Re: [R] slow dget
dput/dget were not intended to save/restore large objects. Understand what is happening in the use of dput/dget. dput is creating a text file that can reconstitute the object with dget. dget is having to read the file in and then parse it: dget function (file) eval(parse(file = file)) environment: namespace:base This can be a complex process if there object is large and complex. save/load basically take the binary object and save it with little additional processing and the load is just as fast. In general, most of the functions can be used both correctly and incorrectly. So should a warning for every potential condition/criteria be put in the help file? Probably not. It is hard to protect the user against him/herself. So what you are doing in seeing how long alternatives take is a good learning tool and will help you improve your use of the features. On Fri, Nov 5, 2010 at 11:16 PM, Jack Tanner i...@hotmail.com wrote: I have a data structure that is fast to dput(), but very slow to dget(). On disk, the file is about 35MB. system.time(dget(r.txt)) user system elapsed 142.93 1.27 192.84 The same data structure is fast to save() and fast to load(). The .RData file on disk is about 12MB. system.time(load(r.RData)) user system elapsed 4.89 0.08 7.82 I imagine that this is a known speed issue with dget, and that the recommended solution is to use load, which is fine with me. If so, perhaps a note to this effect could be added to the dget help page. All timings above using R version 2.12.0 (2010-10-15) Platform: i386-pc-mingw32/i386 (32-bit) __ 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 Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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 required to remove \\N
Dear All, I have .csv file it looks like this : rawdata - read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t' , header=FALSE) head(rawdata,n=5) TenantDomain Owner Current State 1\\N ROOTadmin Running 2\\N ROOTadmin Stopped 3\\N ROOTadmin Running 4\\N ROOTadmin Running 5\\N ROOTadmin Running 20 DEMO ROOTadmin Stopped 21 DEMOROOT admin Stopped 22 Demo ROOT admin Stopped The first column contain the \\n up to 19 row. I need to replace the \\N value to Blankspace . Any help will really appreciated. Thanks for your time. Thanks Rg Mohan L __ 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] About 5.1 Arrays
Hi Richard,
## for an array with
## dim(a) == c(3,4,2)
## a[i,j,k] means select the element in position
##i + (j-1)*3 + (k-1)*3*4
My understanding;
e.g.
1)
dim(a) == c(3,4,2)
3 + (4-1)*3 + (2-1)*3*4
3+9+12=24
2)
## dim(a) == c(1,2,1)
1 + (2-1)*3 + (1-1)*3*4
1+3+0=4
3)
## dim(a) == c(2,3,1)
2 + (3-1)*3 + (1-1)*3*4
2+6+0=8
etc.
It is NOT always the product of i*j*k as I thought before. Thanks for your
explanation.
What are the value of 3 and 4? The values of position and dimension?
e.g.
a - sample(24)
a
[1] 22 18 17 10 24 1 11 13 9 19 20 8 2 21 23 16 7 14 12 15 4 5 3 6
dim(a) - c(3,4,2)
a
, , 1
[,1] [,2] [,3] [,4] -- positions 1, 2, 3, 4 of the second dimension ?
[1,] 22 10 11 19
[2,] 18 24 13 20
[3,] 17198
^ the first dimension ?
, , 2
[,1] [,2] [,3] [,4]
[1,]2 16 125
[2,] 217 153
[3,] 23 1446
? If I'm wrong pls correct me. TIA
Now I'm going to digest Joshua's advice.
B.R.
Stephen L
From: RICHARD M. HEIBERGER r...@temple.edu
Cc: Daniel Nordlund djnordl...@frontier.com; r-help@r-project.org
Sent: Sat, November 6, 2010 12:48:35 AM
Subject: Re: [R] About 5.1 Arrays
Continuing with Daniel's example, but with different data values
a - sample(24)
a
dim(a) - c(3,4,2)
a
as.vector(a)
## for an array with
## dim(a) == c(3,4,2)
## a[i,j,k] means select the element in position
##i + (j-1)*3 + (k-1)*3*4
index - function(i,j,k) {
i + (j-1)*3 + (k-1)*3*4
}
## find the vector position described by row 2, column 1, layer 2
index(2,1,2)## this is the position in the original vector
a[2,1,2]## this is the value in that position with 3D indexing
a[index(2,1,2)] ## this is the same value with 1D vector indexing
a[14] ## this is the same value with 1D vector indexing
## find the position in row 3, column 4, layer 1
index(3,4,1)## this is the position in the original vector
a[3,4,1]## this is the value in that position with 3D indexing
a[index(3,4,1)] ## this is the same value with 1D vector indexing
a[12] ## this is the same value with 1D vector indexing
index(1,1,1)## this is the position in the original vector
index(2,1,1)## this is the position in the original vector
index(3,1,1)## this is the position in the original vector
index(1,2,1)## this is the position in the original vector
index(2,2,1)## this is the position in the original vector
index(3,2,1)## this is the position in the original vector
index(1,3,1)## this is the position in the original vector
index(2,3,1)## this is the position in the original vector
index(3,3,1)## this is the position in the original vector
index(1,4,1)## this is the position in the original vector
index(2,4,1)## this is the position in the original vector
index(3,4,1)## this is the position in the original vector
index(1,1,2)## this is the position in the original vector
index(2,1,2)## this is the position in the original vector
index(3,1,2)## this is the position in the original vector
index(1,2,2)## this is the position in the original vector
index(2,2,2)## this is the position in the original vector
index(3,2,2)## this is the position in the original vector
index(1,3,2)## this is the position in the original vector
index(2,3,2)## this is the position in the original vector
index(3,3,2)## this is the position in the original vector
index(1,4,2)## this is the position in the original vector
index(2,4,2)## this is the position in the original vector
index(3,4,2)## this is the position in the original vector
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[R] anova(lme.model)
Dear R users Topic: Linear effect model fitting using the nlme package (recomended by Pinheiro et al. 2008 for unbalanced data set). The R help provides much info about the controversy to use the anova(lme.model) function to present numerator df and F values. Additionally different p-values calculated by lme and anova are reported. However, I come across the same problem, and I would very much appreciate some R help to fit an anova function to get similar p-values compared to the lme function and additionally to provide corresponding F-values. I tried to use contrasts and to deal with the ‚unbalanced data set’. Thanks Sibylle Kaltenborn-read.table(Kaltenborn_YEARS.txt, na.strings=*, header=TRUE) library(nlme) model5c-lme(asin(sqrt(PropMortality))~Diversity+ Management+Species+Height+Height*Diversity, data=Kaltenborn, random=~1|Plot/SubPlot, na.action=na.omit, weights=varPower(form=~Diversity), subset=Kaltenborn$ADDspecies!=1, method=ML) summary(model5c) Linear mixed-effects model fit by maximum likelihood Data: Kaltenborn Subset: Kaltenborn$ADDspecies != 1 AIC BIC logLik -249.3509 -205.4723 137.6755 Random effects: Formula: ~1 | Plot (Intercept) StdDev: 0.06162279 Formula: ~1 | SubPlot %in% Plot (Intercept) Residual StdDev: 0.03942785 0.05946185 Variance function: Structure: Power of variance covariate Formula: ~Diversity Parameter estimates: power 0.7302087 Fixed effects: asin(sqrt(PropMortality)) ~ Diversity + Management + Species + Height + Height * Diversity Value Std.Error DF t-value p-value (Intercept) 0.5422893 0.05923691 163 9.154585 0. Diversity-0.0734688 0.02333159 14 -3.148896 0.0071 Managementm+ 0.0217734 0.02283375 30 0.953562 0.3479 Managementu -0.0557160 0.02286694 30 -2.436532 0.0210 SpeciesPab -0.2058763 0.02763737 163 -7.449198 0. SpeciesPm 0.0308005 0.02827782 163 1.089210 0.2777 SpeciesQp 0.0968051 0.02689327 163 3.599602 0.0004 Height -0.0017579 0.00031667 163 -5.551251 0. Diversity:Height 0.0005122 0.00014443 163 3.546270 0.0005 Correlation: (Intr) Dvrsty Mngmn+ Mngmnt SpcsPb SpcsPm SpcsQp Height Diversity-0.867 Managementm+ -0.173 -0.019 Managementu -0.206 0.005 0.499 SpeciesPab -0.253 0.085 0.000 0.035 SpeciesPm-0.239 0.058 0.001 0.064 0.521 SpeciesQp-0.250 0.041 -0.001 0.032 0.502 0.506 Height -0.518 0.532 -0.037 -0.004 0.038 0.004 0.033 Diversity:Height 0.492 -0.581 0.031 -0.008 -0.149 -0.099 -0.069 -0.904 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -2.99290873 -0.60522612 -0.05756772 0.62163049 2.80811502 Number of Observations: 216 Number of Groups: Plot SubPlot %in% Plot 1648 anova(model5c) numDF denDF F-value p-value (Intercept) 1 163 244.67887 .0001 Diversity114 1.53025 0.2364 Management 230 6.01972 0.0063 Species 3 163 51.86699 .0001 Height 1 163 30.08090 .0001 Diversity:Height 1 163 12.57603 0.0005 -- __ 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] calculate probability
Dear All I have some problem with calculate probability. Assume I have data with normal distribution with mean = 5 sd = 2. I want to approximate probability = 2.4. I used pnorm(2.4, 5, 2) - pnorm(2.4, 5, 2, lower.tail = FLASE), correct or not. Many Thanks Jumlong __ 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 required to remove \\N
Try this: gsub(N, BlankSpace, rawdata$Tenant) On Sat, Nov 6, 2010 at 8:15 AM, Mohan L l.mohanphys...@gmail.com wrote: Dear All, I have .csv file it looks like this : rawdata - read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t' , header=FALSE) head(rawdata,n=5) TenantDomain Owner Current State 1\\N ROOTadmin Running 2\\N ROOTadmin Stopped 3\\N ROOTadmin Running 4\\N ROOTadmin Running 5\\N ROOTadmin Running 20 DEMO ROOTadmin Stopped 21 DEMOROOT admin Stopped 22 Demo ROOT admin Stopped The first column contain the \\n up to 19 row. I need to replace the \\N value to Blankspace . Any help will really appreciated. Thanks for your time. Thanks Rg Mohan L __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] calculate probability
On 06-Nov-10 11:16:28, Jumlong Vongprasert wrote: Dear All I have some problem with calculate probability. Assume I have data with normal distribution with mean = 5 sd = 2. I want to approximate probability = 2.4. I used pnorm(2.4, 5, 2) - pnorm(2.4, 5, 2, lower.tail = FLASE), correct or not. Many Thanks Jumlong [A] Not correct because FLASE should be FALSE. [B] Not correct because a) pnorm(2.4, 5, 2) is the probability that a value from a Normal distribution with mean 5 will be less than 2.4, and this is less than 1/2 (since 2.4 is less than the mean). b) pnorm(2.4, 5, 2, lower.tail = FALSE) is the probability that such a value will be greater than 2.4, and this is greater than 1/2 (for the same reason. c) The difference will therefore be ( 1/2) - ( 1/2) which will be less than 0, so cannot be a probability. [C] Not correct because a value sampled from a Normal distribution (which is a continuous distribution) has probability 0 of being exactly equal to any given value (e.g. 2.4); so I think your question does not express what you want to know. One possibility which could make your question realistic is that the value 2.4 that you are interested in is a value sampled from Normal(mean=5, sd=2) **that has been rounded to 1 decimal place** and so could have been any value between 2.35 and 2.45; in that case it makes sense to ask what is the probability of a value in this range from Normal(mean=5, sd=2). This would be pnorm(2.45, 5, 2) - pnorm(2.35, 5, 2) = 0.008569045 Of course, the degree of rounding may be different -- for example rounding to *even* values of the first decimal place, i.e. to values ... , 2.0, 2.2, 2.4, 2.6, 2.8, ... in which case the event whose probability you want is that the sampled value is between 2.3 and 2.5, whose probability is pnorm(2.5, 5, 2) - pnorm(2.3, 5, 2) = 0.01714178 Since the question you are really interested in cannot be identified from what you have asked (see examples above), you should try to make your question clear and definite! Hoping this helps, Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 06-Nov-10 Time: 12:06:17 -- 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.
[R] Extracting elements of a particular slot from S4 object
Hi there, can anyone tell me how to extract to values of a particular slot for some S4 object? Let take following example: library(fOptions) val -GBSOption(TypeFlag = c, S = 60, X = 65, Time = 1/4, r = 0.08, b = 0.08, sigma = 0.30) val Title: Black Scholes Option Valuation Call: GBSOption(TypeFlag = c, S = 60, X = 65, Time = 1/4, r = 0.08, b = 0.08, sigma = 0.3) Parameters: Value: TypeFlag c S60 X65 Time 0.25 r0.08 b0.08 sigma0.3 Option Price: 2.133372 Description: Sat Nov 06 19:25:39 2010 Here I have tried with following however slapped with some error: val@Option Price Error: no slot of name Option Price for this object of class fOPTION What is the ideal way to do that? Thanks, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting elements of a particular slot from S4 object
Try this: v...@price On Sat, Nov 6, 2010 at 11:41 AM, Megh Dal megh700...@yahoo.com wrote: Hi there, can anyone tell me how to extract to values of a particular slot for some S4 object? Let take following example: library(fOptions) val -GBSOption(TypeFlag = c, S = 60, X = 65, Time = 1/4, r = 0.08, b = 0.08, sigma = 0.30) val Title: Black Scholes Option Valuation Call: GBSOption(TypeFlag = c, S = 60, X = 65, Time = 1/4, r = 0.08, b = 0.08, sigma = 0.3) Parameters: Value: TypeFlag c S60 X65 Time 0.25 r0.08 b0.08 sigma0.3 Option Price: 2.133372 Description: Sat Nov 06 19:25:39 2010 Here I have tried with following however slapped with some error: val@Option Price Error: no slot of name Option Price for this object of class fOPTION What is the ideal way to do that? Thanks, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] Removing NA in ggplot
Hi list, I just got stuck with this one: In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F and NA). Then getting some nice plots using ggplot(data, aes(age[na.exclude(gender)])) + geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = lightblue ) + facet_grid( gender ~ .) I am trying to get a faceted graph of age distribution excluding the NA data for gender Unfortunately I end up with the error message: Error in data.frame(..., check.names = FALSE) : arguments imply differing number of rows: 206, 219 How do i Wash out NA's in this situation? Regards ottar [[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] Removing NA in ggplot
Create a subset of your data that excludes the NAs before you feed it to ggplot. Ottar Kvindesland ottar.kvindesl...@gmail.com wrote: Hi list, I just got stuck with this one: In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F and NA). Then getting some nice plots using ggplot(data, aes(age[na.exclude(gender)])) + geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = lightblue ) + facet_grid( gender ~ .) I am trying to get a faceted graph of age distribution excluding the NA data for gender Unfortunately I end up with the error message: Error in data.frame(..., check.names = FALSE) : arguments imply differing number of rows: 206, 219 How do i Wash out NA's in this situation? Regards ottar [[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. --- 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. __ 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] About 5.1 Arrays
Hi Joshua, Thanks for your advice. 1) Re your advice:-[quote] a3d , , 1 --- this is the first position of the third dimension [,1] [,2] [,3] [,4] --- positions 1, 2, 3, 4 of the second dimension [1,]147 10 [2,]258 11 [3,]369 12 ^ the first dimension , , 2 --- the second position of the third dimension ... [/quote] Where is the third dimension? 2) Re your advice:-[quote] so you can think that in the original vector a: 1 maps to a[1, 1, 1] in the 3d array 2 maps to a[2, 1, 1]. 3 maps to a[3, 1, 1] 4 maps to a[1, 2, 1] 12 maps to a[3, 4, 1] 20 maps to a[2, 3, 2] 24 maps to a[3, 4, 2] [/quote] My finding; # 1 maps to a[1, 1, 1] in the 3d array a3d - array(a, dim = c(1, 1, 1)) a3d , , 1 [,1] [1,]1 Correct # 2 maps to a[2, 1, 1]. a3d - array(a, dim = c(2, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 Correct # 3 maps to a[3, 1, 1] a3d - array(a, dim = c(3, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 [3,]3 Correct # 4 maps to a[1, 2, 1] a3d - array(a, dim = c(1, 2, 1)) a3d , , 1 [,1] [,2] [1,]12 Incorrect. It is 2 # 12 maps to a[3, 4, 1] a3d - array(a, dim = c(3, 4, 1)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 Correct # 20 maps to a[2, 3, 2] a3d - array(a, dim = c(2, 3, 2)) a3d , , 1 [,1] [,2] [,3] [1,]135 [2,]246 , , 2 [,1] [,2] [,3] [1,]79 11 [2,]8 10 12 Incorrect. It is 12 # 24 maps to a[3, 4, 2] a3d - array(a, dim = c(3, 4, 2)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 , , 2 [,1] [,2] [,3] [,4] [1,] 13 16 19 22 [2,] 14 17 20 23 [3,] 15 18 21 24 Correct. If I'm wrong, pls correct me. Thanks B.R. Stephen - Original Message From: Joshua Wiley jwiley.ps...@gmail.com To: Stephen Liu sati...@yahoo.com Cc: r-help@r-project.org Sent: Sat, November 6, 2010 12:48:27 AM Subject: Re: [R] About 5.1 Arrays On Fri, Nov 5, 2010 at 9:17 AM, Stephen Liu sati...@yahoo.com wrote: Hi Daniel, Thanks for your detail advice. I completely understand your explain. But I can't resolve what does a stand for there? the a just represents some vector. It is the name of the object that stores your data. Like you might tell someone to go look in a book to find some information. a[1,1,1] is 1 * 1 * 1 = 1 a[2,1,1] is 2 * 1 * 1 = 2 a[2,4,2] is 2 * 4 * 2 = 16 a[3,4,2] is 3 * 4 * 2 = 24 That is the basic idea, but it may not be the most helpful way to think of it because it depends on the length of the each dimension. For example a[1, 2, 1] is not 1 * 2 * 1 = 2 a[1, 1, 2] is not 1 * 1 * 2 = 2 in the little 3d array I show below, it would actually be a[1, 2, 1] = 4 a[1, 1, 2] = 13 ? B.R. Stephen L - Original Message From: Daniel Nordlund djnordl...@frontier.com To: r-help@r-project.org Sent: Fri, November 5, 2010 11:54:15 PM Subject: Re: [R] About 5.1 Arrays -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Stephen Liu Sent: Friday, November 05, 2010 7:57 AM To: Steve Lianoglou Cc: r-help@r-project.org Subject: Re: [R] About 5.1 Arrays Hi Steve, It's not clear what you're having problems understanding. By setting the dim attribute of your (1d) vector, you are changing itsdimenensions. I'm following An Introduction to R to learn R On 5.1 Arrays http://cran.r-project.org/doc/manuals/R-intro.html#Vectors-and-assignment It mentions:- ... For example if the dimension vector for an array, say a, is c(3,4,2) then there are 3 * 4 * 2 = 24 entries in a and the data vector holds them in the order a[1,1,1], a[2,1,1], ..., a[2,4,2], a[3,4,2]. I don't understand on =24 entries in a and the data vector holds them in the order a[1,1,1], a[2,1,1], ..., a[2,4,2], a[3,4,2]. the order a[1,1,1], a[2,1,1], ..., a[2,4,2], a[3,4,2]? What does it mean the order a[1,1,1], a[2,1,1], ..., a[2,4,2], a[3,4,2]? because it is actually stored as a 1 dimensional vector, it is just telling you the order. For example, given some vector a that contains the numbers 1 through 24, you could reshape this into a three dimensional object. It would be stored like: # make a vector a and an array (built from a) called a3d a - 1:24 a3d - array(a, dim = c(3, 4, 2)) a [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 a3d , , 1 --- this is the first position of the third dimension [,1] [,2] [,3] [,4] --- positions 1, 2, 3, 4 of the second dimension [1,]147 10 [2,]258 11 [3,]369 12 ^ the first dimension , , 2 --- the second position of the third dimension [,1] [,2] [,3] [,4] [1,] 13 16 19 22 [2,] 14 17 20 23 [3,] 15 18 21 24 a[1, 1, 1] is the first element of dimension
Re: [R] anova(lme.model)
Sounds to me like you should really be seeking help from your local statistician, not this list. What you request probably cannot be done. What is wrong with what you get from lme, whose results seem fairly clear whether the P values are accurate or not? Cheers, Bert On Sat, Nov 6, 2010 at 4:04 AM, Sibylle Stöckli sibylle.stoec...@gmx.ch wrote: Dear R users Topic: Linear effect model fitting using the nlme package (recomended by Pinheiro et al. 2008 for unbalanced data set). The R help provides much info about the controversy to use the anova(lme.model) function to present numerator df and F values. Additionally different p-values calculated by lme and anova are reported. However, I come across the same problem, and I would very much appreciate some R help to fit an anova function to get similar p-values compared to the lme function and additionally to provide corresponding F-values. I tried to use contrasts and to deal with the ‚unbalanced data set’. Thanks Sibylle Kaltenborn-read.table(Kaltenborn_YEARS.txt, na.strings=*, header=TRUE) library(nlme) model5c-lme(asin(sqrt(PropMortality))~Diversity+ Management+Species+Height+Height*Diversity, data=Kaltenborn, random=~1|Plot/SubPlot, na.action=na.omit, weights=varPower(form=~Diversity), subset=Kaltenborn$ADDspecies!=1, method=ML) summary(model5c) Linear mixed-effects model fit by maximum likelihood Data: Kaltenborn Subset: Kaltenborn$ADDspecies != 1 AIC BIC logLik -249.3509 -205.4723 137.6755 Random effects: Formula: ~1 | Plot (Intercept) StdDev: 0.06162279 Formula: ~1 | SubPlot %in% Plot (Intercept) Residual StdDev: 0.03942785 0.05946185 Variance function: Structure: Power of variance covariate Formula: ~Diversity Parameter estimates: power 0.7302087 Fixed effects: asin(sqrt(PropMortality)) ~ Diversity + Management + Species + Height + Height * Diversity Value Std.Error DF t-value p-value (Intercept) 0.5422893 0.05923691 163 9.154585 0. Diversity -0.0734688 0.02333159 14 -3.148896 0.0071 Managementm+ 0.0217734 0.02283375 30 0.953562 0.3479 Managementu -0.0557160 0.02286694 30 -2.436532 0.0210 SpeciesPab -0.2058763 0.02763737 163 -7.449198 0. SpeciesPm 0.0308005 0.02827782 163 1.089210 0.2777 SpeciesQp 0.0968051 0.02689327 163 3.599602 0.0004 Height -0.0017579 0.00031667 163 -5.551251 0. Diversity:Height 0.0005122 0.00014443 163 3.546270 0.0005 Correlation: (Intr) Dvrsty Mngmn+ Mngmnt SpcsPb SpcsPm SpcsQp Height Diversity -0.867 Managementm+ -0.173 -0.019 Managementu -0.206 0.005 0.499 SpeciesPab -0.253 0.085 0.000 0.035 SpeciesPm -0.239 0.058 0.001 0.064 0.521 SpeciesQp -0.250 0.041 -0.001 0.032 0.502 0.506 Height -0.518 0.532 -0.037 -0.004 0.038 0.004 0.033 Diversity:Height 0.492 -0.581 0.031 -0.008 -0.149 -0.099 -0.069 -0.904 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -2.99290873 -0.60522612 -0.05756772 0.62163049 2.80811502 Number of Observations: 216 Number of Groups: Plot SubPlot %in% Plot 16 48 anova(model5c) numDF denDF F-value p-value (Intercept) 1 163 244.67887 .0001 Diversity 1 14 1.53025 0.2364 Management 2 30 6.01972 0.0063 Species 3 163 51.86699 .0001 Height 1 163 30.08090 .0001 Diversity:Height 1 163 12.57603 0.0005 -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics __ 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] Where to get rcom for Linux
Hi folks, Debian 600 64-bit Is rcom for Linux available? rcom rcom: R COM Client Interface and internal COM Server http://cran.r-project.org/web/packages/rcom/index.html If YES please advise where to get it. TIA B.R. Stephen L __ 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] Where to get rcom for Linux
isn't COM a Windows-only technology? Shige On Sat, Nov 6, 2010 at 12:12 PM, Stephen Liu sati...@yahoo.com wrote: Hi folks, Debian 600 64-bit Is rcom for Linux available? rcom rcom: R COM Client Interface and internal COM Server http://cran.r-project.org/web/packages/rcom/index.html If YES please advise where to get it. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] compare and replace
Hello R Users, I'm wondering if there exists any elegant and simple way to do the following: I have a data.frame X fills in numbers. I have a vector y with numbers as well. Every value in X that is equal to any values in y should be replaced by e.g. 1. Any idea? Robert __ 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 and replace
Robert Ruser robert.ruser at gmail.com writes: Hello R Users, I'm wondering if there exists any elegant and simple way to do the following: I have a data.frame X fills in numbers. I have a vector y with numbers as well. Every value in X that is equal to any values in y should be replaced by e.g. 1. Any idea? Robert If the data frames are completely filled with numbers you should be able to operate on them as matrices (?as.matrix, ?as.data.frame). X - matrix(1:16,nrow=4) Y - matrix(rep(2:5,4),nrow=4) (X[X %in% c(Y)] - 1) __ 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 and replace
On 11/06/2010 11:36 AM, Robert Ruser wrote: Hello R Users, I'm wondering if there exists any elegant and simple way to do the following: I have a data.frame X fills in numbers. I have a vector y with numbers as well. Every value in X that is equal to any values in y should be replaced by e.g. 1. Any idea? Keep in mind FAQ 7.31 when programming this... __ 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] table with values as dots in increasing sizes
This is a tableplot, available on R-Forge at https://r-forge.r-project.org/projects/tableplot/ install.packages(tableplot, repos=http://R-Forge.R-project.org;) will install, as long as you are using R 2.12.x; otherwise, you'll have to download the source package and install from source. -Michael On 11/5/2010 4:45 AM, fugelpitch wrote: I was just thinking of a way to present data and if it is possible in R. I have a data frame that looks as follows (this is just mockup data). df location,species1,species2,species3,species4,species5 loc1,0.44,0.28,0.37,-0.24,0.41 loc2,0.54,0.62,0.34,0.52,0.71 loc3,-0.33,0.75,-0.34,0.48,0.61 location is a factor while all the species are numerical vectors. I would like to present this as a table (or something that looks like a table) but instead of the numbers I would like to present circles (pch = 19) that increases in size with increasing number. Is it also possible to make it change color if the value is negative. (E.g. larger blue circles represent larger +values while larger red circles represent larger -values)? Jonas __ 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 and replace
Thank you vary much Ben and Erik. It's exactly what I want. Below is my a little modified example. set.seed(12345) X = sample(c(40:60),40,replace=TRUE) x = matrix(X,nc=5) y = c(40,43,55,60) x[x %in% y] - -1 __ 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] About 5.1 Arrays
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Stephen Liu Sent: Saturday, November 06, 2010 7:38 AM To: Joshua Wiley Cc: r-help@r-project.org Subject: Re: [R] About 5.1 Arrays Hi Joshua, Thanks for your advice. 1) Re your advice:-[quote] a3d , , 1 --- this is the first position of the third dimension [,1] [,2] [,3] [,4] --- positions 1, 2, 3, 4 of the second dimension [1,]147 10 [2,]258 11 [3,]369 12 ^ the first dimension , , 2 --- the second position of the third dimension ... [/quote] Where is the third dimension? 2) Re your advice:-[quote] so you can think that in the original vector a: 1 maps to a[1, 1, 1] in the 3d array 2 maps to a[2, 1, 1]. 3 maps to a[3, 1, 1] 4 maps to a[1, 2, 1] 12 maps to a[3, 4, 1] 20 maps to a[2, 3, 2] 24 maps to a[3, 4, 2] [/quote] My finding; # 1 maps to a[1, 1, 1] in the 3d array a3d - array(a, dim = c(1, 1, 1)) a3d , , 1 [,1] [1,]1 Correct # 2 maps to a[2, 1, 1]. a3d - array(a, dim = c(2, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 Correct # 3 maps to a[3, 1, 1] a3d - array(a, dim = c(3, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 [3,]3 Correct # 4 maps to a[1, 2, 1] a3d - array(a, dim = c(1, 2, 1)) a3d , , 1 [,1] [,2] [1,]12 Incorrect. It is 2 # 12 maps to a[3, 4, 1] a3d - array(a, dim = c(3, 4, 1)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 Correct # 20 maps to a[2, 3, 2] a3d - array(a, dim = c(2, 3, 2)) a3d , , 1 [,1] [,2] [,3] [1,]135 [2,]246 , , 2 [,1] [,2] [,3] [1,]79 11 [2,]8 10 12 Incorrect. It is 12 # 24 maps to a[3, 4, 2] a3d - array(a, dim = c(3, 4, 2)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 , , 2 [,1] [,2] [,3] [,4] [1,] 13 16 19 22 [2,] 14 17 20 23 [3,] 15 18 21 24 Correct. If I'm wrong, pls correct me. Thanks B.R. Stephen Stephen, I am correcting you. :-) You are using dim() incorrectly, and not accessing the array correctly. In all of your examples you should be using dim(3,4,2). Then you need to specify the indexes of the array element you want to look at. So, to use your example a-1:24 a3d - array(a, dim = c(3,4,2)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 , , 2 [,1] [,2] [,3] [,4] [1,] 13 16 19 22 [2,] 14 17 20 23 [3,] 15 18 21 24 # 1 maps to a[1, 1, 1] in the 3d array a3d[1, 1, 1] [1] 1 # 2 maps to a[2, 1, 1]. a3d[2, 1, 1] [1] 2 # 3 maps to a[3, 1, 1] a3d[3, 1, 1] [1] 3 # 4 maps to a[1, 2, 1] a3d[1, 2, 1] [1] 4 Hope this is helpful, Dan Daniel Nordlund Bothell, WA USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] anova(lme.model)
Date: Sat, 6 Nov 2010 07:45:26 -0700 From: gunter.ber...@gene.com To: sibylle.stoec...@gmx.ch CC: r-help@r-project.org Subject: Re: [R] anova(lme.model) Sounds to me like you should really be seeking help from your local statistician, not this list. What you request probably cannot be done. I'm still bringing my install up to speed so I can't immediately read the cited R stuff below but it sounds like the OP mentions a controversy documented in the R packages. Is there a list for discussing these topics? Offhand that seems legitimate for a user help list unless you want people to believe that it came out of a computer so it must be right, whatever a P value is. What is wrong with what you get from lme, whose results seem fairly clear whether the P values are accurate or not? Cheers, Bert On Sat, Nov 6, 2010 at 4:04 AM, Sibylle Stöckli wrote: Dear R users Topic: Linear effect model fitting using the nlme package (recomended by Pinheiro et al. 2008 for unbalanced data set). The R help provides much info about the controversy to use the anova(lme.model) function to present numerator df and F values. Additionally different p-values calculated by lme and anova are reported. However, I come across the same problem, and I would very much appreciate some R help to fit an anova function to get similar p-values compared to the lme function and additionally to provide corresponding F-values. I tried to use contrasts and to deal with the ‚unbalanced data set’. Thanks Sibylle Kaltenborn-read.table(Kaltenborn_YEARS.txt, na.strings=*, header=TRUE) library(nlme) model5c-lme(asin(sqrt(PropMortality))~Diversity+ Management+Species+Height+Height*Diversity, data=Kaltenborn, random=~1|Plot/SubPlot, na.action=na.omit, weights=varPower(form=~Diversity), subset=Kaltenborn$ADDspecies!=1, method=ML) summary(model5c) Linear mixed-effects model fit by maximum likelihood Data: Kaltenborn Subset: Kaltenborn$ADDspecies != 1 AIC BIC logLik -249.3509 -205.4723 137.6755 Random effects: Formula: ~1 | Plot (Intercept) StdDev: 0.06162279 Formula: ~1 | SubPlot %in% Plot (Intercept) Residual StdDev: 0.03942785 0.05946185 Variance function: Structure: Power of variance covariate Formula: ~Diversity Parameter estimates: power 0.7302087 Fixed effects: asin(sqrt(PropMortality)) ~ Diversity + Management + Species + Height + Height * Diversity Value Std.Error DF t-value p-value (Intercept) 0.5422893 0.05923691 163 9.154585 0. Diversity-0.0734688 0.02333159 14 -3.148896 0.0071 Managementm+ 0.0217734 0.02283375 30 0.953562 0.3479 Managementu -0.0557160 0.02286694 30 -2.436532 0.0210 SpeciesPab -0.2058763 0.02763737 163 -7.449198 0. SpeciesPm 0.0308005 0.02827782 163 1.089210 0.2777 SpeciesQp 0.0968051 0.02689327 163 3.599602 0.0004 Height -0.0017579 0.00031667 163 -5.551251 0. Diversity:Height 0.0005122 0.00014443 163 3.546270 0.0005 Correlation: (Intr) Dvrsty Mngmn+ Mngmnt SpcsPb SpcsPm SpcsQp Height Diversity-0.867 Managementm+ -0.173 -0.019 Managementu -0.206 0.005 0.499 SpeciesPab -0.253 0.085 0.000 0.035 SpeciesPm-0.239 0.058 0.001 0.064 0.521 SpeciesQp-0.250 0.041 -0.001 0.032 0.502 0.506 Height -0.518 0.532 -0.037 -0.004 0.038 0.004 0.033 Diversity:Height 0.492 -0.581 0.031 -0.008 -0.149 -0.099 -0.069 -0.904 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -2.99290873 -0.60522612 -0.05756772 0.62163049 2.80811502 Number of Observations: 216 Number of Groups: Plot SubPlot %in% Plot 1648 anova(model5c) numDF denDF F-value p-value (Intercept) 1 163 244.67887 .0001 Diversity114 1.53025 0.2364 Management 230 6.01972 0.0063 Species 3 163 51.86699 .0001 Height 1 163 30.08090 .0001 Diversity:Height 1 163 12.57603 0.0005 -- Bert Gunter Genentech Nonclinical Biostatistics Mike Marchywka | V.P. Technology 415-264-8477 marchy...@phluant.com Online Advertising and Analytics for Mobile http://www.phluant.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] 3-way interaction simple slopes
Can anyone show me how to test for significant simple slopes of a 3-way interaction, with covariates. my equation tmod-(glm(PCL~ rank.f + gender.f + MONTHS + CEXPOSE.M + bf.m + MONTHS*CEXPOSE.M*bf.m, data=mhatv, family=gaussian ,na.action=na.omit)) Thank you Mike [[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] Using changing names in loop in R
Hello everybody,
I have usually solved this problem by repeating lines of codes instead of a
loop, but it's such a waste of time, I thought I should really learn how to
do it with loops:
What I want to do:
Say, I have several data files that differ only in a number, e.g. data
points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
manipulate them
e.g. a simple sum of several data points
data - c(NA,n)
for (i in 1:n){
data[i] - Data_i + Data_[i-1]
}
I know that the above code doesn't work, and I don't want to combine the
files into one vector to solve the problem etc. - I would just like to know
who make sure R recognizes the extension _i. I have the same problem for
say, reading in datafiles that only differ by one digit in the extension,
and I want to (instead of repeating code) combine the process in a loop.
I hope I made myself clear to what my problem is.
Thanks for your help,
//F
--
View this message in context:
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Re: [R] Using changing names in loop in R
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Tuatara
Sent: Saturday, November 06, 2010 9:22 AM
To: r-help@r-project.org
Subject: [R] Using changing names in loop in R
Hello everybody,
I have usually solved this problem by repeating lines of codes instead of
a
loop, but it's such a waste of time, I thought I should really learn how
to
do it with loops:
What I want to do:
Say, I have several data files that differ only in a number, e.g. data
points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
manipulate them
e.g. a simple sum of several data points
data - c(NA,n)
for (i in 1:n){
data[i] - Data_i + Data_[i-1]
}
I know that the above code doesn't work, and I don't want to combine the
files into one vector to solve the problem etc. - I would just like to
know
who make sure R recognizes the extension _i. I have the same problem for
say, reading in datafiles that only differ by one digit in the extension,
and I want to (instead of repeating code) combine the process in a loop.
I hope I made myself clear to what my problem is.
Thanks for your help,
This is one of those cases where a commented, self-contained, reproducible
example would be very helpful in helping you. You mention you have several
data files, but I see no reference to data files in your code. Did you mean
data frames? What is Data_i? A data frame or something else?
You said you normally do this by repeating lines of code. Can you show us a
simple example? Someone should be able to show you how to optimize it.
Dan
Daniel Nordlund
Bothell, WA USA
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and provide commented, minimal, self-contained, reproducible code.
[R] Prettier axis labels when using log scales in Lattice
Hello, I am trying to alter the way in which lattice functions (specifically xyplot) print the axis labels when one uses the 'scales' parameter. I can obtain the effect I want by using scales=list(y=list(log=10, labels=expression(yvalues))) where yvalues are the values that would have been printed as the y-axis labels if the labels argument had not been present. To help clarify what I am looking for, compare the first of the following plots with the second: data(iris) xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10))) xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10, labels=expression(10^0.65,10^0.7,10^0.75,10^0.8,10^0.85,10^0.85,10^0.9 The second is the effect I am trying to achieve. Is there a way to do this without explicitly entering the expressions to be printed on the y-axis? thanks, Marc Paterno __ 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] Prettier axis labels when using log scales in Lattice
On Sat, Nov 6, 2010 at 3:02 PM, Marc Paterno pate...@fnal.gov wrote: Hello, I am trying to alter the way in which lattice functions (specifically xyplot) print the axis labels when one uses the 'scales' parameter. I can obtain the effect I want by using scales=list(y=list(log=10, labels=expression(yvalues))) where yvalues are the values that would have been printed as the y-axis labels if the labels argument had not been present. To help clarify what I am looking for, compare the first of the following plots with the second: data(iris) xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10))) xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10, labels=expression(10^0.65,10^0.7,10^0.75,10^0.8,10^0.85,10^0.85,10^0.9 The second is the effect I am trying to achieve. Is there a way to do this without explicitly entering the expressions to be printed on the y-axis? Try: library(latticeExtra) xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10)), yscale.components = yscale.components.logpower) -- 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] Using changing names in loop in R
On Sat, Nov 6, 2010 at 5:22 PM, Tuatara franziskabro...@gmail.com wrote:
Hello everybody,
I have usually solved this problem by repeating lines of codes instead of a
loop, but it's such a waste of time, I thought I should really learn how to
do it with loops:
Would the following construct help?
for(i in 1:10) assign(paste('x', i, sep=''), c(i:10))
ls()
[1] ipkg tbbt x1 x10 x2 x3 x4 x5 x6
[11] x7 x8 x9
for(i in 1:10) print(get(paste('x', i, sep='')))
[1] 1 2 3 4 5 6 7 8 9 10
[1] 2 3 4 5 6 7 8 9 10
[1] 3 4 5 6 7 8 9 10
[1] 4 5 6 7 8 9 10
[1] 5 6 7 8 9 10
[1] 6 7 8 9 10
[1] 7 8 9 10
[1] 8 9 10
[1] 9 10
[1] 10
Read ?assign, ?get, but also this fortune:
fortune('assign')
The only people who should use the assign function are those who fully
understand why you should never use the assign function.
-- Greg Snow
R-help (July 2009)
I haven't yet figured out why I should heed this. Regards
Liviu
What I want to do:
Say, I have several data files that differ only in a number, e.g. data
points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
manipulate them
e.g. a simple sum of several data points
data - c(NA,n)
for (i in 1:n){
data[i] - Data_i + Data_[i-1]
}
I know that the above code doesn't work, and I don't want to combine the
files into one vector to solve the problem etc. - I would just like to know
who make sure R recognizes the extension _i. I have the same problem for
say, reading in datafiles that only differ by one digit in the extension,
and I want to (instead of repeating code) combine the process in a loop.
I hope I made myself clear to what my problem is.
Thanks for your help,
//F
--
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http://www.alienetworks.com/srtest.cfm
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Do you know how to write?
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Re: [R] anova(lme.model)
Mike Marchywka marchywka at hotmail.com writes: Date: Sat, 6 Nov 2010 07:45:26 -0700 From: gunter.berton at gene.com To: sibylle.stoeckli at gmx.ch CC: r-help at r-project.org Subject: Re: [R] anova(lme.model) Sounds to me like you should really be seeking help from your local statistician, not this list. What you request probably cannot be done. I'm still bringing my install up to speed so I can't immediately read the cited R stuff below but it sounds like the OP mentions a controversy documented in the R packages. Is there a list for discussing these topics? Offhand that seems legitimate for a user help list unless you want people to believe that it came out of a computer so it must be right, whatever a P value is. It's not documented within the packages themselves, it's documented in the r-sig-mixed-models archives, and at http://rwiki.sciviews.org/doku.php?id=guides:lmer-tests, and in the R FAQ http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-displayed-when-using-lmer_0028_0029_003f and here https://stat.ethz.ch/pipermail/r-help/2006-May/094765.html cheers Ben Bolker __ 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] Hashing and environments
Hi,
I'm trying to write a general-purpose lexicon class and associated methods
for storing and accessing information about large numbers of specific words
(e.g., their frequencies in different genres). Crucial to making such a class
practically useful is to get hashing working correctly so that information
about specific words can be accessed quickly. But I've never really understood
very well how hashing works, so I'm having trouble.
Here is an example of what I've done so far:
***
setClass(Lexicon,representation(e=environment))
setMethod(initialize,Lexicon,function(.Object,wfreqs) {
.obj...@e - new.env(hash=T,parent=emptyenv())
assign(wfreqs,wfreqs,envir=.obj...@e)
return(.Object)
})
## function to access word frequencies
wfreq - function(lexicon,word) {
return(get(wfreqs,envir=lexi...@e)[word])
}
## example of use
my.lexicon - new(Lexicon,wfreqs=c(the=2,person=1))
wfreq(my.lexicon,the)
***
However, testing indicates that the way I have set this up does not achieve the
intended benefits of having the environment hashed:
***
sample.wfreqs - trunc(runif(1e5,max=100))
names(sample.wfreqs) - as.character(1:length(sample.wfreqs))
lex - new(Lexicon,wfreqs=sample.wfreqs)
words.to.lookup - trunc(runif(100,min=1,max=1e5))
## look up the words directly from the sample.wfreqs vector
system.time({
for(i in words.to.lookup)
sample.wfreqs[as.character(i)]
},gcFirst=TRUE)
## look up the words through the wfreq() function; time approx the same
system.time({
for(i in words.to.lookup)
wfreq(lex,as.character(i))
},gcFirst=TRUE)
***
I'm guessing that the problem is that the indexing of the wfreqs vector in my
wfreq() function is not happening inside the actual lexicon's environment.
However, I have not been able to figure out the proper call to get the lookup
to happen inside the lexicon's environment. I've tried
wfreq1 - function(lexicon,word) {
return(eval(wfreqs[word],envir=lexi...@e))
}
which I'd thought should work, but this gives me an error:
wfreq1(my.lexicon,'the')
Error in eval(wfreqs[word], envir = lexi...@e) :
object 'wfreqs' not found
Any advice would be much appreciated!
Best many thanks in advance,
Roger
--
Roger Levy Email: rl...@ucsd.edu
Assistant Professor Phone: 858-534-7219
Department of Linguistics Fax: 858-534-4789
UC San DiegoWeb: http://ling.ucsd.edu/~rlevy
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[R] SMATR common slopes test
Hi All, I am confused with SMATR's test for common slope. My null hypothesis here is that all slopes are parallel (common slopes?), right? So if I get a p value 0.05 means that we can have confidence to reject it? That slopes are different? Or the other way around? it means that we have statistical confidence that the slopes are parallel? thanks -- Eugenio Larios PhD Student University of Arizona. Ecology Evolutionary Biology. (520) 481-2263 elari...@email.arizona.edu [[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] saddle points in optim
Hi, I've been trying to use optim to minimise least squares for a function, and then get a guess at the error using the hessian matrix (calculated from numDeriv::hessian, which I read in some other r-help post was meant to be more accurate than the hessian given in optim). To get the standard error estimates, I'm calculating sqrt(diag(solve(x))), hope that's correct. I've found that using numDeriv's hessian gets me some NaNs for errors, whereas the one from optim gets me numbers for all parameters. If I look for eigenvalues for numDeriv::hessian, I get two negative numbers (and six positive - I'm fitting to eight parameters), so does this mean that optim hasn't converged correctly, and has hit a saddle point? If so, is there any way I could assist it to find the minimum? Thanks, Jon Phillips __ 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] I want to call some R function in C code
Hi, all: I want the user to give a function g as the parameter of our function f. In function f, we use user's function g to compute something. Since our function f is implemented in C++, how do I further pass this function to C++ code? Thank you very much! best Liping LIU [[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] likelyhood maximization problem with polr
blackscorpio wrote: Dear community, I am currently trying to fit an ordinal logistic regression model with the polr function. I often get the same error message : attempt to find suitable starting values failed, for example with : ... Does anyone have a clue ? Yes. The code that generates a starting value for the optimization is flakey. lrm in the Design library appear to be more robust. -- View this message in context: http://r.789695.n4.nabble.com/likelyhood-maximization-problem-with-polr-tp2528818p3030397.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 generate multivariate uniform distribution random numbers?
I wish to generate 100 by 1 vector of x1 and x2 both are uniform distributed with covariance matrix \Sigma. Thanks, Michael [[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] bugs and misfeatures in polr(MASS).... fixed!
The start value generation code in polr is also known to fail quite
frequently. For example, against the Iris data as recently posted to this
list by blackscorpio ( Sep 6, 2010).
polr(Species~Sepal_Length+Sepal_Width+Petal_Length+Petal_Width,data=iris)
Error in polr(Species ~ Sepal_Length + Sepal_Width + Petal_Length +
Petal_Width, :
attempt to find suitable starting values failed
In addition: Warning messages:
1: In glm.fit(X, y1, wt, family = binomial(), offset = offset) :
algorithm did not converge
2: In glm.fit(X, y1, wt, family = binomial(), offset = offset) :
fitted probabilities numerically 0 or 1 occurred
I suggest that simply setting the coefficients beta to zero and the
cutpoints zeta to sensible values will always produce a feasible starting
point for non-pathological data. Here is my code that does this:
if(missing(start)) {
# try something that should always work -tjb
u - as.integer(table(y))
u - (cumsum(u)/sum(u))[1:q]
zetas -
switch(method,
logistic= qlogis(u),
probit= qnorm(u),
cauchit= qcauchy(u),
cloglog= -log(-log(u)) )
s0 - c(rep(0,pc),zetas[1],log(diff(zetas)))
Using this start code the problem is not manifested.
source('fixed-polr.R')
polr(Species~Sepal_Length+Sepal_Width+Petal_Length+Petal_Width,data=iris)
Call:
polr(formula = Species ~ Sepal_Length + Sepal_Width + Petal_Length +
Petal_Width, data = iris)
Coefficients:
Sepal_Length Sepal_Width Petal_Length Petal_Width
-2.466724-6.671515 9.43168918.270058
Intercepts:
setosa|versicolor versicolor|virginica
4.08018942.639320
Residual Deviance: 11.89857
AIC: 23.89857
My change would also likely fix the problem reported by Kevin Coombes on May
6, 2010.
--
View this message in context:
http://r.789695.n4.nabble.com/bugs-and-misfeatures-in-polr-MASS-fixed-tp3024677p3030405.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.
Re: [R] bugs and misfeatures in polr(MASS).... fixed!
Note that that the enhancements in my original post solve the unresolved
problem of Chaehyung Ahn (22 Mar 2005) whose data I reproduce:
y,x,lx
0,3.2e-02,-1.49485
0,3.2e-02,-1.49485
0,1.0e-01,-1.0
0,1.0e-01,-1.0
0,3.2e-01,-0.49485
0,3.2e-01,-0.49485
1,1.0e+00,0.0
0,1.0e+00,0.0
1,3.2e+00,0.50515
1,3.2e+00,0.50515
0,1.0e+01,1.0
1,1.0e+01,1.0
1,3.2e+01,1.50515
2,3.2e+01,1.50515
2,1.0e+02,2.0
1,1.0e+02,2.0
2,3.2e+02,2.50515
1,3.2e+02,2.50515
2,1.0e+03,3.0
2,1.0e+03,3.0
Using the MASS version we get
ahn$y-as.factor(ahn$y)
summary(polr(y~lx,data=ahn))
Re-fitting to get Hessian
Error in optim(s0, fmin, gmin, method = BFGS, hessian = Hess, ...) :
initial value in 'vmmin' is not finite
Whereas,
source('fixed-polr.R')
summary(polr(y~lx,data=ahn))
Re-fitting to get Hessian
Call:
polr(formula = y ~ lx, data = ahn)
Coefficients:
Value Std. Error t value
lx 2.421 0.8146 2.971
Intercepts:
Value Std. Error t value
0|1 0.5865 0.8118 0.7224
1|2 4.8966 1.7422 2.8106
Residual Deviance: 20.43286
AIC: 26.43286
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Hashing and environments
I would make make an environemnt called wfreqsEnv
whose entry names are your words and whose entry
values are the information about the words. I find
it convenient to use [[ to make it appear to be
a list (instead of using exists(), assign(), and get()).
E.g., the following enters the 100,000 words from a
list of 17,576 and records their id numbers and the
number of times each is found in the sample.
wfreqsEnv - new.env(hash=TRUE, parent = emptyenv())
words - do.call(paste, c(list(sep=), expand.grid(LETTERS,
letters, letters)))
# length(words) == 17576
set.seed(1)
samp - sample(seq_along(words), size=10, replace=TRUE)
system.time(for(i in samp) {
+word - words[i]
+if (is.null(wfreqsEnv[[word]])) { # new entry
+wfreqsEnv[[word]] - list(Count=1, EntryNo=i)
+} else { # update existing entry
+wfreqsEnv[[word]]$Count - wfreqsEnv[[word]]$Count + 1
+}
+})
user system elapsed
2.280.002.14
(The time, in seconds, is from an ancient Windows laptop, c. 2002.)
Here is a small check that we are getting what we expect:
words[14736]
[1] Tuv
wfreqsEnv[[Tuv]]
$Count
[1] 8
$EntryNo
[1] 14736
sum(samp==14736)
[1] 8
If we do this with a non-hashed environment we get the same
answers but the elapsed time is now 34.81 seconds instead of
2.14. If you make wfreqEnv be a list instead of an environment
then that time is 74.12 seconds (and the answers are the same).
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 Levy, Roger
Sent: Saturday, November 06, 2010 1:39 PM
To: r-help@r-project.org
Subject: [R] Hashing and environments
Hi,
I'm trying to write a general-purpose lexicon class and
associated methods for storing and accessing information
about large numbers of specific words (e.g., their
frequencies in different genres). Crucial to making such a
class practically useful is to get hashing working correctly
so that information about specific words can be accessed
quickly. But I've never really understood very well how
hashing works, so I'm having trouble.
Here is an example of what I've done so far:
***
setClass(Lexicon,representation(e=environment))
setMethod(initialize,Lexicon,function(.Object,wfreqs) {
.obj...@e - new.env(hash=T,parent=emptyenv())
assign(wfreqs,wfreqs,envir=.obj...@e)
return(.Object)
})
## function to access word frequencies
wfreq - function(lexicon,word) {
return(get(wfreqs,envir=lexi...@e)[word])
}
## example of use
my.lexicon - new(Lexicon,wfreqs=c(the=2,person=1))
wfreq(my.lexicon,the)
***
However, testing indicates that the way I have set this up
does not achieve the intended benefits of having the
environment hashed:
***
sample.wfreqs - trunc(runif(1e5,max=100))
names(sample.wfreqs) - as.character(1:length(sample.wfreqs))
lex - new(Lexicon,wfreqs=sample.wfreqs)
words.to.lookup - trunc(runif(100,min=1,max=1e5))
## look up the words directly from the sample.wfreqs vector
system.time({
for(i in words.to.lookup)
sample.wfreqs[as.character(i)]
},gcFirst=TRUE)
## look up the words through the wfreq() function; time
approx the same
system.time({
for(i in words.to.lookup)
wfreq(lex,as.character(i))
},gcFirst=TRUE)
***
I'm guessing that the problem is that the indexing of the
wfreqs vector in my wfreq() function is not happening inside
the actual lexicon's environment. However, I have not been
able to figure out the proper call to get the lookup to
happen inside the lexicon's environment. I've tried
wfreq1 - function(lexicon,word) {
return(eval(wfreqs[word],envir=lexi...@e))
}
which I'd thought should work, but this gives me an error:
wfreq1(my.lexicon,'the')
Error in eval(wfreqs[word], envir = lexi...@e) :
object 'wfreqs' not found
Any advice would be much appreciated!
Best many thanks in advance,
Roger
--
Roger Levy Email: rl...@ucsd.edu
Assistant Professor Phone: 858-534-7219
Department of Linguistics Fax: 858-534-4789
UC San DiegoWeb: http://ling.ucsd.edu/~rlevy
__
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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] Using changing names in loop in R
Hi,
If you have data that is similar enough to warrant only changing the
extension (i.e., 1, 2, etc.) and that you (at least at times) wish to
perform operations on together, it is time to start thinking of a more
flexible framework. Fortunately, such a framework already exists in
lists. Lists let you keep diverse data structures (i.e., you do not
have to combine everything into a simple vector). Lists make it
tremendously easy to do many tasks (including the two you mentioned).
For example, suppose I want to read in 10 files and then do some
manipulations:
## initialize an empty list
dat - vector(mode = list, length = 10)
for(i in 1:10) {
dat[[i]] - read.table(paste(myfilename, i, sep = ''), header = TRUE, etc.)
}
# presumably these are data frames at this point
Now everything is nicely stored, dat[[1]] is even intuitively similar
to Data_1, Data_2, etc. At this point, suppose I want to create some
new stuff:
dat[[11]] - matrix(1, ncol = 1, nrow = 5) # add in a matrix
for(i in 1:n) {
dat[[i]] - dat[[i]] + dat[[i - 1]]
}
dat[[12]] - 5 # just a little vector
Another great advantage of lists is that it is possible to name their
elements. This can make things more meaningful or aid the memory.
However, even when an element is named, it can still be accessed by
its index
# name the first element 'price'
names(dat)[[1]] - price
now I could access it as any of these:
dat$price
dat[[price]]
dat[[1]]
If that weren't enough, you can easily use functions on every element
of a list with constructs such as lapply(), no for loop required!
lapply(X = dat, FUN = mean, na.rm = TRUE)
It is possible to not use lists and still do what you are after, but
frankly it is messier, more prone to error, and less effective in many
cases. It is generally a very nice feature of the assignment operator
that it is aware of its environment and does not go assigning or
overwriting things where you do not expect. You're left to the wolves
and your own wits with assign().
HTH,
Josh
On Sat, Nov 6, 2010 at 9:22 AM, Tuatara franziskabro...@gmail.com wrote:
Hello everybody,
I have usually solved this problem by repeating lines of codes instead of a
loop, but it's such a waste of time, I thought I should really learn how to
do it with loops:
What I want to do:
Say, I have several data files that differ only in a number, e.g. data
points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
manipulate them
e.g. a simple sum of several data points
data - c(NA,n)
for (i in 1:n){
data[i] - Data_i + Data_[i-1]
}
I know that the above code doesn't work, and I don't want to combine the
files into one vector to solve the problem etc. - I would just like to know
who make sure R recognizes the extension _i. I have the same problem for
say, reading in datafiles that only differ by one digit in the extension,
and I want to (instead of repeating code) combine the process in a loop.
I hope I made myself clear to what my problem is.
Thanks for your help,
//F
--
View this message in context:
http://r.789695.n4.nabble.com/Using-changing-names-in-loop-in-R-tp3030132p3030132.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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] SMATR common slopes test
Eugenio - I am confused with SMATR's test for common slope. My null hypothesis here is that all slopes are parallel (common slopes?), right? So if I get a p value 0.05 means that we can have confidence to reject it? That slopes are different? Or the other way around? it means that we have statistical confidence that the slopes are parallel? Try this: set.seed(5) n - 20 x - rnorm(n) y1 - 2 * x + rnorm(n) y2 - 2 * x + rnorm(n) y3 - 4 * x + rnorm(n) # Slopes approximately equal slope.com(x = c(x, x), y = c(y1, y2), groups = rep(c(1,2), each = n)) #$p #[1] 0.4498037 # Slopes of 2 and 4 slope.com(x = c(x, x), y = c(y1, y3), groups = rep(c(1,2), each = n)) #$p #[1] 0.0003850332 Cheers, Kevin - Kevin M. Middleton Department of Biology California State University San Bernardino __ 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 generate multivariate uniform distribution random
On 06-Nov-10 21:54:41, michael wrote: I wish to generate 100 by 1 vector of x1 and x2 both are uniform distributed with covariance matrix \Sigma. Thanks, Michael First, some comments. 1. I don't think you mean a 100 by 1 vector of x1 and x2 since you have two variables. 100 by 2 perhaps? 2. Given the range over which x1 is to be uniformly distributed (say of length A) and the range for x2 (say of length B), then by virtue of the uniform distribution the variance of x1 will be (A^2)/12 and the variance of x2 will be (B^2)/12, so you don't have a choice about these. So your only free parameter is the covariance (or the correlation) between x1 and x2. That said, it is a curious little problem. Here is one simple solution (though it may have properties that you do not want). Using X and Y for x1 and x2 (for simplicity), and using ranges (-1/2,1/2) for the ranges of X and Y (you could rescale later): 1. Generate X uniformly distributed on (-1/2,1/2). 2. With probability p (0 p 1) let Y=X. Otherwise, let Y be independently uniform on (-1/2,1/2). This can be implemented by the following code: n - 100 p - 0.6# (say) X - runif(n)-1/2 Z - runif(n)-1/2 U - rbinom(n,1,p) # =1 where Y=X, = 0 when Y independent of X Y - X*U + Z*(1-U) XY - cbind(X,Y) # your n by 2 matrix of bivariate (X,Y) Now the marginal distributions of X and Y are both uniform. var(X) = 1/12 and var(Y) = 1/12. Since the means are 0, cov(X,Y) = Exp(X*Y) = p*Exp(X^2) = p/12 cor(X,Y) = cov(X,Y)/sqrt(var(X)*var(Y)) = (p/12)/(1/12) = p. So all you need to do to get a desired correlation rho between X and Y is to set p = rho. The one thing you may not like about this is the diagonal line you will get for the cases where X=Y: plot(X,Y) Test: n - 10 p - 0.6 # (say) X - runif(n)-1/2 Z - runif(n)-1/2 U - rbinom(n,1,p) Y - X*U + Z*(1-U) var(X) # [1] 0.08340525 # theory: var = 1/12 = 0.0833 var(Y) [1] 0.08318914 # theory: var = 1/12 = 0.0833 cov(X,Y) [1] 0.04953733 # theory: cov = p/12 = 0.6/12 = 0.05 cor(X,Y) [1] 0.5947063 # theory: cor = p= 0.6 It would be interesting to see a solution which did not involve having cases with X=Y! Hoping this helps, Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 06-Nov-10 Time: 23:07:26 -- 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] Hashing and environments
some of this can be automated using the CRAN package
hash.
Kjetil
On Sat, Nov 6, 2010 at 10:43 PM, William Dunlap wdun...@tibco.com wrote:
I would make make an environemnt called wfreqsEnv
whose entry names are your words and whose entry
values are the information about the words. I find
it convenient to use [[ to make it appear to be
a list (instead of using exists(), assign(), and get()).
E.g., the following enters the 100,000 words from a
list of 17,576 and records their id numbers and the
number of times each is found in the sample.
wfreqsEnv - new.env(hash=TRUE, parent = emptyenv())
words - do.call(paste, c(list(sep=), expand.grid(LETTERS,
letters, letters)))
# length(words) == 17576
set.seed(1)
samp - sample(seq_along(words), size=10, replace=TRUE)
system.time(for(i in samp) {
+ word - words[i]
+ if (is.null(wfreqsEnv[[word]])) { # new entry
+ wfreqsEnv[[word]] - list(Count=1, EntryNo=i)
+ } else { # update existing entry
+ wfreqsEnv[[word]]$Count - wfreqsEnv[[word]]$Count + 1
+ }
+})
user system elapsed
2.28 0.00 2.14
(The time, in seconds, is from an ancient Windows laptop, c. 2002.)
Here is a small check that we are getting what we expect:
words[14736]
[1] Tuv
wfreqsEnv[[Tuv]]
$Count
[1] 8
$EntryNo
[1] 14736
sum(samp==14736)
[1] 8
If we do this with a non-hashed environment we get the same
answers but the elapsed time is now 34.81 seconds instead of
2.14. If you make wfreqEnv be a list instead of an environment
then that time is 74.12 seconds (and the answers are the same).
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 Levy, Roger
Sent: Saturday, November 06, 2010 1:39 PM
To: r-help@r-project.org
Subject: [R] Hashing and environments
Hi,
I'm trying to write a general-purpose lexicon class and
associated methods for storing and accessing information
about large numbers of specific words (e.g., their
frequencies in different genres). Crucial to making such a
class practically useful is to get hashing working correctly
so that information about specific words can be accessed
quickly. But I've never really understood very well how
hashing works, so I'm having trouble.
Here is an example of what I've done so far:
***
setClass(Lexicon,representation(e=environment))
setMethod(initialize,Lexicon,function(.Object,wfreqs) {
.obj...@e - new.env(hash=T,parent=emptyenv())
assign(wfreqs,wfreqs,envir=.obj...@e)
return(.Object)
})
## function to access word frequencies
wfreq - function(lexicon,word) {
return(get(wfreqs,envir=lexi...@e)[word])
}
## example of use
my.lexicon - new(Lexicon,wfreqs=c(the=2,person=1))
wfreq(my.lexicon,the)
***
However, testing indicates that the way I have set this up
does not achieve the intended benefits of having the
environment hashed:
***
sample.wfreqs - trunc(runif(1e5,max=100))
names(sample.wfreqs) - as.character(1:length(sample.wfreqs))
lex - new(Lexicon,wfreqs=sample.wfreqs)
words.to.lookup - trunc(runif(100,min=1,max=1e5))
## look up the words directly from the sample.wfreqs vector
system.time({
for(i in words.to.lookup)
sample.wfreqs[as.character(i)]
},gcFirst=TRUE)
## look up the words through the wfreq() function; time
approx the same
system.time({
for(i in words.to.lookup)
wfreq(lex,as.character(i))
},gcFirst=TRUE)
***
I'm guessing that the problem is that the indexing of the
wfreqs vector in my wfreq() function is not happening inside
the actual lexicon's environment. However, I have not been
able to figure out the proper call to get the lookup to
happen inside the lexicon's environment. I've tried
wfreq1 - function(lexicon,word) {
return(eval(wfreqs[word],envir=lexi...@e))
}
which I'd thought should work, but this gives me an error:
wfreq1(my.lexicon,'the')
Error in eval(wfreqs[word], envir = lexi...@e) :
object 'wfreqs' not found
Any advice would be much appreciated!
Best many thanks in advance,
Roger
--
Roger Levy Email: rl...@ucsd.edu
Assistant Professor Phone: 858-534-7219
Department of Linguistics Fax: 858-534-4789
UC San Diego Web: http://ling.ucsd.edu/~rlevy
__
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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] SMATR common slopes test
great thanks a lot! On Sat, Nov 6, 2010 at 3:54 PM, Kevin Middleton k...@csusb.edu wrote: Eugenio - I am confused with SMATR's test for common slope. My null hypothesis here is that all slopes are parallel (common slopes?), right? So if I get a p value 0.05 means that we can have confidence to reject it? That slopes are different? Or the other way around? it means that we have statistical confidence that the slopes are parallel? Try this: set.seed(5) n - 20 x - rnorm(n) y1 - 2 * x + rnorm(n) y2 - 2 * x + rnorm(n) y3 - 4 * x + rnorm(n) # Slopes approximately equal slope.com(x = c(x, x), y = c(y1, y2), groups = rep(c(1,2), each = n)) #$p #[1] 0.4498037 # Slopes of 2 and 4 slope.com(x = c(x, x), y = c(y1, y3), groups = rep(c(1,2), each = n)) #$p #[1] 0.0003850332 Cheers, Kevin - Kevin M. Middleton Department of Biology California State University San Bernardino -- Eugenio Larios PhD Student University of Arizona. Ecology Evolutionary Biology. (520) 481-2263 elari...@email.arizona.edu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to generate multivariate uniform distribution random
Ted, Thanks for your help, it is right on the money! for your comments: 1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1. 2. The correlation is the only free parameter. Michael On Sat, Nov 6, 2010 at 7:07 PM, Ted Harding ted.hard...@wlandres.netwrote: On 06-Nov-10 21:54:41, michael wrote: I wish to generate 100 by 1 vector of x1 and x2 both are uniform distributed with covariance matrix \Sigma. Thanks, Michael First, some comments. 1. I don't think you mean a 100 by 1 vector of x1 and x2 since you have two variables. 100 by 2 perhaps? 2. Given the range over which x1 is to be uniformly distributed (say of length A) and the range for x2 (say of length B), then by virtue of the uniform distribution the variance of x1 will be (A^2)/12 and the variance of x2 will be (B^2)/12, so you don't have a choice about these. So your only free parameter is the covariance (or the correlation) between x1 and x2. That said, it is a curious little problem. Here is one simple solution (though it may have properties that you do not want). Using X and Y for x1 and x2 (for simplicity), and using ranges (-1/2,1/2) for the ranges of X and Y (you could rescale later): 1. Generate X uniformly distributed on (-1/2,1/2). 2. With probability p (0 p 1) let Y=X. Otherwise, let Y be independently uniform on (-1/2,1/2). This can be implemented by the following code: n - 100 p - 0.6# (say) X - runif(n)-1/2 Z - runif(n)-1/2 U - rbinom(n,1,p) # =1 where Y=X, = 0 when Y independent of X Y - X*U + Z*(1-U) XY - cbind(X,Y) # your n by 2 matrix of bivariate (X,Y) Now the marginal distributions of X and Y are both uniform. var(X) = 1/12 and var(Y) = 1/12. Since the means are 0, cov(X,Y) = Exp(X*Y) = p*Exp(X^2) = p/12 cor(X,Y) = cov(X,Y)/sqrt(var(X)*var(Y)) = (p/12)/(1/12) = p. So all you need to do to get a desired correlation rho between X and Y is to set p = rho. The one thing you may not like about this is the diagonal line you will get for the cases where X=Y: plot(X,Y) Test: n - 10 p - 0.6 # (say) X - runif(n)-1/2 Z - runif(n)-1/2 U - rbinom(n,1,p) Y - X*U + Z*(1-U) var(X) # [1] 0.08340525 # theory: var = 1/12 = 0.0833 var(Y) [1] 0.08318914 # theory: var = 1/12 = 0.0833 cov(X,Y) [1] 0.04953733 # theory: cov = p/12 = 0.6/12 = 0.05 cor(X,Y) [1] 0.5947063 # theory: cor = p= 0.6 It would be interesting to see a solution which did not involve having cases with X=Y! Hoping this helps, Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 06-Nov-10 Time: 23:07:26 -- XFMail -- [[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] Removing NA in ggplot
OK, any reason why ggplot2 does not allow filtering of NA? ottar On 6 November 2010 15:23, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: Create a subset of your data that excludes the NAs before you feed it to ggplot. Ottar Kvindesland ottar.kvindesl...@gmail.com wrote: Hi list, I just got stuck with this one: In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F and NA). Then getting some nice plots using ggplot(data, aes(age[na.exclude(gender)])) + geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = lightblue ) + facet_grid( gender ~ .) I am trying to get a faceted graph of age distribution excluding the NA data for gender Unfortunately I end up with the error message: Error in data.frame(..., check.names = FALSE) : arguments imply differing number of rows: 206, 219 How do i Wash out NA's in this situation? Regards ottar [[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. --- 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. [[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] Removing NA in ggplot
On Sat, Nov 6, 2010 at 4:43 PM, Ottar Kvindesland ottar.kvindesl...@gmail.com wrote: OK, any reason why ggplot2 does not allow filtering of NA? It is not so much that ggplot2 does not allow the filtering of NA values, it is that you need to use data from the dataset you specified. By subsetting in aes() rather than in data, ggplot2 has differing datasets that it is being told to work with, so it returns an error (I'm sure that is a simplification, but the general point). Do your exclusion in the data argument. I imagine something like this, but untested since I have nothing to test it on. ggplot(data[na.exclude(gender), ], aes(age)) + geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = lightblue ) + facet_grid( gender ~ .) HTH, Josh ottar On 6 November 2010 15:23, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote: Create a subset of your data that excludes the NAs before you feed it to ggplot. Ottar Kvindesland ottar.kvindesl...@gmail.com wrote: Hi list, I just got stuck with this one: In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F and NA). Then getting some nice plots using ggplot(data, aes(age[na.exclude(gender)])) + geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = lightblue ) + facet_grid( gender ~ .) I am trying to get a faceted graph of age distribution excluding the NA data for gender Unfortunately I end up with the error message: Error in data.frame(..., check.names = FALSE) : arguments imply differing number of rows: 206, 219 How do i Wash out NA's in this situation? Regards ottar [[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. --- Jeff Newmiller The . . Go Live... DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. [[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 University of California, Los Angeles http://www.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] Using changing names in loop in R
A more detailed example: Say I would like to read in data files that are set-up identically and have identical (but somewhat) different text names (see below): data_1 - read.csv(data1.txt) data_2 - read.csv(data2.txt) data_3 - read.csv(data3.txt) How do I automate this process? (I assume the way I make R understand that the data file extension is to be read as a number rather than a string is the same for things like applying functions to matrices with different extensions, e.g. data_i, i = 1,2,3) -- View this message in context: http://r.789695.n4.nabble.com/Using-changing-names-in-loop-in-R-tp3030132p3030412.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to generate multivariate uniform distribution random
Dear Michael, On Sat, Nov 6, 2010 at 7:27 PM, michael tufemich...@gmail.com wrote: Ted, Thanks for your help, it is right on the money! for your comments: 1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1. 2. The correlation is the only free parameter. Michael I like Ted's solution. If all you are looking for is unif(0,1), you could use the Probability Integral Transform; something like this: set.seed(1) library(MASS) S - matrix(c(1, 0.9, 0.9, 1), nrow = 2) X - mvrnorm(100, mu = c(0,0), Sigma = S) Y - pnorm(X) var(Y) cor(Y) You could also use copulas, but those depend on contributed packages (and you can read more about them on the CRAN Task View for probability distributions). Hope this helps, Jay __ G. Jay Kerns, Ph.D. Youngstown State University http://people.ysu.edu/~gkerns/ __ 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] Using changing names in loop in R
On Sat, Nov 6, 2010 at 3:13 PM, Tuatara franziskabro...@gmail.com wrote:
A more detailed example:
Say I would like to read in data files that are set-up identically and have
identical (but somewhat) different text names (see below):
data_1 - read.csv(data1.txt)
data_2 - read.csv(data2.txt)
data_3 - read.csv(data3.txt)
How do I automate this process?
Well, the (nearly) verbatim automated duplicate would be:
for(i in 1:3) {
assign(x = paste(data, i, sep = _), value =
read.csv(paste(data, i, .txt, sep = '')))
}
but the preferred way would be:
dat - lapply(1:3, function(x) {read.csv(paste(data, x, .txt, sep = ''))})
which would read in and store all three files in one convenient list.
Josh
(I assume the way I make R understand that the data file extension is to be
read as a number rather than a string is the same for things like applying
functions to matrices with different extensions, e.g. data_i, i = 1,2,3)
--
View this message in context:
http://r.789695.n4.nabble.com/Using-changing-names-in-loop-in-R-tp3030132p3030412.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to generate multivariate uniform distribution random
Jay, Yes I'm looking for unif(0,1) and your method works just fine. I suppose your method should work for dimensions greater than 2, am I right? Michael On Sat, Nov 6, 2010 at 8:05 PM, G. Jay Kerns gke...@ysu.edu wrote: Dear Michael, On Sat, Nov 6, 2010 at 7:27 PM, michael tufemich...@gmail.com wrote: Ted, Thanks for your help, it is right on the money! for your comments: 1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1. 2. The correlation is the only free parameter. Michael I like Ted's solution. If all you are looking for is unif(0,1), you could use the Probability Integral Transform; something like this: set.seed(1) library(MASS) S - matrix(c(1, 0.9, 0.9, 1), nrow = 2) X - mvrnorm(100, mu = c(0,0), Sigma = S) Y - pnorm(X) var(Y) cor(Y) You could also use copulas, but those depend on contributed packages (and you can read more about them on the CRAN Task View for probability distributions). Hope this helps, Jay __ G. Jay Kerns, Ph.D. Youngstown State University http://people.ysu.edu/~gkerns/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to generate multivariate uniform distribution random
On Sat, Nov 6, 2010 at 8:22 PM, michael tufemich...@gmail.com wrote: Jay, Yes I'm looking for unif(0,1) and your method works just fine. I suppose your method should work for dimensions greater than 2, am I right? Michael Yes, but it gets that much more tricky to specify the covariance matrix. Two ways around this are to suppose that Sigma has a simplified correlation structure, or again, to use copulas. Jay __ 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] Where to get rcom for Linux
- Original Message From: Shige Song shiges...@gmail.com To: Stephen Liu sati...@yahoo.com Cc: r-help@r-project.org Sent: Sun, November 7, 2010 12:17:42 AM Subject: Re: [R] Where to get rcom for Linux isn't COM a Windows-only technology? Hi Shige Thanks. I see. I was surprised for unable to find it after having turned over the whole Internet World. Which Windows version Win7/Vista/Win Server 2008 will be more suitable running RBloomberg? B.R. Stephen L On Sat, Nov 6, 2010 at 12:12 PM, Stephen Liu sati...@yahoo.com wrote: Hi folks, Debian 600 64-bit Is rcom for Linux available? rcom rcom: R COM Client Interface and internal COM Server http://cran.r-project.org/web/packages/rcom/index.html If YES please advise where to get it. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] postscript window size
Dear R users, simple figure: postscript(file=~/Desktop/figure.ps, horizontal=T, width=20, height=10) par(mfcol=c(2,5)) plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') ##- plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') dev.off() Does not resize when saved on Desktop. I could ignore 'width' and 'height', and the same result. Particular figures plot(rnorm()) are too rectangular, when I want to have them more like squares (stretched horizontally). Working on Ubuntu and need to paste 'figure.ps' into latex (Kile). Thanks a lot, robert -- View this message in context: http://r.789695.n4.nabble.com/postscript-window-size-tp3030514p3030514.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] postscript window size
Hi Robert, You need to add paper = special to the postscript() call. postscript(file=~/Desktop/figure.ps, horizontal=TRUE, width=20, height=10, paper = special) plot() ... plot() dev.off() Otherwise it is reset because you are specifying a size outside of what can fit on the default paper size. Once that is changed, it should work (tested both ways on Debian :)) Cheers, Josh On Sat, Nov 6, 2010 at 6:23 PM, threshold r.kozar...@gmail.com wrote: Dear R users, simple figure: postscript(file=~/Desktop/figure.ps, horizontal=T, width=20, height=10) par(mfcol=c(2,5)) plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') ##- plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') plot(rnorm(100), type='l') dev.off() Does not resize when saved on Desktop. I could ignore 'width' and 'height', and the same result. Particular figures plot(rnorm()) are too rectangular, when I want to have them more like squares (stretched horizontally). Working on Ubuntu and need to paste 'figure.ps' into latex (Kile). Thanks a lot, robert -- View this message in context: http://r.789695.n4.nabble.com/postscript-window-size-tp3030514p3030514.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.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] can't load nlme on windoze 7
Hi, I've got a problem that sounds a lot like this, http://r.789695.n4.nabble.com/Re-R-R-2-12-0-hangs-while-loading-RGtk2-on-FreeBSD-td3005929.html under windoze 7. but it seems to hang with this stack trace, #0 0x77830190 in ntdll!LdrFindResource_U () from /cygdrive/c/Windows/system32/ntdll.dll building goes as follows, $ ./R CMD INSTALL --no-test-load nlme_3.1-97.tar.gz * installing to library 'C:/pfs/R/R-2.11.1/library' * installing *source* package 'nlme' ... ** libs making DLL ... gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c corStruct.c - o corStruct.o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c gnls.c -o gnl s.o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c init.c -o ini t.o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c matrix.c -o m atrix.o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c nlOptimizer.c -o nlOptimizer.o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c nlme.c -o nlm e.o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c nlmefit.c -o nlmefit.o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c nls.c -o nls. o gcc -IC:/pfs/R/R-2.11.1/include -O3 -Wall -std=gnu99 -c pdMat.c -o pd Mat.o gcc -shared -s -static-libgcc -o nlme.dll tmp.def corStruct.o gnls.o init.o matr ix.o nlOptimizer.o nlme.o nlmefit.o nls.o pdMat.o -LC:/pfs/R/R-2.11.1/bin -lR installing to C:/pfs/R/R-2.11.1/library/nlme/libs ... done ** R ** data ** moving datasets to lazyload DB ** inst ** preparing package for lazy loading ** help *** installing help indices ** building package indices ... * DONE (nlme) $ gcc --version gcc (GCC) 4.3.4 20090804 (release) 1 Copyright (C) 2008 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. $ gdb GNU gdb 6.8.0.20080328-cvs (cygwin-special) Copyright (C) 2008 Free Software Foundation, Inc. License GPLv3+: GNU GPL version 3 or later http://gnu.org/licenses/gpl.html This is free software: you are free to change and redistribute it. There is NO WARRANTY, to the extent permitted by law. Type show copying and show warranty for details. This GDB was configured as i686-pc-cygwin. (gdb) target exec R.exe (gdb) run Starting program: /cygdrive/c/pfs/R/R-2.11.1/bin/R.exe [New thread 20844.0x5368] Error: dll starting at 0x7742 not found. Error: dll starting at 0x769c not found. Error: dll starting at 0x7742 not found. Error: dll starting at 0x7754 not found. (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) (no debugging symbols found) Error: dll starting at 0x4a0b not found. R version 2.11.1 (2010-05-31) Copyright (C) 2010 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. library(nlme) [New thread 20844.0x5154] [Switching to thread 20844.0x5154] Quit (gdb) bt #0 0x77830190 in ntdll!LdrFindResource_U () from /cygdrive/c/Windows/system32/ntdll.dll (gdb) Mike Marchywka | V.P. Technology 415-264-8477 marchy...@phluant.com Online Advertising and Analytics for Mobile http://www.phluant.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] 3-way interaction simple slopes
On Nov 6, 2010, at 9:06 AM, Michael Wood wrote: Can anyone show me how to test for significant simple slopes of a 3- way interaction, with covariates. You might start by defining what you mean by simple slopes when discussing a model with a three way interaction. You might also include a description of the variables involved. my equation tmod-(glm(PCL~ rank.f + gender.f + MONTHS + CEXPOSE.M + bf.m + MONTHS*CEXPOSE.M*bf.m, data=mhatv, family=gaussian ,na.action=na.omit)) Thank you Mike 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] About 5.1 Arrays
Hi Daniel, I am correcting you. :-) You are using dim() incorrectly, and not accessing the array correctly. In all of your examples you should be using dim(3,4,2). Then you need to specify the indexes of the array element you want to look at. So, to use your example Thanks for your correction. So the index to be used in my example should be (3,4,2) only. But still I'm not very clear re your advice on follows a-1:24 a3d - array(a, dim = c(3,4,2)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 , , 2 [,1] [,2] [,3] [,4] [1,] 13 16 19 22 [2,] 14 17 20 23 [3,] 15 18 21 24 # 1 maps to a[1, 1, 1] in the 3d array a3d[1, 1, 1] [1] 1 # 2 maps to a[2, 1, 1]. a3d[2, 1, 1] [1] 2 # 3 maps to a[3, 1, 1] a3d[3, 1, 1] [1] 3 # 4 maps to a[1, 2, 1] a3d[1, 2, 1] [1] 4 What does it mean; [1] 1 [1] 2 [1] 3 [1] 4 as mentioned ? Anyway I'll move/continue on the manual to see what will happen. B.R. Stephen L - Original Message From: Daniel Nordlund djnordl...@frontier.com To: r-help@r-project.org Sent: Sun, November 7, 2010 2:08:04 AM Subject: Re: [R] About 5.1 Arrays -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Stephen Liu Sent: Saturday, November 06, 2010 7:38 AM To: Joshua Wiley Cc: r-help@r-project.org Subject: Re: [R] About 5.1 Arrays Hi Joshua, Thanks for your advice. 1) Re your advice:-[quote] a3d , , 1 --- this is the first position of the third dimension [,1] [,2] [,3] [,4] --- positions 1, 2, 3, 4 of the second dimension [1,]147 10 [2,]258 11 [3,]369 12 ^ the first dimension , , 2 --- the second position of the third dimension ... [/quote] Where is the third dimension? 2) Re your advice:-[quote] so you can think that in the original vector a: 1 maps to a[1, 1, 1] in the 3d array 2 maps to a[2, 1, 1]. 3 maps to a[3, 1, 1] 4 maps to a[1, 2, 1] 12 maps to a[3, 4, 1] 20 maps to a[2, 3, 2] 24 maps to a[3, 4, 2] [/quote] My finding; # 1 maps to a[1, 1, 1] in the 3d array a3d - array(a, dim = c(1, 1, 1)) a3d , , 1 [,1] [1,]1 Correct # 2 maps to a[2, 1, 1]. a3d - array(a, dim = c(2, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 Correct # 3 maps to a[3, 1, 1] a3d - array(a, dim = c(3, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 [3,]3 Correct # 4 maps to a[1, 2, 1] a3d - array(a, dim = c(1, 2, 1)) a3d , , 1 [,1] [,2] [1,]12 Incorrect. It is 2 # 12 maps to a[3, 4, 1] a3d - array(a, dim = c(3, 4, 1)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 Correct # 20 maps to a[2, 3, 2] a3d - array(a, dim = c(2, 3, 2)) a3d , , 1 [,1] [,2] [,3] [1,]135 [2,]246 , , 2 [,1] [,2] [,3] [1,]79 11 [2,]8 10 12 Incorrect. It is 12 # 24 maps to a[3, 4, 2] a3d - array(a, dim = c(3, 4, 2)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 , , 2 [,1] [,2] [,3] [,4] [1,] 13 16 19 22 [2,] 14 17 20 23 [3,] 15 18 21 24 Correct. If I'm wrong, pls correct me. Thanks B.R. Stephen Stephen, I am correcting you. :-) You are using dim() incorrectly, and not accessing the array correctly. In all of your examples you should be using dim(3,4,2). Then you need to specify the indexes of the array element you want to look at. So, to use your example a-1:24 a3d - array(a, dim = c(3,4,2)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 , , 2 [,1] [,2] [,3] [,4] [1,] 13 16 19 22 [2,] 14 17 20 23 [3,] 15 18 21 24 # 1 maps to a[1, 1, 1] in the 3d array a3d[1, 1, 1] [1] 1 # 2 maps to a[2, 1, 1]. a3d[2, 1, 1] [1] 2 # 3 maps to a[3, 1, 1] a3d[3, 1, 1] [1] 3 # 4 maps to a[1, 2, 1] a3d[1, 2, 1] [1] 4 Hope this is helpful, Dan Daniel Nordlund Bothell, WA USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] Hashing and environments
Wow, that is perfect: the hash package is exactly what I needed. Thank you!
Roger
On Nov 6, 2010, at 4:09 PM, Kjetil Halvorsen wrote:
some of this can be automated using the CRAN package
hash.
Kjetil
On Sat, Nov 6, 2010 at 10:43 PM, William Dunlap wdun...@tibco.com wrote:
I would make make an environemnt called wfreqsEnv
whose entry names are your words and whose entry
values are the information about the words. I find
it convenient to use [[ to make it appear to be
a list (instead of using exists(), assign(), and get()).
E.g., the following enters the 100,000 words from a
list of 17,576 and records their id numbers and the
number of times each is found in the sample.
wfreqsEnv - new.env(hash=TRUE, parent = emptyenv())
words - do.call(paste, c(list(sep=), expand.grid(LETTERS,
letters, letters)))
# length(words) == 17576
set.seed(1)
samp - sample(seq_along(words), size=10, replace=TRUE)
system.time(for(i in samp) {
+word - words[i]
+if (is.null(wfreqsEnv[[word]])) { # new entry
+wfreqsEnv[[word]] - list(Count=1, EntryNo=i)
+} else { # update existing entry
+wfreqsEnv[[word]]$Count - wfreqsEnv[[word]]$Count + 1
+}
+})
user system elapsed
2.280.002.14
(The time, in seconds, is from an ancient Windows laptop, c. 2002.)
Here is a small check that we are getting what we expect:
words[14736]
[1] Tuv
wfreqsEnv[[Tuv]]
$Count
[1] 8
$EntryNo
[1] 14736
sum(samp==14736)
[1] 8
If we do this with a non-hashed environment we get the same
answers but the elapsed time is now 34.81 seconds instead of
2.14. If you make wfreqEnv be a list instead of an environment
then that time is 74.12 seconds (and the answers are the same).
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 Levy, Roger
Sent: Saturday, November 06, 2010 1:39 PM
To: r-help@r-project.org
Subject: [R] Hashing and environments
Hi,
I'm trying to write a general-purpose lexicon class and
associated methods for storing and accessing information
about large numbers of specific words (e.g., their
frequencies in different genres). Crucial to making such a
class practically useful is to get hashing working correctly
so that information about specific words can be accessed
quickly. But I've never really understood very well how
hashing works, so I'm having trouble.
Here is an example of what I've done so far:
***
setClass(Lexicon,representation(e=environment))
setMethod(initialize,Lexicon,function(.Object,wfreqs) {
.obj...@e - new.env(hash=T,parent=emptyenv())
assign(wfreqs,wfreqs,envir=.obj...@e)
return(.Object)
})
## function to access word frequencies
wfreq - function(lexicon,word) {
return(get(wfreqs,envir=lexi...@e)[word])
}
## example of use
my.lexicon - new(Lexicon,wfreqs=c(the=2,person=1))
wfreq(my.lexicon,the)
***
However, testing indicates that the way I have set this up
does not achieve the intended benefits of having the
environment hashed:
***
sample.wfreqs - trunc(runif(1e5,max=100))
names(sample.wfreqs) - as.character(1:length(sample.wfreqs))
lex - new(Lexicon,wfreqs=sample.wfreqs)
words.to.lookup - trunc(runif(100,min=1,max=1e5))
## look up the words directly from the sample.wfreqs vector
system.time({
for(i in words.to.lookup)
sample.wfreqs[as.character(i)]
},gcFirst=TRUE)
## look up the words through the wfreq() function; time
approx the same
system.time({
for(i in words.to.lookup)
wfreq(lex,as.character(i))
},gcFirst=TRUE)
***
I'm guessing that the problem is that the indexing of the
wfreqs vector in my wfreq() function is not happening inside
the actual lexicon's environment. However, I have not been
able to figure out the proper call to get the lookup to
happen inside the lexicon's environment. I've tried
wfreq1 - function(lexicon,word) {
return(eval(wfreqs[word],envir=lexi...@e))
}
which I'd thought should work, but this gives me an error:
wfreq1(my.lexicon,'the')
Error in eval(wfreqs[word], envir = lexi...@e) :
object 'wfreqs' not found
Any advice would be much appreciated!
Best many thanks in advance,
Roger
--
Roger Levy Email: rl...@ucsd.edu
Assistant Professor Phone: 858-534-7219
Department of Linguistics Fax: 858-534-4789
UC San DiegoWeb: http://ling.ucsd.edu/~rlevy
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
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[R] is this matrix symmetric
Hi, I have this symmetric matrix, at least I think so. col1 col2 col3 [1,] 0.20 0.05 0.06 [2,] 0.05 0.10 0.03 [3,] 0.06 0.03 0.08 or structure(c(0.2, 0.05, 0.06, 0.05, 0.1, 0.03, 0.06, 0.03, 0.08 ), .Dim = c(3L, 3L), .Dimnames = list(NULL, c(var1, var2, var3))) But isSymmetric() doesn't agree. Any comment? I am on R 2.10.1 Thanks. Jun [[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.
