Re: [R] Bus stop sequence matching problem
Try dtw. First convert ref to numeric since dtw does not handle
character input. Then align using dtw and NA out repeated values in
the alignment. Finally zap ugly row names and calculate loading:
library(dtw)
s1 - as.numeric(stop_sequence$ref)
s2 - as.numeric(factor(as.character(stop_onoff$ref),
levels(stop_sequence$ref)))
a - dtw(s1, s2)
DF - cbind(stop_sequence,
stop_onoff[replace(a$index2, c(FALSE, diff(a$index2) == 0), NA), ])[-3]
rownames(DF) - NULL
transform(DF, loading = cumsum(ifelse(is.na(on), 0, on)) -
cumsum(ifelse(is.na(off), 0, off)))
giving:
seq ref on off loading
1 10 A 5 0 5
2 20 B NA NA 5
3 30 C NA NA 5
4 40 D 0 2 3
5 50 B 10 2 11
6 60 A 0 6 5
You will need to test this with more data and tweak it if necessary
via the various dtw arguments.
On Fri, Aug 29, 2014 at 8:46 PM, Adam Lawrence alaw...@gmail.com wrote:
I am hoping someone can help me with a bus stop sequencing problem in R,
where I need to match counts of people getting on and off a bus to the
correct stop in the bus route stop sequence. I have tried looking
online/forums for sequence matching but seems to refer to numeric sequences
or DNA matching and over my head. I am after a simple example if anyone can
please help.
I have two data series as per below (from database), that I want to
combine. In this example “stop_sequence” includes the equence (seq) of bus
stops and “stop_onoff” is a count of people getting on and off at certain
stops (there is no entry if noone gets on or off).
stop_sequence - data.frame(seq=c(10,20,30,40,50,60),
ref=c('A','B','C','D','B','A'))
## seq ref
## 1 10 A
## 2 20 B
## 3 30 C
## 4 40 D
## 5 50 B
## 6 60 A
stop_onoff -
data.frame(ref=c('A','D','B','A'),on=c(5,0,10,0),off=c(0,2,2,6))
## ref on off
## 1 A 5 0
## 2 D 0 2
## 3 B 10 2
## 4 A 0 6
I need to match the stop_onoff numbers in the right sto sequence, with the
correctly matched output as follows (load is a cumulative count of on and
off)
desired_output - data.frame(seq=c(10,20,30,40,50,60),
ref=c('A','B','C','D','B','A'),
on=c(5,'-','-',0,10,0),off=c(0,'-','-',2,2,6), load=c(5,0,0,3,11,5))
## seq ref on off load
## 1 10 A 5 05
## 2 20 B - -0
## 3 30 C - -0
## 4 40 D 0 23
## 5 50 B 10 2 11
## 6 60 A 0 65
In this example the stop “B” is matched to the second stop “B” in the stop
sequence and not the first because the onoff data is after stop “D”.
Any guidance much appreciated.
Regards
Adam
[[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.
--
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] Bus stop sequence matching problem
Homework? The list has a no homework policy - but perhaps I'll be forgiven por
posting hints.
In general terms, this is how I appraoched the problem:
* Loop through the rows of stop_onoff - for (idx in ...someething...) {...
* For each row, find the first of ref in a suitably filtered subset of
stop_sequence, and keep track of these row numbers
* Update columns on and off
* Use cumsum to calculate the number of passengers on the bus
Note the loop. Someone cleverer than I might be able to vectorise that step,
but I couldn't see how.
By the way, if this is homework...
Are you sure you're desired_output is correct? I would expect someething like
seq ref on off load
1 10 A 5 05
2 20 B 0 05
3 30 C 0 05
4 40 D 0 23
5 50 B 10 2 11
6 60 A 0 65
Are you aware that you're ref ccolumns are factors and not characters? If
you use stringsAsFactors = FALSE or
stop_onoff -
data.frame(ref=factor(c('A','D','B','A'), levels =
levels(stop_sequence$ref)),on=c(5,0,10,0),off=c(0,2,2,6))
it will simplify your'e analysis (or at least reduce some typing).
Type the following in an R console
?data.frame
?factor
and have a read.
Now, if this ain't homework, or you just want someone to do it for you, e-mail
me offline and I'll send you my appraoch. If it is homework, let me know - I'm
happy to help anyway, but I will be trying to help you solve this for
yourself.
Cheers,
DMcP
On Sat, 30 Aug 2014 12:46:17 +1200 Adam Lawrence alaw...@gmail.com wrote
I am hoping someone can help me with a bus stop sequencing problem in R,
where I need to match counts of people getting on and off a bus to the
correct stop in the bus route stop sequence. I have tried looking
online/forums for sequence matching but seems to refer to numeric sequences
or DNA matching and over my head. I am after a simple example if anyone can
please help.
I have two data series as per below (from database), that I want to
combine. In this example “stop_sequence” includes the equence (seq) of
bus
stops and “stop_onoff” is a count of people getting on and off at
certain
stops (there is no entry if noone gets on or off).
stop_sequence - data.frame(seq=c(10,20,30,40,50,60),
ref=c('A','B','C','D','B','A'))
## seq ref
## 1 10 A
## 2 20 B
## 3 30 C
## 4 40 D
## 5 50 B
## 6 60 A
stop_onoff -
data.frame(ref=c('A','D','B','A'),on=c(5,0,10,0),off=c(0,2,2,6))
## ref on off
## 1 A 5 0
## 2 D 0 2
## 3 B 10 2
## 4 A 0 6
I need to match the stop_onoff numbers in the right sto sequence, with the
correctly matched output as follows (load is a cumulative count of on and
off)
desired_output - data.frame(seq=c(10,20,30,40,50,60),
ref=c('A','B','C','D','B','A'),
on=c(5,'-','-',0,10,0),off=c(0,'-','-',2,2,6), load=c(5,0,0,3,11,5))
## seq ref on off load
## 1 10 A 5 05
## 2 20 B - -0
## 3 30 C - -0
## 4 40 D 0 23
## 5 50 B 10 2 11
## 6 60 A 0 65
In this example the stop “B” is matched to the second stop “B” in
the stop
sequence and not the first because the onoff data is after stop “D”.
Any guidance much appreciated.
Regards
Adam
[[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.
South Africas premier free email service - www.webmail.co.za
Cheapest Insurance Quotes!
https://www.outsurance.co.za/insurance-quote/personal/?source=msncr=Postit14_468x60_gifcid=322
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with sapa package and spectral density function (SDF)
Hello, Did you find a solution to this problem in the R mailing list ? I am having the same problem but there were apparently no replies to your question or didnt find them ? Thanks Anusha [[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] new error with QuantMod getSymbols
You didn't provide the file lista.csv, so it's not possible to
reproduce any of these errors. And there's no call to getSymbols in
your code. You use loadSymbols, and I am not familiar with that
function.
That said, this sounds like an issue with some of the data being sent
by Yahoo Finance.
--
Joshua Ulrich | about.me/joshuaulrich
FOSS Trading | www.fosstrading.com
On Thu, Aug 28, 2014 at 11:12 PM, Adolfo Yanes adolfoya...@gmail.com wrote:
Hello,
I use getSymbols function daily to run some models with stock data. Today
when I tried to update the stock info i get this error
Error in charToDate(x) :
character string is not in a standard unambiguous format
Sometimes I get it after 2 symbols, other times after 150 symbols, another
time after 40 symbols, then after 203 symbols.
The code for the symbol list is:
lista-read.csv(lista.csv, header=FALSE)
lista.list.ana-vector('list',nrow(lista))
names(lista.list.ana) - lista[,1]
lista.sum-as.vector(lista[,1])
##actualizar la lista
lista_simbolos-download_symbols(lista.sum, lista.list.ana)
*The code for the function download_symbols is:*
download_symbols- function(lista.sum.,lista.list.ana..){
newnames.- c(Open, High, Low, Close, Volume, Adjusted)
for (m in 1:length(lista.sum.))
{
print(paste(c(Downloading symbol , lista.sum.[m], . , length(lista.sum.
)-m, symbols missing), sep=, collapse=))
temp-get(loadSymbols(lista.sum.[m]))
names(temp)-newnames.
#lista.list.ana..[[m]]-loadSymbols(lista.sum.[m])
lista.list.ana..[[m]]-temp
}
return(lista.list.ana..)
}
Is it something wrong with yahoo? I tried google and got another error
Error in `colnames-`(`*tmp*`, value = c(Open, High, Low, Close, :
length of 'dimnames' [2] not equal to array extent
Thanks for your help
--
Adolfo Yanes Musetti
[[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] Bus stop sequence matching problem
Adam Lawrence alaw005 at gmail.com writes:
I am hoping someone can help me with a bus stop sequencing problem in R,
where I need to match counts of people getting on and off a bus to the
correct stop in the bus route stop sequence. I have tried looking
online/forums for sequence matching but seems to refer to numeric
sequences
or DNA matching and over my head. I am after a simple example if anyone
can
please help.
Adam,
Yet another way...
See inline code. BTW, you should have mentioned that you are
a transit planner or included a signature block so folks would know this
is not a homework question.
As others have noted/hinted, there are some unstated assumptions, so
you need to try some test cases to be sure any solution always works.
You only have one outbound/inbound cycle in stop_onoff, right??
If not, I think almost any approach can fail given the right
sequence of 'seq's.
I have two data series as per below (from database), that I want to
combine. In this example “stop_sequence” includes the equence (seq) of bus
stops and “stop_onoff” is a count of people getting on and off at certain
stops (there is no entry if noone gets on or off).
stop_sequence - data.frame(seq=c(10,20,30,40,50,60),
ref=c('A','B','C','D','B','A'))
## seq ref
## 1 10 A
## 2 20 B
## 3 30 C
## 4 40 D
## 5 50 B
## 6 60 A
stop_onoff -
data.frame(ref=c('A','D','B','A'),on=c(5,0,10,0),off=c(0,2,2,6))
## ref on off
## 1 A 5 0
## 2 D 0 2
## 3 B 10 2
## 4 A 0 6
I need to match the stop_onoff numbers in the right sto sequence, with the
correctly matched output as follows (load is a cumulative count of on and
off)
desired_output - data.frame(seq=c(10,20,30,40,50,60),
ref=c('A','B','C','D','B','A'),
on=c(5,'-','-',0,10,0),off=c(0,'-','-',2,2,6), load=c(5,0,0,3,11,5))
## seq ref on off load
## 1 10 A 5 05
## 2 20 B - -0
## 3 30 C - -0
## 4 40 D 0 23
## 5 50 B 10 2 11
## 6 60 A 0 65
Start here:
stop_onoff$load - with(stop_onoff,cumsum(on)-cumsum(off))
split.ref - with(stop_sequence,split(seq,ref))
split.ref.onoff - split.ref[as.character(stop_onoff$ref)]
stop.mat - sapply(split.ref.onoff,rep,length=2)
inout - cbind(stop.mat,c(0,Inf))cbind(c(0,Inf),stop.mat)
stop_onoff$seq - head(stop.mat[inout],-1)
merge(stop_sequence[c(ref,seq)],stop_onoff[-1],by=seq,all.x=T)
seq ref on off load
1 10 A 5 05
2 20 B NA NA NA
3 30 C NA NA NA
4 40 D 0 23
5 50 B 10 2 11
6 60 A 0 65
You can take care of turning the NA's to zeroes or '-'s, I think.
HTH,
Chuck
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] clean email addresses?
Does anyone have suggestions for cleaning a list of email addresses? I ask, because I can read into R data on registered voters that includes an email address field. I wondered if anyone had experience doing such, especially in R. (I found an article on How to Clean Large Email Contact Lists for Email Marketing Campaigns; www.wikihow.com/Clean-Large-Email-Contact-Lists-for-Email-Marketing-Campaigns. Before I went further with this, I felt a need to ask. Thanks, Spencer Graves __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ddply question
I apologize about cross posting but my question keeps bouncing back from the list How come pct doesn't work in this ddply call? I am trying to get a percent of 'TotalCount' by SampleDate and Age library(plyr) b - structure(list(SampleDate = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = 5/8/1996, class = factor), TotalCount = c(1L, 2L, 1L, 1L, 4L, 3L, 1L, 10L, 3L), ForkLength = c(61L, 22L, NA, NA, 72L, 34L, 100L, 23L, 25L), TotalSalvage = c(12L, 24L, 12L, 12L, 17L, 23L, 31L, 12L, 15L), Age = c(1L, 0L, NA, NA, 1L, 0L, 1L, 0L, 0L)), .Names = c(SampleDate, TotalCount, ForkLength, TotalSalvage, Age), class = data.frame, row.names = c(NA, -9L)) b ddply(b,.(SampleDate,Age),summarise,salvage=sum(TotalSalvage),pct=TotalCount/sum(TotalCount)) Error: expecting result of length one, got : 4 #Computing TotalCount inside ddply works but the pct seems wrong... ddply(b,.(SampleDate,Age),summarise,salvage=sum(TotalSalvage),Count=sum(TotalCount),pct=Count/sum(Count)) [[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] Error in inDL(x, as.logical(local), as.logical(now), ...)
Dear Group, I get this error when loadin RCurl. What could be the reason? My configuration: R-version : 3.1.1 Windows- 32, Error in inDL(x, as.logical(local), as.logical(now), ...) : unable to load shared object 'C:/Program Files/R/R-3.1.1/library/RCurl/libs/i386/RCurl.dll': LoadLibrary failure: The specified procedure could not be found. how do i solve this. Since RCurl seems mandatory to install Rgbif. -- :) Smile is my Style :) [[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] new error with QuantMod getSymbols
Thank you very much Joshua. Pardon me for confusing loadSymbols with
getSymbols and not sending the file lista.csv .. Apparently the issue was with
Yahoo Finance that day. The next day it worked perfectlty.
Best Regards,
Adolfo Yanes
Enviado desde mi BlackBerry de Movistar
-Original Message-
From: Joshua Ulrich josh.m.ulr...@gmail.com
Date: Sat, 30 Aug 2014 10:45:01
To: Adolfo Yanesadolfoya...@gmail.com
Cc: R-Helpr-help@r-project.org
Subject: Re: [R] new error with QuantMod getSymbols
You didn't provide the file lista.csv, so it's not possible to
reproduce any of these errors. And there's no call to getSymbols in
your code. You use loadSymbols, and I am not familiar with that
function.
That said, this sounds like an issue with some of the data being sent
by Yahoo Finance.
--
Joshua Ulrich | about.me/joshuaulrich
FOSS Trading | www.fosstrading.com
On Thu, Aug 28, 2014 at 11:12 PM, Adolfo Yanes adolfoya...@gmail.com wrote:
Hello,
I use getSymbols function daily to run some models with stock data. Today
when I tried to update the stock info i get this error
Error in charToDate(x) :
character string is not in a standard unambiguous format
Sometimes I get it after 2 symbols, other times after 150 symbols, another
time after 40 symbols, then after 203 symbols.
The code for the symbol list is:
lista-read.csv(lista.csv, header=FALSE)
lista.list.ana-vector('list',nrow(lista))
names(lista.list.ana) - lista[,1]
lista.sum-as.vector(lista[,1])
##actualizar la lista
lista_simbolos-download_symbols(lista.sum, lista.list.ana)
*The code for the function download_symbols is:*
download_symbols- function(lista.sum.,lista.list.ana..){
newnames.- c(Open, High, Low, Close, Volume, Adjusted)
for (m in 1:length(lista.sum.))
{
print(paste(c(Downloading symbol , lista.sum.[m], . , length(lista.sum.
)-m, symbols missing), sep=, collapse=))
temp-get(loadSymbols(lista.sum.[m]))
names(temp)-newnames.
#lista.list.ana..[[m]]-loadSymbols(lista.sum.[m])
lista.list.ana..[[m]]-temp
}
return(lista.list.ana..)
}
Is it something wrong with yahoo? I tried google and got another error
Error in `colnames-`(`*tmp*`, value = c(Open, High, Low, Close, :
length of 'dimnames' [2] not equal to array extent
Thanks for your help
--
Adolfo Yanes Musetti
[[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] posterior probabilities from lda.predict
Function predict.lda() is just answering a different question from the one you are posing. It is answering the question, given the values on this object what is the probability of membership in each of the groups used to construct the discriminant functions in the first place. Those probabilities sum to 1 and are generally called the posterior probabilities. Your question is somewhat different, if this object was a member of group x, what is the probability that it would have values like these. These are typicality probabilities (how typical is this observation in this group). There are two ways to compute typicality probabilities. One is to use the reduced space defined by the discriminant functions and measure the distance of a new observation to the centroid of the group. This is the approach taken by SPSS which provides the typicality for the group which has the highest posterior probability. Huberty and Olejink recommend this procedure on the grounds that the probability distribution is known. The alternate approach which is used commonly in compositional analysis is to use Mahalanobis distance with the probability assumed to follow a chi square distribution. I am not aware of a package that has a function to produce either of these. Huberty, Carl J. and Stephen Olejink. 2006. Applied Manova and Discriminant Analysis. Second Edition. Wiley-Interscience. David L. Carlson Department of Anthropology Texas AM University -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Fraser D. Neiman Sent: Friday, August 29, 2014 4:14 PM To: r-help@r-project.org Subject: [R] posterior probabilities from lda.predict Dear All, I have used the lda() function in the MASS library to estimate a set of discriminant functions to assign samples from a training set to one of six groups. The cross validation generates nearly perfect predictions for samples in the training set. Hooray! Now I want to use lda.predict() to estimate both discriminant function scores and probabilities of group membership for a second set of samples whose group membership is unknown. For each unknown sample, lda.predict() produces a six probabilities. These probabilities sum to one. So lda.predict() seems to assume that the unknown samples do, in fact, belong to one of the six groups. The problem is that it is nearly certain that some of the unknown samples in the second set do not belong to any of the six groups. For those samples, probabilities of group membership should be close to zero for all six groups. In fact, identifying which samples are unlikely to belong to any of the six groups is a major goal of the analysis. So the question is, what is lda.predict() doing behind the scenes to force the group membership probabilities to sum to one? How do I get it to not do this and produce probabilities that accurately reflect the large Mahalanobis distances of some of the unknown sample from any group centroid?\ I have searched the R-list archive on this and have found several folks asking similar questions, but no helpful answers. Thanks very much! Fraser __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Unexpected behavior when giving a value to a new variable basedon the value of another variable
On Aug 29, 2014, at 8:54 PM, David McPearson wrote: On Fri, 29 Aug 2014 06:33:01 -0700 Jeff Newmiller jdnew...@dcn.davis.ca.us wrote One clue is the help file for $... ? $ In particular there see the discussion of character indices and the exact argument. ...snip... On August 29, 2014 1:53:47 AM PDT, Angel Rodriguez angel.rodrig...@matiainstituto.net wrote: Dear subscribers, I've found that if there is a variable in the dataframe with a name ...sip... N - structure(list(V1 = c(67, 62, 74, 61, 60, 55, 60, 59, 58), V2 = c(NA, 1, 1, 1, 1,1,1,1,NA)), + .Names = c(age,samplem), row.names = c(NA, -9L), class = data.frame) N$sample[N$age = 65] - 1 N age samplem sample 1 67 NA 1 2 62 1 1 3 74 1 1 4 61 1 1 5 60 1 1 6 55 1 1 7 60 1 1 8 59 1 1 9 58 NA NA ...snip... Having seen all the responses about partial matching I almost understand. I've also replicated the behaviour on R 2.11.1 so it's been around awhile. This tells me it ain't a bug - so if any of the cognoscenti have the time and inclination can someone give me a brief (and hopefully simple) explanation of what is going on under the hood? It looks (to me) like N$sample[N$age = 65] - 1 copies N$samplem to N$sample and then does the assignment. If partial matching is the problem (which it clearly is) my expectation is that the output should look like age samplem 1 67 1 2 62 1 3 74 1 4 61 1 5 60 1 6 55 1 7 60 1 8 59 1 9 58 NA That is - no new column. (and I just hate it when the world doesn't live up to my expectations!) Not sure what you are seeing. I am seeing what you expected: test - data.frame(age=1:10, sample=1) test$sample[test$age5] - 2 test age sample 11 2 22 2 33 2 44 2 55 1 66 1 77 1 88 1 99 1 10 10 1 -- David Bewildered and confused, DMcP South Africas premier free email service - www.webmail.co.za Cotlands - Shaping tomorrows Heroes http://www.cotlands.org.za/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, 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.
Re: [R] Split PVClust plot
Hi Tom,
There is a as.dendrogram.pvclust function in the package dendextend.
(it is on CRAN: http://cran.r-project.org/web/packages/dendextend/)
You can run:
install.packages('dendextend')
library(dendextend)
result2 - as.dendrogram(result)
# You can then also use the prune function in dendextend, to get the
subtree you are interested in.
Also, there is an example of pvclust in the package vignette (just
search pvclust
here):
http://cran.r-project.org/web/packages/dendextend/vignettes/introduction.html
The example shows how to highlight significant branches (with line width
and color).
With regards,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com |
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Tue, Jul 29, 2014 at 12:40 AM, Worthington, Thomas A
thomas.worthing...@okstate.edu wrote:
Dear All
I'm using PVClust to perform hierarchical clustering, for the output plot
I can control most of the graphical I need, however the plot is large and I
would like to split it vertically into two panels one above the other. Is
there a way to plot only part of a PVClust plot, I tried to convert it to a
dendrogram with
result2 = as.dendrogram(result)
however I get the error message no applicable method for 'as.dendrogram'
applied to an object of class pvclust. I also wondered whether it would
be possible to convert to a phylogenetic tree and use the functions in the
'ape' package?
Any suggestion on how to split up a PVclust plot would be greatly
appreciated (code for the plot below)
Thanks
Tom
result - pvclust(df.1, method.dist=uncentered,
method.hclust=average,nboot=10)
par(mar=c(0,0,0,0))
par(oma=c(0,0,0,0))
plot(result, print.pv =FALSE, col.pv=c(red,,), print.num=FALSE,
float = 0.02, font=1,
axes=T, cex =0.85, main=, sub=, xlab=, ylab= ,
labels=NULL, hang=-1)
pvrect(result, alpha=0.95)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unexpected behavior when giving a value to a new variable basedon the value of another variable
On Aug 30, 2014, at 7:38 PM, David Winsemius wrote: On Aug 29, 2014, at 8:54 PM, David McPearson wrote: On Fri, 29 Aug 2014 06:33:01 -0700 Jeff Newmiller jdnew...@dcn.davis.ca.us wrote One clue is the help file for $... ? $ In particular there see the discussion of character indices and the exact argument. ...snip... On August 29, 2014 1:53:47 AM PDT, Angel Rodriguez angel.rodrig...@matiainstituto.net wrote: Dear subscribers, I've found that if there is a variable in the dataframe with a name ...sip... N - structure(list(V1 = c(67, 62, 74, 61, 60, 55, 60, 59, 58), V2 = c(NA, 1, 1, 1, 1,1,1,1,NA)), + .Names = c(age,samplem), row.names = c(NA, -9L), class = data.frame) N$sample[N$age = 65] - 1 N age samplem sample 1 67 NA 1 2 62 1 1 3 74 1 1 4 61 1 1 5 60 1 1 6 55 1 1 7 60 1 1 8 59 1 1 9 58 NA NA ...snip... Having seen all the responses about partial matching I almost understand. I've also replicated the behaviour on R 2.11.1 so it's been around awhile. This tells me it ain't a bug - so if any of the cognoscenti have the time and inclination can someone give me a brief (and hopefully simple) explanation of what is going on under the hood? It looks (to me) like N$sample[N$age = 65] - 1 copies N$samplem to N$sample and then does the assignment. If partial matching is the problem (which it clearly is) my expectation is that the output should look like age samplem 1 67 1 2 62 1 3 74 1 4 61 1 5 60 1 6 55 1 7 60 1 8 59 1 9 58 NA That is - no new column. (and I just hate it when the world doesn't live up to my expectations!) Not sure what you are seeing. I am seeing what you expected: test - data.frame(age=1:10, sample=1) test$sample[test$age5] - 2 test age sample 11 2 22 2 33 2 44 2 55 1 66 1 77 1 88 1 99 1 10 10 1 I realized later that I had not constructed a test of you behavior and that when I did I see the creation of a third column. The answer is to read the help page: ?`[-` Character indices can in some circumstances be partially matched (see pmatch) to the names or dimnames of the object being subsetted (but never for subassignment). Note the caveat in parentheses. -- 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.
[R] what happened when copying a function definition into R prompt then press Enter?
Dear expeRts, That's to say,what happened when loading source code into memory? what's the difference between it and loading installed code into memory? Do they related with .Rdata? -- PO SU mail: desolato...@163.com Majored in Statistics from SJTU __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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-es] help: shiny leer ficheros desde google drive
Hola buenas, Un compañero y yo estamos haciendo una aplicación shiny. Nos ha quedado bastante aparente y en sevidor local (con R) funciona bastante bien. El problema es que cuando cargamos los ficheros en la web deja de fucionar ¿Por qué? Pues porque al principio de la aplicacion cargamos unos datos de nuestro ordenador y esto no es posible a la hora de poner los datos ShinyApps.io (una de las multiples soluciones que se nos ofrece en la red) [1, 2,a p 3] Una de las posibles soluciones que se nos había ocurrido era subir los datos en formato csv a goolge drive y una vez permitido a cualquiera acceder a ellos leerlos con un read table a través del link. La pregunta es ¿Alguien sabe como leer unos datos colgados en csv desde google drive? ¿Podría poner un ejemplo práctico? los que hemos encontrado no hemos sabido reproducirlos. Otra pregunta es ¿alguna otra solución gratuita para leer datos colgados on-line? Como siempre, gracias por adelantado y un cordial saludo Javier Bibliografía [1] https://github.com/rstudio/shinyapps/blob/master/guide/guide.md [2] http://shiny.rstudio.com/tutorial/lesson7/ [3] https://www.shinyapps.io/ -- [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R-es] help: shiny leer ficheros desde google drive
Hola, Hay un paquete en Bioconductor para esto RGoogleDocs. Referencias adicionales: http://www.statsravingmad.com/a-tiny-rcurl-headache/ Y si no, puedes hacerlo a través de un fichero en el Public de un Dropbox... Saludos, Carlos Ortega www.qualityexcellence.es El 30 de agosto de 2014, 13:58, Javier Villacampa González javier.villacampa.gonza...@gmail.com escribió: Hola buenas, Un compañero y yo estamos haciendo una aplicación shiny. Nos ha quedado bastante aparente y en sevidor local (con R) funciona bastante bien. El problema es que cuando cargamos los ficheros en la web deja de fucionar ¿Por qué? Pues porque al principio de la aplicacion cargamos unos datos de nuestro ordenador y esto no es posible a la hora de poner los datos ShinyApps.io (una de las multiples soluciones que se nos ofrece en la red) [1, 2,a p 3] Una de las posibles soluciones que se nos había ocurrido era subir los datos en formato csv a goolge drive y una vez permitido a cualquiera acceder a ellos leerlos con un read table a través del link. La pregunta es ¿Alguien sabe como leer unos datos colgados en csv desde google drive? ¿Podría poner un ejemplo práctico? los que hemos encontrado no hemos sabido reproducirlos. Otra pregunta es ¿alguna otra solución gratuita para leer datos colgados on-line? Como siempre, gracias por adelantado y un cordial saludo Javier Bibliografía [1] https://github.com/rstudio/shinyapps/blob/master/guide/guide.md [2] http://shiny.rstudio.com/tutorial/lesson7/ [3] https://www.shinyapps.io/ -- [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es -- Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
