Re: [R] uniform number
On 05/05/14 17:05, Ragia Ibrahim wrote: Dear group, How to generate uniform probability choosing p to be 2% and 5%, in separate trials for 100 times. No idea WTF you are talking about. Can you formulate a question that is comprehensible to the human mind? 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] uniform number
Maybe you thought the binomial distribution? see ?rbinom best, kd Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ; meghatalmaz#243;: Rolf Turner [r.tur...@auckland.ac.nz] Küldve: 2014. május 5. 9:27 To: Ragia Ibrahim Cc: r-help@r-project.org Tárgy: Re: [R] uniform number On 05/05/14 17:05, Ragia Ibrahim wrote: Dear group, How to generate uniform probability choosing p to be 2% and 5%, in separate trials for 100 times. No idea WTF you are talking about. Can you formulate a question that is comprehensible to the human mind? cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] DESeq vs DESeq2 different DEGs results
Hi, I want to compare DESeq vs DESeq2 and I am getting different number of DEGs which I will expect to be normal. However, when I compare the 149 genes ID that I get with DESeq with the 869 from DESeq2 there are only ~10 genes that are in common which I donât understand (using FDR 0.05 for both). I want to block the Subject effect for which I am including the reduced formula of ~1. Shouldnât these two methods output similar results? Because at the moment I could interpret my results in different ways. Thanks for your help, Catalina This the DESeq script that I am using: DESeq library(DESeq) co=as.matrix(read.table(2014_04_01_6h_LP.csv,header=T, sep=,, row.names=1)) Subject=c(1,2,3,4,5,1,2,4,5) Treatment=c(rep(co,5),rep(c2,4)) a.con=cbind(Subject,Treatment) cds=newCountDataSet(co,a.con) cds - estimateSizeFactors( cds) cds - estimateDispersions(cds,method=pooled-CR, modelFormula=count~Subject+Treatment) #filtering rs = rowSums ( counts ( cds )) theta = 0.2 use = (rs quantile(rs, probs=theta)) table(use) cdsFilt= cds[ use, ] fit0 - fitNbinomGLMs (cdsFilt, count~1) fit1 - fitNbinomGLMs (cdsFilt, count~Treatment) pvals - nbinomGLMTest (fit1, fit0) padj - p.adjust( pvals, method=BH ) padj - data.frame(padj) row.names(padj)=row.names(cdsFilt) padj_fil - subset (padj,padj 0.05 ) dim (padj_fil) [1] 149 1 ââââââ library (DESeq2) countdata=as.matrix(read.table(2014_04_01_6h_LP.csv,header=T, sep=,, row.names=1)) coldata= read.table (targets.csv, header = T, sep=,,row.names=1) coldata Subject Treatment F1 1co F2 2co F3 3co F4 4co F5 5co H1 1c2 H2 2c2 H4 4c2 H5 5c2 dds - DESeqDataSetFromMatrix( countData = countdata, colData = coldata, design = ~ Subject + Treatment) dds dds$Treatment - relevel (dds$Treatment, co) dds - estimateSizeFactors( dds) dds - estimateDispersions(dds) rs = rowSums ( counts ( dds )) theta = 0.2 use = (rs quantile(rs, probs=theta)) table(use) ddsFilt= dds[ use, ] dds - nbinomLRT(ddsFilt, full = design(dds), reduced = ~ 1) resLRT - results(dds) sum( resLRT$padj 0.05, na.rm=TRUE ) #[1] 869 [[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] uniform number
WTF? Is that a R package from you? On 5 May 2014 09:27, Rolf Turner r.tur...@auckland.ac.nz wrote: On 05/05/14 17:05, Ragia Ibrahim wrote: Dear group, How to generate uniform probability choosing p to be 2% and 5%, in separate trials for 100 times. No idea WTF you are talking about. Can you formulate a question that is comprehensible to the human mind? cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] function to compare lengths of objects?
On 05/05/2014, 1:02 AM, Spencer Graves wrote:
Is there a standard function to generate messages like longer
object length is not a multiple of shorter object length but providing
more information like the names of the objects?
No, not really.
At the C level objects don't really have names. The same object may be
linked to several different names, or none at all. Most messages like
the length warning are generated there, so they don't include name
information.
At the R level you can use substitute() to find the expression that was
used to supply a function argument, so you can do a little better. For
example, I would write a simpler version of your compareLengths function
something like this:
compareLengths - function(x, y, name.x = deparse(substitute(x)),
name.y = deparse(substitute(y))) {
if (length(x) == length(y))
return( c(equal, paste(objects, dQuote(name.x), and,
dQuote(name.y), are the same length.))
else ...
}
Then a function could call it like this:
f - function(a, b) {
cmp1 - compareLengths(a, b) # Message talks about a and b
cmp2 - compareLengths(a, b, deparse(substitute(a)),
deparse(substitute(b))) # Message talks about names
# in call to f
...
}
Duncan Murdoch
For example, a/b issues, Warning message: In a/b : longer
object length is not a multiple of shorter object length if the length
of one is not a multiple of the other. I want a utility to produce
messages that include the lengths of a and b and other information.
If you were to write such a function, what would you include in
it? Arguments? Values? Options?
I started creating compareLengths.Rd with
\usage{compareLengths(x, y, name.x=NULL, name.y=NULL, Source='',
compFun=c('length', 'NROW'), action=c(comparable='',
incomparable='warning'), maxChar=20, ...)} and \value{A character vector
of length 2. The first element is either 'equal', 'compatible' or
'incompatible'. The second element is the message composed for the
desired action.
Thanks,
Spencer Graves
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Re: [R] uniform number
That paper you cite is about Social networks. You may want to use igraph or sna packages On 5 May 2014 10:54, Ragia Ibrahim ragi...@hotmail.com wrote: thanks for replying in the following paper http://www.cs.cornell.edu/home/kleinber/kdd03-inf.pdf page 6 third paragraph the author writes: assigned a uniform probability of p to each edge of the graph, choosing p to be 1% and 10% in separate trials. how to use R function to get such probability ? Regards Date: Mon, 5 May 2014 10:12:49 +0200 Subject: Re: [R] uniform number From: msu...@gmail.com To: r.tur...@auckland.ac.nz CC: ragi...@hotmail.com; r-help@r-project.org WTF? Is that a R package from you? On 5 May 2014 09:27, Rolf Turner r.tur...@auckland.ac.nz wrote: On 05/05/14 17:05, Ragia Ibrahim wrote: Dear group, How to generate uniform probability choosing p to be 2% and 5%, in separate trials for 100 times. No idea WTF you are talking about. Can you formulate a question that is comprehensible to the human mind? cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] uniform number
Please keep responses on-list unless there are compelling reasons not to. It is still not clear what you want to do. You *might* want to assign probabilities to the edges of a graph where these probabilities are uniformly (and independently) distributed on the interval [0.01, 0.10]. This could be done by probs - runif(n,0.01,0.10) where n is the number of edges. See ?runif. You *might* want to assign probabilities either 0.01 or 0.10 to each edge of the graph, each probability being chosen with probability 0.5 (???). This could be done by probs - sample(c(0.01,0.10),n,TRUE). You really need to learn something about R if you are going to use R. Start with An Introduction to R available (under Manuals) from the R web site. You also need to learn to express yourself clearly and unambiguously. Do not expect your readers to be telepathic. cheers, Rolf Turner On 05/05/14 19:41, Ragia Ibrahim wrote: thanks for replying in the following paper http://www.cs.cornell.edu/home/kleinber/kdd03-inf.pdf page 6 third paragraph the author writes: assigned a uniform probability of p to each edge of the graph, choosing p to be 1% and 10% in separate trials. how to use R function to get such probability ? Regards Date: Mon, 5 May 2014 19:27:43 +1200 From: r.tur...@auckland.ac.nz To: ragi...@hotmail.com CC: r-help@r-project.org Subject: Re: [R] uniform number On 05/05/14 17:05, Ragia Ibrahim wrote: Dear group, How to generate uniform probability choosing p to be 2% and 5%, in separate trials for 100 times. No idea WTF you are talking about. Can you formulate a question that is comprehensible to the human mind? __ 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] metaMDS in vegan: zero stress
Dear R and vegan package users, I have been experiencing problems with the metaMDS function when working on a dataset (euk) consisting of 9 sites (RNA extracts of 9 biofilms samples) and 340 species (microbial taxa based on rRNA sequences). The problem is that I get nMDS ordinations with overlapping points, so that it looks like 6 samples are identical, while the remaining 3 are well separated. The dataset is rather large, and quite complex so I do not think that this is a correct representation of dissimilarities. Every time this happens, I get the warning message: In metaMDS(t(euk_red), distance = bray) : Stress is (nearly) zero - you may have insufficient data Removing relatively rare species using: euk_red-euk[rowSums(euk)100,] results in an nMDS ordination with scattered points that looks more reasonable. However, removing even more rare species (1000) results in a different, but similarly uncomplex ordination and the same warning message. Changing the distance metric to euclidean seems even more sensitive to this problem, yielding uncomplex ordinations with almost any rare species cutoff. The code I am using is: euk_MDS-metaMDS(t(euk_red), distance=bray) ordiplot(euk_MDS, display=sites) I have tried the arguments engine=isoMDS, and changing maxit and trymax without noticeable effect. Generating the distance matrices separately in vegdist yields distance matrices that look normal to my eye. It seems like a similar problem to a recent thread in the R-help (Fabian Boetzl, Feb 2014), but my dataset has many species that are shared between all sites. I would be happy to provide the dataset if this may help in figuring out what the problem is. I am using R version 3.1.0 for MacOS, coupled to R-studio (version 0.98.510). Thanks in advance and best regards! Mia M. Bengtsson, University of Vienna [[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] uniform number
... or this could mean to accept or reject an edge with p = 0.01 or 0.1 in which case you *might* use rbinom() as in n - 100 # number of edges p - 0.27 # desired probability rbinom(n,1,p) [1] 0 1 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 1 ... etc. B. On 2014-05-05, at 5:19 AM, Rolf Turner wrote: Please keep responses on-list unless there are compelling reasons not to. It is still not clear what you want to do. You *might* want to assign probabilities to the edges of a graph where these probabilities are uniformly (and independently) distributed on the interval [0.01, 0.10]. This could be done by probs - runif(n,0.01,0.10) where n is the number of edges. See ?runif. You *might* want to assign probabilities either 0.01 or 0.10 to each edge of the graph, each probability being chosen with probability 0.5 (???). This could be done by probs - sample(c(0.01,0.10),n,TRUE). You really need to learn something about R if you are going to use R. Start with An Introduction to R available (under Manuals) from the R web site. You also need to learn to express yourself clearly and unambiguously. Do not expect your readers to be telepathic. cheers, Rolf Turner On 05/05/14 19:41, Ragia Ibrahim wrote: thanks for replying in the following paper http://www.cs.cornell.edu/home/kleinber/kdd03-inf.pdf page 6 third paragraph the author writes: assigned a uniform probability of p to each edge of the graph, choosing p to be 1% and 10% in separate trials. how to use R function to get such probability ? Regards Date: Mon, 5 May 2014 19:27:43 +1200 From: r.tur...@auckland.ac.nz To: ragi...@hotmail.com CC: r-help@r-project.org Subject: Re: [R] uniform number On 05/05/14 17:05, Ragia Ibrahim wrote: Dear group, How to generate uniform probability choosing p to be 2% and 5%, in separate trials for 100 times. No idea WTF you are talking about. Can you formulate a question that is comprehensible to the human mind? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotting a Regression holding some exogeneous constant.
Dear R-community, I am totally lost and need help. For a Visualization I need to plot two regressions in the same plot. The intention is to provide a visual basis for a synthesized theoretical discussion. 1.)fixed.1-plm(CSmean~ FCRlong, data = data.plm, index = c(countrynr,quartal),model = within) 2.)fixed.2-plm(FCRlong ~ GDP+ GDPpCapita+ budget+ percentDebt+ primSurplus +Inflation+lagBY, data = data.plm, index = c(countrynr,quartal),model = within Usually I would do plot(data.plm$ FCRlong,data.plm$BY,ylim = c(0,20)) abline(fixed.1) abline(fixed.2) For Regression Number one, fixed.1, it´s easy, because I have only two Dimensions, but here I need to do a parallel shift by adding an average Value, say 4, (to solve for CSmean to BY) So loosely speaking, I want to plot fixed.1+4 ? What shall I do in this case? Second, in Regression fixed.2 I have a Hyperplane, but for a synthesized general discussion the exogeneous variables, except lagBY, shall be set to their average values in the panel-dataset. With this restriction I am able to plot Regression 2 in the two-dimensional-plot as well. But I am not even sure how to reconstruct this by a code? I unfortunately can´t give you an reproduceable example, But any suggestions or hints or just a link to a similar problem would be helpful Thank you! Katie [[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] R Package making error
Hi all ,
I have a GUI package function like below ,
trader-function()
{
install.packages(Rgtk2)
install.packages(cluster)
install.packages(gWidgets)
install.packages(gWidgetsRGtk2)
install.packages(scales)
install.packages(RGtk2Extras)
install.packages(FuzzyToolkitUoN)
install.packages(splines)
install.packages(plyr)
install.packages(cairo)
install.packages(cairoDevices)
install.packages(e1071)
install.packages(quantmod)
install.packages(TTR)
install.packages(xts)
library(RGtk2)
library(RGtk2Extras)
library(FuzzyToolkitUoN)
library(splines)
library(plyr)
library(Cairo)
library(cairoDevice)
library(gWidgets2)
library(gWidgetsRGtk2)
library(e1071)
library(quantmod)
library(TTR)
library(xts)
}
Ididn't put the whole code here ,
My problem is I had converted thisfunction into an
package , it works good ,but every time it has to
install all the packages, is there something by which only once
it will install the packages and next time it is not needed .
Thanks
ASHIS
[[alternative HTML version deleted]]
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Re: [R] R Package making error
While I don't particularly like the idea of a package automatically
making changes to the system, you can use
installed.packages()
to see which of your needed packages are already available, and only
install the ones that are missing.
Better would be to put the whole list in Depends, and let R manage the loading.
http://stackoverflow.com/questions/8637993/better-explanation-of-when-to-use-imports-depends
Sarah
On Mon, May 5, 2014 at 9:46 AM, Ashis Deb ashisde...@gmail.com wrote:
Hi all ,
I have a GUI package function like below ,
trader-function()
{
install.packages(Rgtk2)
install.packages(cluster)
install.packages(gWidgets)
install.packages(gWidgetsRGtk2)
install.packages(scales)
install.packages(RGtk2Extras)
install.packages(FuzzyToolkitUoN)
install.packages(splines)
install.packages(plyr)
install.packages(cairo)
install.packages(cairoDevices)
install.packages(e1071)
install.packages(quantmod)
install.packages(TTR)
install.packages(xts)
library(RGtk2)
library(RGtk2Extras)
library(FuzzyToolkitUoN)
library(splines)
library(plyr)
library(Cairo)
library(cairoDevice)
library(gWidgets2)
library(gWidgetsRGtk2)
library(e1071)
library(quantmod)
library(TTR)
library(xts)
}
Ididn't put the whole code here ,
My problem is I had converted thisfunction into an
package , it works good ,but every time it has to
install all the packages, is there something by which only once
it will install the packages and next time it is not needed .
Thanks
ASHIS
---
Sarah Goslee
http://www.functionaldiversity.org
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Re: [R] auglag store parameters
Just guessing, but something like
pars - numeric(nloops)
for (i in 1:nloops) {
## use auglag
pars[i] - {the $par part of the answer}
}
might be what you're looking for.
All of which has nothing to do with auglag in particular, just simple R
programming.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 5/3/14 5:02 AM, Delger Enkhbayar d.enkhba...@lse.ac.uk wrote:
Hi,
I am minimizing a non-linear function subject to linear and nonlinear
equalities and inequalities. I managed to understand how to use auglag for
this case. Now, I'm using loops to get solutions for different
assumptions,
however, I'm facing a difficulty in storing just $par part of the answer
to
a vector form. Could someone point me to how to do this in auglag?
Thank you for all your great work!
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[R] Tick marks at the 0:0 point in a plot ()
Dear R usres, Sorry to bother you with my basic question. I have a quick question about tick mark in R. A reviewer wants the tick marks of the graph to start at 0:0 point. My current code below produces the tick mark a little far from the corner; plot(Simu2~Irrig, data=Ahmed,pch=17, col=1,subset=ET==40,tck=-0.01, xlab = (Applied Irrigation (mm)), ylab = expression(paste(Cumulative ET (mm))) ,cex.lab=1.2, cex.axis=0.8,cex=0.9,font.lab=1,xlim = c(0,1000), xaxp=c(0,1000,10), yaxp=c(100,800,7), ylim=c(100,800)) Any ideas. Thank you Ahmed M. Attia Research Assistant Dept. of SoilCrop Sciences Texas AM University ahmed.at...@ag.tamu.edu Cell phone: 001-979-248-5215 FAX: 001-308-455-4024 __ 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] Parsing XML file to data frame
I didn't find an attached XML file. Maybe the list removes attachments? You might try posting to StackOverflow.com if this is the case. On Fri, May 2, 2014 at 2:17 PM, starcraz chan_will...@email.com wrote: Hi all - I am trying to parse out the attached XML file into a data frame. The file is extracted from Hadoop File Services (HFS). I am new in using the XML package so need some help in parsing out the data. Below is some code that I explore to get the attribute into a data frame. Any help is appreciated. library(XML) temp - xmlParseDoc(sample.xml) temp.root - xmlRoot(temp) xmlName(temp.root) xmlSize(temp.root) #21 child nodes temp.root[[2]] #headers temp.root[[2]][[2]] #extracts just the revision temp.2 - xmlToList(temp.root[[2]]) #extracts the info in temp.root[[2]] into a list temp.2 temp.2.df - xmlToDataFrame(temp.root[[2]]) #data frame of the list temp.2.df xmlValue(temp.root[[2]]) #string the values of the node inside [[2]] temp.revision - xmlValue(temp.root[[2]][[Revision]]) temp.revision test - xmlTreeParse(sample.xml) test -- View this message in context: http://r.789695.n4.nabble.com/Parsing-XML-file-to-data-frame-tp4689883.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. -- Timothy Cook LinkedIn Profile:http://www.linkedin.com/in/timothywaynecook MLHIM http://www.mlhim.org [[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] Tick marks at the 0:0 point in a plot ()
I don't understand your question - you want a tick mark at 0,0 but you explicitly ask for the y axis to start at 100 (and to go upwards from there). Do you mean you want the initial tick mark on an axis to be at the lower left corner of the plot? If so, see if adding xaxs=i, yaxs=i to your plot command helps. E.g., compare plot(0:10,0:10, xaxs=i, yaxs=i) and plot(0:10,0:10) Bill Dunlap TIBCO Software wdunlap tibco.com On Mon, May 5, 2014 at 11:27 AM, Ahmed Attia ahmedati...@gmail.com wrote: Dear R usres, Sorry to bother you with my basic question. I have a quick question about tick mark in R. A reviewer wants the tick marks of the graph to start at 0:0 point. My current code below produces the tick mark a little far from the corner; plot(Simu2~Irrig, data=Ahmed,pch=17, col=1,subset=ET==40,tck=-0.01, xlab = (Applied Irrigation (mm)), ylab = expression(paste(Cumulative ET (mm))) ,cex.lab=1.2, cex.axis=0.8,cex=0.9,font.lab=1,xlim = c(0,1000), xaxp=c(0,1000,10), yaxp=c(100,800,7), ylim=c(100,800)) Any ideas. Thank you Ahmed M. Attia Research Assistant Dept. of SoilCrop Sciences Texas AM University ahmed.at...@ag.tamu.edu Cell phone: 001-979-248-5215 FAX: 001-308-455-4024 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] function to compare lengths?
Is there a function someplace to compare lengths?
For example, if longer object length is not a multiple of
shorter object length, a/b issues a warning. I'd like something with
more flexibility in what is reported.
Hadley Wickham's testthat and assertthat packages provide
tools for testing and comparing objects, but I didn't see this.
I'm starting to write compareLengths(x, y, name.x=NULL,
name.y=NULL, Source='', compFun=c('length', 'NROW'),
action=c(comparable='', incomparable='warning'), maxChar=20, ...).
However, I'd prefer to use an existing function if I can find one that I
feel is adequate.
Thanks,
Spencer
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Re: [R] function to compare lengths?
On 05/05/2014, 3:39 PM, Spencer Graves wrote:
Is there a function someplace to compare lengths?
For example, if longer object length is not a multiple of
shorter object length, a/b issues a warning. I'd like something with
more flexibility in what is reported.
Hadley Wickham's testthat and assertthat packages provide
tools for testing and comparing objects, but I didn't see this.
I'm starting to write compareLengths(x, y, name.x=NULL,
name.y=NULL, Source='', compFun=c('length', 'NROW'),
action=c(comparable='', incomparable='warning'), maxChar=20, ...).
However, I'd prefer to use an existing function if I can find one that I
feel is adequate.
You posted this question (with a different subject line) yesterday.
What is wrong with the response I sent then?
Duncan Murdoch
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[R] multiply of two expressions
hello, i want to differentiate of L with respect to b when: L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln likelihood) and ln(mu/(mu+k)) = a+bx #link function how can i do it in R? thank you. _ Best Regards Niloofar.Javanrouh Ph.D Student of BioStatistics [[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] Lattice Histogram with Normal Curve - Y axis as percentages
Hello,
This may seem like a simple problem, but it's frustrating me immensely. I'm
trying to overlay a normal curve (dnorm) on top of a histogram using the
code below. This works find when the type = density, but the person for
whom I'm making the plot wants the y axis in percent of total rather than
density. When I change type to percent, I get the histogram scale I'm
after, but the dnorm plot is greatly reduced. How could I scale the density
plot to the percent of total axis. Alternatively, perhaps there is a way to
add density to a secondary y axis?
Thanks in advance for your help.
Jimdare
plot-histogram(~rdf[,j]|Year,nint=20, data=rdf,main = i,strip =
my.strip,xlab = j,
type = percent,layout=c(2,1),
panel=function(x, ...) {
panel.histogram(x, ...)
panel.mathdensity(dmath=dnorm, col=black,
# Add na.rm = TRUE to mean() and sd()
args=list(mean=mean(x, na.rm = TRUE),
sd=sd(x, na.rm = TRUE)), ...)
})
--
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[R] Sciplot: Increasing the width of bargraph and decreasing the sapce b/n groups
Hello, I am trying to plot bargraphs susing Sciplot. Is there a way to increase the width of the bar graphs and decrease the space b/n the groups? I am pasting the script as well as attaching the graph. Bio6 - read.csv(Data/Plin1.csv,na.strings=,header=T) attach(Bio6) head(Bio6) par(family=serif, font=11) Bio6$Sps - factor(Bio6$Sps, levels = c(FFA1, FFA2,FFA3)) Bio6$Gp - factor(Bio6$Gp, levels = c(N-FFA1, FU1,FA1,N-FFA2,FU2,FA2,N-FFA3,FU3,FA3)) bargraph.CI(Sps, O.D, group = Gp, data = Bio6,ylab = Relative expression levels, cex.lab = 1.5, y.leg = 6,cex.leg = 0.82,cex=1.5, axisnames=TRUE, col = c(red,blue,grey),space=c(0, 0.5), ylim=c(0,7),cex.names = 1.0,density = c(30,30,30), legend = TRUE, main=PLIN1) detach(Bio6) O.D Gp Sps 1 1.00 N-FFA1 FFA1 2 2.996432FU1 FFA1 3 3.223413FU1 FFA1 4 3.524465FU1 FFA1 5 1.311971FA1 FFA1 6 6.755860FA1 FFA1 7 1.566000FA1 FFA1 8 1.00 N-FFA2 FFA2 9 2.741612FU2 FFA2 10 2.800644FU2 FFA2 11 3.569509FU2 FFA2 12 4.141500FA2 FFA2 13 7.049476FA2 FFA2 14 4.694674FA2 FFA2 15 1.00 N-FFA3 FFA3 16 4.163601FU3 FFA3 17 3.903986FU3 FFA3 18 4.73FU3 FFA3 19 0.00FA3 FFA3 20 0.00FA3 FFA3 21 0.00FA3 FFA3 Any help would be appreciated. Thanks in advance, Roopa __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Degrees of Freedom GAM/GAMM
Dear All I've fitted a GAMM to relate water temperature to the day of the year (DOY) at three different sites. I used summary(MFinal$gam) and anova(MFinal$gam) to produce the output of my model. I'm confused on how to report the degrees of freedom for the smother and the factor. I currently have The final model suggested water temperature was significant related to DOY (6.46 estimated degrees of freedom, F = 224.2, p 0.001) and was significantly different between sites (F2 = 82.14, p 0.001) But think there should be more to the degrees of freedom particularly for the sites factor. If anyone could advise on the correct way to display the results I would greatly appreciate it . The output from the model is below Best wishes Tom summary(MFinal$gam) Family: gaussian Link function: identity Formula: Temperature ~ s(DOY) + Site Parametric coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 12.0366 0.2411 49.92 2e-16 *** SiteG 4.2460 0.3979 10.67 2e-16 *** SiteH 4.1439 0.4069 10.19 2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Approximate significance of smooth terms: edf Ref.df F p-value s(DOY) 6.457 6.457 224.2 2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 R-sq.(adj) = 0.914 Scale est. = 0.098602 n = 137 anova(MFinal$gam) Family: gaussian Link function: identity Formula: Temperature ~ s(DOY) + Site Parametric Terms: df F p-value Site 2 82.14 2e-16 Approximate significance of smooth terms: edf Ref.df F p-value s(DOY) 6.457 6.457 224.2 2e-16 __ 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] list.files accessing subdirectory as relative path?
Dear list members, I would like to access a subdirectory given where the work directory has been set. So I have: getwd() [1] C:/Users/Lord Adellus/Dropbox/I8child1/Data list.files() # give three folders [1] Least Developed Countries [4] Low middle income grouping More advanced developing countries and territories list.files(path = ../Least Developed Countries) # I want to access now one of the subdirectories character(0) But as you can see R does not want to look into the specified subdirectory. I have tried several combination and searched the list but without any great success. Actually: list.files(path = ../...) #goes up one level in the folder structure so I cannot see what the problem is. Thanks in advance. Adel -- View this message in context: http://r.789695.n4.nabble.com/list-files-accessing-subdirectory-as-relative-path-tp4689997.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] function to compare lengths?
On 5/5/2014 1:19 PM, Duncan Murdoch wrote:
On 05/05/2014, 3:39 PM, Spencer Graves wrote:
Is there a function someplace to compare lengths?
For example, if longer object length is not a multiple of
shorter object length, a/b issues a warning. I'd like something with
more flexibility in what is reported.
Hadley Wickham's testthat and assertthat packages provide
tools for testing and comparing objects, but I didn't see this.
I'm starting to write compareLengths(x, y, name.x=NULL,
name.y=NULL, Source='', compFun=c('length', 'NROW'),
action=c(comparable='', incomparable='warning'), maxChar=20, ...).
However, I'd prefer to use an existing function if I can find one that I
feel is adequate.
You posted this question (with a different subject line) yesterday.
What is wrong with the response I sent then?
Your earlier response was great, but I didn't see it until I
received this reply.
I apologize for sending essentially the same question twice. It
was not my intent. I missed Duncan's reply and couldn't find my
original email in my sent folder. I concluded erroneously that I may
have deleted the draft rather than sending it.
Thanks again,
Spencer Graves
Duncan Murdoch
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Re: [R] list.files accessing subdirectory as relative path?
If 'foo' is a subdirectory of the current working directory, then
list.files('foo')
lists the files in foo.
.. goes up one level
But you did not want to go up one level. You wanted to stay at the same
level.
(also, '...' has no meaning in pathnames)
Try either of
list.files(path = ./Least Developed Countries)
list.files(path = Least Developed Countries)
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 5/5/14 12:29 PM, Adel adel.da...@sociology.gu.se wrote:
Dear list members,
I would like to access a subdirectory given where the work directory has
been set. So I have:
getwd()
[1] C:/Users/Lord Adellus/Dropbox/I8child1/Data
list.files() # give three folders
[1] Least Developed Countries
[4] Low middle income grouping More advanced
developing countries and territories
list.files(path = ../Least Developed Countries) # I want to access now
one of the subdirectories
character(0)
But as you can see R does not want to look into the specified
subdirectory.
I have tried several combination and searched the list but without any
great
success.
Actually:
list.files(path = ../...) #goes up one level in the folder structure so
I
cannot see what the problem is.
Thanks in advance.
Adel
--
View this message in context:
http://r.789695.n4.nabble.com/list-files-accessing-subdirectory-as-relativ
e-path-tp4689997.html
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Re: [R] Tick marks at the 0:0 point in a plot ()
Dear William, Worded beautifully. Yes, I wanted to get the tick marks at the left corner. I had to just add xaxs=i and yaxs=i Thanks Ahmed Ahmed M. Attia Research Assistant Dept. of SoilCrop Sciences Texas AM University ahmed.at...@ag.tamu.edu Cell phone: 001-979-248-5215 FAX: 001-308-455-4024 On Mon, May 5, 2014 at 12:13 PM, William Dunlap wdun...@tibco.com wrote: I don't understand your question - you want a tick mark at 0,0 but you explicitly ask for the y axis to start at 100 (and to go upwards from there). Do you mean you want the initial tick mark on an axis to be at the lower left corner of the plot? If so, see if adding xaxs=i, yaxs=i to your plot command helps. E.g., compare plot(0:10,0:10, xaxs=i, yaxs=i) and plot(0:10,0:10) Bill Dunlap TIBCO Software wdunlap tibco.com On Mon, May 5, 2014 at 11:27 AM, Ahmed Attia ahmedati...@gmail.com wrote: Dear R usres, Sorry to bother you with my basic question. I have a quick question about tick mark in R. A reviewer wants the tick marks of the graph to start at 0:0 point. My current code below produces the tick mark a little far from the corner; plot(Simu2~Irrig, data=Ahmed,pch=17, col=1,subset=ET==40,tck=-0.01, xlab = (Applied Irrigation (mm)), ylab = expression(paste(Cumulative ET (mm))) ,cex.lab=1.2, cex.axis=0.8,cex=0.9,font.lab=1,xlim = c(0,1000), xaxp=c(0,1000,10), yaxp=c(100,800,7), ylim=c(100,800)) Any ideas. Thank you Ahmed M. Attia Research Assistant Dept. of SoilCrop Sciences Texas AM University ahmed.at...@ag.tamu.edu Cell phone: 001-979-248-5215 FAX: 001-308-455-4024 __ 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] SQL vs R
On May 5, 2014, at 11:44 AM, Dr Eberhard Lisse wrote: I do not wish to prolong this metadiscussion but I remain confused by your advice: 1) You don't understand what I asked (ie would have to parse two simple SQL statements) Correct ... at least for me. I could have guessed at what that statement might have meant, but why should I need to guess? Why not use a shared naturallanguage rather than restricting your audience to the more limited group that understands both languages? 2) The Original Post is understood enough, however, to point me to the Introduction to R (where I have not found something to help me) That means my guess would have been wrong, since like Jeff Newmiller, I thought a simple call to `table` would have succeeded. (Section 5.10 although the desire for ordering suggested by my guess regarding 3) My name is not Pete. I'm actually not sure who Pete was. It's a local expression of astonishment directed, not at you, but at Satish. That was a prelude to my effort explain to Satish why the other respondents to the list have not seen fit to be more expansive in their responses. I thought Satish's comment was gratuitous (and likewise unhelpful). If you don't want to help me, don't. Several people are trying to help. You are remaining obdurate in failing to explain what is desired in natural language or posting an example in R code with desired output, as well as in failing to heed multiple other bits of advice in the Posting Guide. The accepted practice in responses is to include any context that might further the conversation. To my mind that would have required that you include the original request: How do I do something like this without using sqldf? a - sqldf(SELECT COUNT(*) FROM b WHERE c = 'd') or e - sqldf(SELECT f, COUNT(*) FROM b GROUP BY f ORDER BY f) How does the earlier suggestion to look at the 'table' function fail to address the first alternative? ( It appears it might satisfy the second one as well.) -- David. Nobody is forcing you to reply. el On 2014-05-04, 06:56 , David Winsemius wrote: On May 3, 2014, at 9:10 PM, Satish Anupindi Rao wrote: By making the effort to learn R?? very constructive and not condescending at all. We, lesser beings, are indebted to you, sir. For Pete's sake. The OP didn't even express his original request in natural language or offer a working example. Those of us who are not regular SQL users would have needed to parse out the SQL code in order to figure out what was intended. (My guess is that it would have been quite easy to solve if those were what were offered.) But making the effort to divine the intent didn't seem justified by the level of courtesy offered by the questioner. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Max characters from deparse(substitute(.))?
Is there a standard or or a standard utility to limit the size of
deparse(substitute(.))?
Below please find an example of the problem plus one solution.
If another solution already exists, I might prefer to use it.
Thanks,
Spencer
##
## Problem
##
deparse.x0 - function(x)deparse(substitute(x))
deparse.a - do.call(deparse.x0, list(letters))
nchar(deparse.a) # unacceptable
[1] 62 65 4
##
## Better
##
deparse.x - function(x, maxChar=20){
name.x - deparse(substitute(x))
nch.x - nchar(name.x)
name2 - name.x[nch.x0]
nch2 - nch.x[nch.x0]
if((length(name2)1)){
name2 - name2[1]
}
if(nch2[1]maxChar){
name2 - paste0(substring(name2, 1, maxChar), '...')
}
name2
}
do.call(deparse.x, list(letters)) # better
[1] c(\a\, \b\, \c\, \d\...
[[alternative HTML version deleted]]
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Re: [R] Sciplot: Increasing the width of bargraph and decreasing the sapce b/n groups
On 05/06/2014 05:02 AM, Roopa Subbaiaih wrote: Hello, I am trying to plot bargraphs susing Sciplot. Is there a way to increase the width of the bar graphs and decrease the space b/n the groups? I am pasting the script as well as attaching the graph. Bio6- read.csv(Data/Plin1.csv,na.strings=,header=T) attach(Bio6) head(Bio6) par(family=serif, font=11) Bio6$Sps- factor(Bio6$Sps, levels = c(FFA1, FFA2,FFA3)) Bio6$Gp- factor(Bio6$Gp, levels = c(N-FFA1, FU1,FA1,N-FFA2,FU2,FA2,N-FFA3,FU3,FA3)) bargraph.CI(Sps, O.D, group = Gp, data = Bio6,ylab = Relative expression levels, cex.lab = 1.5, y.leg = 6,cex.leg = 0.82,cex=1.5, axisnames=TRUE, col = c(red,blue,grey),space=c(0, 0.5), ylim=c(0,7),cex.names = 1.0,density = c(30,30,30), legend = TRUE, main=PLIN1) detach(Bio6) O.D Gp Sps 1 1.00 N-FFA1 FFA1 2 2.996432FU1 FFA1 3 3.223413FU1 FFA1 4 3.524465FU1 FFA1 5 1.311971FA1 FFA1 6 6.755860FA1 FFA1 7 1.566000FA1 FFA1 8 1.00 N-FFA2 FFA2 9 2.741612FU2 FFA2 10 2.800644FU2 FFA2 11 3.569509FU2 FFA2 12 4.141500FA2 FFA2 13 7.049476FA2 FFA2 14 4.694674FA2 FFA2 15 1.00 N-FFA3 FFA3 16 4.163601FU3 FFA3 17 3.903986FU3 FFA3 18 4.73FU3 FFA3 19 0.00FA3 FFA3 20 0.00FA3 FFA3 21 0.00FA3 FFA3 Hi Roopa, bargraph.CI does something with the space argument that I can't quite work out. I can get a reasonable plot like this: Bmeans-matrix(by(Bio6$O.D,Bio6$Gp],FUN=mean),ncol=3) barpos-barplot(Bmeans,beside=TRUE, ylim=c(0,7),col=c(red,blue,grey),space=c(0.1,1),main=PLIN1) legend(8.5,7.1, c(N-FFA1,FU1,FA1,N-FFA2,FU2,FA2,N-FFA3,FU3,FA3), fill=c(red,blue,grey),bty=n) library(plotrix) Bse-matrix(by(Bio6$O.D,Bio6$Gp,FUN=std.error),ncol=3) dispersion(barpos,Bmeans,Bse,display.na=FALSE) 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] Max characters from deparse(substitute(.))?
Have you looked at the width.cutoff and nlines arguments to deparse?
width.cutoff controls how long a line can be and it quits producing
output (saving time and space) after nlines of output are produced.
You want essentially the following
f0 - function(x, maxChar=20) deparse(x, width.cutoff=maxChar, nlines=1)
or, to add the '...' to the end if something was cut off
f1 - function(x, maxChar=20) {
# deparse warns and uses 65 if maxChar20, so avoid that.
retval - deparse(x, width.cutoff=max(20, maxChar+1), nlines=1)
if (nchar(retval) maxChar) {
retval - paste0(substring(retval, 1, maxChar), ...)
}
retval
}
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, May 5, 2014 at 2:44 PM, Spencer Graves
spencer.gra...@structuremonitoring.com wrote:
Is there a standard or or a standard utility to limit the size of
deparse(substitute(.))?
Below please find an example of the problem plus one solution.
If another solution already exists, I might prefer to use it.
Thanks,
Spencer
##
## Problem
##
deparse.x0 - function(x)deparse(substitute(x))
deparse.a - do.call(deparse.x0, list(letters))
nchar(deparse.a) # unacceptable
[1] 62 65 4
##
## Better
##
deparse.x - function(x, maxChar=20){
name.x - deparse(substitute(x))
nch.x - nchar(name.x)
name2 - name.x[nch.x0]
nch2 - nch.x[nch.x0]
if((length(name2)1)){
name2 - name2[1]
}
if(nch2[1]maxChar){
name2 - paste0(substring(name2, 1, maxChar), '...')
}
name2
}
do.call(deparse.x, list(letters)) # better
[1] c(\a\, \b\, \c\, \d\...
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] SQL vs R
On Fri, May 2, 2014 at 5:23 PM, Dr Eberhard Lisse nos...@lisse.na wrote: Hi, How do I do something like this without using sqldf? a - sqldf(SELECT COUNT(*) FROM b WHERE c = 'd') or e - sqldf(SELECT f, COUNT(*) FROM b GROUP BY f ORDER BY f) In the examples section at the bottom of ?sqldf are a number of SQL statements and the corresponding R statements. -- 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] Max characters from deparse(substitute(.))?
On 5/5/2014 3:42 PM, William Dunlap wrote:
Have you looked at the width.cutoff and nlines arguments to deparse?
width.cutoff controls how long a line can be and it quits producing
output (saving time and space) after nlines of output are produced.
You want essentially the following
f0 - function(x, maxChar=20) deparse(x, width.cutoff=maxChar, nlines=1)
or, to add the '...' to the end if something was cut off
f1 - function(x, maxChar=20) {
# deparse warns and uses 65 if maxChar20, so avoid that.
retval - deparse(x, width.cutoff=max(20, maxChar+1), nlines=1)
if (nchar(retval) maxChar) {
retval - paste0(substring(retval, 1, maxChar), ...)
}
retval
}
Hi, Bill: Thanks very much. I could have solved it myself if I had
known which FMTR. Thanks again. Spencer
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, May 5, 2014 at 2:44 PM, Spencer Graves
spencer.gra...@structuremonitoring.com wrote:
Is there a standard or or a standard utility to limit the size of
deparse(substitute(.))?
Below please find an example of the problem plus one solution.
If another solution already exists, I might prefer to use it.
Thanks,
Spencer
##
## Problem
##
deparse.x0 - function(x)deparse(substitute(x))
deparse.a - do.call(deparse.x0, list(letters))
nchar(deparse.a) # unacceptable
[1] 62 65 4
##
## Better
##
deparse.x - function(x, maxChar=20){
name.x - deparse(substitute(x))
nch.x - nchar(name.x)
name2 - name.x[nch.x0]
nch2 - nch.x[nch.x0]
if((length(name2)1)){
name2 - name2[1]
}
if(nch2[1]maxChar){
name2 - paste0(substring(name2, 1, maxChar), '...')
}
name2
}
do.call(deparse.x, list(letters)) # better
[1] c(\a\, \b\, \c\, \d\...
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] multiply of two expressions
On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh javanrou...@yahoo.com wrote: hello, i want to differentiate of L with respect to b when: L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln likelihood) and ln(mu/(mu+k)) = a+bx #link function how can i do it in R? Try this. First we solve for 'mu' in terms of the other variables using the link equation: library(Ryacas) k - Sym(k) mu - Sym(mu) y - Sym(y) L - Sym(L) a - Sym(a) b - Sym(b) x - Sym(x) sumy - Sym(sumy) Solve(log(mu/(mu+k)) == a+b*x, mu) expression(list(mu == k * exp(a + b * x)/(1 - exp(a + b * x Now in 'k*log(k/(k+mu)) + sumy * log(1-(k/mu+k))' substitute 'k * exp(a + b * x)/(1 - exp(a + b * x)))' for 'mu' using 'Subst' and take the derivative with respect to 'a' using 'deriv': s - Subst(k*log(k/(k+mu)) + sumy * log(1-(k/mu+k)), mu, + k * exp(a + b * x)/(1 - exp(a + b * x))) deriv(s, a) expression(sumy * (k * ((1 - exp(a + b * x)) * (k * exp(a + b * x)) + k * exp(a + b * x)^2))/((1 - exp(a + b * x))^2 * (k * exp(a + b * x)/(1 - exp(a + b * x)))^2 * (1 - (k * (1 - exp(a + b * x))/(k * exp(a + b * x)) + k))) - k * ((k + k * exp(a + b * x)/(1 - exp(a + b * x))) * (k * ((1 - exp(a + b * x)) * (k * exp(a + b * x)) + k * exp(a + b * x)^2)))/((1 - exp(a + b * x))^2 * ((k + k * exp(a + b * x)/(1 - exp(a + b * x)))^2 * k))) __ 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] multiply of two expressions
On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote: On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh javanrou...@yahoo.com wrote: hello, i want to differentiate of L with respect to b when: L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln likelihood) and ln(mu/(mu+k)) = a+bx #link function how can i do it in R? Try this. First we solve for 'mu' in terms of the other variables using the link equation: library(Ryacas) k - Sym(k) mu - Sym(mu) y - Sym(y) L - Sym(L) a - Sym(a) b - Sym(b) x - Sym(x) sumy - Sym(sumy) Solve(log(mu/(mu+k)) == a+b*x, mu) expression(list(mu == k * exp(a + b * x)/(1 - exp(a + b * x Now in 'k*log(k/(k+mu)) + sumy * log(1-(k/mu+k))' substitute 'k * exp(a + b * x)/(1 - exp(a + b * x)))' for 'mu' using 'Subst' and take the derivative with respect to 'a' using 'deriv': s - Subst(k*log(k/(k+mu)) + sumy * log(1-(k/mu+k)), mu, + k * exp(a + b * x)/(1 - exp(a + b * x))) deriv(s, a) expression(sumy * (k * ((1 - exp(a + b * x)) * (k * exp(a + b * x)) + k * exp(a + b * x)^2))/((1 - exp(a + b * x))^2 * (k * exp(a + b * x)/(1 - exp(a + b * x)))^2 * (1 - (k * (1 - exp(a + b * x))/(k * exp(a + b * x)) + k))) - k * ((k + k * exp(a + b * x)/(1 - exp(a + b * x))) * (k * ((1 - exp(a + b * x)) * (k * exp(a + b * x)) + k * exp(a + b * x)^2)))/((1 - exp(a + b * x))^2 * ((k + k * exp(a + b * x)/(1 - exp(a + b * x)))^2 * k))) But can we maybe get the Taylor series approximation to first or second order? -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multiply of two expressions
On Mon, May 5, 2014 at 10:16 PM, David Winsemius dwinsem...@comcast.net wrote: On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote: On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh javanrou...@yahoo.com wrote: hello, i want to differentiate of L with respect to b when: L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln likelihood) and ln(mu/(mu+k)) = a+bx #link function how can i do it in R? Try this. First we solve for 'mu' in terms of the other variables using the link equation: library(Ryacas) k - Sym(k) mu - Sym(mu) y - Sym(y) L - Sym(L) a - Sym(a) b - Sym(b) x - Sym(x) sumy - Sym(sumy) Solve(log(mu/(mu+k)) == a+b*x, mu) expression(list(mu == k * exp(a + b * x)/(1 - exp(a + b * x Now in 'k*log(k/(k+mu)) + sumy * log(1-(k/mu+k))' substitute 'k * exp(a + b * x)/(1 - exp(a + b * x)))' for 'mu' using 'Subst' and take the derivative with respect to 'a' using 'deriv': s - Subst(k*log(k/(k+mu)) + sumy * log(1-(k/mu+k)), mu, + k * exp(a + b * x)/(1 - exp(a + b * x))) deriv(s, a) expression(sumy * (k * ((1 - exp(a + b * x)) * (k * exp(a + b * x)) + k * exp(a + b * x)^2))/((1 - exp(a + b * x))^2 * (k * exp(a + b * x)/(1 - exp(a + b * x)))^2 * (1 - (k * (1 - exp(a + b * x))/(k * exp(a + b * x)) + k))) - k * ((k + k * exp(a + b * x)/(1 - exp(a + b * x))) * (k * ((1 - exp(a + b * x)) * (k * exp(a + b * x)) + k * exp(a + b * x)^2)))/((1 - exp(a + b * x))^2 * ((k + k * exp(a + b * x)/(1 - exp(a + b * x)))^2 * k))) But can we maybe get the Taylor series approximation to first or second order? Taylor(d, a, 0, 2) expression(sumy * (k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2))/((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (1 - (k * (1 - exp(b * x))/(k * exp(b * x)) + k))) - k * ((k + k * exp(b * x)/(1 - exp(b * x))) * (k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2)))/((1 - exp(b * x))^2 * ((k + k * exp(b * x)/(1 - exp(b * x)))^2 * k)) + (((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (1 - (k * (1 - exp(b * x))/(k * exp(b * x)) + k)) * (sumy * (k * ((1 - exp(b * x)) * (k * exp(b * x)) - k * exp(b * x)^2 + k * (2 * exp(b * x)^2 - sumy * (k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2)) * ((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2))/((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2) + ((1 - exp(b * x))^2 * (k * exp(b * x) * (2 * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2)))/(1 - exp(b * x))^3 + -2 * exp(b * x) * (1 - exp(b * x)) * (k * exp(b * x)/(1 - exp(b * x)))^2) * (1 - (k * (1 - exp(b * x))/(k * exp(b * x)) + k/((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (1 - (k * (1 - exp(b * x))/(k * exp(b * x)) + k)))^2 - ((1 - exp(b * x))^2 * ((k + k * exp(b * x)/(1 - exp(b * x)))^2 * k) * (k * ((k + k * exp(b * x)/(1 - exp(b * x))) * (k * ((1 - exp(b * x)) * (k * exp(b * x)) - k * exp(b * x)^2 + k * (2 * exp(b * x)^2))) + k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2)^2/(1 - exp(b * x))^2)) - k * ((k + k * exp(b * x)/(1 - exp(b * x))) * (k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2))) * ((1 - exp(b * x))^2 * (k * ((k + k * exp(b * x)/(1 - exp(b * x))) * (2 * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2/(1 - exp(b * x))^2 + -2 * exp(b * x) * (1 - exp(b * x)) * ((k + k * exp(b * x)/(1 - exp(b * x)))^2 * k)))/((1 - exp(b * x))^2 * ((k + k * exp(b * x)/(1 - exp(b * x)))^2 * k))^2) * a + a^2 * 1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (1 - (k * (1 - exp(b * x))/(k * exp(b * x)) + k)))^2 * ((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (1 - (k * (1 - exp(b * x))/(k * exp(b * x)) + k)) * (sumy * (k * ((1 - exp(b * x)) * (k * exp(b * x)) - k * exp(b * x)^2 - k * (2 * exp(b * x)^2) + k * (4 * exp(b * x)^2 + ((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2))/((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2) + ((1 - exp(b * x))^2 * (k * exp(b * x) * (2 * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2)))/(1 - exp(b * x))^3 + -2 * exp(b * x) * (1 - exp(b * x)) * (k * exp(b * x)/(1 - exp(b * x)))^2) * (1 - (k * (1 - exp(b * x))/(k * exp(b * x)) + k))) * (sumy * (k * ((1 - exp(b * x)) * (k * exp(b * x)) - k * exp(b * x)^2 + k * (2 * exp(b * x)^2 - (sumy * (k * ((1 - exp(b * x)) * (k * exp(b * x)) + k * exp(b * x)^2)) * (((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * ((1 - exp(b * x))^2 * (k * exp(b * x)/(1 - exp(b * x)))^2 * (k * ((1 - exp(b * x)) *
Re: [R] Problem with products in R ?
Gabor,
Can you confirm that the bc function is supposed to be current.
The bc package works with my Mac, but not with Windows.
I keep getting the message
Error in system(cmd, input = input, intern = TRUE) : -l
The FAQ on your https://code.google.com/p/r-bc/ page didn't get me past that
problem. I tried both the download bc.zip from the page and also the cygwin bc.
A secondary issue is that placing your bc.exe into c:/Program
Files/R/R-3.1.0/library/bc/bcdir/bc.exe
gives the message
one - bc(1)
Error in system(cmd, input = input, intern = TRUE) :
'C:/Program' not found
Working around that is possible with
bc.cmd - C:/Progra~1/R/R-3.1.0/library/bc/bcdir/bc.exe -l
one - bc(1, cmd=bc.cmd)
but the next line gives the same problem
one
Error in system(cmd, input = input, intern = TRUE) :
'C:/Program' not found
Thanks
Rich
On Sun, May 4, 2014 at 1:10 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Checking this with the bc R package (https://code.google.com/p/r-bc/),
the Ryacas package (CRAN), the gmp package (CRAN) and the Windows 8.1
calculator all four give the same result:
library(bc)
bc(168988580159 * 36662978)
[1] 6195624596620653502
library(Ryacas)
yacas(168988580159 * 36662978, retclass = character)
6195624596620653502
library(gmp)
as.bigz(168988580159) * as.bigz(36662978)
Big Integer ('bigz') :
[1] 6195624596620653502
On Sun, May 4, 2014 at 12:50 PM, Ted Harding ted.hard...@wlandres.net wrote:
On 04-May-2014 14:13:27 Jorge I Velez wrote:
Try
options(digits = 22)
168988580159 * 36662978
# [1] 6195624596620653568
HTH,
Jorge.-
Err, not quite ... !
I hitch my horses to my plough (with help from R):
options(digits=22)
168988580159*8 = 1351908641272 (copy down)
168988580159*7 = 1182920061113 ( )
168988580159*9 = 1520897221431 ( )
168988580159*2 = 337977160318 ( )
168988580159*6 = 1013931480954 ( )^3
168988580159*3 = 506965740477 ( )
1351908641272
11829200611130
152089722143100
337977160318000
1013931480954
10139314809540
101393148095400
506965740477000
==
6195624596620653502
[after adding up mentally]
compared with Jorge's:
6195624596620653568
(02 vs 68 in the final two digits).
Alternatively, if using a unixoid system with 'bc' present,
one can try interfacing R with 'bc'. 'bc' is an calculating
engine which works to arbitrary precision.
There certainly used to be a utility in which R can evoke 'bc',
into which one can enter a 'bc' command and get the result
returned as a string, but I can't seem to find it on CRAN now.
In any case, the raw UNIX command line for this calculation
with 'bc' (with result) is:
$ bc -l
[...]
168988580159 * 36662978
6195624596620653502
quit
which agrees with my horse-drawn working.
Best wishes to all,
Ted.
On Sun, May 4, 2014 at 10:44 PM, ARTENTOR Diego Tentor
diegotento...@gmail.com wrote:
Trying algorithm for products with large numbers i encountered a difference
between result of 168988580159 * 36662978 in my algorithm and r product.
The Microsoft calculator confirm my number.
Thanks.
--
*Gráfica ARTENTOR *
de Diego L. Tentor
Echagüe 558
Tel.:0343 4310119
Paraná - Entre Ríos
-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 04-May-2014 Time: 17:50:54
This message was sent by XFMail
__
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.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with products in R ?
1. Firstly your library should be in user space. Don't use the one in
C:\Program Files\... . That is the case not only for bc but for other
packages too. Reinstall R and the first time you install a package it
should ask you to allow it to create a user library. Be sure you
allow that.
2. As indicated in the FAQ which bc you use matters. I am using the
one on the Downloads tab, i.e. here:
https://code.google.com/p/r-bc/downloads/list
If you are using a different bc then try this one.
3. Also ensure bc on your path. From within R: Sys.which(bc) should
find it if its on your path.
4. I usually just source the bc.R file and that seems to work for me.
I just added a NAMESPACE file to the svn repo so that the package will
build and install on more recent versions of R.
This works for me on R 3.1 patched under Windows 8.1:
source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;)
one - bc(1)
one
[1] 1
and with the version of the repo that I just updated a few minutes ago
and building and installing bc in a separate session this works too:
library(bc)
bc(1)
[1] 1
On Mon, May 5, 2014 at 11:09 PM, Richard M. Heiberger r...@temple.edu wrote:
Gabor,
Can you confirm that the bc function is supposed to be current.
The bc package works with my Mac, but not with Windows.
I keep getting the message
Error in system(cmd, input = input, intern = TRUE) : -l
The FAQ on your https://code.google.com/p/r-bc/ page didn't get me past that
problem. I tried both the download bc.zip from the page and also the cygwin
bc.
A secondary issue is that placing your bc.exe into c:/Program
Files/R/R-3.1.0/library/bc/bcdir/bc.exe
gives the message
one - bc(1)
Error in system(cmd, input = input, intern = TRUE) :
'C:/Program' not found
Working around that is possible with
bc.cmd - C:/Progra~1/R/R-3.1.0/library/bc/bcdir/bc.exe -l
one - bc(1, cmd=bc.cmd)
but the next line gives the same problem
one
Error in system(cmd, input = input, intern = TRUE) :
'C:/Program' not found
Thanks
Rich
On Sun, May 4, 2014 at 1:10 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Checking this with the bc R package (https://code.google.com/p/r-bc/),
the Ryacas package (CRAN), the gmp package (CRAN) and the Windows 8.1
calculator all four give the same result:
library(bc)
bc(168988580159 * 36662978)
[1] 6195624596620653502
library(Ryacas)
yacas(168988580159 * 36662978, retclass = character)
6195624596620653502
library(gmp)
as.bigz(168988580159) * as.bigz(36662978)
Big Integer ('bigz') :
[1] 6195624596620653502
On Sun, May 4, 2014 at 12:50 PM, Ted Harding ted.hard...@wlandres.net
wrote:
On 04-May-2014 14:13:27 Jorge I Velez wrote:
Try
options(digits = 22)
168988580159 * 36662978
# [1] 6195624596620653568
HTH,
Jorge.-
Err, not quite ... !
I hitch my horses to my plough (with help from R):
options(digits=22)
168988580159*8 = 1351908641272 (copy down)
168988580159*7 = 1182920061113 ( )
168988580159*9 = 1520897221431 ( )
168988580159*2 = 337977160318 ( )
168988580159*6 = 1013931480954 ( )^3
168988580159*3 = 506965740477 ( )
1351908641272
11829200611130
152089722143100
337977160318000
1013931480954
10139314809540
101393148095400
506965740477000
==
6195624596620653502
[after adding up mentally]
compared with Jorge's:
6195624596620653568
(02 vs 68 in the final two digits).
Alternatively, if using a unixoid system with 'bc' present,
one can try interfacing R with 'bc'. 'bc' is an calculating
engine which works to arbitrary precision.
There certainly used to be a utility in which R can evoke 'bc',
into which one can enter a 'bc' command and get the result
returned as a string, but I can't seem to find it on CRAN now.
In any case, the raw UNIX command line for this calculation
with 'bc' (with result) is:
$ bc -l
[...]
168988580159 * 36662978
6195624596620653502
quit
which agrees with my horse-drawn working.
Best wishes to all,
Ted.
On Sun, May 4, 2014 at 10:44 PM, ARTENTOR Diego Tentor
diegotento...@gmail.com wrote:
Trying algorithm for products with large numbers i encountered a
difference
between result of 168988580159 * 36662978 in my algorithm and r product.
The Microsoft calculator confirm my number.
Thanks.
--
*Gráfica ARTENTOR *
de Diego L. Tentor
Echagüe 558
Tel.:0343 4310119
Paraná - Entre Ríos
-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 04-May-2014 Time: 17:50:54
This message was sent by XFMail
__
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
Re: [R] list.files accessing subdirectory as relative path?
Hi .. is the parent directory whereas . is the current directory, so you probably want list.files(path = ./Least Developed Countries) instead of list.files(path = ../Least Developed Countries) Yours sincerely / Med venlig hilsen Frede Aakmann Tøgersen Specialist, M.Sc., Ph.D. Plant Performance Modeling Technology Service Solutions T +45 9730 5135 M +45 2547 6050 fr...@vestas.com http://www.vestas.com Company reg. name: Vestas Wind Systems A/S This e-mail is subject to our e-mail disclaimer statement. Please refer to www.vestas.com/legal/notice If you have received this e-mail in error please contact the sender. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Adel Sent: 5. maj 2014 21:30 To: r-help@r-project.org Subject: [R] list.files accessing subdirectory as relative path? Dear list members, I would like to access a subdirectory given where the work directory has been set. So I have: getwd() [1] C:/Users/Lord Adellus/Dropbox/I8child1/Data list.files() # give three folders [1] Least Developed Countries [4] Low middle income grouping More advanced developing countries and territories list.files(path = ../Least Developed Countries) # I want to access now one of the subdirectories character(0) But as you can see R does not want to look into the specified subdirectory. I have tried several combination and searched the list but without any great success. Actually: list.files(path = ../...) #goes up one level in the folder structure so I cannot see what the problem is. Thanks in advance. Adel -- View this message in context: http://r.789695.n4.nabble.com/list-files- accessing-subdirectory-as-relative-path-tp4689997.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Making a package works on any R version
Hi all I had made a package in R-3.0.3 , and its running well , my issue is it is not running in other versions or R like R-3.0.2/3.0.1 it is showing error like --- Error: This is R 3.0.2, package âxxxâ needs = 3.0.3 Does anybody have the solution on how to make this package run on any versions . i know its not a big problem , please forgive my ignorance if it sounds silly . Thanks , ASHIS [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice Histogram with Normal Curve - Y axis as percentages
Hi Jim
Without going to the histogram function to study it in detail it appears
when there is density used in the histogram function the data has to be in
the original scale to to send to the panel function.
As you example is not reproducible use the singer data to get the ylimits
for the histogram and density plot separately
For density
d -
histogram( ~ height | voice.part, data = singer,
xlab = Height (inches), type = density,
panel = function(x, ...) {
# panel.histogram(x, ...)
panel.abline(v= 70)
panel.mathdensity(dmath = dnorm, col = black,
args = list(mean=mean(x),sd=sd(x)))
} )
d$y.limits ...
and the same for the histogram with type = percent
if you divide the last by the former there is a difference of 192 (ie
scaling factor)
Using the full dataset as it is quicker to demonstrate
histogram( ~ height, data = singer,
type = percent, border = transparent, col.line = grey60,
xlab = Height (inches),
ylab = Density Histogram\n with Normal Fit )
trellis.focus(panel, 1, 1, clip.off=F, highlight = FALSE)
llines(density(singer$height)$x, density(singer$height)$y*192)
trellis.unfocus()
This will give you the idea that it is possible. y axes labels etc are all
funny now
By using a user defined panel function plotting the histogram and the
density output plotted as panel.lines after scaling within the function you
should be able to get a plot
It may be easier to start off with an xyplot to send the data to the panel
function as you will need 2 types of data as original and that for the
histogram.
Whether you need to use panel.groups is another mater
HTH
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet.com.au
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of jimdare
Sent: Tuesday, 6 May 2014 06:23
To: r-help@r-project.org
Subject: [R] Lattice Histogram with Normal Curve - Y axis as percentages
Hello,
This may seem like a simple problem, but it's frustrating me immensely. I'm
trying to overlay a normal curve (dnorm) on top of a histogram using the
code below. This works find when the type = density, but the person for
whom I'm making the plot wants the y axis in percent of total rather than
density. When I change type to percent, I get the histogram scale I'm
after, but the dnorm plot is greatly reduced. How could I scale the density
plot to the percent of total axis. Alternatively, perhaps there is a way to
add density to a secondary y axis?
Thanks in advance for your help.
Jimdare
plot-histogram(~rdf[,j]|Year,nint=20, data=rdf,main = i,strip =
my.strip,xlab = j,
type = percent,layout=c(2,1),
panel=function(x, ...) {
panel.histogram(x, ...)
panel.mathdensity(dmath=dnorm, col=black,
# Add na.rm = TRUE to mean() and sd()
args=list(mean=mean(x, na.rm = TRUE),
sd=sd(x, na.rm = TRUE)), ...)
})
--
View this message in context:
http://r.789695.n4.nabble.com/Lattice-Histogram-with-Normal-Curve-Y-axis-as-
percentages-tp469.html
Sent from the R help mailing list archive at Nabble.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.
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Re: [R-es] Mapa de quantiles con spplot
No resulta fácil contestarte con la información que proporcionas. Sobre la base de las variables que definiste, intuyó que lo siguiente debería funcionar: zm$clases - class spplot(zm, clases , col.regions=plotclr) Un saludo. Olivier Hola, Intento representar en un mapa participaciones porcentuales de los sectores económicos y no logro hacerlo con spplot. He intentado con spplot(zm, c(part88, part93), cuts=4, col.regions=brewer.pal(4, Set3)) Lo pude hacer utilizando el base graphics de R, definiendo: plotvar88 - zm$part88 nclr - 8 plotclr - brewer.pal(nclr, PuOr) plotclr - plotclr[nclr:1] class - classIntervals(plotvar88, nclr, style=quantile) colcode - findColours(class, plotclr) plot(zm) plot(zm, col=colcode, add=T) legend(3377189,2249615, legend=names(attr(colcode, table)), fill=attr(colcode, palette), cex=0.6, bty=n) Después intenté meter algunos de esos argumentos utilizando spplot spplot(zm, c(part88, part93), col.regions=plotclr, cuts=nclr) y tampoco funcionó. Tengo algunas nociones de cómo hacerlo en spplot, pero no sé cómo, pues desconozco la forma en utilizar los argumentos por ejemplo usando cuts, at. He revisado la ayuda que provee el paquete y sigo sin poder concretarlo. Agradecería un poco de orientación. Saludos cordiales. Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es -- Olivier G. Nuñez Email: onu...@unex.es http://kolmogorov.unex.es/~onunez Tel : +34 663 03 69 09 Departamento de Matemáticas Universidad de Extremadura ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R-es] Mapa de quantiles con spplot
No me sale ningún error, simplemente el resultado no es el esperado. Les comparto la carpeta con la capa y los datos que estoy usando. https://www.dropbox.com/s/umy0evt3qm1wr4d/dissolve.zip Esto es lo que estoy haciendo: library(rgdal) library(maptools) library(sp) library(classInt) library(RColorBrewer) zm - readOGR(“.”, “zmdis”) data - read.csv(“part_pot.csv”, header=T, sep=“,”, dec=“.”) zm@data=data.frame(data) plotvar88 - zm$part88 nclr - 8 plotclr - brewer.pal(nclr, “Blues”) class - classIntervals(plotvar88, nclr, style=“quantile”) colcode - findColours(class, plotclr) plot(zm) plot(zm, col=colcode, add=T) legend(3777189,2249615, legend=names(attar(colcode, “table”)), fill=attr(colcode, “palette”), cex=0.6, bty=“n”) Entonces obtengo el mapa como lo quiero (Me falta la división política, pero para este ejemplo la omití): https://www.dropbox.com/s/w3ps61o4jr1ciy5/part88.tiff Cuando lo hago con spplot: spplot(zm, part88, col.regions=plotclr, cuts=nclr, key.space=bottom”) Me regresa lo siguiente: https://www.dropbox.com/s/2cgiua5ra2tx4a0/part88_spplot.tiff Entonces lo que quiero es usar el spplot para pintar dos variables c(“part88”, “part93”), que el mismo ‘legend’ aplique para ambas. El 05/05/2014, a las 05:44, Oscar Perpiñan oscar.perpi...@upm.es escribió: Hola Rolando, Por el código que muestras debiera funcionar, pero falta información para poder responderte. Cuando dices que no funcionó, ¿que significa? ¿Obtienes un error? ¿El resultado no es el esperado? Por favor, envía los datos que estás utilizando, al menos una muestra, o algo que sirva para poder reproducir tu código. Saludos. Oscar. - Oscar Perpiñán Lamigueiro Dpto. Ingeniería Eléctrica (ETSIDI-UPM) Grupo de Sistemas Fotovoltaicos (IES-UPM) URL: http://oscarperpinan.github.io Twitter: @oscarperpinan El día 3 de mayo de 2014, 21:39, Rolando Valdez rvald...@gmail.com escribió: Hola, Intento representar en un mapa participaciones porcentuales de los sectores económicos y no logro hacerlo con spplot. He intentado con spplot(zm, c(“part88”, “part93”), cuts=4, col.regions=brewer.pal(4, “Set3”)) Lo pude hacer utilizando el base graphics de R, definiendo: plotvar88 - zm$part88 nclr - 8 plotclr - brewer.pal(nclr, PuOr) plotclr - plotclr[nclr:1] class - classIntervals(plotvar88, nclr, style=quantile) colcode - findColours(class, plotclr) plot(zm) plot(zm, col=colcode, add=T) legend(3377189,2249615, legend=names(attr(colcode, table)), fill=attr(colcode, palette), cex=0.6, bty=n”) Después intenté meter algunos de esos argumentos utilizando spplot spplot(zm, c(“part88”, “part93”), col.regions=plotclr, cuts=nclr) y tampoco funcionó. Tengo algunas nociones de cómo hacerlo en spplot, pero no sé cómo, pues desconozco la forma en utilizar los argumentos por ejemplo usando “cuts”, “at”. He revisado la ayuda que provee el paquete y sigo sin poder concretarlo. Agradecería un poco de orientación. Saludos cordiales. Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R-es] Mapa de quantiles con spplot
Hola, no me funciona: Te contesto lo mismo que le contesté a otro compañero de la lista, espero ser más claro. Les comparto la carpeta con la capa y los datos que estoy usando. https://www.dropbox.com/s/umy0evt3qm1wr4d/dissolve.zip Esto es lo que estoy haciendo: library(rgdal) library(maptools) library(sp) library(classInt) library(RColorBrewer) zm - readOGR(“.”, “zmdis”) data - read.csv(“part_pot.csv”, header=T, sep=“,”, dec=“.”) zm@data=data.frame(data) plotvar88 - zm$part88 nclr - 8 plotclr - brewer.pal(nclr, “Blues”) class - classIntervals(plotvar88, nclr, style=“quantile”) colcode - findColours(class, plotclr) plot(zm) plot(zm, col=colcode, add=T) legend(3777189,2249615, legend=names(attar(colcode, “table”)), fill=attr(colcode, “palette”), cex=0.6, bty=“n”) Entonces obtengo el mapa como lo quiero (Me falta la división política, pero para este ejemplo la omití): https://www.dropbox.com/s/w3ps61o4jr1ciy5/part88.tiff Cuando lo hago con spplot: spplot(zm, part88, col.regions=plotclr, cuts=nclr, key.space=bottom”) Me regresa lo siguiente: https://www.dropbox.com/s/2cgiua5ra2tx4a0/part88_spplot.tiff Entonces lo que quiero es usar el spplot para pintar dos variables c(“part88”, “part93”), que el mismo ‘legend’ aplique para ambas. Saludos cordiales El 05/05/2014, a las 10:39, Olivier Nuñez onu...@unex.es escribió: No resulta fácil contestarte con la información que proporcionas. Sobre la base de las variables que definiste, intuyó que lo siguiente debería funcionar: zm$clases - class spplot(zm, clases , col.regions=plotclr) Un saludo. Olivier Hola, Intento representar en un mapa participaciones porcentuales de los sectores económicos y no logro hacerlo con spplot. He intentado con spplot(zm, c(“part88”, “part93”), cuts=4, col.regions=brewer.pal(4, “Set3”)) Lo pude hacer utilizando el base graphics de R, definiendo: plotvar88 - zm$part88 nclr - 8 plotclr - brewer.pal(nclr, PuOr) plotclr - plotclr[nclr:1] class - classIntervals(plotvar88, nclr, style=quantile) colcode - findColours(class, plotclr) plot(zm) plot(zm, col=colcode, add=T) legend(3377189,2249615, legend=names(attr(colcode, table)), fill=attr(colcode, palette), cex=0.6, bty=n”) Después intenté meter algunos de esos argumentos utilizando spplot spplot(zm, c(“part88”, “part93”), col.regions=plotclr, cuts=nclr) y tampoco funcionó. Tengo algunas nociones de cómo hacerlo en spplot, pero no sé cómo, pues desconozco la forma en utilizar los argumentos por ejemplo usando “cuts”, “at”. He revisado la ayuda que provee el paquete y sigo sin poder concretarlo. Agradecería un poco de orientación. Saludos cordiales. Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es -- Olivier G. Nuñez Email: onu...@unex.es http://kolmogorov.unex.es/~onunez Tel : +34 663 03 69 09 Departamento de Matemáticas Universidad de Extremadura Rolando Valdez ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R-es] Realizar pruebas unitarias al codigo de R
Estimada Daliana Ramos Yo preste un libro que tenía sobre esos temas y que nunca me devolvieron. No recuerdo el nombre o tengo la posibilidad de buscar un ejemplo para enviarle. Pero yo buscaría por estos lados: http://www.stats.uwo.ca/faculty/murdoch/software/debuggingR/ Javier Marcuzzi El 05/05/2014 03:44 p.m., Daliana Ramos Garcia escribió: hola a todos, nuevamente tengo una duda y me gustarÃa compartirla con ustedes para ver si pueden ayudarme. Estoy haciendo un proyecto en R, usando la metodologÃa XP que como saben en la fase de pruebas hay que hacer pruebas unitaria al código, pero no se como realizarlas en R. Si alguien sabe como se pueden hacer le agradecerÃa su ayuda. ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-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
