance, by a long shot.
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Re: [R] Creating rectangular plots with x and y coordinates and treatments from a matrix for a randomized block design
On 09/23/13 07:52, Laz wrote: SNIP I was wondering if we could draw some horizontal and vertical lines to separate the blocks. Is it possible using R? SNIP See fortune(Yoda). :-) cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Creating rectangular plots with x and y coordinates and treatments from a matrix for a randomized block design
Is Yoda an R package? I tried: install.packages(Yoda) Warning in install.packages : package 'Yoda' is not available (for R version 3.0.1) Warning in install.packages : package 'Yoda' is not available (for R version 3.0.1) How do I proceed from here? Regards, Laz On 9/22/2013 4:21 PM, Rolf Turner wrote: On 09/23/13 07:52, Laz wrote: SNIP I was wondering if we could draw some horizontal and vertical lines to separate the blocks. Is it possible using R? SNIP See fortune(Yoda). :-) cheers, Rolf Turner [[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] Creating rectangular plots with x and y coordinates and treatments from a matrix for a randomized block design
Hello, No, 'Yoda' is not a package, the package is 'fortunes': library(fortunes) # load it in the R session fortune(Yoda) Or, if you don't want to load the package, fortunes::fortune(Yoda) Rui Barradas Em 22-09-2013 21:32, Laz escreveu: Is Yoda an R package? I tried: install.packages(Yoda) Warning in install.packages : package 'Yoda' is not available (for R version 3.0.1) Warning in install.packages : package 'Yoda' is not available (for R version 3.0.1) How do I proceed from here? Regards, Laz On 9/22/2013 4:21 PM, Rolf Turner wrote: On 09/23/13 07:52, Laz wrote: SNIP I was wondering if we could draw some horizontal and vertical lines to separate the blocks. Is it possible using R? SNIP See fortune(Yoda). :-) cheers, Rolf Turner [[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] Creating rectangular plots with x and y coordinates and treatments from a matrix for a randomized block design
Hi, fortune(Yoda) returns a quote of fortunes. My problem was to draw horizontal and vertical lines to separate the blocks in my design. Each block has 4 elements. The first and second rows and columns are in block 1, 3rd and 4th rows and 1st and 2nd columns are in 3 etc [,1] [,2] [,3] [,4] [1,]2423 [2,]1341 [3,]3223 [4,]1441 On 9/22/2013 4:38 PM, Rui Barradas wrote: fortune(Yoda). :-) __ 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] Coding several dummy variables into a single categorical variable
Hi Mosi, The easiest approach would be be generate native from re and usborn directly, rather than generating dummy codes as an intermediate step: n$native - interaction(n[c(re, usborn)]) This has the advantage of preserving the meanings of the levels of native in the resulting factor, so that you don't have to remember what 0,1,2,3 mean. If for some reason you prefer the numeric levels you can then change them using the levels() function. If for some reason you don't have the original variables (re and usborn in your example), you can construct a factor from the dummy codes as follows: n$native - factor( with(n, paste(native0, native1, native2, native3, sep = )), levels = c(1000, 0100, 0010, 0001), labels = c(0, 1, 2, 3)) Best, Ista On Sun, Sep 22, 2013 at 2:40 PM, Mosi Ifatunji ifatu...@gmail.com wrote: Colleagues, I have generated several dummy variables: n$native0 - 1 * (n$re==white n$usborn==yes) n$native1 - 1 * (n$re==afam n$usborn==yes) n$native2 - 1 * (n$re==carib n$usborn==yes) n$native3 - 1 * (n$re==carib n$usborn==no) I would now like to combine these into a single categorical variable where the new variable would be n$native. And values of native would be 0 through 3, where n$native0 would be a 0 value on n$native, n$native1 would be a 1 value on n$native etc. Any help would be greatly appreciated. -- Mosi __ 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] Conditioning plots (wth coplot function) with logistic regression curves
On 9/21/2013 11:12 PM, Kiyoshi Sasaki wrote: I have been trying to produce a conditional plot using coplot function (http://stat.ethz.ch/R-manual/R-devel/library/graphics/html/coplot.html) for a binary response (Presence in my case) variable and one continuous variable (Overstory) given a specific levels of the other continuous variable (Ivy). But, my codes produces an overlapping graph. Also, I want to use three equal intervals for Ivy (i.e.,33.3 each), but I could not figure out how. Here is my data and codes I used: If you feel it hurts because you are banging your head into a wall trying to get coplot to do this, the answer is: Don't do that! Instead, you might consider using ggplot2, which handles this case nicely, as far as I can tell from your description. But first a due diligence caveat: Say you come to me for consulting on this little plotting question. I look at your data frame, dat, and I see there are a number of other variables that might explain Presence, so maybe the marginal plot that ignores them could be misleading, e.g., any of Moist, Leaf, Prey, ... could moderate the relation between overstory and presence, but you won't see that in a marginal plot. Here are a couple of quick ggplot examples, plotting classes of Ivy in the same plot frame, and on the probability scale. library(ggplot2) ggplot(dat, aes(Overstory, Presence), color=Ivy50 ) + stat_smooth(method=glm, family=binomial, formula= y~x, alpha=0.3, aes(fill=Ivy50)) dat$Ivy3 - factor(cut(dat$Ivy,3)) plt - ggplot(dat, aes(Overstory, Presence), color=Ivy3 ) + stat_smooth(method=glm, family=binomial, formula= y~x, alpha=0.3, aes(fill=Ivy3)) plt If you want separate panels, try something like plt + facet_grid(. ~ Ivy3) For plots on the logit scale, try something like logit - function(p) log(p)/log(1-p), then plt + coord_trans(y=logit) -- 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 __ 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] Arcsine transformation
I am tryin to perform an arcsine transformation on my data containig percentages as the dep. variable. Does anyone have a code that I could use to do that? I am relatively new to R. Thanks for your help! -- View this message in context: http://r.789695.n4.nabble.com/Arcsine-transformation-tp4676706.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] Creating rectangular plots with x and y coordinates and treatments from a matrix for a randomized block design
CHeck out the 'tables' package if you want to create pretty outputs of your tables. Exactly where do you plan to use them? You can also use Sweave/Latex to create such tables. Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. On Sun, Sep 22, 2013 at 4:45 PM, Laz lmra...@ufl.edu wrote: Hi, fortune(Yoda) returns a quote of fortunes. My problem was to draw horizontal and vertical lines to separate the blocks in my design. Each block has 4 elements. The first and second rows and columns are in block 1, 3rd and 4th rows and 1st and 2nd columns are in 3 etc [,1] [,2] [,3] [,4] [1,]2423 [2,]1341 [3,]3223 [4,]1441 On 9/22/2013 4:38 PM, Rui Barradas wrote: fortune(Yoda). :-) __ 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] Conditioning plots (wth coplot function) with logistic regression curves
Thank you! Your codes produced plots that was fairly close to what I wanted. I am not familiar with ggplot2, so I need to study what exactly each code is doing. Actually, I also want to plot the observed (raw) data points (not just confidence intervals shown by ggplot, assuming that colored portion is confidence intervals). I actually produced the one I wanted before, but my hard drive died and lost all R codes I wrote are lost. I have posted an example plot I created before using coplot function (downloadable in PowerPoint format from http://laurentian.academia.edu/KiyoshiSasaki/Miscellaneous ). Could you please show me how to plot raw data points in those ggplots? And, can anyone help me reproduce that plot I made before using coplot? I just cannot figure out how I did to make that plot (I spent several days...). Thank you for taking your time to help me out. Sincerely, Kiyoshi From: Michael Friendly frien...@yorku.ca Cc: r-help@r-project.org r-help@r-project.org Sent: Sunday, September 22, 2013 7:46:04 PM Subject: Re: Conditioning plots (wth coplot function) with logistic regression curves On 9/21/2013 11:12 PM, Kiyoshi Sasaki wrote: I have been trying to produce a conditional plot using coplot function (http://stat.ethz.ch/R-manual/R-devel/library/graphics/html/coplot.html) for a binary response (Presence in my case) variable and one continuous variable (Overstory) given a specific levels of the other continuous variable (Ivy). But, my codes produces an overlapping graph. Also, I want to use three equal intervals for Ivy (i.e.,33.3 each), but I could not figure out how. Here is my data and codes I used: If you feel it hurts because you are banging your head into a wall trying t[[elided Yahoo spam]] Instead, you might consider using ggplot2, which handles this case nicely, as far as I can tell from your description. But first a due diligence caveat: Say you come to me for consulting on this little plotting question. I look at your data frame, dat, and I see there are a number of other variables that might explain Presence, so maybe the marginal plot that ignores them could be misleading, e.g., any of Moist, Leaf, Prey, ... could moderate the relation between overstory and presence, but you won't see that in a marginal plot. Here are a couple of quick ggplot examples, plotting classes of Ivy in the same plot frame, and on the probability scale. library(ggplot2) ggplot(dat, aes(Overstory, Presence), color=Ivy50 ) + stat_smooth(method=glm, family=binomial, formula= y~x, alpha=0.3, aes(fill=Ivy50)) dat$Ivy3 - factor(cut(dat$Ivy,3)) plt - ggplot(dat, aes(Overstory, Presence), color=Ivy3 ) + stat_smooth(method=glm, family=binomial, formula= y~x, alpha=0.3, aes(fill=Ivy3)) plt If you want separate panels, try something like plt + facet_grid(. ~ Ivy3) For plots on the logit scale, try something like logit - function(p) log(p)/log(1-p), then plt + coord_trans(y=logit) -- 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 Street Web: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA [[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] Arcsine transformation
peake peake.19 at osu.edu writes: I am tryin to perform an arcsine transformation on my data containig percentages as the dep. variable. Does anyone have a code that I could use to do that? I am relatively new to R. Thanks for your help! asin(x/100) ? or asin(x/100)*2/pi if you want the results rescaled to (0,1) curve(asin(x/100)*2/pi,from=0,to=100) __ 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] Arcsine transformation
Tena koe
I think you'll find the arcsine transformation is asin(sqrt(x/100)) where × is
the percentage. However, it might be better to ask whether the data wouldn't
be better analysed using generalised models (e.g., glm).
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Ben Bolker
Sent: Monday, 23 September 2013 12:54 p.m.
To: r-h...@stat.math.ethz.ch
Subject: Re: [R] Arcsine transformation
peake peake.19 at osu.edu writes:
I am tryin to perform an arcsine transformation on my data containig
percentages as the dep. variable. Does anyone have a code that I could
use to do that? I am relatively new to R. Thanks for your help!
asin(x/100)
? or
asin(x/100)*2/pi if you want the results rescaled to (0,1)
curve(asin(x/100)*2/pi,from=0,to=100)
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The contents of this e-mail are confidential and may be ...{{dropped:14}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Arcsine transformation
On 13-09-22 09:25 PM, Peter Alspach wrote: Tena koe I think you'll find the arcsine transformation is asin(sqrt(x/100)) where × is the percentage. However, it might be better to ask whether the data wouldn't be better analysed using generalised models (e.g., glm). HTH Good point about arcsine-sqrt, and about GLMs: specifically see Warton, David I., and Francis K. C. Hui. 2011. “The Arcsine Is Asinine: The Analysis of Proportions in Ecology.” Ecology 92 (1) (January): 3–10. doi:10.1890/10-0340.1. http://www.esajournals.org/doi/full/10.1890/10-0340.1. I think the title is a little silly, but it's worth reading. Peter Alspach -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Ben Bolker Sent: Monday, 23 September 2013 12:54 p.m. To: r-h...@stat.math.ethz.ch Subject: Re: [R] Arcsine transformation peake peake.19 at osu.edu writes: I am tryin to perform an arcsine transformation on my data containig percentages as the dep. variable. Does anyone have a code that I could use to do that? I am relatively new to R. Thanks for your help! asin(x/100) ? or asin(x/100)*2/pi if you want the results rescaled to (0,1) curve(asin(x/100)*2/pi,from=0,to=100) __ 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. The contents of this e-mail are confidential and may be subject to legal privilege. If you are not the intended recipient you must not use, disseminate, distribute or reproduce all or any part of this e-mail or attachments. If you have received this e-mail in error, please notify the sender and delete all material pertaining to this e-mail. Any opinion or views expressed in this e-mail are those of the individual sender and may not represent those of The New Zealand Institute for Plant and Food Research Limited. __ 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] PLS1 NIPALS question: error with chemometrics package?
I am doing a self study, trying to understand PLS.
I have run across the following and hope that someone here can clarify as
to what is going on.
These are the data from Wold et al. (1984)
dat - structure(list(t = 1:15, y = c(4.39, 4.42, 5, 5.85, 4.35, 4.51,
6.33, 6.37, 4.68, 5.04, 7.1, 5.04, 6, 5.48, 7.1), x1 = c(4.55,
4.74, 5.07, 5.77, 4.62, 4.41, 6.17, 6.17, 4.33, 4.62, 7.22, 4.64,
5.62, 6.19, 7.85), x2 = c(8.93, 8.93, 9.29, 9.9, 9.9, 9.93, 9.19,
9.19, 10.03, 10.29, 9.29, 10.22, 9.94, 9.77, 9.29), x3 = c(1.14,
1.14, 1.14, 1.14, 1.14, 1.14, 1.14, 1.14, 1.14, 1.14, 1.14, 1.14,
-0.07, -0.07, -0.07), x4 = c(0.7, 1.23, 0.19, 0.19, 1.23, 1.23,
0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19, 0.19), x5 = c(0.19,
0.19, 0.7, 1.64, 1.64, 2.35, 2.83, 2.56, 2.42, 3.36, 2.43, 2.95,
1.64, 1.64, 3.8), x6 = c(0.49, 0.49, 0, -0.1, -0.1, -0.2, -0.13,
-0.13, -0.08, -0.13, -0.3, -0.08, -0.19, -0.19, -0.3), x7 = c(1.24,
1.24, 0, -0.47, -0.47, -0.51, -0.93, -0.93, -0.38, -0.93, -1.6,
-0.38, -0.47, -0.47, -1.6), x8 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 0L, 0L, 1L, 0L, 1L, 1L, 1L)), .Names = c(t, y, x1,
x2, x3, x4, x5, x6, x7, x8), class = data.frame, row.names
= c(NA,
-15L))
In Wold et al. (1984) (pg 741, Table 2) beta coefficients are given for PLS
(1) and PLS (2) where (1) and (2) correspond to the number of PLS
components retained. The coefficients are not the same as those when I run
the following:
library(chemometrics)
# retaining 1 component:
pls1_nipals(X=dat[,c(3:10)], y=dat[2], a=1, scale=TRUE)$b
# retaining 2 components:
pls1_nipals(X=dat[,c(3:10)], y=dat[2], a=2, scale=TRUE)$b
However, if I modify the source code like this:
pls1_nipals_mod - function(X, y, a, it = 50, tol = 1e-08, scale = FALSE)
{
Xh - scale(X, center = TRUE, scale = scale)
yh - scale(y, center = TRUE, scale = scale)
T - NULL
P - NULL
C - NULL
W - NULL
for (h in 1:a) {
wh - t(Xh) %*% yh/sum(yh^2) # modified: / SS
wh - wh/as.vector(sqrt(t(wh) %*% wh))
th - Xh %*% wh
ch - as.numeric(t(yh) %*% th)/sum(th^2) # modified: / SS
# ch - ch/as.vector(sqrt(t(th) %*% th)) # modified: removed
normalization of ch
ph - t(Xh) %*% th/as.vector(t(th) %*% th)
Xh - Xh - th %*% t(ph)
yh - yh - th * ch
T - cbind(T, th)
P - cbind(P, ph)
C - c(C, ch)
W - cbind(W, wh)
}
b - W %*% solve(t(P) %*% W) %*% C
list(P = P, T = T, W = W, C = C, b = b)
}
pls1_nipals_mod(X=dat[,c(3:10)], y=dat[2], a=1, scale=TRUE)$b
pls1_nipals_mod(X=dat[,c(3:10)], y=dat[2], a=2, scale=TRUE)$b
These beta coefficients are exactly the same as in Wold et al. (1984)
Furthermore, if I do a leave-one-out CV, my modified version has a PRESS =
1.27, whereas the original pls1_nipals() function has a PRESS = 18.11!
That's not good, right? Here is my LOOCV code, 1:1 lines added in plots:
### LOOCV for original function
out.j - vector(list, length=nrow(dat))
for(j in c(2:nrow(dat), 1)){
b - pls1_nipals(X=dat[-j,c(3:10)], y=dat[-j,2], a=2, scale=TRUE)$b
dats - scale(dat)
y.est - dats[j,c(3:10)] %*% b
y.obs - dats[j,2]
out.j[[j]] - data.frame(y.obs, y.est)
}
out - do.call(rbind, out.j)
sqrt(sum((out[,1]-out[,2])^2) )
plot(out[,2]~ out[,1], ylab=pred, xlab=obs)
abline(0,1, col=grey)
### LOOCV for modified function
out.j - vector(list, length=nrow(dat))
for(j in c(2:nrow(dat), 1)){
b - pls1_nipals_mod(X=dat[-j,c(3:10)], y=dat[-j,2], a=2, scale=TRUE)$b
dats - scale(dat)
y.est - dats[j,c(3:10)] %*% b
y.obs - dats[j,2]
out.j[[j]] - data.frame(y.obs, y.est)
}
out - do.call(rbind, out.j)
sqrt(sum((out[,1]-out[,2])^2) )
plot(out[,2]~ out[,1], ylab=pred, xlab=obs)
abline(0,1, col=grey)
Is this an error with the chemometrics function; or am I simply
understanding something incorrectly?
Thank you.
Citation: Wold, S., A. Ruhe, H. Wold, W. Dunn. (1984) The collinearity
problem in linear regression. The partial least squares (PLS) approach to
generalized inverses* SIAM J. Sci. Stat. Comput.
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Re: [R] Arcsine transformation
Thanks for the advice. Like I said, I am still pretty new to R, and stats in general. I tried wading through the search results on google, but I didn't really find anything that I could understand. I am working with percentages as my dependent variable, and I am trying to get my distribution as close to normal as possible. Is an arcsine transformation even the correct choice for my data? Querying R for help usually leaves me even more confused, so I was hoping someone could help me out by walking me through the process. On Sun, Sep 22, 2013 at 8:00 PM, chuck.01 [via R] ml-node+s789695n4676708...@n4.nabble.com wrote: ?asin also, try Googling anything you might want to do in R... it is there also, google... R cheatsheet you will find several helpful sheets of useful functions. peake wrote I am tryin to perform an arcsine transformation on my data containig percentages as the dep. variable. Does anyone have a code that I could use to do that? I am relatively new to R. Thanks for your help! -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Arcsine-transformation-tp4676706p4676708.html To unsubscribe from Arcsine transformation, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4676706code=cGVha2UuMTlAb3N1LmVkdXw0Njc2NzA2fDE4MjEzMDk3MTU= . NAMLhttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewerid=instant_html%21nabble%3Aemail.namlbase=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespacebreadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml -- View this message in context: http://r.789695.n4.nabble.com/Arcsine-transformation-tp4676706p4676712.html Sent from the R help mailing list archive at Nabble.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] xlim with barplot
Hi, I want to compare to barplots with same horizontal axis limits. x=c(55,56,57,58,59,60,60,60,61,62,63,64,65) y=c(35,40,45,50,55,60,60,60,65,70,75,80,85) par(mfrow=c(2,1)) barplot(table(x),xlim=c(35,85)) barplot(table(y),xlim=c(35,85)) par(mfrow=c(1,1)) But the bars disappear. http://r.789695.n4.nabble.com/file/n4676717/Rplot.png Any suggestions? -- View this message in context: http://r.789695.n4.nabble.com/xlim-with-barplot-tp4676717.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] Correlate rows of 2 matrices
Hi, You may try: set.seed(49) m1 = matrix(rnorm(30), nrow = 3) m2 = matrix(rnorm(30), nrow = 3) corsP-vector() for(i in 1:3) corsP[i] = cor(m1[i,], m2[i,]) corsP #[1] 0.58411274 -0.02382329 0.03760757 diag(cor(t(m1),t(m2))) #[1] 0.58411274 -0.02382329 0.03760757 #or mNew- rbind(m1,m2) indx-rep(seq(nrow(mNew)/2),2) sapply(split(seq_len(nrow(mNew)),indx),function(x) cor(t(mNew[x,]),t(mNew[x,]))[2]) # 1 2 3 #0.58411274 -0.02382329 0.03760757 #or tapply(seq_along(indx),list(indx),FUN=function(x) cor(t(mNew[x,]),t(mNew[x,]))[2]) # 1 2 3 #0.58411274 -0.02382329 0.03760757 A.K. From: Ira Sharenow irasharenow...@yahoo.com To: arun smartpink...@yahoo.com Sent: Sunday, September 22, 2013 9:57 PM Subject: Correlate rows of 2 matrices Arun, I have a new problem for you. I have two data frames (or matrices) and row by row I want to take the correlations. So if I have a 3 row by 10 column matrix, I would produce 3 correlations. Is there a way to merge the matrices and then use some sort of split? Ideas/solutions much appreciated. m1 = matrix(rnorm(30), nrow = 3) m2 = matrix(rnorm(30), nrow = 3) set.seed(22) m1 = matrix(rnorm(30), nrow = 3) m2 = matrix(rnorm(30), nrow = 3) for(i in 1:3) corsP[i] = cor(m1[i,], m2[i,]) corsP [1] -0.50865019 -0.27760046 0.01423144 Thanks. Ira __ 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] xlim with barplot
On 09/23/2013 12:55 PM, David Arnold wrote: Hi, I want to compare to barplots with same horizontal axis limits. x=c(55,56,57,58,59,60,60,60,61,62,63,64,65) y=c(35,40,45,50,55,60,60,60,65,70,75,80,85) par(mfrow=c(2,1)) barplot(table(x),xlim=c(35,85)) barplot(table(y),xlim=c(35,85)) par(mfrow=c(1,1)) But the bars disappear. http://r.789695.n4.nabble.com/file/n4676717/Rplot.png Any suggestions? Hi David, The first suggestion is: par(usr) [1] -0.32 13.72 -0.03 3.00 Specifying the x limits as above means that the bars are floating somewhere off to the left of the plot and thus not visible. You are mistaking the labels of the bars for their position. Now, having admonished you like some grumpy old school teacher, I suppose I should do something useful: barplot(tabulate(x,nbins=85)[35:85]) barplot(tabulate(y,nbins=85)[35:85]) This is an underhanded trick to line up the bars as I think you want them. I suppose you want x labels as well: barpos-barplot(tabulate(x,nbins=85)[35:85],names.arg=xylabels) axis(1,at=barpos[c(6,16,26,36,46)],labels=c(40,50,60,70,80)) barplot(tabulate(y,nbins=85)[35:85],names.arg=xylabels) axis(1,at=barpos[c(6,16,26,36,46)],labels=c(40,50,60,70,80)) 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.
[R] legend for the plot with type = b
Hi there, I plot a simple plot with the following code: plot (rnorm(1:10), type = b) legend(top, test, lty = 1, pch = 21) The result is something wired for the line crosses the point in the legend while the line does not cross the point in the main plot. Is there possibility to draw the legend that line does not cross the point, i.e., like the pattern in the main plot? Any help is really appreciated. Best regards, Jinsong __ 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] xlim with barplot
You could try ggplot() as well. library(ggplot2) library(gridExtra) library(plyr) x1- count(x) y1- count(y) p1-ggplot(x1,aes(x=x,y=freq))+geom_bar(stat=identity,colour=gray,fill=red)+xlim(c(35,85))+ theme_bw()+ theme(axis.line=element_line(colour=black), panel.grid.major=element_blank(), panel.grid.minor=element_blank(), panel.border=element_blank(),panel.background=element_blank()) p2-ggplot(y1,aes(x=x,y=freq))+geom_bar(stat=identity,colour=gray,fill=blue)+xlim(c(35,85)) +theme_bw()+ theme(axis.line=element_line(colour=black), panel.grid.major=element_blank(), panel.grid.minor=element_blank(), panel.border=element_blank(),panel.background=element_blank()) grid.arrange(p1,p2,nrow=2) A.K. - Original Message - From: David Arnold dwarnol...@suddenlink.net To: r-help@r-project.org Cc: Sent: Sunday, September 22, 2013 10:55 PM Subject: [R] xlim with barplot Hi, I want to compare to barplots with same horizontal axis limits. x=c(55,56,57,58,59,60,60,60,61,62,63,64,65) y=c(35,40,45,50,55,60,60,60,65,70,75,80,85) par(mfrow=c(2,1)) barplot(table(x),xlim=c(35,85)) barplot(table(y),xlim=c(35,85)) par(mfrow=c(1,1)) But the bars disappear. http://r.789695.n4.nabble.com/file/n4676717/Rplot.png Any suggestions? -- View this message in context: http://r.789695.n4.nabble.com/xlim-with-barplot-tp4676717.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] xlim with barplot
On 09/23/2013 01:56 PM, arun wrote: You could try ggplot() as well. library(ggplot2) library(gridExtra) library(plyr) x1- count(x) y1- count(y) p1-ggplot(x1,aes(x=x,y=freq))+geom_bar(stat=identity,colour=gray,fill=red)+xlim(c(35,85))+ theme_bw()+ theme(axis.line=element_line(colour=black), panel.grid.major=element_blank(), panel.grid.minor=element_blank(), panel.border=element_blank(),panel.background=element_blank()) p2-ggplot(y1,aes(x=x,y=freq))+geom_bar(stat=identity,colour=gray,fill=blue)+xlim(c(35,85)) +theme_bw()+ theme(axis.line=element_line(colour=black), panel.grid.major=element_blank(), panel.grid.minor=element_blank(), panel.border=element_blank(),panel.background=element_blank()) grid.arrange(p1,p2,nrow=2) A.K. Hi arun, Okay, if we're allowed to use packages, challenge taken: library(plotrix) barp(tabulate(x,nbins=85)[35:85],names.arg=35:85) barp(tabulate(y,nbins=85)[35:85],names.arg=35:85) 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] legend for the plot with type = b
Hi, May be this helps. set.seed(55) x-rnorm(1:10) plot(x,type=n,xaxt=n,yaxt=n) legend1- legend(top,test,lty=1,pch=21) range1- range(x) range1[2]- 1.05* (range1[2]+ legend1$rect$h) plot(x,ylim=range1,type=b) legend1- legend(top,test,lty=1,pch=21) A.K. - Original Message - From: Jinsong Zhao jsz...@yeah.net To: R help r-help@r-project.org Cc: Sent: Sunday, September 22, 2013 11:54 PM Subject: [R] legend for the plot with type = b Hi there, I plot a simple plot with the following code: plot (rnorm(1:10), type = b) legend(top, test, lty = 1, pch = 21) The result is something wired for the line crosses the point in the legend while the line does not cross the point in the main plot. Is there possibility to draw the legend that line does not cross the point, i.e., like the pattern in the main plot? Any help is really appreciated. Best regards, Jinsong __ 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] basic matrix function
Hi, Use `drop=FALSE`. b- matrix(c(2,1,-1,-2),ncol=1) b[1:3,1] #[1] 2 1 -1 b[1:3,1,drop=FALSE] #or b[1:3,,drop=FALSE] # [,1] #[1,] 2 #[2,] 1 #[3,] -1 A.K. hi all, i got a small question tonight. matrix(b,4)[] [,1] [1,] 2 [2,] 1 [3,] -1 [4,] -2 dim(matrix(betan,4)) [1] 4 1 As shown, b is a 4X1 matrix. matrix(betan,4)[1:3,1] [1] 2 1 -1 However, I think the result should be [,1] [1,] 2 [2,] 1 [3,] -1 How could I get the result above? Many thanks, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] legend for the plot with type = b
On Sep 22, 2013, at 10:54 PM, Jinsong Zhao wrote: Hi there, I plot a simple plot with the following code: plot (rnorm(1:10), type = b) legend(top, test, lty = 1, pch = 21) ?par plot (rnorm(1:10), type = b) legend(top, test, lty = 69, pch = 21) The result is something wired for the line crosses the point in the legend while the line does not cross the point in the main plot. Is there possibility to draw the legend that line does not cross the point, i.e., like the pattern in the main plot? Any help is really appreciated. -- David Winsemius, MD 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.
