Re: [R] R-bash beginneR
Dear Jeff, thanks ever so much for your comment. Please excuse me, but is it really not an R question?? well, my IT support team at the lab said it is not an informatics question either, so what shall I do? Complain to my boss because they don't speak R?. I found information for an R function(?) called Rscript in the R utils package that deals with this in some way (that I am not understanding completely), so somehow I think this is not a bad forum to ask. All the best, Dave Subject: Re: [R] R-bash beginneR From: jdnew...@dcn.davis.ca.us Date: Mon, 7 Nov 2011 08:29:43 -0800 To: dasol...@hotmail.com; r-help@r-project.org This is not an R question. Use the print function in R and use backticks in bash. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. David A. dasol...@hotmail.com wrote: Hi, I am trying to run some R commands into my bash scripts and want to use shell variables in the R commands and store the output of R objects into shell variables for further usage in downstream analyses. So far I have managed the first, but how to get values out of R script? I am using here documents (as a starter, maybe something else is simpler or better; suggestions greatly appreciated). A basic random example: #!/bin/sh MYVAR=2 R --slave --quiet --no-save EEE x-5 z-x/$MYVAR zz-x*$MYVAR EEE How get the values of z and zz into shell variables? Thanks D [[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. [[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] Sampling with conditions
[Yet another correction -- this one is important.
I start from scratch this time]
On 07-Nov-11 22:22:54, SarahJoyes wrote:
Hey everyone,
I am at best, an amateur user of R, but I am stuck on how
to set-up the following situation.
I am trying to select a random sample of numbers from 0 to 10
and insert them into the first column of a matrix (which will
used later in a loop).
However, I need to have those numbers add up to 10. How can
I set those conditions?
So far I have:
n-matrix(0,nr=5,ncol=10)
for(i in 1:10){n[i,1]-sample(0:10,1)}
How do I set-up the BUT sum(n[i,1])=10?
Thanks
SarahJ
Sarah, your example is confusing because you have set up a
matrix 'n' with 5 rows and 10 columns. But your loop cycles
through 10 rows!
However, assuming that your basic requirement is to sample
10 integers which add up to 10, consider rmultinom():
### Instead of: rmultinom(n=1,size=10,prob=(1:10)/10) ###
rmultinom(n=1,size=10,prob=rep(1,10)/10)
# [,1]
# [1,]1
# [2,]0
# [3,]2
# [4,]3
# [5,]1
# [6,]1
# [7,]0
# [8,]0
# [9,]1
#[10,]1
rmultinom(n=1,size=10,prob=rep(1,10)/10)
# [,1]
# [1,]2
# [2,]0
# [3,]1
# [4,]1
# [5,]2
# [6,]2
# [7,]1
# [8,]0
# [9,]1
#[10,]0
This gives a uniform distribution over the positions in
the sample vector for the sampled integers, so that all
permutations are equally likely. For a non-uniform
distribution, vary 'prob'.
Ted.
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Fax-to-email: +44 (0)870 094 0861
Date: 08-Nov-11 Time: 08:13:36
-- XFMail --
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Re: [R] rearrange set of items randomly
Sorry, but I dont think that I get what you mean (I am a quite new user though) I hae a data file of a sample, so I cannot use random numbers, the only thing I want to do is to randomly reassign the order of the items. -- View this message in context: http://r.789695.n4.nabble.com/rearrange-set-of-items-randomly-tp4013723p4015161.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] rearrange set of items randomly
On 08-Nov-11 07:46:15, flokke wrote: Sorry, but I dont think that I get what you mean (I am a quite new user though) I hae a data file of a sample, so I cannot use random numbers, the only thing I want to do is to randomly reassign the order of the items. The simplest method of randomly re-ordering is to use sample() to re-arrange (1:N) randomly, where N is the number of items in the data. Then use the result to access the items. Example: D - data.frame(X1=c(1.1,2.1,3.1,4.1), X2=c(1.2,2.2,3.2,4.2)) N - nrow(D) ix - sample((1:N)) D #X1 X2 # 1 1.1 1.2 # 2 2.1 2.2 # 3 3.1 3.2 # 4 4.1 4.2 N # [1] 4 ix # [1] 3 2 4 1 D[ix,] #X1 X2 # 3 3.1 3.2 # 2 2.1 2.2 # 4 4.1 4.2 # 1 1.1 1.2 Note that the defaults for sample() are (see ?sample): For 'sample' the default for 'size' is the number of items inferred from the first argument, so that 'sample(x)' generates a random permutation of the elements of 'x' (or '1:x'). and the default for the 'replace' option is FALSE, so sample((1:N)) samples N from (1:N) without replacement, i.e. a random permutation. Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 08-Nov-11 Time: 08:59:35 -- 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.
Re: [R] ordination in vegan: what does downweight() do?
kelsmith kelsmith at usgs.gov writes: Can anyone point me in the right direction of figuring out what downweight() is doing? I am using vegan to perform CCA on diatom assemblage data. I have a lot of rare species, so I want to reduce the influence of rare species in my CCA. I have read that some authors reduce rare species by only including species with an abundance of at least 1% in at least one sample (other authors use 5% as a rule, but this removes at least half my species). If I code it as follows: cca(downweight(diatoms, fraction=5) ~ ., env) It is clearly not removing these species entirely from analysis, as some authors suggest. So I am wondering: what is downweight() doing exactly? I assume it is somehow ranking the species and reducing their abundance values based on their rank, but I'm not entirely sure and can't seem to figure out how to look at the code (R novice here). Nor can I find a clear description within the documentation (although I may be looking in all the wrong places). So, my inclination is to remove species that are very rare (max abundance 1%) prior to the CCA and then use the downweight function (fraction = 5?) in my CCA (as above). This way, I can include most of my species, but overall still reduce the impact of rare species. Dear kelsmith, First a question: how do you *know* that rare species influence the result? I know many people *believe* that rare species have an unduly high influence in CCA, but that is only ecological folklore with no empirical basis. You could have a look at the influence: run ordinations with all species and without rare species and compare resulting ordinations using procrustes() function in vegan. If the site/lake/sample/core/river results are very different, then rare species indeed were influential (use plot() so that you don't get misled by numbers). Rare species often are extreme in CA-family results, but that does not mean that they influenced the results. CA-family methods are weighted ordination methods, and for species the weights are their total abundances (marginal totals). For rare species the weights are low. The species are blown to the outskirts of the ordination after the ordination rotation, and they do not often influence that rotation very much. I see that I ignored explaining downweighting in detail in vegan documentation. You must read the reference cited there (Hill Gauch): they explain the procedure (at least vaguely: you must read the code to see how it is actually implemented). However, the principle is simple: like I wrote above, rare species have low weights, and downweighting makes these weights even smaller. It does not remove the species, but it makes them even less influential. In vegan, downweighting was implemented for decorana, because it was the part of Mark Hill's original decorana code. However, it was implemented in R (instead of the original FORTRAN) and made independent of decorana so that it could be used with other functions. This was done to serve those people who wanted to use it outside its original context (greetings to Oslo!). If you do not know what downweighting does, I suggest you don't use it. I also suggest that you check the influence of rare species before you remove them from your data. Cheers, Jari Oksanen __ 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] rearrange set of items randomly
On 08-Nov-11 08:59:38, Ted Harding wrote: On 08-Nov-11 07:46:15, flokke wrote: Sorry, but I dont think that I get what you mean (I am a quite new user though) I hae a data file of a sample, so I cannot use random numbers, the only thing I want to do is to randomly reassign the order of the items. The simplest method of randomly re-ordering is to use sample() to re-arrange (1:N) randomly, where N is the number of items in the data. Then use the result to access the items. Example: D - data.frame(X1=c(1.1,2.1,3.1,4.1), X2=c(1.2,2.2,3.2,4.2)) N - nrow(D) ix - sample((1:N)) D #X1 X2 # 1 1.1 1.2 # 2 2.1 2.2 # 3 3.1 3.2 # 4 4.1 4.2 N # [1] 4 ix # [1] 3 2 4 1 D[ix,] #X1 X2 # 3 3.1 3.2 # 2 2.1 2.2 # 4 4.1 4.2 # 1 1.1 1.2 Note that the defaults for sample() are (see ?sample): For 'sample' the default for 'size' is the number of items inferred from the first argument, so that 'sample(x)' generates a random permutation of the elements of 'x' (or '1:x'). and the default for the 'replace' option is FALSE, so sample((1:N)) samples N from (1:N) without replacement, i.e. a random permutation. Ted. While I am at it, an alternative to this use of sample() is to use order() to find the permutation which re-arranges a set of random numbers into increasing order. This in effect returns a random permutation of (1:N). Hence, instead of ix - sample(1:N)) in the above, you could use: ix - order(runif(N)) Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 08-Nov-11 Time: 09:30:33 -- 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.
Re: [R] ordination in vegan: what does downweight() do?
On Mon, 2011-11-07 at 10:24 -0800, kelsmith wrote: Can anyone point me in the right direction of figuring out what downweight() is doing? I am using vegan to perform CCA on diatom assemblage data. I have a lot of rare species, so I want to reduce the influence of rare species in my CCA. I have read that some authors reduce rare species by only including species with an abundance of at least 1% in at least one sample (other authors use 5% as a rule, but this removes at least half my species). That is not what downweight() is for. If you want this sort of selection, see chooseTaxa() in my analogue package diat.sel - chooseTaxa(diatoms, max.abun = 1, type = AND) or, if proportions not percent diat.sel - chooseTaxa(diatoms, max.abun = 0.01, type = AND) This sort of indexing is trivial (but I made a typo in the current version so type = OR won't work) so you can study the code of analogue:::chooseTaxa.default once I fix the CRAN version or on R-Forge now: https://r-forge.r-project.org/scm/viewvc.php/pkg/R/chooseTaxa.R?view=markuproot=analogue Jari has addressed the other part of your question. Jari also mentioned the issue about whether you should be removing or downweighting rare species. Many people, especially diatomists, do this for practical purposes in a routine fashion because their data sets are especially speciose and have a large proportion of low abundance taxa. As a general matter of routine practice, I don't think this is a very good way of working, especially as we have no good ecological grounds for doing so and who knows what information these species could be telling us if we just listened to them instead of discarding them. HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ 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] Reading a specific column of a csv file in a loop
Dear all:
I have two larges files with 2000 columns. For each file I am
performing a loop to extract the ith element of each file and create
a data frame with both ith elements in order to perform further
analysis. I am not extracting all the ith elements but only certain
which I am indicating on a vector called d.
See an example of my code below
### generate an example for the CSV files, the original files contain
more than 2000 columns, here for the sake of simplicity they have only
10 columns
M1-matrix(rnorm(1000), nrow=100, ncol=10,
dimnames=list(seq(1:100),letters[1:10]))
M2-matrix(rnorm(1000), nrow=100, ncol=10,
dimnames=list(seq(1:100),letters[1:10]))
write.table(M1, file=M1.csv, sep=,)
write.table(M2, file=M2.csv, sep=,)
### the vector containing the i elements to be read
d-c(1,4,7,8)
P1-read.table(M1.csv, header=TRUE)
P2-read.table(M1.csv, header=TRUE)
for (i in d) {
M-data.frame(P1[i],P2[i])
rm(list=setdiff(ls(),d))
}
As the files are quite large, I want to include read.table within
the loop so as it only read the ith element. I know that there is
the option colClasses for which I have to create a vector with zeros
for all the columns I do not want to load. Nonetheless I have no idea
how to make this vector to change in the loop, so as the only element
with no zeros is the ith element following the vector d. Any ideas
how to do this? Or is there anz other approach to load only an
specific element?
best regards,
Sergio René
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Re: [R] error message
Ok, I got it, I found a stupid typing error. Just got the genind object.
Tks
J
2011/11/7 JulianaMF j...@ib.usp.br
As I mentioned, the archives with the exact thread were back at the lab.
Here they are:
read.structure(file = SimStru.str)
How many genotypes are there?
820
How many markers are there?
10
Which column contains labels for genotypes ('0' if absent)?
0
Which column contains the population factor ('0' if absent)?
2
Which other optional columns should be read (press 'return' when done)?
1:
Which row contains the marker names ('0' if absent)?
1
Are genotypes coded by a single row (y/n)?
n
Converting data from a STRUCTURE .stru file to a genind object...
Error in txt[(lastline - n + 1):lastline] :
only 0's may be mixed with negative subscripts
Errr, what are you referring to? You also forgot to read the posting
guide of this mailing list that asks you to cite the original thread,
Uwe Ligges
Juliana
--
Juliana Machado Ferreira
TED Senior Fellow 2010-2012
Laboratory of Evolutionary Biology and Vertebrate Conservation
(LABEC, USP)
www.sosfauna.org/
--
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http://r.789695.n4.nabble.com/error-message-tp3223412p3998488.html
Sent from the R help mailing list archive at Nabble.com.
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--
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TED Senior Fellow 2010-2012
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www.sosfauna.org/
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Re: [R] Reading a specific column of a csv file in a loop
2011/11/8 Sergio René Araujo Enciso araujo.enc...@gmail.com:
Dear all:
I have two larges files with 2000 columns. For each file I am
performing a loop to extract the ith element of each file and create
a data frame with both ith elements in order to perform further
analysis. I am not extracting all the ith elements but only certain
which I am indicating on a vector called d.
See an example of my code below
### generate an example for the CSV files, the original files contain
more than 2000 columns, here for the sake of simplicity they have only
10 columns
M1-matrix(rnorm(1000), nrow=100, ncol=10,
dimnames=list(seq(1:100),letters[1:10]))
M2-matrix(rnorm(1000), nrow=100, ncol=10,
dimnames=list(seq(1:100),letters[1:10]))
write.table(M1, file=M1.csv, sep=,)
write.table(M2, file=M2.csv, sep=,)
### the vector containing the i elements to be read
d-c(1,4,7,8)
P1-read.table(M1.csv, header=TRUE)
P2-read.table(M1.csv, header=TRUE)
for (i in d) {
M-data.frame(P1[i],P2[i])
rm(list=setdiff(ls(),d))
}
As the files are quite large, I want to include read.table within
the loop so as it only read the ith element. I know that there is
the option colClasses for which I have to create a vector with zeros
for all the columns I do not want to load. Nonetheless I have no idea
how to make this vector to change in the loop, so as the only element
with no zeros is the ith element following the vector d. Any ideas
how to do this? Or is there anz other approach to load only an
specific element?
Its a bit messy if there are row names so lets generate M1.csv like this:
write.csv(M1, file = M1.csv, row.names = FALSE)
Then we can do this:
nc - ncol(read.csv(M1.csv, nrows = 1))
colClasses - replace(rep(NULL, nc), d, NA)
M1.subset - read.csv(M1.csv, colClasses = colClasses)
or using the same M1.csv that we just generated try this which uses
sqldf with the H2 backend:
library(sqldf)
library(RH2)
M0 - read.csv(M1.csv, nrows = 1)[0L, ]
M1.subset.h2 - sqldf(c(insert into M0 (select * from csvread('M1.csv')),
select a, d, g, h from M0))
This is referred to as Alternative 3 in FAQ#10 Example 6a on the sqldf
home page:
http://sqldf.googlecode.com
Alternative 1 and Alternative 2 listed there could also be tried.
(Note that although sqldf has a read.csv.sql command we did not use it
here since that command only works with the sqlite back end and the
RSQLite driver has a max of 999 columns.)
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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[R] skip on error
Dear all,
I have a different data sets and I am doing some calculations over time,
For that every data set is split into junks based on the time stamps
so one data set has like 10 timestamps.
There is also the case that one data set has less than 10 timestamps.
In my code I was doing the following
lapply(Datasource,analysis_for_one_data_source)
lapply(TimeFrames,analysis_for_one_data_source_and_one_time_frame)
as you can imagine there are times where the second lapply will explode
analysis_for_one_data_source_and_one_time_frame- function( DataSource,
TimeFrame,) {
return( do_analysis(DataSource,TimeFrame))
}
as you can understand this will return an error for a given Datasource that
does not have a timestamp. I was looking though if I can ask from R to handle
the error by continuing to the next one. So the lapply that returns and error
can for example to store the error message to the list it returns and continue
to the next element of the lapply list
Would that be possible in R?
B.R
Alex
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Re: [R] Reading a specific column of a csv file in a loop
have you considered reading in the data and then creating objects for each
column and then saving (save) each to disk. That way you incur the expense of
the read once and now have quick access (?load) to the column as you need them.
You could also use a database for this.
On Nov 8, 2011, at 5:04, Sergio René Araujo Enciso araujo.enc...@gmail.com
wrote:
Dear all:
I have two larges files with 2000 columns. For each file I am
performing a loop to extract the ith element of each file and create
a data frame with both ith elements in order to perform further
analysis. I am not extracting all the ith elements but only certain
which I am indicating on a vector called d.
See an example of my code below
### generate an example for the CSV files, the original files contain
more than 2000 columns, here for the sake of simplicity they have only
10 columns
M1-matrix(rnorm(1000), nrow=100, ncol=10,
dimnames=list(seq(1:100),letters[1:10]))
M2-matrix(rnorm(1000), nrow=100, ncol=10,
dimnames=list(seq(1:100),letters[1:10]))
write.table(M1, file=M1.csv, sep=,)
write.table(M2, file=M2.csv, sep=,)
### the vector containing the i elements to be read
d-c(1,4,7,8)
P1-read.table(M1.csv, header=TRUE)
P2-read.table(M1.csv, header=TRUE)
for (i in d) {
M-data.frame(P1[i],P2[i])
rm(list=setdiff(ls(),d))
}
As the files are quite large, I want to include read.table within
the loop so as it only read the ith element. I know that there is
the option colClasses for which I have to create a vector with zeros
for all the columns I do not want to load. Nonetheless I have no idea
how to make this vector to change in the loop, so as the only element
with no zeros is the ith element following the vector d. Any ideas
how to do this? Or is there anz other approach to load only an
specific element?
best regards,
Sergio René
__
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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] skip on error
?try
Sent from my iPad
On Nov 8, 2011, at 5:49, Alaios ala...@yahoo.com wrote:
Dear all,
I have a different data sets and I am doing some calculations over time,
For that every data set is split into junks based on the time stamps
so one data set has like 10 timestamps.
There is also the case that one data set has less than 10 timestamps.
In my code I was doing the following
lapply(Datasource,analysis_for_one_data_source)
lapply(TimeFrames,analysis_for_one_data_source_and_one_time_frame)
as you can imagine there are times where the second lapply will explode
analysis_for_one_data_source_and_one_time_frame- function( DataSource,
TimeFrame,) {
return( do_analysis(DataSource,TimeFrame))
}
as you can understand this will return an error for a given Datasource that
does not have a timestamp. I was looking though if I can ask from R to handle
the error by continuing to the next one. So the lapply that returns and error
can for example to store the error message to the list it returns and
continue to the next element of the lapply list
Would that be possible in R?
B.R
Alex
[[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.
[R] splitting by the last occurance of a dot
Dear R-helpers, I want to split the following vector into 2 vectors by the last occurance of a . dput(rownames(sensext)) c(pat, cash_bank_bal, invest_abroad, pat.1, cash_bank_bal.1, invest_abroad.1, pat.2, cash_bank_bal.2, invest_abroad.2, pat.3, cash_bank_bal.3, invest_abroad.3, pat.4, cash_bank_bal.4, invest_abroad.4, Market.Capitalisation, Market.Capitalisation.1, Market.Capitalisation.2, Market.Capitalisation.3, Market.Capitalisation.4 ) My attempt : I tried strsplit(rownames(sensext),\\.) but that splits it into 3 parts sometimes,the logic of which I can see,since there are 2 dots sometimes. Can someone tell me how to split this ? Many thanks, Ashim [[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] help to ... import the data from Excel
Sarah, it seems your current directory is not where your LTS.xls file is
located.
What I would is the following:
always use the xlsReadWrite functions in the current directory where your
excel file is located
Also, just reaad the ENTIR excel sheet, rather than referencing columns.
This the simplest way and it has ALWAYS worked for me.
e.g. STI-read.xls('STI_CASES.xls',sheet=1)
And I have my data, now I have to work a bit to get what I want (e.g. cases
for a certain year ect), but that is doable.
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Re: [R] R-bash beginner
On 11/07/2011 05:17 PM, David A. wrote: Hi, I am trying to run some R commands into my bash scripts and want to use shell variables in the R commands and store the output of R objects into shell variables for further usage in downstream analyses. So far I have managed the first, but how to get values out of R script? I am using here documents (as a starter, maybe something else is simpler or better; suggestions greatly appreciated). hi, maybe this can be helpful: http://rwiki.sciviews.org/doku.php?id=tips:scriptingr I personally didn't manage to successfully write a 'fully functional' script, though, because of quoting troubles and so on. best regards, -- Matteo Giani __ 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] Estimate of intercept in loglinear model
Sorry about that. However I have solved the problem by declaring the
explanatory variables as factors.
An unresolved problem is: what does R do when the explanatory factors
are not defined as factors when it obtains a different value for the
intercept but the correct value for the fitted value?
A description of the data and the R code and output is attached for
anyone interested.
Best wishes,
Colin Aitken
---
David Winsemius wrote:
On Nov 7, 2011, at 12:59 PM, Colin Aitken wrote:
How does R estimate the intercept term \alpha in a loglinear
model with Poisson model and log link for a contingency table of counts?
(E.g., for a 2-by-2 table {n_{ij}) with \log(\mu) = \alpha + \beta_{i}
+ \gamma_{j})
I fitted such a model and checked the calculations by hand. I agreed
with the main effect terms but not the intercept. Interestingly, I
agreed with the fitted value provided by R for the first cell {11} in
the table.
If my estimate of intercept = \hat{\alpha}, my estimate of the fitted
value for the first cell = exp(\hat{\alpha}) but R seems to be doing
something else for the estimate of the intercept.
However if I check the R $fitted_value for n_{11} it agrees with my
exp(\hat{\alpha}).
I would expect that with the corner-point parametrization, the
estimates for a 2 x 2 table would correspond to expected frequencies
exp(\alpha), exp(\alpha + \beta), exp(\alpha + \gamma), exp(\alpha +
\beta + \gamma). The MLE of \alpha appears to be log(n_{.1} *
n_{1.}/n_{..}), but this is not equal to the intercept given by R in
the example I tried.
With thanks in anticipation,
Colin Aitken
--
Professor Colin Aitken,
Professor of Forensic Statistics,
Do you suppose you could provide a data-corpse for us to dissect?
Noting the tag line for every posting
and provide commented, minimal, self-contained, reproducible code.
--
Professor Colin Aitken,
Professor of Forensic Statistics,
School of Mathematics, King’s Buildings, University of Edinburgh,
Mayfield Road, Edinburgh, EH9 3JZ.
Tel:0131 650 4877
E-mail: c.g.g.ait...@ed.ac.uk
Fax : 0131 650 6553
http://www.maths.ed.ac.uk/~cgga
The University of Edinburgh is a charitable body, registered in
Scotland, with registration number SC005336.
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Re: [R] iplots problem
I also have the same problem. Can you send me the link to download a proper Sun Java application (I have Microsot Windows XP Profesional V.2002)? Anyway, this is what I tried: library(JGR) Loading required package: JavaGD Loading required package: iplots Error : .onLoad failed in loadNamespace() for 'iplots', details: call: .jnew(org/rosuda/iplots/Framework) error: java.awt.HeadlessException Error: package ‘iplots’ could not be loaded install.packages(JavaGD, dependencies= TRUE) Aviso: package ‘JavaGD’ is in use and will not be installed install.packages(iplots, dependencies= TRUE) probando la URL 'http://cran.es.r-project.org/bin/windows/contrib/2.14/iplots_1.1-4.zip' Content type 'application/zip' length 448623 bytes (438 Kb) URL abierta downloaded 438 Kb package ‘iplots’ successfully unpacked and MD5 sums checked Aviso: cannot remove prior installation of package ‘iplots’ The downloaded packages are in C:\Documents and Settings\cristina.pascual\Configuración local\Temp\RtmpGuRMKp\downloaded_packages library(iplots) Error en library(iplots) : there is no package called ‘iplots’ library(JGR) Error: package ‘iplots’ required by ‘JGR’ could not be found u...@host.com -- View this message in context: http://r.789695.n4.nabble.com/iplots-problem-tp825990p4015425.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] Aggregate or extract function ?
Thanks for your help, I still have a little problem with this function because I don't have the same number of line in my two datarame so when I try to apply the dataframe function, I obtain this response ; that I have a different number of lines. Erreur dans data.frame(train, test[rowz, ]) : les arguments impliquent des nombres de lignes différents : 50327, 66592 Do you know how I could solve this problem ? Thanks, Céline -- View this message in context: http://r.789695.n4.nabble.com/Aggregate-or-extract-function-tp4013673p4015263.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] variance explained by each predictor in GAM
Dear Prof. Wood, I read your methods of extracting the variance explained by each predictor in different places. My question is: using the method you suggested, the sum of the deviance explained by all terms is not equal to the deviance explained by the full model. Could you tell me what caused such problem? set.seed(0) n-400 x1 - runif(n, 0, 1) ## to see problem with not fixing smoothing parameters ## remove the `##' from the next line, and the `sp' ## arguments from the `gam' calls generating b1 and b2. x2 - runif(n, 0, 1) ## *.1 + x1 f1 - function(x) exp(2 * x) f2 - function(x) 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10 f - f1(x1) + f2(x2) e - rnorm(n, 0, 2) y - f + e ## fit full and reduced models... b - gam(y~s(x1)+s(x2)) b1 - gam(y~s(x1),sp=b$sp[1]) b2 - gam(y~s(x2),sp=b$sp[2]) b0 - gam(y~1) ## calculate proportions deviance explained... dev.1 - (deviance(b1)-deviance(b))/deviance(b0) ## prop explained by s(x2) dev.2 - (deviance(b2)-deviance(b))/deviance(b0) ## prop explained by s(x1) dev.1 + dev.2 [1] 0.6974949 summary(b)$dev.expl [1] 0.7298136 I checked the two models (b1 b2), found the model coefficients are different with model b, so I feel it could be the problem. wish to hear your comments. Huidong Tian -- View this message in context: http://r.789695.n4.nabble.com/Re-variance-explained-by-each-predictor-in-GAM-tp896222p4015368.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] Selecting 3 different hours in a day
attach another column to you dataframe with the hour of the day and then use
'subset' to collect the three hours of interest.
Sent from my iPad
On Nov 8, 2011, at 1:12, jck13 jennake...@hotmail.com wrote:
Hello,
I have a csv with 5months of hourly data for 4 years. I would like to get
9am, 12pm and 3pm from each day and create a subset or a new data frame
that I can analyze. The time are from hour 0-23 for each day.
I am not sure how to create a loop which will take out each of these hours
and create a subset.
I was thinking of doing a for loop using the row number since:
9am= row 10
12pm= row 13
3pm= row 16
trying to loop through to extract these 3 times each day.
n=length(date_stamp)
for (i in i:n) {
m= 10
i= 1
new1= mv[m,]
i= i+1
m= m+3
##m+18 at row 16?
}
I need some help creating a loop through this! Thank you!
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Re: [R] How much time a process need?
could be that the process was waiting for the user to select a file from a list, or some other input before proceeding. Would have to see what the overall performance of the system was at that point. It could also have been that the process was low on physical memory and there was a lot of paging going on. Sent from my iPad On Nov 7, 2011, at 21:50, Rolf Turner rolf.tur...@xtra.co.nz wrote: On 08/11/11 02:40, Joshua Wiley wrote: On Mon, Nov 7, 2011 at 5:32 AM, Alaiosala...@yahoo.com wrote: So I just need to get the user system elapsed 0.460 0.048 67.366 user value and convert the seconds to days and then to hours ? Right? What about this elapsed field? It's all in seconds. Convert whatever fields you want. That being said, doesn't having an elapsed time of over 67 seconds, when the actual calculation takes less than half a second, indicate that something weird is going on? At the very least the R calculations are fighting for resources with some other very greedy processes. 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] splitting by the last occurance of a dot
On Tue, Nov 8, 2011 at 6:06 AM, Ashim Kapoor ashimkap...@gmail.com wrote: Dear R-helpers, I want to split the following vector into 2 vectors by the last occurance of a . dput(rownames(sensext)) c(pat, cash_bank_bal, invest_abroad, pat.1, cash_bank_bal.1, invest_abroad.1, pat.2, cash_bank_bal.2, invest_abroad.2, pat.3, cash_bank_bal.3, invest_abroad.3, pat.4, cash_bank_bal.4, invest_abroad.4, Market.Capitalisation, Market.Capitalisation.1, Market.Capitalisation.2, Market.Capitalisation.3, Market.Capitalisation.4 ) My attempt : I tried strsplit(rownames(sensext),\\.) but that splits it into 3 parts sometimes,the logic of which I can see,since there are 2 dots sometimes. Can someone tell me how to split this ? Assuming we want to split off the number at the end try this which splits on those dots which are followed by a digit: strsplit(r, \\.(?=\\d), perl = TRUE) -- 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] splitting by the last occurance of a dot
Assuming we want to split off the number at the end try this which splits on those dots which are followed by a digit: strsplit(r, \\.(?=\\d), perl = TRUE) Dear Gabor, Thank you very much. That works very well. I don't completely understand it though. A few words on what the (?=\\d) is doing would be nice. Regards, Ashim [[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] ggplot2 reorder factors for faceting
Dear List
I am trying to draw a heatmap using ggplot2. In this heatmap I have faceted my
data by 'infection' of which I have four. These four infections break down into
two types and I would like to reorder the 'infection' column of my data to
reflect this.
Toy example below:
library(ggplot2)
# test data for ggplot reordering
genes - (rep (c(rep('a',4), rep('b',4), rep('c',4), rep('d',4), rep('e',4),
rep('f',4)) ,4))
fcData - rnorm(96)
times - rep(rep(c(2,6,24,48),6),4)
infection - c(rep('InfA', 24), rep('InfB', 24), rep('InfC', 24), rep('InfD',
24))
infType - c(rep('M', 24), rep('D',24), rep('M', 24), rep('D', 24))
# data is long format for ggplot2
plotData - as.data.frame(cbind(genes, as.numeric(fcData), as.numeric(times),
infection, infType))
hp2 - ggplot(plotData, aes(factor(times), genes)) + geom_tile(aes(fill =
scale(as.numeric(fcData + facet_wrap(~infection, ncol=4)
# set scale
hp2 - hp2 + scale_fill_gradient2(name=NULL, low=#0571B0, mid=#F7F7F7,
high=#CA0020, midpoint=0, breaks=NULL, labels=NULL, limits=NULL,
trans=identity)
# set up text (size, colour etc etc)
hp2 - hp2 + labs(x = Time, y = ) + scale_y_discrete(expand = c(0, 0)) +
opts(axis.ticks = theme_blank(), axis.text.x = theme_text(size = 10, angle =
360, hjust = 0, colour = grey25), axis.text.y = theme_text(size=10, colour =
'gray25'))
hp2 - hp2 + theme_bw()
In the resulting plot I would like infections infA and infC plotted next to
each other and likewise for infB and infD. I have a column in the data -
infType - which I could use to reorder the infection column but so far I have
no luck getting this to work.
Could someone give me a pointer to the best way to reorder the infection factor
and accompanying data into the order I would like?
Best
iain
sessionInfo()
R version 2.13.2 (2011-09-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_GB.utf8 LC_NUMERIC=C
[3] LC_TIME=en_GB.utf8 LC_COLLATE=en_GB.utf8
[5] LC_MONETARY=C LC_MESSAGES=en_GB.utf8
[7] LC_PAPER=en_GB.utf8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_GB.utf8 LC_IDENTIFICATION=C
attached base packages:
[1] grid stats graphics grDevices utils datasets methods
[8] base
other attached packages:
[1] ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.6
loaded via a namespace (and not attached):
[1] digest_0.5.0 tools_2.13.2
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Re: [R] similar package in R like SKEW CALCULATOR?
See the replies given to you two days ago when you asked the same question: http://r.789695.n4.nabble.com/similar-package-in-R-like-quot-SKEW-CALCULATOR-quot-td3993216.html Feel free to follow up though with further questions, Michael On Tue, Nov 8, 2011 at 2:34 AM, Knut Krueger r...@knut-krueger.de wrote: Hi to all is there a similar package like the SKEW CALCULATOR from Peter Nonacs (University of California - Department of Ecology and Evolutionary Biology) http://www.eeb.ucla.edu/Faculty/Nonacs/shareware.htm Kind Regards Knut __ 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] splitting by the last occurance of a dot
On Tue, Nov 8, 2011 at 6:48 AM, Ashim Kapoor ashimkap...@gmail.com wrote: Assuming we want to split off the number at the end try this which splits on those dots which are followed by a digit: strsplit(r, \\.(?=\\d), perl = TRUE) Dear Gabor, Thank you very much. That works very well. I don't completely understand it though. A few words on what the (?=\\d) is doing would be nice. See the info on zero width lookahead assertions on the ?regex page. -- 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] Unexpected behavior when browsing list objects by name, when the object name is a substring (prefix) of an existing object that has a valid value
You've stumbled on one of the reasons use of the $ operator is discouraged in formal programming: it uses partial matching when given names. E.g., a = list(boy = 1:5, cat = 1:10, dog = 1:3) a$d #Exists If you want to require exact matching (and it seems you do), use the [[ operator. a[[d]] # Error a[[dog]] # Works a[[alligator]] - 1:100 #Works Michael On Tue, Nov 8, 2011 at 12:29 AM, Val Athaide vath...@gmail.com wrote: Hi, The program I am writing requires me to append named objects to my existing list objects dynamically. So my list object is retval. retval always has a metadatatabletype object createMetadata=TRUE retval -list() retval$metadatatype =c(normal) retval$metadata=NULL How, depending on certain logic, I create a metadata object if (createMetadata==TRUE) retval$metadata =rbind(retval$metadata,c(1,0) ) The results are retval$metadata [,1] [,2] [1,] normal normal [2,] 1 0 What I expected to see is retval$metadata [,1] [,2] [1,] 1 0 I have been able to reproduce this problem only when the object retval$metadata is NULL and there is an existing object that has a valid value and the NULL object is a sub-string (prefix) of the existing object with a valid value. Also, retval$metadata takes on a value of normal even though it has been explicity set as NULL Your assistance is appreciated. Thanks Vathaid [[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] splitting by the last occurance of a dot
See the info on zero width lookahead assertions on the ?regex page. Thank you again. Best Regards, Ashim [[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] Simple coordinate transformations
Hi, I need to do some simple coordinate transforms between cartesian, cylindrical and spherical coordinates. I can't find any built in functions or packages to do this, is this because they do not exist? Obviously I can write my own code, but don't want to re-invent the wheel if I can avoid it. Cheers, Matt __ 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] Building package problem
Dear R-users I am trying to recompile a CRAN package on Windows 32. Rtools for 2.14 (that is the version I am running) and miktex were sucessfully installed on my machine. Problems: a) hydroGOF is a CRAN package, but R CMD check does not work on it. C:\Users\eduardo\Documents\R_tests2R CMD check hydroGOF * using log directory 'C:/Users/eduardo/Documents/R_tests2/hydroGOF.Rcheck' * using R version 2.14.0 (2011-10-31) * using platform: i386-pc-mingw32 (32-bit) * using session charset: ISO8859-1 * checking for file 'hydroGOF/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'hydroGOF' version '0.3-2' * checking package namespace information ... OK * checking package dependencies ... OK * checking if this is a source package ... OK * checking if there is a namespace ... OK * checking for executable files ... OK * checking whether package 'hydroGOF' can be installed ... OK * checking installed package size ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking whether the package can be unloaded cleanly ... OK * checking whether the namespace can be loaded with stated dependencies ... OK * checking whether the namespace can be unloaded cleanly ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd metadata ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking Rd contents ... OK * checking for unstated dependencies in examples ... OK * checking contents of 'data' directory ... OK * checking data for non-ASCII characters ... OK * checking data for ASCII and uncompressed saves ... OK * checking examples ... ERROR Running examples in 'hydroGOF-Ex.R' failed The error most likely occurred in: ### Name: plot2 ### Title: Plotting 2 Time Series ### Aliases: plot2 ### Keywords: dplot ### ** Examples sim - 2:11 obs - 1:10 ## Not run: ##D plot2(sim, obs) ## End(Not run) ## # Loading daily streamflows of the Ega River (Spain), from 1961 to 1970 require(zoo) Loading required package: zoo Attaching package: 'zoo' The following object(s) are masked from 'package:base': as.Date, as.Date.numeric data(EgaEnEstellaQts) obs - EgaEnEstellaQts # Generating a simulated daily time series, initially equal to the observed se ries sim - obs # Randomly changing the first 2000 elements of 'sim', by using a normal distri bution # with mean 10 and standard deviation equal to 1 (default of 'rnorm'). sim[1:2000] - obs[1:2000] + rnorm(2000, mean=10) # Plotting 'sim' and 'obs' in 2 separate panels plot2(x=obs, y=sim) # Plotting 'sim' and 'obs' in the same window plot2(x=obs, y=sim, plot.type=single) Error in as.POSIXlt.character(x, tz, ...) : character string is not in a standard unambiguous format Calls: plot2 ... as.POSIXct.default - as.POSIXct - as.POSIXlt - as.POSIXlt.ch aracter Execution halted b) option --binary is no longer available, is that so? How can an extension zip can be built on Windows? R CMD build --no-vignettes hydroGOF works. And R CMD INSTALL hydroGOFxx.tar.gz too. Many thanks Ed __ 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] Building package problem
Hi Ed, If the only error is in examples then this should work: R CMD check --no-examples foopkg should not have anything to do with vignettes (although those may also not run, who knows). As far as building a binary, look at: R CMD INSTALL --help which leads you to R CMD INSTALL --build foopkg HTH, Josh On Tue, Nov 8, 2011 at 4:35 AM, Eduardo M. A. M. Mendes emammen...@gmail.com wrote: Dear R-users I am trying to recompile a CRAN package on Windows 32. Rtools for 2.14 (that is the version I am running) and miktex were sucessfully installed on my machine. Problems: a) hydroGOF is a CRAN package, but R CMD check does not work on it. C:\Users\eduardo\Documents\R_tests2R CMD check hydroGOF * using log directory 'C:/Users/eduardo/Documents/R_tests2/hydroGOF.Rcheck' * using R version 2.14.0 (2011-10-31) * using platform: i386-pc-mingw32 (32-bit) * using session charset: ISO8859-1 * checking for file 'hydroGOF/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'hydroGOF' version '0.3-2' * checking package namespace information ... OK * checking package dependencies ... OK * checking if this is a source package ... OK * checking if there is a namespace ... OK * checking for executable files ... OK * checking whether package 'hydroGOF' can be installed ... OK * checking installed package size ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking whether the package can be unloaded cleanly ... OK * checking whether the namespace can be loaded with stated dependencies ... OK * checking whether the namespace can be unloaded cleanly ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd metadata ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking Rd contents ... OK * checking for unstated dependencies in examples ... OK * checking contents of 'data' directory ... OK * checking data for non-ASCII characters ... OK * checking data for ASCII and uncompressed saves ... OK * checking examples ... ERROR Running examples in 'hydroGOF-Ex.R' failed The error most likely occurred in: ### Name: plot2 ### Title: Plotting 2 Time Series ### Aliases: plot2 ### Keywords: dplot ### ** Examples sim - 2:11 obs - 1:10 ## Not run: ##D plot2(sim, obs) ## End(Not run) ## # Loading daily streamflows of the Ega River (Spain), from 1961 to 1970 require(zoo) Loading required package: zoo Attaching package: 'zoo' The following object(s) are masked from 'package:base': as.Date, as.Date.numeric data(EgaEnEstellaQts) obs - EgaEnEstellaQts # Generating a simulated daily time series, initially equal to the observed se ries sim - obs # Randomly changing the first 2000 elements of 'sim', by using a normal distri bution # with mean 10 and standard deviation equal to 1 (default of 'rnorm'). sim[1:2000] - obs[1:2000] + rnorm(2000, mean=10) # Plotting 'sim' and 'obs' in 2 separate panels plot2(x=obs, y=sim) # Plotting 'sim' and 'obs' in the same window plot2(x=obs, y=sim, plot.type=single) Error in as.POSIXlt.character(x, tz, ...) : character string is not in a standard unambiguous format Calls: plot2 ... as.POSIXct.default - as.POSIXct - as.POSIXlt - as.POSIXlt.ch aracter Execution halted b) option --binary is no longer available, is that so? How can an extension zip can be built on Windows? R CMD build --no-vignettes hydroGOF works. And R CMD INSTALL hydroGOFxx.tar.gz too. Many thanks Ed __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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] ggplot2 reorder factors for faceting
Hi, Iain-
You might want to have a look at
?relevel
i.e. plotData$infection-relevel(plotData$infection, ref = 'InfC')
Ken
On 11/08/11, Iain Gallagher wrote:
Dear List
I am trying to draw a heatmap using ggplot2. In this heatmap I have faceted
my data by 'infection' of which I have four. These four infections break down
into two types and I would like to reorder the 'infection' column of my data
to reflect this.
Toy example below:
library(ggplot2)
# test data for ggplot reordering
genes - (rep (c(rep('a',4), rep('b',4), rep('c',4), rep('d',4), rep('e',4),
rep('f',4)) ,4))
fcData - rnorm(96)
times - rep(rep(c(2,6,24,48),6),4)
infection - c(rep('InfA', 24), rep('InfB', 24), rep('InfC', 24), rep('InfD',
24))
infType - c(rep('M', 24), rep('D',24), rep('M', 24), rep('D', 24))
# data is long format for ggplot2
plotData - as.data.frame(cbind(genes, as.numeric(fcData), as.numeric(times),
infection, infType))
hp2 - ggplot(plotData, aes(factor(times), genes)) + geom_tile(aes(fill =
scale(as.numeric(fcData + facet_wrap(~infection, ncol=4)
# set scale
hp2 - hp2 + scale_fill_gradient2(name=NULL, low=#0571B0, mid=#F7F7F7,
high=#CA0020, midpoint=0, breaks=NULL, labels=NULL, limits=NULL,
trans=identity)
# set up text (size, colour etc etc)
hp2 - hp2 + labs(x = Time, y = ) + scale_y_discrete(expand = c(0, 0)) +
opts(axis.ticks = theme_blank(), axis.text.x = theme_text(size = 10, angle =
360, hjust = 0, colour = grey25), axis.text.y = theme_text(size=10, colour
= 'gray25'))
hp2 - hp2 + theme_bw()
In the resulting plot I would like infections infA and infC plotted next to
each other and likewise for infB and infD. I have a column in the data -
infType - which I could use to reorder the infection column but so far I have
no luck getting this to work.
Could someone give me a pointer to the best way to reorder the infection
factor and accompanying data into the order I would like?
Best
iain
sessionInfo()
R version 2.13.2 (2011-09-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_GB.utf8 LC_NUMERIC=C
[3] LC_TIME=en_GB.utf8 LC_COLLATE=en_GB.utf8
[5] LC_MONETARY=C LC_MESSAGES=en_GB.utf8
[7] LC_PAPER=en_GB.utf8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_GB.utf8 LC_IDENTIFICATION=C
attached base packages:
[1] grid stats graphics grDevices utils datasets methods
[8] base
other attached packages:
[1] ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.6
loaded via a namespace (and not attached):
[1] digest_0.5.0 tools_2.13.2
__
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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] ggplot2 reorder factors for faceting
Dear Iain,
RSiteSearch(ggplot2 reorder, restrict = c(Rhelp10, Rhelp08)) gives you
the solution.
Best regards,
Thierry
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Iain Gallagher
Verzonden: dinsdag 8 november 2011 12:51
Aan: r-help
Onderwerp: [R] ggplot2 reorder factors for faceting
Dear List
I am trying to draw a heatmap using ggplot2. In this heatmap I have faceted my
data by 'infection' of which I have four. These four infections break down
into
two types and I would like to reorder the 'infection' column of my data to
reflect
this.
Toy example below:
library(ggplot2)
# test data for ggplot reordering
genes - (rep (c(rep('a',4), rep('b',4), rep('c',4), rep('d',4), rep('e',4),
rep('f',4)) ,4))
fcData - rnorm(96) times - rep(rep(c(2,6,24,48),6),4) infection -
c(rep('InfA',
24), rep('InfB', 24), rep('InfC', 24), rep('InfD', 24)) infType - c(rep('M',
24),
rep('D',24), rep('M', 24), rep('D', 24))
# data is long format for ggplot2
plotData - as.data.frame(cbind(genes, as.numeric(fcData), as.numeric(times),
infection, infType))
hp2 - ggplot(plotData, aes(factor(times), genes)) + geom_tile(aes(fill =
scale(as.numeric(fcData + facet_wrap(~infection, ncol=4)
# set scale
hp2 - hp2 + scale_fill_gradient2(name=NULL, low=#0571B0, mid=#F7F7F7,
high=#CA0020, midpoint=0, breaks=NULL, labels=NULL, limits=NULL,
trans=identity)
# set up text (size, colour etc etc)
hp2 - hp2 + labs(x = Time, y = ) + scale_y_discrete(expand = c(0, 0)) +
opts(axis.ticks = theme_blank(), axis.text.x = theme_text(size = 10, angle =
360,
hjust = 0, colour = grey25), axis.text.y = theme_text(size=10, colour =
'gray25'))
hp2 - hp2 + theme_bw()
In the resulting plot I would like infections infA and infC plotted next to
each
other and likewise for infB and infD. I have a column in the data - infType -
which
I could use to reorder the infection column but so far I have no luck getting
this
to work.
Could someone give me a pointer to the best way to reorder the infection
factor
and accompanying data into the order I would like?
Best
iain
sessionInfo()
R version 2.13.2 (2011-09-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_GB.utf8 LC_NUMERIC=C
[3] LC_TIME=en_GB.utf8 LC_COLLATE=en_GB.utf8
[5] LC_MONETARY=C LC_MESSAGES=en_GB.utf8
[7] LC_PAPER=en_GB.utf8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C [11]
LC_MEASUREMENT=en_GB.utf8 LC_IDENTIFICATION=C
attached base packages:
[1] grid stats graphics grDevices utils datasets methods [8]
base
other attached packages:
[1] ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.6
loaded via a namespace (and not attached):
[1] digest_0.5.0 tools_2.13.2
__
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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.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Export interactive linked graphs?
Hi I am falling in love with iplots, especially the linked graphs, i.e. selecting points in one will highlight the same points in the other graph as well. But I would like to be able to export the graphs, so that I can give them to somebody else to be able to look at the graphs and explore the data. I could write a script in R, but that would require installing the packages. Is there a way of exporting these graphs, so that the user can select points and that they are highlighted in the other graph as well? Giving the data along would not be a problem. Any help appreciated, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax (F): +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug [[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] Aggregate or extract function ?
Celine wrote on 11/08/2011 02:28:32 AM: Thanks for your help, I still have a little problem with this function because I don't have the same number of line in my two datarame so when I try to apply the dataframe function, I obtain this response ; that I have a different number of lines. Erreur dans data.frame(train, test[rowz, ]) : les arguments impliquent des nombres de lignes différents : 50327, 66592 Do you know how I could solve this problem ? Thanks, Céline The knn1() function finds, for each row of test (df1), the closest coordinates in train (df2). rowz - knn1(train, test, cl) Thus, rowz should have the same number of elements as the test (df1) has rows, and df1 and df2[rowz, ] should have the same numbers of rows. Without the data in hand, I don't know what's going on. I suggest you look at the rowz variable and see if it makes sense. Is it matching up the rows the way it should? Try looking at a subset of the df1 and df2 data within a certain narrow range of Xs and Ys. Also, when replying to the r-help list, it is helpful for other readers to maintain the history of previous posts. Jean --- previous posts --- Celine wrote on 11/07/2011 02:50:55 PM: Hi R user, I have two dataframe with different variables and coordinates : X Y sp bio3 bio5 bio6 bio13 bio14 1 -70.91667 -45.08333 0 47 194 -274712 2 -86.58333 66.25000 0 16 119 -34542 3 3 -62.58333 -17.91667 0 68 334 152 14428 4 -68.91667 -31.25000 0 54 235 -4525 7 5 55.58333 48.41667 0 23 319 -1722314 6 66.25000 37.75000 0 34 363 -1849 0 and this one : X Y LU1 LU2 LU3 LU4 LU5 LU6 LU7 LU8 LU9 LU10 LU11 LU12 LU13 LU14 LU15 LU16 LU17 LU18 1 -36.5 84 0 0 0 0 0 0 0 0 00000 0.000000 2 -36.0 84 0 0 0 0 0 0 0 0 00000 0.000000 3 -35.5 84 0 0 0 0 0 0 0 0 00000 26.0854680000 4 -35.0 84 0 0 0 0 0 0 0 0 00000 0.000000 5 -34.5 84 0 0 0 0 0 0 0 0 00000 5.2677610000 6 -34.0 84 0 0 0 0 0 0 0 0 00000 105.3710690000 I wouldlike to add to my first dataframe the value of the LU variables at the coordinates of the first dataframe. Of course, the coordinates are not at the same resolution and are different, this is the problem. I wouldlike to decrease the resolution of the first one because the second dataframe have a coarser resolution and obtain something like that : X Y sp bio3 bio5 bio6 bio13 bio14 LU1 LU2 LU3 LU4 ... 1 -70.91667 -45.08333 0 47 194 -274712 0 22.08 76.9 2 -86.58333 66.25000 0 16 119 -34542 3 0 22.08 76.9 3 -62.58333 -17.91667 0 68 334 152 14428 0 22.08 76.9 4 -68.91667 -31.25000 0 54 235 -4525 7 0 22.08 76.9 5 55.58333 48.41667 0 23 319 -1722314 0 22.08 76.9 6 66.25000 37.75000 0 34 363 -1849 0 0 22.08 76.9 Do someone know a function or a way to do obtain that ? Thanks in advance for the help, Céline You could use 1-nearest neighbor classification to find the closest set of coordinates in the second data frame (df2) to each row of coordinates in the first data frame (df1). The function knn1() is in the r package class. For example: library(class) train - df2[, c(X, Y)] test - df1[, c(X, Y)] cl - 1:dim(train)[1] rowz - knn1(train, test, cl) data.frame(df1, df2[rowz, ]) Jean [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How much time a process need?
Actually I want to have a rough approximation. A process that takes one day and a half it is good to send me how many hours it gets. It is not a problem to convert the values of system.error the major is that I am not sure If I should use the user or elapsed time for getting an estimation of how much it takes once I press enter theh process to finish A From: Jim Holtman jholt...@gmail.com To: Rolf Turner rolf.tur...@xtra.co.nz Cc: R-help@r-project.org R-help@r-project.org Sent: Tuesday, November 8, 2011 12:34 PM Subject: Re: [R] How much time a process need? could be that the process was waiting for the user to select a file from a list, or some other input before proceeding. Would have to see what the overall performance of the system was at that point. It could also have been that the process was low on physical memory and there was a lot of paging going on. Sent from my iPad On Nov 7, 2011, at 21:50, Rolf Turner rolf.tur...@xtra.co.nz wrote: On 08/11/11 02:40, Joshua Wiley wrote: So I just need to get the user system elapsed 0.460 0.048 67.366 user value and convert the seconds to days and then to hours ? Right? What about this elapsed field? It's all in seconds. Convert whatever fields you want. That being said, doesn't having an elapsed time of over 67 seconds, when the actual calculation takes less than half a second, indicate that something weird is going on? At the very least the R calculations are fighting for resources with some other very greedy processes. cheers, Rolf Turner __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] save at relative directory
Dear all, I have a variable called thres and before I run a script I set it to a value like thres- -10 at the end of the execution I am issuing a save(variablename,file='Results') which will end up with a file saved at the current directory with the name Results I would like though to use thres value and do the followingg save at the directory called 10 so to get ./10/Results, (yes I want this in a relative order) My question is how I can also check if the directory exists R to create it? I would like to thank you in advance for your help B.R Alex [[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] save at relative directory
Hi Alex, Look at some of these functions: apropos(dir) apropos(exists) Cheers, Josh On Tue, Nov 8, 2011 at 5:36 AM, Alaios ala...@yahoo.com wrote: Dear all, I have a variable called thres and before I run a script I set it to a value like thres- -10 at the end of the execution I am issuing a save(variablename,file='Results') which will end up with a file saved at the current directory with the name Results I would like though to use thres value and do the followingg save at the directory called 10 so to get ./10/Results, (yes I want this in a relative order) My question is how I can also check if the directory exists R to create it? I would like to thank you in advance for your help B.R Alex [[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. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] match first consecutive list of capitalized words in string
Dear R-Helpers,
this is my first post ever to a mailing list, so please feel free to point out
any missunderstandings on my side regarding the conventions of this mailing
list.
My problem:
Assuming the following character vector is given:
names - c(filia Maria, vidua Joh Dirck Kleve (oo 02.02.1732), Bernardus
Engelb Franciscus Linde j.u.Doktor referendarius sereniss Judex et gograven
Rheinensis)
Is there a regular expression matching the first consecutive list of
capitalized words in a single characterstring (Maria, Joh Dirck Kleve,
Bernardus Engelb Franciscus Linde)?
This expression would very reliably seperate the person names from the
additional information in my historic church register transcription.
Thank you very much for your effort,
Jonas
--
This mail has been sent through the MPI for Demographic ...{{dropped:10}}
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and provide commented, minimal, self-contained, reproducible code.
[R] compare linear regressions
Hi, I'm trying to compare two linear regressions. I'm using the following approach: ## xx-1:100 df1 - data.frame(x = xx, y = xx * 2 + 30 + rnorm(n=length(xx),sd=10), g = 1) df2 - data.frame(x = xx, y = xx * 4 + 9 + rnorm(n=length(xx),sd=10), g = 2) dta - rbind(df1, df2) dta$g - factor(dta$g) plot(df2$x,df2$y,type=l,col=red) lines(df1$x,df1$y,col=blue) summary(lm(formula = y ~ x + g + x:g, dta)) ## I learned that the coefficients (g2 and x:g2) tell me about the differences in intercept and slope and the corresponding p-values. Now I'm trying to do the same except that there should be no intercept term: ## xx-1:100 df1 - data.frame(x = xx, y = xx * 2 + rnorm(n=length(xx),sd=10), g = 1) df2 - data.frame(x = xx, y = xx * 4 + rnorm(n=length(xx),sd=10), g = 2) dta - rbind(df1, df2) dta$g - factor(dta$g) plot(df2$x,df2$y,type=l,col=red) lines(df1$x,df1$y,col=blue) summary(lm(formula = y ~ x - 1 + x:g, dta)) ## I assume that the last line is the correct way to specify a linear model without intercept. But I'm not certain about that. Can someone please confirm? Thanks a lot, Holger __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How much time a process need?
The 'user' + 'system' will give you how much CPU is required which is an indication of how many of the CPU cycles you are using. The elapsed time is just how long it spent. If the script is CPU intensive, and there is no paging going on, you should see the CPU time close to the elapsed time. Longer extensions in the elapsed time is an indication of extensive I/O, or it might be an indication that you are calling another process; e.g., you have a PERL script that is parsing some data in preparation to using in R. If you are really interested, then enable the monitoring on your system and see what else is running. Use 'perfmon' on WIndows or a combination of 'ps' and 'vmstat' on UNIX/Linux systems. Run you script on a system that has nothing else running and interfering with resources and then look at the results along with the performance data from the entire system in order to create a model of how long things will take. On Tue, Nov 8, 2011 at 8:31 AM, Alaios ala...@yahoo.com wrote: Actually I want to have a rough approximation. A process that takes one day and a half it is good to send me how many hours it gets. It is not a problem to convert the values of system.error the major is that I am not sure If I should use the user or elapsed time for getting an estimation of how much it takes once I press enter theh process to finish A From: Jim Holtman jholt...@gmail.com To: Rolf Turner rolf.tur...@xtra.co.nz Cc: R-help@r-project.org R-help@r-project.org Sent: Tuesday, November 8, 2011 12:34 PM Subject: Re: [R] How much time a process need? could be that the process was waiting for the user to select a file from a list, or some other input before proceeding. Would have to see what the overall performance of the system was at that point. It could also have been that the process was low on physical memory and there was a lot of paging going on. Sent from my iPad On Nov 7, 2011, at 21:50, Rolf Turner rolf.tur...@xtra.co.nz wrote: On 08/11/11 02:40, Joshua Wiley wrote: On Mon, Nov 7, 2011 at 5:32 AM, Alaiosala...@yahoo.com wrote: So I just need to get the user system elapsed 0.460 0.048 67.366 user value and convert the seconds to days and then to hours ? Right? What about this elapsed field? It's all in seconds. Convert whatever fields you want. That being said, doesn't having an elapsed time of over 67 seconds, when the actual calculation takes less than half a second, indicate that something weird is going on? At the very least the R calculations are fighting for resources with some other very greedy processes. 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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ 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] Rename a directory in R
Hi, I want to be able to rename a folder using R, similar to file.rename. I want to paste - done onto the folder name. The reason for this is we run a loop on a large number of folders on a server and it would be nice for people to be able to log in and instantly see if their data has been processed so they can remove it. I have searched the help list but have not found anything and was wondering if this is possible? I realise I could probably do this through the R/system command line interface but I thought there may be a simpler way to do this through R. Many thanks, Gavin. Dr. Gavin Blackburn SULSA Technologist Strathclyde institute of Pharmacy and Biomedical Science 161 Cathedral Street, Glasgow. G4 0RE Tel: +44 (0)1415483828 ScotMet: The Scottish Metabolomics Facility www.metabolomics.strath.ac.ukhttp://www.metabolomics.strath.ac.uk [[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] compare linear regressions
Yes, adding -1 or +0 both return a lm without an intercept term. Michael On Tue, Nov 8, 2011 at 8:52 AM, Holger Taschenberger holger.taschenber...@mpi-bpc.mpg.de wrote: Hi, I'm trying to compare two linear regressions. I'm using the following approach: ## xx-1:100 df1 - data.frame(x = xx, y = xx * 2 + 30 + rnorm(n=length(xx),sd=10), g = 1) df2 - data.frame(x = xx, y = xx * 4 + 9 + rnorm(n=length(xx),sd=10), g = 2) dta - rbind(df1, df2) dta$g - factor(dta$g) plot(df2$x,df2$y,type=l,col=red) lines(df1$x,df1$y,col=blue) summary(lm(formula = y ~ x + g + x:g, dta)) ## I learned that the coefficients (g2 and x:g2) tell me about the differences in intercept and slope and the corresponding p-values. Now I'm trying to do the same except that there should be no intercept term: ## xx-1:100 df1 - data.frame(x = xx, y = xx * 2 + rnorm(n=length(xx),sd=10), g = 1) df2 - data.frame(x = xx, y = xx * 4 + rnorm(n=length(xx),sd=10), g = 2) dta - rbind(df1, df2) dta$g - factor(dta$g) plot(df2$x,df2$y,type=l,col=red) lines(df1$x,df1$y,col=blue) summary(lm(formula = y ~ x - 1 + x:g, dta)) ## I assume that the last line is the correct way to specify a linear model without intercept. But I'm not certain about that. Can someone please confirm? Thanks a lot, Holger __ 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] Simple coordinate transformations
On 11-11-08 7:24 AM, Matthew Young wrote: Hi, I need to do some simple coordinate transforms between cartesian, cylindrical and spherical coordinates. I can't find any built in functions or packages to do this, is this because they do not exist? Obviously I can write my own code, but don't want to re-invent the wheel if I can avoid it. There are several contributed packages that do at least spherical coordinates. Try RSiteSearch(spherical coordinates) to find them. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] from points in Lon/Lat to physical distance in dist class
Dear R-listers, Here, I would like to hearing helps from you. I have GPS data (multiple points in the geographic scale) in longitude/latitude. I intend to calculate distance (in kilometer) among such points and output the distance matrix in dist class. I have gotten some progress, but I still can not get final goal. Could please give me any directions/advice? This email cc. to Mr. Pierre, the author of BoSSA package. Thanks a lot in advance. Best wishes, Jian-Feng, Mao # What I have gotten are (1) distance among GPS points could be calculated by distGPS() function in BoSSA package. But, it can not output the distance in dist class. And, I do not know how to convert such distance matrix to dist class. (2) dist() function in base R can calculate distance among units and export it in dist class. But, it could not be used to work on points in lon/lat. (3) some dummy codes # (3.1) generate dummy points in lon/lat (in degree) points - data.frame(lon=seq(95, 105),lat=seq(35, 45)) # (3.2) calculate distance between points using distGPS() library(BoSSA) Geodist-distGPS(points) str(Geodist) class(Geodist) [[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] Rekeying value denoting NA
SML wrote on 11/07/2011 09:10:30 PM: I'm trying to rekey values which denote there is no values, i.e., '-999' in a dataset which contains both '-999' and NA entries. When I try the following command I get the following error: data.frame[data.frame$MAR = -9,MAR] - NA missing values are not allowed in subscripted assignments of data frames Example of data: YEAR JAN FEB MAR ... DEC 1931 5 -999NA 3 1932 2 1 -9992 . . . 2010 -999NA 2 1 I've tried to replace the NAs with -999 values first to remove the NA values, but got the same error. I'm quite new to R, and these little issues seem to be a stumbling block. Many thanks for any help you might be able to offer. First of all, you should call your data frame something other than data.frame because data.frame is already a function in the base package of R. Let's call it df, instead, df - data.frame rm(data.frame) Secondly, it looks like the variables in your data frame (JAN, FEB, MAR, ..., DEC) are character not numeric, because their values are left aligned in your example print out. You can test this out by showing the class of each variable in the data frame, lapply(df, class) If the variables are character, you can convert them to numeric, df2 - as.data.frame(lapply(df, as.numeric)) Then you can convert all the -999 values to NAs, df2[df2 -99] - NA YEAR JAN FEB MAR ... DEC 1931 5 NA NA 3 1932 2 NA NA 2 . . . 2010 NA NA 2 1 Jean [[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] Rename a directory in R
apropos('rename')
you will find 'file.rename' which also works on directories.
On Tue, Nov 8, 2011 at 7:45 AM, Gavin Blackburn
gavin.blackb...@strath.ac.uk wrote:
Hi,
I want to be able to rename a folder using R, similar to file.rename. I want
to paste - done onto the folder name. The reason for this is we run a loop
on a large number of folders on a server and it would be nice for people to
be able to log in and instantly see if their data has been processed so they
can remove it.
I have searched the help list but have not found anything and was wondering
if this is possible?
I realise I could probably do this through the R/system command line
interface but I thought there may be a simpler way to do this through R.
Many thanks,
Gavin.
Dr. Gavin Blackburn
SULSA Technologist
Strathclyde institute of Pharmacy and Biomedical Science
161 Cathedral Street,
Glasgow.
G4 0RE
Tel: +44 (0)1415483828
ScotMet: The Scottish Metabolomics Facility
www.metabolomics.strath.ac.ukhttp://www.metabolomics.strath.ac.uk
[[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.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
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] from points in Lon/Lat to physical distance in dist class
Look at distm() in package geosphere. -- David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77843-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Mao Jianfeng Sent: Tuesday, November 08, 2011 8:22 AM To: r-help@r-project.org Cc: bossa.pack...@gmail.com Subject: [R] from points in Lon/Lat to physical distance in dist class Dear R-listers, Here, I would like to hearing helps from you. I have GPS data (multiple points in the geographic scale) in longitude/latitude. I intend to calculate distance (in kilometer) among such points and output the distance matrix in dist class. I have gotten some progress, but I still can not get final goal. Could please give me any directions/advice? This email cc. to Mr. Pierre, the author of BoSSA package. Thanks a lot in advance. Best wishes, Jian-Feng, Mao # What I have gotten are (1) distance among GPS points could be calculated by distGPS() function in BoSSA package. But, it can not output the distance in dist class. And, I do not know how to convert such distance matrix to dist class. (2) dist() function in base R can calculate distance among units and export it in dist class. But, it could not be used to work on points in lon/lat. (3) some dummy codes # (3.1) generate dummy points in lon/lat (in degree) points - data.frame(lon=seq(95, 105),lat=seq(35, 45)) # (3.2) calculate distance between points using distGPS() library(BoSSA) Geodist-distGPS(points) str(Geodist) class(Geodist) [[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] how to use quadrature to integrate some complicated functions
Have you tried wrapping it in a function and using integrate()? R is pretty
good at handling numerical integration. If integrate() isn't good for you, can
you say more as to why?
Michael
On Nov 6, 2011, at 4:15 PM, JeffND zuofeng.shan...@nd.edu wrote:
Hello to all,
I am having trouble with intregrating a complicated uni-dimensional function
of the following form
Phi(x-a_1)*Phi(x-a_2)*...*Phi(x-a_{n-1})*phi(x-a_n).
Here n is about 5000, Phi is the cumulative distribution function of
standard normal,
phi is the density function of standard normal, and x ranges over
(-infty,infty).
My idea is to to use quadrature to handle this integral. But since Phi has
not cloaed form,
I don't know how to do this effeciently. I appreciate very much if someone
has any ideas about it.
Thanks!
Jeff
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-use-quadrature-to-integrate-some-complicated-functions-tp3996765p3996765.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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] compare linear regressions
Thank you. I was confused because the output of the line summary(lm(formula =... reads: Coefficients: (1 not defined because of singularities), which did not look like a normal message (which can safely be ignored) to me. --Holger On Tue, 8 Nov 2011 15:06:28 +0100 Peter Konings peter.l.e.koni...@gmail.com wrote: On Tue, Nov 8, 2011 at 2:52 PM, Holger Taschenberger holger.taschenber...@mpi-bpc.mpg.de wrote: snip summary(lm(formula = y ~ x - 1 + x:g, dta)) ## I assume that the last line is the correct way to specify a linear model without intercept. But I'm not certain about that. Can someone please confirm? Yes, that's true. See chapter 11 of the Introduction To R manual that was installed with R for an overview of model specification in R. HTH Peter. __ 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] Building package problem
On 08.11.2011 13:50, Joshua Wiley wrote: Hi Ed, If the only error is in examples then this should work: R CMD check --no-examples foopkg Disabling the example checks is not the solution - well, it is the one to hide the errors, of course. should not have anything to do with vignettes (although those may also not run, who knows). As far as building a binary, look at: R CMD INSTALL --help which leads you to R CMD INSTALL --build foopkg ... and was the recommended way to build binaries as long as I am maintaining the Windows binaries on CRAN (which is more than 8 years, I think), if I remember correctly. HTH, Josh On Tue, Nov 8, 2011 at 4:35 AM, Eduardo M. A. M. Mendes emammen...@gmail.com wrote: Dear R-users I am trying to recompile a CRAN package on Windows 32. Rtools for 2.14 (that is the version I am running) and miktex were sucessfully installed on my machine. Problems: a) hydroGOF is a CRAN package, but R CMD check does not work on it. It works on it and tells you there is an error. Since the CRAN check summary pages do not show that error, I anticipate your unstated versions of zoo and/or other packages are outdated. Please update all your packages an dtry again. Best, Uwe Ligges C:\Users\eduardo\Documents\R_tests2R CMD check hydroGOF * using log directory 'C:/Users/eduardo/Documents/R_tests2/hydroGOF.Rcheck' * using R version 2.14.0 (2011-10-31) * using platform: i386-pc-mingw32 (32-bit) * using session charset: ISO8859-1 * checking for file 'hydroGOF/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'hydroGOF' version '0.3-2' * checking package namespace information ... OK * checking package dependencies ... OK * checking if this is a source package ... OK * checking if there is a namespace ... OK * checking for executable files ... OK * checking whether package 'hydroGOF' can be installed ... OK * checking installed package size ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking whether the package can be unloaded cleanly ... OK * checking whether the namespace can be loaded with stated dependencies ... OK * checking whether the namespace can be unloaded cleanly ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd metadata ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking Rd contents ... OK * checking for unstated dependencies in examples ... OK * checking contents of 'data' directory ... OK * checking data for non-ASCII characters ... OK * checking data for ASCII and uncompressed saves ... OK * checking examples ... ERROR Running examples in 'hydroGOF-Ex.R' failed The error most likely occurred in: ### Name: plot2 ### Title: Plotting 2 Time Series ### Aliases: plot2 ### Keywords: dplot ### ** Examples sim- 2:11 obs- 1:10 ## Not run: ##D plot2(sim, obs) ## End(Not run) ## # Loading daily streamflows of the Ega River (Spain), from 1961 to 1970 require(zoo) Loading required package: zoo Attaching package: 'zoo' The following object(s) are masked from 'package:base': as.Date, as.Date.numeric data(EgaEnEstellaQts) obs- EgaEnEstellaQts # Generating a simulated daily time series, initially equal to the observed se ries sim- obs # Randomly changing the first 2000 elements of 'sim', by using a normal distri bution # with mean 10 and standard deviation equal to 1 (default of 'rnorm'). sim[1:2000]- obs[1:2000] + rnorm(2000, mean=10) # Plotting 'sim' and 'obs' in 2 separate panels plot2(x=obs, y=sim) # Plotting 'sim' and 'obs' in the same window plot2(x=obs, y=sim, plot.type=single) Error in as.POSIXlt.character(x, tz, ...) : character string is not in a standard unambiguous format Calls: plot2 ... as.POSIXct.default - as.POSIXct - as.POSIXlt - as.POSIXlt.ch aracter Execution halted b) option --binary is no longer available, is that so? How can an extension zip can be built on Windows? R CMD build --no-vignettes hydroGOF works. And R CMD INSTALL hydroGOFxx.tar.gz too. Many thanks Ed __ 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
Re: [R] compare linear regressions
Just because your model specification is valid does not mean your data can be analyzed using that particular model specification. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Holger Taschenberger holger.taschenber...@mpi-bpc.mpg.de wrote: Thank you. I was confused because the output of the line summary(lm(formula =... reads: Coefficients: (1 not defined because of singularities), which did not look like a normal message (which can safely be ignored) to me. --Holger On Tue, 8 Nov 2011 15:06:28 +0100 Peter Konings peter.l.e.koni...@gmail.com wrote: On Tue, Nov 8, 2011 at 2:52 PM, Holger Taschenberger holger.taschenber...@mpi-bpc.mpg.de wrote: snip summary(lm(formula = y ~ x - 1 + x:g, dta)) ## I assume that the last line is the correct way to specify a linear model without intercept. But I'm not certain about that. Can someone please confirm? Yes, that's true. See chapter 11 of the Introduction To R manual that was installed with R for an overview of model specification in R. HTH Peter. __ 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] Building package problem
R CMD check is *not* 'building a package'. Nor is making a Windows binary package. 'Building a package' is creating a source tarball from a source directory. On Tue, 8 Nov 2011, Joshua Wiley wrote: Hi Ed, If the only error is in examples then this should work: R CMD check --no-examples foopkg should not have anything to do with vignettes (although those may also not run, who knows). As far as building a binary, look at: R CMD INSTALL --help which leads you to R CMD INSTALL --build foopkg And as for the hydroGOF issue, my guess is that the problem is your locale or timezone. But despite the posting guide, you failed to tell us. AFAIK CRAN only checks packages in English locales. (One thing we know is that in Columbia there was no midnight on one of the dates in that file. So hydroGOF really ought to be specifying a timezone when reading character datetimes.) HTH, Josh On Tue, Nov 8, 2011 at 4:35 AM, Eduardo M. A. M. Mendes emammen...@gmail.com wrote: Dear R-users I am trying to recompile a CRAN package on Windows 32. Rtools for 2.14 (that is the version I am running) and miktex were sucessfully installed on my machine. Problems: a) hydroGOF is a CRAN package, but R CMD check does not work on it. C:\Users\eduardo\Documents\R_tests2R CMD check hydroGOF * using log directory 'C:/Users/eduardo/Documents/R_tests2/hydroGOF.Rcheck' * using R version 2.14.0 (2011-10-31) * using platform: i386-pc-mingw32 (32-bit) * using session charset: ISO8859-1 * checking for file 'hydroGOF/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'hydroGOF' version '0.3-2' * checking package namespace information ... OK * checking package dependencies ... OK * checking if this is a source package ... OK * checking if there is a namespace ... OK * checking for executable files ... OK * checking whether package 'hydroGOF' can be installed ... OK * checking installed package size ... OK * checking package directory ... OK * checking for portable file names ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... OK * checking whether the package can be loaded with stated dependencies ... OK * checking whether the package can be unloaded cleanly ... OK * checking whether the namespace can be loaded with stated dependencies ... OK * checking whether the namespace can be unloaded cleanly ... OK * checking for unstated dependencies in R code ... OK * checking S3 generic/method consistency ... OK * checking replacement functions ... OK * checking foreign function calls ... OK * checking R code for possible problems ... OK * checking Rd files ... OK * checking Rd metadata ... OK * checking Rd cross-references ... OK * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking Rd contents ... OK * checking for unstated dependencies in examples ... OK * checking contents of 'data' directory ... OK * checking data for non-ASCII characters ... OK * checking data for ASCII and uncompressed saves ... OK * checking examples ... ERROR Running examples in 'hydroGOF-Ex.R' failed The error most likely occurred in: ### Name: plot2 ### Title: Plotting 2 Time Series ### Aliases: plot2 ### Keywords: dplot ### ** Examples sim - 2:11 obs - 1:10 ## Not run: ##D plot2(sim, obs) ## End(Not run) ## # Loading daily streamflows of the Ega River (Spain), from 1961 to 1970 require(zoo) Loading required package: zoo Attaching package: 'zoo' The following object(s) are masked from 'package:base': as.Date, as.Date.numeric data(EgaEnEstellaQts) obs - EgaEnEstellaQts # Generating a simulated daily time series, initially equal to the observed se ries sim - obs # Randomly changing the first 2000 elements of 'sim', by using a normal distri bution # with mean 10 and standard deviation equal to 1 (default of 'rnorm'). sim[1:2000] - obs[1:2000] + rnorm(2000, mean=10) # Plotting 'sim' and 'obs' in 2 separate panels plot2(x=obs, y=sim) # Plotting 'sim' and 'obs' in the same window plot2(x=obs, y=sim, plot.type=single) Error in as.POSIXlt.character(x, tz, ...) : character string is not in a standard unambiguous format Calls: plot2 ... as.POSIXct.default - as.POSIXct - as.POSIXlt - as.POSIXlt.ch aracter Execution halted b) option --binary is no longer available, is that so? How can an extension zip can be built on Windows? R CMD build --no-vignettes hydroGOF works. And R CMD INSTALL hydroGOFxx.tar.gz too. Many thanks Ed __ 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
Re: [R] save at relative directory
Hmm I will try something like that
if (file.exists('threshold')==FALSE)
dir.create(paste('./',abs(threshold)))
save(var,file=paste('./',abs(threshold),'/',DataSource[[4]],sep=)
I just need a bit of confirmation If I am doing soemthing terribly wrong that
might harm my filesystem.
I also did accidentaly
var- -13
dir.create(paste('./',var))
which created a folder called -13 which I do not know how to remove it
rmdir -13
rmdir: invalid option -- '1'
Try `rmdir --help' for more information.
B.R
Alex
From: Joshua Wiley jwiley.ps...@gmail.com
Cc: R-help@r-project.org R-help@r-project.org
Sent: Tuesday, November 8, 2011 2:43 PM
Subject: Re: [R] save at relative directory
Hi Alex,
Look at some of these functions:
apropos(dir)
apropos(exists)
Cheers,
Josh
Dear all,
I have a variable called thres and before I run a script I set it to a value
like
thres- -10
at the end of the execution I am issuing a save(variablename,file='Results')
which will end up with a file saved at the current directory with the name
Results
I would like though to use thres value and do the followingg
save at the directory called 10 so to get ./10/Results, (yes I want this in a
relative order)
My question is how I can also check if the directory exists R to create it?
I would like to thank you in advance for your help
B.R
Alex
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__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
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__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] save at relative directory
On 08.11.2011 16:09, Alaios wrote:
Hmm I will try something like that
if (file.exists('threshold')==FALSE)
dir.create(paste('./',abs(threshold)))
save(var,file=paste('./',abs(threshold),'/',DataSource[[4]],sep=)
I just need a bit of confirmation If I am doing soemthing terribly wrong that
might harm my filesystem.
I also did accidentaly
var- -13
dir.create(paste('./',var))
which created a folder called -13 which I do not know how to remove it
rmdir -13
rmdir: invalid option -- '1'
Try `rmdir --help' for more information.
In R:
unlink(-13, recursive=TRUE)
or in your shell:
I guess your intated OS can do:
rm --help
which points you to:
rm -r -- -13
Uwe Ligges
B.R
Alex
From: Joshua Wileyjwiley.ps...@gmail.com
Cc: R-help@r-project.orgR-help@r-project.org
Sent: Tuesday, November 8, 2011 2:43 PM
Subject: Re: [R] save at relative directory
Hi Alex,
Look at some of these functions:
apropos(dir)
apropos(exists)
Cheers,
Josh
Dear all,
I have a variable called thres and before I run a script I set it to a value
like
thres- -10
at the end of the execution I am issuing a save(variablename,file='Results')
which will end up with a file saved at the current directory with the name
Results
I would like though to use thres value and do the followingg
save at the directory called 10 so to get ./10/Results, (yes I want this in a
relative order)
My question is how I can also check if the directory exists R to create it?
I would like to thank you in advance for your help
B.R
Alex
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] save at relative directory
Hi Alex,
For the R part, I would abstract it a bit:
mydir - paste(./, abs(threshold), sep = )
if (!file.exists(mydir)) dir.create(mydir)
save(var, file = paste(mydir, DataSource[[4]], sep = /))
if you use file.exists('threshold') you are testing for the existence
of threshold, not the value contained in threshold, and anyway, you
seem not not want the value contained in threshold, but the absolute
value of the value in threshold, hence, in part, the value of
abstraction.
In R, see ?unlink for ways to delete things, rmdir looks like you are
using the command prompt, and for that I will refer you to the help
for your OS/shell on how to go about removing unwanted directories
(hint, rmdir --help is not a bad place to start ;)
Cheers,
Josh
On Tue, Nov 8, 2011 at 7:09 AM, Alaios ala...@yahoo.com wrote:
Hmm I will try something like that
if (file.exists('threshold')==FALSE)
dir.create(paste('./',abs(threshold)))
save(var,file=paste('./',abs(threshold),'/',DataSource[[4]],sep=)
I just need a bit of confirmation If I am doing soemthing terribly wrong
that might harm my filesystem.
I also did accidentaly
var- -13
dir.create(paste('./',var))
which created a folder called -13 which I do not know how to remove it
rmdir -13
rmdir: invalid option -- '1'
Try `rmdir --help' for more information.
B.R
Alex
From: Joshua Wiley jwiley.ps...@gmail.com
To: Alaios ala...@yahoo.com
Cc: R-help@r-project.org R-help@r-project.org
Sent: Tuesday, November 8, 2011 2:43 PM
Subject: Re: [R] save at relative directory
Hi Alex,
Look at some of these functions:
apropos(dir)
apropos(exists)
Cheers,
Josh
On Tue, Nov 8, 2011 at 5:36 AM, Alaios ala...@yahoo.com wrote:
Dear all,
I have a variable called thres and before I run a script I set it to a
value
like
thres- -10
at the end of the execution I am issuing a
save(variablename,file='Results')
which will end up with a file saved at the current directory with the name
Results
I would like though to use thres value and do the followingg
save at the directory called 10 so to get ./10/Results, (yes I want this
in a relative order)
My question is how I can also check if the directory exists R to create
it?
I would like to thank you in advance for your help
B.R
Alex
[[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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.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] window?
Can someone enlighten me on why the following doesn't work?
setwd('C:/Temp/R')
d - rep(1:53,2)
(s - ts(d, frequency=53, start=c(2000,10)))
n - length(s)
k - n%/%3
for(i in (n-k):n)
{
st - c(start(s)[1] + (start(s)[2] + i)%/%frequency(s), (start(s)[2] +
i) %% frequency(s))
ed - c(start(s)[1] + (start(s)[2]+k+i)%/%frequency(s),
(start(s)[2]+i+k) %% frequency(s))
xshort - window(s, start=st, end=ed)
cat(Start , st, End , ed, \n)
cat(Length , length(xshort), start , start(xshort),
end , end(xshort), \n)
}
I get a bunch of warnings like:
36: In window.default(x, ...) : 'end' value not changed
Thank you.
Kevin
rkevinbur...@charter.net
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] window?
On 08.11.2011 16:26, Kevin Burton wrote:
Can someone enlighten me on why the following doesn't work?
setwd('C:/Temp/R')
d- rep(1:53,2)
(s- ts(d, frequency=53, start=c(2000,10)))
n- length(s)
k- n%/%3
for(i in (n-k):n)
{
st- c(start(s)[1] + (start(s)[2] + i)%/%frequency(s), (start(s)[2] +
i) %% frequency(s))
ed- c(start(s)[1] + (start(s)[2]+k+i)%/%frequency(s),
(start(s)[2]+i+k) %% frequency(s))
xshort- window(s, start=st, end=ed)
cat(Start , st, End , ed, \n)
cat(Length , length(xshort), start , start(xshort),
end , end(xshort), \n)
}
I get a bunch of warnings like:
36: In window.default(x, ...) : 'end' value not changed
Yes, your original s has
End = c(2002, 9)
and you try to set to, e.g.,
c(2002, 45)
in your last iteration which is later and hence cannot be changed.
Uwe ligges
Thank you.
Kevin
rkevinbur...@charter.net
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] RpgSQL row names
This is great, thanks!
I have another unrelated question. I'll create a new email for that one.
ben
On Mon, Nov 7, 2011 at 4:16 PM, Gabor Grothendieck
ggrothendi...@gmail.comwrote:
On Mon, Nov 7, 2011 at 5:34 PM, Ben quant ccqu...@gmail.com wrote:
Hello,
Using the RpgSQL package, there must be a way to get the row names into
the
table automatically. In the example below, I'm trying to get rid of the
cbind line, yet have the row names of the data frame populate a column.
bentest = matrix(1:4,2,2)
dimnames(bentest) = list(c('ra','rb'),c('ca','cb'))
bentest
ca cb
ra 1 3
rb 2 4
bentest = cbind(item_name=rownames(bentest),bentest)
dbWriteTable(con, r.bentest, bentest)
[1] TRUE
dbGetQuery(con, SELECT * FROM r.bentest)
item_name ca cb
1ra 1 3
2rb 2 4
The RJDBC based drivers currently don't support that. You can create a
higher level function that does it.
dbGetQuery2 - function(...) {
out - dbGetQuery(...)
i - match(row_names, names(out), nomatch = 0)
if (i 0) {
rownames(out) - out[[i]]
out - out[-1]
}
out
}
rownames(BOD) - letters[1:nrow(BOD)]
dbWriteTable(con, BOD, cbind(row_names = rownames(BOD), BOD))
dbGetQuery2(con, select * from BOD)
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rekeying value denoting NA
On Nov 8, 2011, at 8:57 AM, Jean V Adams wrote: SML wrote on 11/07/2011 09:10:30 PM: I'm trying to rekey values which denote there is no values, i.e., '-999' in a dataset which contains both '-999' and NA entries. When I try the following command I get the following error: data.frame[data.frame$MAR = -9,MAR] - NA missing values are not allowed in subscripted assignments of data frames Example of data: YEAR JAN FEB MAR ... DEC 1931 5 -999NA 3 1932 2 1 -9992 . . . 2010 -999NA 2 1 I've tried to replace the NAs with -999 values first to remove the NA values, but got the same error. I'm quite new to R, and these little issues seem to be a stumbling block. Many thanks for any help you might be able to offer. First of all, you should call your data frame something other than data.frame because data.frame is already a function in the base package of R. Let's call it df, instead, df - data.frame rm(data.frame) Secondly, it looks like the variables in your data frame (JAN, FEB, MAR, ..., DEC) are character not numeric, because their values are left aligned in your example print out. You can test this out by showing the class of each variable in the data frame, lapply(df, class) If the variables are character, you can convert them to numeric, df2 - as.data.frame(lapply(df, as.numeric)) Then you can convert all the -999 values to NAs, df2[df2 -99] - NA Agreed this is what _should_ be done. YEAR JAN FEB MAR ... DEC 1931 5 NA NA 3 1932 2 NA NA 2 . . . 2010 NA NA 2 1 But ... I thought she wanted (unwisely in my opinion) to go the other way, NA's - -999. In R the replacement of NA's is a bit convoluted because nothing =='s NA. You might need use the `is.na` function in this manner. df[is.na(df[[MAR]]), MAR] - -999 -- David Winsemius, MD West Hartford, CT __ 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] Rename a directory in R
Thanks Jim,
I didn't think it worked on directories but I've got it working now.
Cheers,
Gavin.
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: 08 November 2011 14:18
To: Gavin Blackburn
Cc: r-help@r-project.org
Subject: Re: [R] Rename a directory in R
apropos('rename')
you will find 'file.rename' which also works on directories.
On Tue, Nov 8, 2011 at 7:45 AM, Gavin Blackburn
gavin.blackb...@strath.ac.uk wrote:
Hi,
I want to be able to rename a folder using R, similar to file.rename. I want
to paste - done onto the folder name. The reason for this is we run a loop
on a large number of folders on a server and it would be nice for people to
be able to log in and instantly see if their data has been processed so they
can remove it.
I have searched the help list but have not found anything and was wondering
if this is possible?
I realise I could probably do this through the R/system command line
interface but I thought there may be a simpler way to do this through R.
Many thanks,
Gavin.
Dr. Gavin Blackburn
SULSA Technologist
Strathclyde institute of Pharmacy and Biomedical Science
161 Cathedral Street,
Glasgow.
G4 0RE
Tel: +44 (0)1415483828
ScotMet: The Scottish Metabolomics Facility
www.metabolomics.strath.ac.ukhttp://www.metabolomics.strath.ac.uk
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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Re: [R] how to use quadrature to integrate some complicated functions
On Nov 8, 2011, at 9:43 AM, R. Michael Weylandt wrote:
Have you tried wrapping it in a function and using integrate()? R is
pretty good at handling numerical integration. If integrate() isn't
good for you, can you say more as to why?
Michael
On Nov 6, 2011, at 4:15 PM, JeffND zuofeng.shan...@nd.edu wrote:
Hello to all,
I am having trouble with intregrating a complicated uni-dimensional
function
of the following form
Phi(x-a_1)*Phi(x-a_2)*...*Phi(x-a_{n-1})*phi(x-a_n).
Here n is about 5000, Phi is the cumulative distribution function of
standard normal,
phi is the density function of standard normal, and x ranges over
(-infty,infty).
The density of a standard normal is very tractable mathematically. Why
wouldn't you extract the arguments and sum them before submitting to
integration ... which also might not be needed since pnorm could
economically provide the answer. Perhaps with limits a, b:
suitable_norm_factor*
pnorm(
dnorm( sum(x-a_1, x-a_2, ..., x-a_{n-1}, x-a_n) ), a,
lower.tail=FALSE) ) -
pnorm(
dnorm( sum(x-a_1, x-a_2, ..., x-a_{n-1}, x-a_n) ), b,
lower.tail=FALSE) ) )
My idea is to to use quadrature to handle this integral. But since
Phi has
not cloaed form,
I don't know how to do this effeciently. I appreciate very much if
someone
has any ideas about it.
If efficiency is desired ... use mathematical theory to maximum extent
before resorting to pickaxes.
--
David Winsemius, MD
West Hartford, CT
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Re: [R] similar package in R like SKEW CALCULATOR?
Am 08.11.2011 13:01, schrieb R. Michael Weylandt: See the replies given to you two days ago when you asked the same Hi Michael, thank you for your second answer. I did not get my first question and I did not get your answer via mail - strange Knut __ 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] multi-line query
Hello, I'm using package RpgSQL. Is there a better way to create a multi-line query/character string? I'm looking for less to type and readability. This is not very readable for large queries: s - 'create table r.BOD(id int primary key,name varchar(12))' I write a lot of code, so I'm looking to type less than this, but it is more readable from and SQL standpoint: s - gsub(\n, , 'create table r.BOD( id int primary key ,name varchar(12)) ') How it is used: dbSendUpdate(con, s) Regards, Ben [[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] rearrange set of items randomly
Thanks for the replies! Indeed, I could use order() instead of sample() but that it wouldnt be random anymore, as it sorts data points in increasing(!) order. Second, I have to use 1000 different samples. I got the hint that I have to do somethin with indeces, but still cant figure out what this should be. Maybe someone of you knows? -- View this message in context: http://r.789695.n4.nabble.com/rearrange-set-of-items-randomly-tp4013723p4016613.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] Sampling with conditions
Sorry about being confusing, I have so many loops in loops and ifelses that I get mixed up sometimes, it was just a typo, it was supposed to be for(i in 1:5) Sorry, Thanks for you help! SJ -- View this message in context: http://r.789695.n4.nabble.com/Sampling-with-conditions-tp4014036p4016058.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] Error message after updating pkg spatstat
Hello,
I recently downloaded an updated version of the {spatstat} package 1.24-1. But
now, when I do help() to look up a certain function in the spatstat library,
I get an error message. It only seems to be for spatstat fns. The following error:
Error in fetch(key) : internal error -3 in R_decompress1
Can anyone help?
Thanks greatly, my session info is below. I am running R 2.13.1 on a Mac.
Robin
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] gpclib_1.5-1rgdal_0.6-33maptools_0.8-10 lattice_0.19-33
foreign_0.8-45 spatstat_1.24-1
[7] deldir_0.0-15 mgcv_1.7-6 sp_0.9-91
loaded via a namespace (and not attached):
[1] grid_2.13.1Matrix_0.999375-50 nlme_3.1-102 tools_2.13.1
--
Robin Hunnewell
PhD student
Department of Biology
University of New Brunswick
Fredericton, NB
Canada
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Re: [R] Sampling with conditions
That is exactly what I want, and it's so simple! Thanks so much! -- View this message in context: http://r.789695.n4.nabble.com/Sampling-with-conditions-tp4014036p4016050.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 with SEM package: Error message
Hello. I started using the sem package in R and after a lot of searching and trying things I am still having difficulty. I get the following error message when I use the sem() function: Warning message: In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars, : Could not compute QR decomposition of Hessian. Optimization probably did not converge. I started with a simple example using the specify.model() function, but it is really straight forward. I uploaded my specify.model script and my data covariance matrix here too so I wouldn't clutter this email with the entire model (20 observed variables, 5 factors). Could this error message be from the data itself and not from my path model? I have my observed variables X and my unobserved variables F. I have ONLY exogenous latent variables (i.e. they never appear on the right side of the single head arrow -). I include all possible factor covariances FjFk, and the only constraints I've made was to restrict the Factor variances to 1. My model follows in this basic format (as you can see from my uploaded file): # Factors (where I specify which observed variables load on to which factors) # I have only exogenous latent variables F.i - X.j, lamj.i, NA . . . # Observed variable variances X.j - X.j, ej, NA . . . # Factor variances (I fixed all factor variances to 1) F.i - F.i, NA, 1 . . . # Factor covariances (I represent all factor covariances, i.e. the upper or lower triangle of a covariance matrix) F.i - F.k, FiFk, NA . . . Did I do something wrong here? Here are my uploaded files: CFA script: http://r.789695.n4.nabble.com/file/n4016569/CFA_script.txt CFA_script.txt Covariance matrix: http://r.789695.n4.nabble.com/file/n4016569/covariance_matrix.RData covariance_matrix.RData Thank you so much for any and all of your help. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Help-with-SEM-package-Error-message-tp4016569p4016569.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] Question
Hi My name is Rocky and I am trying to use the org.Dm.eg.db library. When I am using the org.Dm.egFLYBASE2EG[fb_ids] it is stopping at a point where it cannot find any value for a given ID such as the following: Error in .checkKeys(value, Rkeys(x), x@ifnotfound) : value for FBgn0004461 not found Then the whole thing stops. I cannot retrieve any information on the values that has been found until this ID was reached. Can anyone help in retrieving all the IDs without stopping at a certain ID or if the value for the ID not found just generate an NA for that ID but do not stop. Another thing when I am using FlybaseID converter FBgn0004461 ID shows a value. I am not sure why R is not able to retrieve it. Please comment on this. Waiting for your reply. Thanking you Rocky [[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] lapply to list of variables
Hi
Can someone help me with this?
How can I apply a function to a list of variables.
something like this
listvar=list(Monday,Tuesday,Wednesday)
func=function(x){x[which(x=10)]=NA}
lapply(listvar, func)
were
Monday=[213,56,345,33,34,678,444]
Tuesday=[213,56,345,33,34,678,444]
...
in my case I have a neverending list of vectors.
Thanks!
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[R] determine length of bivariate polynomial
Dear R-community,
I have a fitted bivariate polynomial, i.e:
fit = lm(cbind(x, y)~poly(t, 15))
and I would like to determine the length of the line in the interval t =
[a, b]. Obviously, I could use predict and go through all the points, i.e.
for (t in a:(b-1)) {
length = length + sqrt((x.pred[t] - x.pred[t+1])^2 + (y.pred[t] -
y.pred[t+1])^2)
}
but that would take very long given the amount of data I have. Do you know
of any better solutions?
Many thanks!
Nicolas
[[alternative HTML version deleted]]
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[R] GAM
Hi R community! I am analyzing the data set motorins in the package faraway by using the generalized additive model. it shows the following error. Can some one suggest me the right way? library(faraway) data(motorins) motori - motorins[motorins$Zone==1,] library(mgcv) amgam - gam(log(Payment) ~ offset(log(Insured))+ s(as.numeric(Kilometres)) + s(Bonus) + Make + s(Claims),family = gaussian, data = motori) Error in smooth.construct.tp.smooth. spec(object, dk$data, dk$knots) : A term has fewer unique covariate combinations than specified maximum degrees of freedom summary(amgam) Error in summary(amgam) : object 'amgam' not found Gyan [[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] rearrange set of items randomly
On Nov 8, 2011, at 11:31 AM, flokke wrote: Thanks for the replies! Indeed, I could use order() instead of sample() but that it wouldnt be random anymore, as it sorts data points in increasing(!) order. Second, I have to use 1000 different samples. Please re-read the help page for `order` _and_ the one for `sort`, examine their differences, re-read the replies you have gotten so far, work through _all_ of the examples, and if illumination still does not arrive, then repeat the above process until it does. I got the hint that I have to do somethin with indeces, but still cant figure out what this should be. Maybe someone of you knows? Many (if not most) of us do. We try to avoid having r-help becoming a homework tutoring site, however. You should be using your academic resources for this effort. -- David Winsemius, MD West Hartford, CT __ 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] Question
I suspect you are lost. This is almost certainly more appropriate on the BioConductor list. There are relatively few people on this list who will know what (un-named) packages you might be using. Read the Posting Guide, note the link to the Bioconductor mailing list page, configure your mailer for plain text, and subscribe to the help list at Bioc. Then construct a post that shows your code, and mentions all the packages being required. -- David. On Nov 8, 2011, at 11:22 AM, Parida, Mrutyunjaya wrote: Hi My name is Rocky and I am trying to use the org.Dm.eg.db library. When I am using the org.Dm.egFLYBASE2EG[fb_ids] it is stopping at a point where it cannot find any value for a given ID such as the following: Error in .checkKeys(value, Rkeys(x), x@ifnotfound) : value for FBgn0004461 not found Then the whole thing stops. I cannot retrieve any information on the values that has been found until this ID was reached. Can anyone help in retrieving all the IDs without stopping at a certain ID or if the value for the ID not found just generate an NA for that ID but do not stop. Another thing when I am using FlybaseID converter FBgn0004461 ID shows a value. I am not sure why R is not able to retrieve it. Please comment on this. Waiting for your reply. Thanking you Rocky [[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. David Winsemius, MD West Hartford, CT __ 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] dbWriteTable with field data type
Hello, When I do: dbWriteTable(con, r.BOD, cbind(row_names = rownames(BOD), BOD)) ...can I specify the data types such as varchar(12), float, double precision, etc. for each of the fields/columns? If not, what is the best way to create a table with specified field data types (with the RpgSQL package/R)? Regards, Ben [[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] lapply to list of variables
On 08.11.2011 17:59, Ana wrote:
Hi
Can someone help me with this?
How can I apply a function to a list of variables.
something like this
listvar=list(Monday,Tuesday,Wednesday)
This is a list of length one character vectors rather than a list of
variables.
func=function(x){x[which(x=10)]=NA}
To make it work, redefine:
func -function(x){
x - get(x)
is.na(x[x=10]) - TRUE
x
}
lapply(listvar, func)
were
Monday=[213,56,345,33,34,678,444]
Tuesday=[213,56,345,33,34,678,444]
This is not R syntax.
...
in my case I have a neverending list of vectors.
Then your function will take an infinite amount of time - or you will
get amazing reputation in computer sciences.
Uwe Ligges
Thanks!
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Re: [R] why NA coefficients
Sure it does, but still struggling with what is going on... Thanks John From:David Winsemius dwinsem...@comcast.net To:David Winsemius dwinsem...@comcast.net oject.org Sent:Monday, November 7, 2011 10:27 PM Subject:Re: [R] why NA coefficients But this output suggests there may be alligators in the swamp: predict(lmod, newdata=data.frame(treat=1, group=2)) 1 0.09133691 Warning message: In predict.lm(lmod, newdata = data.frame(treat = 1, group = 2)) : prediction from a rank-deficient fit may be misleading --David. --David. Thanks John From: David Winsemius dwinsem...@comcast.net Cc: r-help@r-project.org r-help@r-project.org Sent: Monday, November 7, 2011 5:13 PM Subject: Re: [R] why NA coefficients On Nov 7, 2011, at 7:33 PM, array chip wrote: Hi, I am trying to run ANOVA with an interaction term on 2 factors (treat has 7 levels, group has 2 levels). I found the coefficient for the last interaction term is always 0, see attached dataset and the code below: test-read.table(test.txt,sep='\t',header=T,row.names=NULL) lm(y~factor(treat)*factor(group),test) Call: lm(formula = y ~ factor(treat) * factor(group), data = test) Coefficients: (Intercept) factor(treat)2 factor(treat)3 0.429244 0.499982 0.352971 factor(treat)4 factor(treat)5 factor(treat)6 -0.204752 0.142042 0.044155 factor(treat)7 factor(group)2 factor(treat)2:factor(group)2 -0.007775 -0.337907 -0.208734 factor(treat)3:factor(group)2 factor(treat)4:factor(group)2 factor(treat)5:factor(group)2 -0.195138 0.800029 0.227514 factor(treat)6:factor(group)2 factor(treat)7:factor(group)2 0.331548 NA I guess this is due to model matrix being singular or collinearity among the matrix columns? But I can't figure out how the matrix is singular in this case? Can someone show me why this is the case? Because you have no cases in one of the crossed categories. --David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ R-help@r-project.orgmailing 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 West Hartford, CT [[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] why NA coefficients
true, why it has to omit treat 7-group 2 Thanks again From: David Winsemius dwinsem...@comcast.net Cc: r-help@r-project.org r-help@r-project.org Sent: Monday, November 7, 2011 10:19 PM Subject: Re: [R] why NA coefficients On Nov 7, 2011, at 10:07 PM, array chip wrote: Thanks David. The only category that has no cases is treat 1-group 2: with(test,table(treat,group)) group treat 1 2 1 8 0 2 1 5 3 5 5 4 7 3 5 7 4 6 3 3 7 8 2 But why the coefficient for treat 7-group 2 is not estimable? Well, it had to omit one of them didn't it? (But I don't know why that level was chosen.) --David. Thanks John From: David Winsemius dwinsem...@comcast.net Cc: r-help@r-project.org r-help@r-project.org Sent: Monday, November 7, 2011 5:13 PM Subject: Re: [R] why NA coefficients On Nov 7, 2011, at 7:33 PM, array chip wrote: Hi, I am trying to run ANOVA with an interaction term on 2 factors (treat has 7 levels, group has 2 levels). I found the coefficient for the last interaction term is always 0, see attached dataset and the code below: test-read.table(test.txt,sep='\t',header=T,row.names=NULL) lm(y~factor(treat)*factor(group),test) Call: lm(formula = y ~ factor(treat) * factor(group), data = test) Coefficients: (Intercept) factor(treat)2 factor(treat)3 0.429244 0.499982 0.352971 factor(treat)4 factor(treat)5 factor(treat)6 -0.204752 0.142042 0.044155 factor(treat)7 factor(group)2 factor(treat)2:factor(group)2 -0.007775 -0.337907 -0.208734 factor(treat)3:factor(group)2 factor(treat)4:factor(group)2 factor(treat)5:factor(group)2 -0.195138 0.800029 0.227514 factor(treat)6:factor(group)2 factor(treat)7:factor(group)2 0.331548 NA I guess this is due to model matrix being singular or collinearity among the matrix columns? But I can't figure out how the matrix is singular in this case? Can someone show me why this is the case? Because you have no cases in one of the crossed categories. --David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT [[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] multi-line query
Why not just send it in as is. I use SQLite (via sqldf) and here is
the way I write my SQL statements:
inRange - sqldf('
select t.*
, r.start
, r.end
from total t, commRange r
where t.comm = r.comm and
t.loc between r.start and r.end and
t.loc != t.new
')
On Tue, Nov 8, 2011 at 11:43 AM, Ben quant ccqu...@gmail.com wrote:
Hello,
I'm using package RpgSQL. Is there a better way to create a multi-line
query/character string? I'm looking for less to type and readability.
This is not very readable for large queries:
s - 'create table r.BOD(id int primary key,name varchar(12))'
I write a lot of code, so I'm looking to type less than this, but it is
more readable from and SQL standpoint:
s - gsub(\n, , 'create table r.BOD(
id int primary key
,name varchar(12))
')
How it is used:
dbSendUpdate(con, s)
Regards,
Ben
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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Re: [R] Sampling with conditions
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of SarahJoyes
Sent: Tuesday, November 08, 2011 5:57 AM
To: r-help@r-project.org
Subject: Re: [R] Sampling with conditions
That is exactly what I want, and it's so simple!
Thanks so much!
Sarah,
I want to point out that my post was qualified by something like. I am not
sure it is exactly what you want. Since you didn't quote my post, let me show
my suggestion and then express my concern.
n - matrix(0,nrow=5, ncol=10)
repeat{
c1 - sample(0:10, 4, replace=TRUE)
if(sum(c1) = 10) break
}
n[,1] - c(c1,10-sum(c1))
n
This nominally meets your criteria, but it will tend to result in larger digits
being under-represented. For example, you unlikely to get a result like
c(0,8,0,0,2) or (9,0,0,1,0).
That may be OK for your purposes, but I wanted to point it out.
You could use something like
n - matrix(0,nrow=5, ncol=10)
c1 - rep(0,4)
for(i in 1:4){
upper - 10-sum(c1)
c1[i] - sample(0:upper, 1, replace=TRUE)
if(sum(c1) == 10) break
}
n[,1] - c(c1,10-sum(c1))
n
if that would suit your purposes better.
Good luck,
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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Re: [R] multi-line query
Because I don't know anything about sqldf. :)
Here is what happens, but Im sure it is happening because I didn't read
the manual yet:
s - sqldf('create table r.dat(id int primary key,val int)')
Error in ls(envir = envir, all.names = private) :
invalid 'envir' argument
Error in !dbPreExists : invalid argument type
ben
On Tue, Nov 8, 2011 at 10:41 AM, jim holtman jholt...@gmail.com wrote:
Why not just send it in as is. I use SQLite (via sqldf) and here is
the way I write my SQL statements:
inRange - sqldf('
select t.*
, r.start
, r.end
from total t, commRange r
where t.comm = r.comm and
t.loc between r.start and r.end and
t.loc != t.new
')
On Tue, Nov 8, 2011 at 11:43 AM, Ben quant ccqu...@gmail.com wrote:
Hello,
I'm using package RpgSQL. Is there a better way to create a multi-line
query/character string? I'm looking for less to type and readability.
This is not very readable for large queries:
s - 'create table r.BOD(id int primary key,name varchar(12))'
I write a lot of code, so I'm looking to type less than this, but it is
more readable from and SQL standpoint:
s - gsub(\n, , 'create table r.BOD(
id int primary key
,name varchar(12))
')
How it is used:
dbSendUpdate(con, s)
Regards,
Ben
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
[[alternative HTML version deleted]]
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[R] passing dataframe col name through cbind()
Hi all --- I note that the column name of the first column in a dataframe does not necessarily get passed on when using cbind (example below)… I'm looking for help in clarifying why this behavior occurs, and how I can get all col names, including the first, passed on to the result…while I suspect it's obvious and documented to the cognoscenti, it's puzzling me… Many thanks for any help on this... Eric scores - data.frame(name=c(Bob,Ron,Bud),round1=c(40,30,20),round2=c(5,6,4)) #some toy data scores name round1 round2 1 Bob 40 5 2 Ron 30 6 3 Bud 20 4 cbind(scores[,1],total=rowSums(scores[,2:3]),scores[,2:3]) scores[, 1] total round1 round2 1 Bob45 40 5 2 Ron36 30 6 3 Bud24 20 4 ...first column renamed... …yet this passes all column names: cbind(scores[,1:3]) name round1 round2 1 Bob 40 5 2 Ron 30 6 3 Bud 20 4 …but this doesn't: cbind(scores[,1],scores[,2:3]) scores[, 1] round1 round2 1 Bob 40 5 2 Ron 30 6 3 Bud 20 4 -- Eric Rupley University of Michigan, Museum of Anthropology 1109 Geddes Ave, Rm. 4013 Ann Arbor, MI 48109-1079 erup...@umich.edu +1.734.276.8572 __ 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] why NA coefficients
On Nov 8, 2011, at 12:36 PM, array chip wrote: Sure it does, but still struggling with what is going on... Have you considered redefining the implicit base level for treat so it does not create the missing crossed-category? test$treat2_ - factor(test$treat, levels=c(2:7, 1) ) lm(y~treat2_*factor(group),test) Call: lm(formula = y ~ treat2_ * factor(group), data = test) Coefficients: (Intercept) treat2_3 treat2_4 0.9292256 -0.1470106 -0.7047343 treat2_5 treat2_6 treat2_7 -0.3579398 -0.4558269 -0.5077571 treat2_1 factor(group)2 treat2_3:factor(group)2 -0.4999820 -0.5466405 0.0135963 treat2_4:factor(group)2 treat2_5:factor(group)2 treat2_6:factor(group)2 1.00876280.4362479 0.5402821 treat2_7:factor(group)2 treat2_1:factor(group)2 0.2087338 NA All the group-less coefficients are for group1 , so now get a prediction for group=1:treat=2 == Intercept, group=1:treat=3 , a total of 7 values. And there are 6 predictions for group2. The onus is obviously on you to check the predictions against the data. 'aggregate' is a good function for that purpose. -- david. Thanks John From: David Winsemius dwinsem...@comcast.net To: David Winsemius dwinsem...@comcast.net Cc: array chip arrayprof...@yahoo.com; r-help@r-project.org r-help@r-project.org Sent: Monday, November 7, 2011 10:27 PM Subject: Re: [R] why NA coefficients But this output suggests there may be alligators in the swamp: predict(lmod, newdata=data.frame(treat=1, group=2)) 1 0.09133691 Warning message: In predict.lm(lmod, newdata = data.frame(treat = 1, group = 2)) : prediction from a rank-deficient fit may be misleading --David. --David. Thanks John From: David Winsemius dwinsem...@comcast.net To: array chip arrayprof...@yahoo.com Cc: r-help@r-project.org r-help@r-project.org Sent: Monday, November 7, 2011 5:13 PM Subject: Re: [R] why NA coefficients On Nov 7, 2011, at 7:33 PM, array chip wrote: Hi, I am trying to run ANOVA with an interaction term on 2 factors (treat has 7 levels, group has 2 levels). I found the coefficient for the last interaction term is always 0, see attached dataset and the code below: test-read.table(test.txt,sep='\t',header=T,row.names=NULL) lm(y~factor(treat)*factor(group),test) Call: lm(formula = y ~ factor(treat) * factor(group), data = test) Coefficients: (Intercept) factor(treat)2factor(treat)3 0.429244 0.499982 0.352971 factor(treat)4 factor(treat)5factor(treat)6 -0.204752 0.142042 0.044155 factor(treat)7factor(group)2 factor(treat)2:factor(group)2 -0.007775 -0.337907 -0.208734 factor(treat)3:factor(group)2 factor(treat)4:factor(group)2 factor(treat)5:factor(group)2 -0.195138 0.800029 0.227514 factor(treat)6:factor(group)2 factor(treat)7:factor(group)2 0.331548NA I guess this is due to model matrix being singular or collinearity among the matrix columns? But I can't figure out how the matrix is singular in this case? Can someone show me why this is the case? Because you have no cases in one of the crossed categories. --David Winsemius, MD West Hartford, CT David Winsemius, MD West Hartford, CT __ 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 West Hartford, CT David Winsemius, MD West Hartford, CT __ 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] why NA coefficients
The cell mean mu_{12} is non-estimable because it has no data in the
cell. How can you estimate something that's not there (at least
without imputation :)? Every parametric function that involves mu_{12}
will also be non-estimable - in particular, the interaction term and
the population marginal means . That's why you get the NA estimates
and the warning. All this follows from the linear model theory
described in, for example, Milliken and Johnson (1992), Analysis of
Messy Data, vol. 1, ch. 13.
Here's an example from Milliken and Johnson (1992) to illustrate:
B1 B2 B3
T1 2, 6 8, 6
T23 14 12, 9
T36 9
Assume a cell means model E(Y_{ijk}) = \mu_{ij}, where the cell means
are estimated by the cell averages.
From M J (p. 173, whence this example is taken):
Whenever treatment combinations are missing, certain
hypotheses cannot be tested without making some
additional assumptions about the parameters in the model.
Hypotheses involving parameters corresponding to the
missing cells generally cannot be tested. For example,
for the data [above] it is not possible to estimate any
linear combinations (or to test any hypotheses) that
involve parameters \mu_{12} and \mu_{33} unless one
is willing to make some assumptions about them.
They continue:
One common assumption is that there is no
interactions between the levels of T and the levels of B.
In our opinion, this assumption should not be made
without some supporting experimental evidence.
In other words, removing the interaction term makes the
non-estimability problem disappear, but it's a copout unless there is
some tangible scientific justification for an additive rather than an
interaction model.
For the above data, M J note that it is not possible to estimate all
of the expected marginal means - in particular, one cannot estimate
the population marginal means $\bar{\mu}_{1.}$, $\bar{\mu}_{3.}$,
$\bar{\mu}_{.2}$ or $\bar{\mu}_{.3}$. OTOH, $\bar{\mu}_{2.}$ and
$\bar{\mu}_{.1}$ since these functions of the parameters involve terms
associated with the means of the missing cells. Moreover, any
hypotheses involving parametric functions that contain non-estimable
cell means are not testable. In this example, the test of equal row
population marginal means is not testable because $\bar{\mu}_{1.}$ and
$\bar{\mu}_{3.}$ are not estimable.
[Aside: if the term parametric function is not familiar, in this
context it refers to linear combinations of model parameters. In the
M J example, $\bar{\mu}_{1.} = \mu_{11} + \mu_{12} + \mu_{13}$ is a
parametric function.]
Hopefully this sheds some light on the situation.
Dennis
On Mon, Nov 7, 2011 at 10:17 PM, array chip arrayprof...@yahoo.com wrote:
Hi Dennis,
The cell mean mu_12 from the model involves the intercept and factor 2:
Coefficients:
(Intercept) factor(treat)2
factor(treat)3
0.429244
0.499982 0.352971
factor(treat)4 factor(treat)5
factor(treat)6
-0.204752
0.142042 0.044155
factor(treat)7 factor(group)2
factor(treat)2:factor(group)2
-0.007775
-0.337907 -0.208734
factor(treat)3:factor(group)2 factor(treat)4:factor(group)2
factor(treat)5:factor(group)2
-0.195138
0.800029 0.227514
factor(treat)6:factor(group)2 factor(treat)7:factor(group)2
0.331548 NA
So mu_12 = 0.429244-0.337907 = 0.091337. This can be verified by:
predict(fit,data.frame(list(treat=1,group=2)))
1
0.09133691
Warning message:
In predict.lm(fit, data.frame(list(treat = 1, group = 2))) :
prediction from a rank-deficient fit may be misleading
But as you can see, it gave a warning about rank-deficient fit... why this
is a rank-deficient fit?
Because treat 1_group 2 has no cases, so why it is still estimable while
on the contrary, treat 7_group 2 which has 2 cases is not?
Thanks
John
From: Dennis Murphy djmu...@gmail.com
To: array chip arrayprof...@yahoo.com
Sent: Monday, November 7, 2011 9:29 PM
Subject: Re: [R] why NA coefficients
Hi John:
What is the estimate of the cell mean \mu_{12}? Which model effects
involve that cell mean? With this data arrangement, the expected
population marginal means of treatment 1 and group 2 are not estimable
either, unless you're willing to assume a no-interaction model.
Chapters 13 and 14 of Milliken and Johnson's Analysis of Messy Data
(vol. 1) cover this topic in some detail, but it assumes you're
familiar with the matrix form of a linear statistical model. Both
chapters cover the two-way model with interaction - Ch.13 from the
cell means model approach and Ch. 14 from the model effects approach.
Because this was written in
[R] rpanel package - retrieve data from panel
Dear Co-Forumeees
Does anybody have experience with using rpanel..or how to retrieve data from
created panel.
For example my panel draws some interactive graph and computes something
inside the panel.
Question : is there a way to retrieve those data ?
For illustration:
if (interactive()) {
plot.hist - function(panel) {
with(panel, {
xlim - range(c(x, mean(x) + c(-3, 3) * sd(x)))
if (panel$cbox[3])
clr - lightblue else clr - NULL
hist(x, freq = FALSE, col = clr, xlim = xlim)
y-x+2
if (panel$cbox[1]) {
xgrid - seq(xlim[1], xlim[2], length = 50)
dgrid - dnorm(xgrid, mean(x), sd(x))
lines(xgrid, dgrid, col = red, lwd = 3)
}
if (panel$cbox[2])
box()
})
panel
}
x - rnorm(50)
panel - rp.control(x = x)
rp.checkbox(panel, cbox, plot.hist,
labels = c(normal density, box, shading), title = Options)
rp.do(panel, plot.hist)
}
and I want to retrieve y in my further computations outside the panel.
Thanks and regards
Mike
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Re: [R] geoR, variofit/likfit
up -- View this message in context: http://r.789695.n4.nabble.com/geoR-variofit-likfit-tp4013734p4016970.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] passing dataframe col name through cbind()
Try this: cbind(scores[,1,drop = FALSE], scores[,2:3]) name round1 round2 1 Bob 40 5 2 Ron 30 6 3 Bud 20 4 Then do ?'[' to learn about 'drop' On Tue, Nov 8, 2011 at 1:06 PM, Eric Rupley erup...@umich.edu wrote: Hi all --- I note that the column name of the first column in a dataframe does not necessarily get passed on when using cbind (example below)… I'm looking for help in clarifying why this behavior occurs, and how I can get all col names, including the first, passed on to the result…while I suspect it's obvious and documented to the cognoscenti, it's puzzling me… Many thanks for any help on this... Eric scores - data.frame(name=c(Bob,Ron,Bud),round1=c(40,30,20),round2=c(5,6,4)) #some toy data scores name round1 round2 1 Bob 40 5 2 Ron 30 6 3 Bud 20 4 cbind(scores[,1],total=rowSums(scores[,2:3]),scores[,2:3]) scores[, 1] total round1 round2 1 Bob 45 40 5 2 Ron 36 30 6 3 Bud 24 20 4 ...first column renamed... …yet this passes all column names: cbind(scores[,1:3]) name round1 round2 1 Bob 40 5 2 Ron 30 6 3 Bud 20 4 …but this doesn't: cbind(scores[,1],scores[,2:3]) scores[, 1] round1 round2 1 Bob 40 5 2 Ron 30 6 3 Bud 20 4 -- Eric Rupley University of Michigan, Museum of Anthropology 1109 Geddes Ave, Rm. 4013 Ann Arbor, MI 48109-1079 erup...@umich.edu +1.734.276.8572 __ 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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ 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] geoR, variofit/likfit
On Nov 8, 2011, at 1:20 PM, katiab81 wrote: up down ... for a variety of reasons. And... PLEASE do read the posting guide http://www.R-project.org/posting-guide.html It's very possible that you are not communicating with the package authors by posting to rhelp (by way of Nabble) , and since you are asking in your initial posting if there is an error, that would seem to be the proper avenue. -- David Winsemius, MD West Hartford, CT __ 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] why NA coefficients
Hi Dennis, Thanks very much for the details. All those explanations about
non-estimable mu_{12} when it has no data make sense to me!
Regarding my specific example data where mu_{12} should NOT be estimable in a
linear model with interaction because it has no data, yet the linear model I
created by using lm() in R still CAN estimate the mean mu_{12}, while on the
other hand, mu_{72} is instead NOT estimable from lm() even this category does
have data. Does this contradiction to the theory imply that the linear model by
lm() in R on my specific example data is NOT reliable/trustable and should not
be used?
Thanks
John
From: Dennis Murphy djmu...@gmail.com
Cc: r-help@r-project.org r-help@r-project.org
Sent: Tuesday, November 8, 2011 10:22 AM
Subject: Re: [R] why NA coefficients
The cell mean mu_{12} is non-estimable because it has no data in the
cell. How can you estimate something that's not there (at least
without imputation :)? Every parametric function that involves mu_{12}
will also be non-estimable - in particular, the interaction term and
the population marginal means . That's why you get the NA estimates
and the warning. All this follows from the linear model theory
described in, for example, Milliken and Johnson (1992), Analysis of
Messy Data, vol. 1, ch. 13.
Here's an example from Milliken and Johnson (1992) to illustrate:
B1 B2 B3
T1 2, 6 8, 6
T2 3 14 12, 9
T3 6 9
Assume a cell means model E(Y_{ijk}) = \mu_{ij}, where the cell means
are estimated by the cell averages.
From M J (p. 173, whence this example is taken):
Whenever treatment combinations are missing, certain
hypotheses cannot be tested without making some
additional assumptions about the parameters in the model.
Hypotheses involving parameters corresponding to the
missing cells generally cannot be tested. For example,
for the data [above] it is not possible to estimate any
linear combinations (or to test any hypotheses) that
involve parameters \mu_{12} and \mu_{33} unless one
is willing to make some assumptions about them.
They continue:
One common assumption is that there is no
interactions between the levels of T and the levels of B.
In our opinion, this assumption should not be made
without some supporting experimental evidence.
In other words, removing the interaction term makes the
non-estimability problem disappear, but it's a copout unless there is
some tangible scientific justification for an additive rather than an
interaction model.
For the above data, M J note that it is not possible to estimate all
of the expected marginal means - in particular, one cannot estimate
the population marginal means $\bar{\mu}_{1.}$, $\bar{\mu}_{3.}$,
$\bar{\mu}_{.2}$ or $\bar{\mu}_{.3}$. OTOH, $\bar{\mu}_{2.}$ and
$\bar{\mu}_{.1}$ since these functions of the parameters involve terms
associated with the means of the missing cells. Moreover, any
hypotheses involving parametric functions that contain non-estimable
cell means are not testable. In this example, the test of equal row
population marginal means is not testable because $\bar{\mu}_{1.}$ and
$\bar{\mu}_{3.}$ are not estimable.
[Aside: if the term parametric function is not familiar, in this
context it refers to linear combinations of model parameters. In the
M J example, $\bar{\mu}_{1.} = \mu_{11} + \mu_{12} + \mu_{13}$ is a
parametric function.]
Hopefully this sheds some light on the situation.
Dennis
Hi Dennis,
The cell mean mu_12 from the model involves the intercept and factor 2:
Coefficients:
(Intercept) factor(treat)2
factor(treat)3
0.429244
0.499982 0.352971
factor(treat)4 factor(treat)5
factor(treat)6
-0.204752
0.142042 0.044155
factor(treat)7 factor(group)2
factor(treat)2:factor(group)2
-0.007775
-0.337907 -0.208734
factor(treat)3:factor(group)2 factor(treat)4:factor(group)2
factor(treat)5:factor(group)2
-0.195138
0.800029 0.227514
factor(treat)6:factor(group)2 factor(treat)7:factor(group)2
0.331548 NA
So mu_12 = 0.429244-0.337907 = 0.091337. This can be verified by:
predict(fit,data.frame(list(treat=1,group=2)))
1
0.09133691
Warning message:
In predict.lm(fit, data.frame(list(treat = 1, group = 2))) :
prediction from a rank-deficient fit may be misleading
But as you can see, it gave a warning about rank-deficient fit... why this
is a rank-deficient fit?
Because treat 1_group 2 has no cases, so why it is still estimable while
on the contrary, treat 7_group 2 which has 2 cases is not?
Thanks
John
From: Dennis Murphy djmu...@gmail.com
Sent:
Re: [R] ggplot2 reorder factors for faceting
Hi:
(1) Here is one way to reorganize the levels of a factor:
plotData[['infection']] - factor(plotData[['infection']],
levels = c('InfA', 'InfC', 'InfB', 'InfD'))
Do this ahead of the call to ggplot(), preferably after plotData is defined.
relevel() resets the baseline category of a factor, but here you want
to make multiple changes.
(2) You probably want a better title for the legend. Assuming you want
'Scale' as the title, you can add the following to labs:
labs(..., fill = 'Scale')
HTH,
Dennis
On Tue, Nov 8, 2011 at 3:51 AM, Iain Gallagher
iaingallag...@btopenworld.com wrote:
Dear List
I am trying to draw a heatmap using ggplot2. In this heatmap I have faceted
my data by 'infection' of which I have four. These four infections break down
into two types and I would like to reorder the 'infection' column of my data
to reflect this.
Toy example below:
library(ggplot2)
# test data for ggplot reordering
genes - (rep (c(rep('a',4), rep('b',4), rep('c',4), rep('d',4), rep('e',4),
rep('f',4)) ,4))
fcData - rnorm(96)
times - rep(rep(c(2,6,24,48),6),4)
infection - c(rep('InfA', 24), rep('InfB', 24), rep('InfC', 24), rep('InfD',
24))
infType - c(rep('M', 24), rep('D',24), rep('M', 24), rep('D', 24))
# data is long format for ggplot2
plotData - as.data.frame(cbind(genes, as.numeric(fcData), as.numeric(times),
infection, infType))
hp2 - ggplot(plotData, aes(factor(times), genes)) + geom_tile(aes(fill =
scale(as.numeric(fcData + facet_wrap(~infection, ncol=4)
# set scale
hp2 - hp2 + scale_fill_gradient2(name=NULL, low=#0571B0, mid=#F7F7F7,
high=#CA0020, midpoint=0, breaks=NULL, labels=NULL, limits=NULL,
trans=identity)
# set up text (size, colour etc etc)
hp2 - hp2 + labs(x = Time, y = ) + scale_y_discrete(expand = c(0, 0)) +
opts(axis.ticks = theme_blank(), axis.text.x = theme_text(size = 10, angle =
360, hjust = 0, colour = grey25), axis.text.y = theme_text(size=10, colour
= 'gray25'))
hp2 - hp2 + theme_bw()
In the resulting plot I would like infections infA and infC plotted next to
each other and likewise for infB and infD. I have a column in the data -
infType - which I could use to reorder the infection column but so far I have
no luck getting this to work.
Could someone give me a pointer to the best way to reorder the infection
factor and accompanying data into the order I would like?
Best
iain
sessionInfo()
R version 2.13.2 (2011-09-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_GB.utf8 LC_NUMERIC=C
[3] LC_TIME=en_GB.utf8 LC_COLLATE=en_GB.utf8
[5] LC_MONETARY=C LC_MESSAGES=en_GB.utf8
[7] LC_PAPER=en_GB.utf8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_GB.utf8 LC_IDENTIFICATION=C
attached base packages:
[1] grid stats graphics grDevices utils datasets methods
[8] base
other attached packages:
[1] ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.6
loaded via a namespace (and not attached):
[1] digest_0.5.0 tools_2.13.2
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to use quadrature to integrate some complicated functions
It's polite to include the whole list in your replies so the threads
get archived properly.
Yes, underflow is occasionally a problem: hence the common use of
log-likelihood in MLE and other applications. Might that help you out
here? You can get log-likelihoods directly from pnorm and dnorm with
the log = TRUE option.
Michael
On Tue, Nov 8, 2011 at 11:35 AM, Zuofeng Shang zuofeng.shan...@nd.edu wrote:
Hi Michael , Thanks for your suggestion!
Integrate() possibly workx for my problem. However my function is a product
of
a sequence with values between zero and one, which could be extremely small
when the length of the sequence is large. Note: my integrand is like
Phi(x-a_1)*Phi(x-a_2)*...*Phi(x-a_{n-1})*phi(x-a_n).
So when n is very small, the above product will be very close to zero.
So my concern is whether integrate() can handle accurately such integrand.
Thanks a lot for your time!
Best wishes,
Zuofeng
于 2011/11/8 8:43, R. Michael Weylandt michael.weyla...@gmail.com 写道:
Have you tried wrapping it in a function and using integrate()? R is
pretty good at handling numerical integration. If integrate() isn't good for
you, can you say more as to why?
Michael
On Nov 6, 2011, at 4:15 PM, JeffNDzuofeng.shan...@nd.edu wrote:
Hello to all,
I am having trouble with intregrating a complicated uni-dimensional
function
of the following form
Phi(x-a_1)*Phi(x-a_2)*...*Phi(x-a_{n-1})*phi(x-a_n).
Here n is about 5000, Phi is the cumulative distribution function of
standard normal,
phi is the density function of standard normal, and x ranges over
(-infty,infty).
My idea is to to use quadrature to handle this integral. But since Phi
has
not cloaed form,
I don't know how to do this effeciently. I appreciate very much if
someone
has any ideas about it.
Thanks!
Jeff
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--
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Re: [R] why NA coefficients
It might make the discussion easier to follow if you used
a smaller dataset that anyone can make and did some experiments
with contrasts. E.g.,
D - data.frame(expand.grid(X1=LETTERS[1:3], X2=letters[24:26])[-1,],
Y=2^(1:8))
D
X1 X2 Y
2 B x 2
3 C x 4
4 A y 8
5 B y 16
6 C y 32
7 A z 64
8 B z 128
9 C z 256
lm(data=D, Y ~ X1 * X2)
Call:
lm(formula = Y ~ X1 * X2, data = D)
Coefficients:
(Intercept) X1B X1C
-188 190 192
X2y X2z X1B:X2y
196 252 -182
X1C:X2y X1B:X2z X1C:X2z
-168 -126 NA
lm(data=D, Y ~ X1 * X2, contrasts=list(X2=contr.SAS))
Call:
lm(formula = Y ~ X1 * X2, data = D, contrasts = list(X2 = contr.SAS))
Coefficients:
(Intercept) X1B X1C
64 64 192
X2x X2y X1B:X2x
-252 -56 126
X1C:X2x X1B:X2y X1C:X2y
NA -56 -168
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of array chip
Sent: Tuesday, November 08, 2011 10:57 AM
To: Dennis Murphy
Cc: r-help@r-project.org
Subject: Re: [R] why NA coefficients
Hi Dennis, Thanks very much for the details. All those explanations about
non-estimable mu_{12} when
it has no data make sense to me!
Regarding my specific example data where mu_{12} should NOT be estimable in a
linear model with
interaction because it has no data, yet the linear model I created by using
lm() in R still CAN
estimate the mean mu_{12}, while on the other hand, mu_{72} is instead NOT
estimable from lm() even
this category does have data. Does this contradiction to the theory imply
that the linear model by
lm() in R on my specific example data is NOT reliable/trustable and should
not be used?
Thanks
John
From: Dennis Murphy djmu...@gmail.com
Cc: r-help@r-project.org r-help@r-project.org
Sent: Tuesday, November 8, 2011 10:22 AM
Subject: Re: [R] why NA coefficients
The cell mean mu_{12} is non-estimable because it has no data in the
cell. How can you estimate something that's not there (at least
without imputation :)? Every parametric function that involves mu_{12}
will also be non-estimable - in particular, the interaction term and
the population marginal means . That's why you get the NA estimates
and the warning. All this follows from the linear model theory
described in, for example, Milliken and Johnson (1992), Analysis of
Messy Data, vol. 1, ch. 13.
Here's an example from Milliken and Johnson (1992) to illustrate:
B1 B2 B3
T1 2, 6 8, 6
T2 3 14 12, 9
T3 6 9
Assume a cell means model E(Y_{ijk}) = \mu_{ij}, where the cell means
are estimated by the cell averages.
From M J (p. 173, whence this example is taken):
Whenever treatment combinations are missing, certain
hypotheses cannot be tested without making some
additional assumptions about the parameters in the model.
Hypotheses involving parameters corresponding to the
missing cells generally cannot be tested. For example,
for the data [above] it is not possible to estimate any
linear combinations (or to test any hypotheses) that
involve parameters \mu_{12} and \mu_{33} unless one
is willing to make some assumptions about them.
They continue:
One common assumption is that there is no
interactions between the levels of T and the levels of B.
In our opinion, this assumption should not be made
without some supporting experimental evidence.
In other words, removing the interaction term makes the
non-estimability problem disappear, but it's a copout unless there is
some tangible scientific justification for an additive rather than an
interaction model.
For the above data, M J note that it is not possible to estimate all
of the expected marginal means - in particular, one cannot estimate
the population marginal means $\bar{\mu}_{1.}$, $\bar{\mu}_{3.}$,
$\bar{\mu}_{.2}$ or $\bar{\mu}_{.3}$. OTOH, $\bar{\mu}_{2.}$ and
$\bar{\mu}_{.1}$ since these functions of the parameters involve terms
associated with the means of the missing cells. Moreover, any
hypotheses involving parametric functions that contain non-estimable
cell means are not testable. In this example, the test of equal row
population marginal means is not testable because $\bar{\mu}_{1.}$ and
$\bar{\mu}_{3.}$ are not estimable.
[Aside: if the term parametric function is not familiar, in this
context it refers to linear combinations of model parameters. In the
M J example, $\bar{\mu}_{1.} = \mu_{11} + \mu_{12} + \mu_{13}$ is a
parametric function.]
Hopefully this sheds some light on the situation.
Dennis
Hi Dennis,
[R] nesting scale_manual caracteristics in ggplot
Hi there,
I am having a little problem with combining three scale_manual commands in a
facet plot. I am not able to combine the three different characteristics,
instead ending up with three different descriptions next to the graph for
the same geom. I would like to see two separate labels (not three); one
describing lines 1-7 and the other 8-14. For each of the treatments (A-B) I
want a combination of color, line type and symbol. How do I do this?
Here are my codes (Feel free to modify the example to make it easier to work
with. I was not able to do this while keeping the problem I wanted help
with)
df -structure(list(year = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), treatment =
structure(c(1L,
1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L,
6L, 7L, 7L, 7L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L,
5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 1L, 2L, 2L, 2L, 3L,
3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L,
1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L,
7L, 7L, 7L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L,
5L, 5L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L,
3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 2L,
2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 7L, 7L,
7L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L,
6L, 6L, 6L, 7L, 7L, 7L), .Label = c(A, B, C, D, E,
F, G), class = factor), total = c(135L, 118L, 121L, 64L,
53L, 49L, 178L, 123L, 128L, 127L, 62L, 129L, 126L, 99L, 183L,
45L, 57L, 45L, 72L, 30L, 71L, 123L, 89L, 102L, 60L, 44L, 59L,
124L, 145L, 126L, 103L, 67L, 97L, 66L, 76L, 108L, 36L, 48L, 41L,
69L, 47L, 57L, 167L, 136L, 176L, 85L, 36L, 82L, 222L, 149L, 171L,
145L, 122L, 192L, 136L, 164L, 154L, 46L, 57L, 57L, 70L, 55L,
102L, 111L, 152L, 204L, 41L, 46L, 103L, 156L, 148L, 155L, 103L,
124L, 176L, 111L, 142L, 187L, 43L, 52L, 75L, 64L, 91L, 78L, 196L,
314L, 265L, 44L, 39L, 98L, 197L, 273L, 274L, 89L, 91L, 74L, 91L,
112L, 98L, 140L, 90L, 121L, 120L, 161L, 83L, 230L, 266L, 282L,
35L, 53L, 57L, 315L, 332L, 202L, 90L, 79L, 89L, 67L, 116L, 109L,
44L, 68L, 75L, 29L, 52L, 52L, 253L, 203L, 87L, 105L, 234L, 152L,
247L, 243L, 144L, 167L, 165L, 95L, 300L, 128L, 125L, 84L, 183L,
88L, 153L, 185L, 175L, 226L, 216L, 118L, 118L, 94L, 224L, 259L,
176L, 175L, 147L, 197L, 141L, 176L, 187L, 87L, 92L, 148L, 86L,
139L, 122L), country = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), .Label = c(high, low), class = factor)), .Names = c(year,
treatment, total, country), class = data.frame, row.names = c(NA,
-167L))
lines - structure(list(`Line #` = 1:14, country = structure(c(2L, 2L, 2L,
2L,2L,2L,2L,1L, 1L,1L,1L,1L,1L,1L), .Label = c(high, low), class =
factor),treatment = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L,1L, 2L,
3L,4L,5L,6L,7L), .Label = c(A, B, C,D, E, F,G,H ), class =
factor), Intercept = c(81.47, 31.809,69.892,82.059,106.392,45.059,38.809,
67.024, 17.357, 105.107,79.191,91.357,5.691,24.357), Slope = c(47.267,
20.234,33.717,14.667,13.434,25.817,21.967, 47.267, 20.234,
33.717,14.667,13.434,25.817,21.967)), .Names = c(Line #, country,
treatment, Intercept, Slope), class = data.frame, row.names = c(NA,
-14L))
ggplot(data = df, aes(x = year, y = total, colour = treatment,
linetype=treatment)) +
geom_point(aes(shape = treatment)) +
facet_wrap(~country) +
scale_colour_manual(breaks=c('A','B','C','D','E','F','G'),
values=c('A'='black','B'='black', 'C'='grey','D'='grey',
'E'='red','F'='grey', 'G'='red'), labels=c('A: Line 1','B: Line 2','C:
Line3','D: Line 4',
'E: Line 5 ','F:Line 6','G:Line 7'))+
scale_linetype_manual(breaks=c('A','B','C','D','E','F','G'),
[R] Question about R mantissa and number of bits
Dear all,
I think that every number x in R can be represented in floating point
arithmetic as:
x = (-1)^s (1+f) 2^(e-1023)
where s is coded on 1 bit, e (positive integer) is coded on 11 bits, and
f (real in [0,1)) is coded on 52 bits.
Am I right?
We have f=\sum_{i=1}^{52} k_i 2^{-i} for some values k_i in {0,1}.
If this is the case (for the 52 bits), we should have:
The number next to 2^150 should be (-1)^02^150(1+2^(-52))=2^150+2^98
I can check this:
a - 2^150; b - a + 2^97; b == a
[1] TRUE
a - 2^150; b - a + 2^98; b == a
[1] FALSE
So it seems that the mantissa is really coded on 52 bits.
But now, if I issue the following commands (using some functions
provided below to translate from decimal to binary):
dec2bin(0.1,52)
[1] 0.0001100110011001100110011001100110011001100110011001
formatC(sum(as.numeric(strsplit(dec2bin(0.1,52),)[[1]][-(1:2)])*2^(-(1:52))),50)
[1] 0.099866773237044981215149164199829101562
formatC(0.1,50)
[1] 0.155511151231257827021181583404541
formatC(sum(as.numeric(strsplit(dec2bin(0.1,55),)[[1]][-(1:2)])*2^(-(1:55))),50)
[1] 0.155511151231257827021181583404541
formatC(0.1,50)
[1] 0.155511151231257827021181583404541
So now, using formatC() it seems that f is coded on 55 bits!
Do you have an explanation for this fact?
Many thanks!
Pierre
dec2bin.ent - function(x) {
as.integer(paste(rev(as.integer(intToBits(x))), collapse=))
}
dec2bin.frac - function(x,prec=52) {
res - rep(NA,prec)
for (i in 1:prec) {
res[i] - as.integer(x*2)
x - (x*2) %% 1
}
return(paste(res,collapse=))
}
dec2bin - function(x,prec=52) {
x - as.character(x)
res - strsplit(x,.,fixed=TRUE)[[1]]
return(paste(dec2bin.ent(as.numeric(res[1])),dec2bin.frac(as.numeric(paste(0.,res[2],sep=)),prec),sep=.))
}
--
Pierre Lafaye de Micheaux
Adresse courrier:
Département de Mathématiques et Statistique
Université de Montréal
CP 6128, succ. Centre-ville
Montréal, Québec H3C 3J7
CANADA
Adresse physique:
Département de Mathématiques et Statistique
Bureau 4249, Pavillon André-Aisenstadt
2920, chemin de la Tour
Montréal, Québec H3T 1J4
CANADA
Tél.: (00-1) 514-343-6607 / Fax: (00-1) 514-343-5700
laf...@dms.umontreal.ca
http://www.biostatisticien.eu
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Re: [R] rpanel package - retrieve data from panel
I found names(y)-y before the end is working ;) some more ideas ? -- View this message in context: http://r.789695.n4.nabble.com/rpanel-package-retrieve-data-from-panel-tp4016953p4017082.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] Estimate of intercept in loglinear model
On Nov 08, 2011 at 11:16am Colin Aitken wrote: An unresolved problem is: what does R do when the explanatory factors are not defined as factors when it obtains a different value for the intercept but the correct value for the fitted value? Colin, I don't think that happens (that the fitted values are identical if predictors are cast as numerical), but the following could (really is answered by my initial answer). Once again, using the example I gave above, but using the second level of outcome as a reference level for a new fit, called glm.D93R. (For this part of the question a corpse would have been nice, though not really needed---yours was unfortunately buried too deeply for me to find it,) ## Dobson (1990) Page 93: Randomized Controlled Trial : counts - c(18,17,15,20,10,20,25,13,12) outcome - gl(3,1,9) treatment - gl(3,3) glm.D93 - glm(counts ~ outcome + treatment, family=poisson()) glm.D93R - glm(counts ~ C(outcome, base=2) + treatment, family=poisson()) ## treat predictor as numeric glm.D93N - glm(counts ~ as.numeric(as.character(outcome)) + as.numeric(as.character(treatment)), family=poisson()) coef(glm.D93) (Intercept) outcome2 outcome3treatment2treatment3 3.044522e+00 -4.542553e-01 -2.929871e-01 1.337909e-15 1.421085e-15 ## Different value for the Intercept but same fitted values (see below) as the earlier fit (above) ## summary(glm.D93R) snipped and edited for clarity Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 2.590e+00 1.958e-01 13.230 2e-16 *** outcome1 4.543e-01 2.022e-01 2.247 0.0246 * outcome3 1.613e-01 2.151e-01 0.750 0.4535 treatment2-3.349e-16 2.000e-01 0.000 1. treatment3-6.217e-16 2.000e-01 0.000 1. snip fitted(glm.D93) 12345678 9 21.0 13.3 15.7 21.0 13.3 15.7 21.0 13.3 15.7 fitted(glm.D93R) 12345678 9 21.0 13.3 15.7 21.0 13.3 15.7 21.0 13.3 15.7 ## if predictors treated as numeric---check summary(glm.D93N) yourself fitted(glm.D93N) 12345678 9 19.40460 16.52414 14.07126 19.40460 16.52414 14.07126 19.40460 16.52414 14.07126 Regards, Mark. - Mark Difford (Ph.D.) Research Associate Botany Department Nelson Mandela Metropolitan University Port Elizabeth, South Africa -- View this message in context: http://r.789695.n4.nabble.com/Estimate-of-intercept-in-loglinear-model-tp4009905p4017091.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] How to handle empty arguments
Dear all,
I am having a data stucture that contains Products and Time Stamps,
I have made also two lists
ProductList=list(c('Example1','Example2'...)
TimeStamp=list(c(1990-02-03 12:57:60),c(1990-02-03 12:57:60),
then I have made few functions that call each other
do_analysis_for_all the data-function(arguments){
.
return(lapply(ProductList,do_analysis_for_one_product_list_for_all_time_stamps)
}
do_analysis_for_one_product_list_for_all_time_stamps-function(arguments){
return(lapply(TimeStamps,do_analysis_for_one_product_list_for_one_time_stamps)
}
(this is mostly just an algorithm to show you the main logic).
as one is getting down to the chain I have described, there are functions that
chop the data as requested. For example for a specific TimeFram one will get
the entries that correspond to the specific dates.
The problem I have though is that sometimes (is not too often) there are no
entries for a specific interval and thus the next function will rely on the
chopped data will explode with an error.
I want to ask you for a clear solution to handle this cases
I have two ideas in mind
a). I change all the function in my code so to check that the input argument is
empty (how to do that?) and in that case I return an empty list
b) I change nothing to the code and I ask kindly from the R in that case to
return a lovely error message (Best is to find it saved in the list that lapply
returns) and continue to the next case. I have tried earlier to add the
try(myfunction,silent=TRUE) but I ended up changing all my code with the
try(..) which of course is a bit of dirty solution. Do you know if I can ask
globally R (perhaps add some directive in the beginning of my file) to handle
all these errors in a silent manner so to skip them.
What should I try to a or b and why?
I would like to thank you in advance for your time spent to read this email
B.R
Alex
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[R] window?
This doesn't seem to work: d - rnorm(2*53) ds - ts(d, frequency=53, start=c(2000,10)) dswin - window(ds, start=c(2001,1), end=c(2001,10), frequency=1) dswin Time Series: Start = 2001 End = 2001 Frequency = 1 [1] 1.779409 dswin - window(ds, start=c(2001,1), end=c(2001,10)) dswin Time Series: Start = c(2001, 1) End = c(2001, 10) Frequency = 53 [1] 1.7794090 0.6916779 -0.6641813 -0.7426889 -0.5584049 -0.2312959 [7] -0.0183454 -1.0026301 0.4534920 0.6058198 The problem is that when the frequency is specified only one value shows up in the window. When no frequency is specified I get all 10 values but now the time series has a frequency that I don't want. Comments? Kevin [[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.
