[R] problems with geom_vline, histograms, scale=free and facets in ggplot
Dear list,
I´m having a little trouble with adding vertical lines to a histogram.
I need to draw a matrix of histograms (using facets), and in each
histogram add a vertical line indicating the mean value of the data in
each facet. According to the last example in the geom_hline help, to
display different lines in each facet I need to provide a data frame.
I modified this example to make a plot similar to what I need:
# BEGIN R CODE
# This example works and does exactly what I need
library(reshape)
library(ggplot2)
# First, create a data frame with the mean value, to add to each facet
mean.data=tapply(mtcars$wt,INDEX=list(mtcars$vs,mtcars$am),mean)
mean.data=melt(mean.data)
colnames(mean.data)=c(vs,am,wt.mean)
# Now do the plot
p - qplot(wt, data=mtcars)
p - p + facet_grid(vs ~ am,scale=free)
p - p + geom_vline(aes(xintercept = wt.mean), mean.data,col=red)
p
###
###
# This one does not work
# Create a test data set
set.seed(100)
my.data=data.frame(code1=rep(c(A,B,C),each=30),
code2=rep(rep(c(D,E,F),each=10),3),
n=runif(90))
# Create a data frame with the mean value, to add to each facet
mean.data=tapply(my.data$n,INDEX=list(my.data$code1,my.data$code2),mean)
mean.data=melt(mean.data)
colnames(mean.data)=c(code1,code2,n.mean)
# Now do the plot
p = qplot(n, data=my.data) +
facet_grid(code2 ~ code1, scales=free)
p + geom_vline(aes(xintercept = n.mean), mean.data,col=red)
#With this I get the following error: Error in if (length(range) == 1
|| diff(range) == 0) { : missing value where TRUE/FALSE needed.
# If I remove the scales=free argument. I get the lines but also a
lot of extra empty facets.
p = qplot(n, data=my.data) +
facet_grid(code2 ~ code1)
p + geom_vline(aes(xintercept = n.mean), mean.data,col=red)
# END R CODE ##
I can´t figure out why the second example does not work when having
the scales=free argument, and why I get the extra facets when
removing it. Is this a bug? Any help will be very welcomed.
Julian
--
Julian Mariano Burgos
Hafrannsóknastofnunin/Marine Research Institute
Skúlagata 4, 121 Reykjavík, Iceland
Sími/Telephone : +354-5752037
Bréfsími/Telefax: +354-5752001
Netfang/Email: jul...@hafro.is, jmbur...@uw.edu
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Re: [R] R-/Text-editor for Windows?
I've tried Emacs Speaks Statistics and found it just confusing, to say the least (by the way, I'm puzzled by why so many people adhere to this one. Maybe it's a matter of getting used to it...?) You are right... it is confusing, and it takes some time to get used to, and it is not for everyone. But, it is totally customizable, and combined with org mode and org babel it becomes a tool as no other. I´ve tried about every editor, including Tinn-R, Notepad++, and Eclipse with StatEt, but after switching to Emacs/ESS there is no turning back. Julian -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.edu __ 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] post-hoc comparisons in GAMs (mgcv) with parametric terms
Dear list, I´m wondering if there is something analogous to the TukeyHSD function that could be used for parametric terms in a GAM. I´m using the mgcv package to fit models that have some continuous predictors (modeled as smooth terms) and a single categorical predictor. I would like to do post hoc test on the categorical predictor in the models where it is significant. Any suggestions? Thanks, Julian -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.edu __ 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 me understand how things work.
Hi Alex, What is happening is that the ´dist´function calculates a distance matrix, and returns an object of the ´dist´ class. temp - rbind (c(10,1),c(99,98)) x=dist(temp) x 1 2 131.6435 class(x) [1] dist You can see a description of the ´dist´class at the end of the function´s help file. Do this: ?dist So when you do dist(temp) you do not just get a number. You get an object that has the distance between the points, plus additional information (i.e. the method used to calculate the distance, if the matrix contains only the lower triangle or also the upper triangle and the diagonal, etc.). If you do sqrt() or 1/ over this object it is still of the ´dist´ class: x=dist(temp) x=sqrt(x) class(x) [1] dist x=1/x class(x) [1] dist Notice that you still the right answer (0.08715662). For some reason R prints the attributes of the object when you do the inverse but not when you do the square root (I´m curious about why...If anyone has an answer please pitch in). If you only want to get a number, do this: x=as.numeric(dist(temp)) class(x) [1] numeric 1/sqrt(as.numeric(dist(temp))) [1] 0.08715662 All the best, Julián Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.edu On Thu, Sep 16, 2010 at 10:02 AM, Alaios ala...@yahoo.com wrote: Hello I have some strange output from R and I try to understand how R works. Could you please help me with that? temp - rbind (c(10,1),c(99,98)) temp [,1] [,2] [1,] 101 [2,] 99 98 dist(temp) 1 2 131.6435 sqrt(dist(temp)) 1 2 11.47360 so far so good. until the nex line: when I try to do what i did before but adding the 1/(what I did before). I was expecting a number as a result of the division but unfortunately I took the following: 1/sqrt(dist(temp)) [1] 0.08715662 attr(,Size) [1] 2 attr(,Diag) [1] FALSE attr(,Upper) [1] FALSE attr(,method) [1] euclidean attr(,call) dist(x = temp) attr(,class) [1] dist Could you please help me understand what is this about? I would like to thank you in advance for your help Best REgards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- [[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] Strange behaviour when using diff with POSIXt and POSIXlt objects
Hi Brian, This is really odd. I keep getting the NA secs answer, only by running these three lines of code in a new session. time3=strptime(2009 06 01 00 47 00,format=%Y %m %d %H %M) time4=strptime(2009 06 01 00 57 00,format=%Y %m %d %H %M) diff(c(time3,time4)) Time difference of NA secs Here is the information you requested: dput(time3) structure(list(sec = 0, min = 47L, hour = 0L, mday = 1L, mon = 5L, year = 109L, wday = 1L, yday = 151L, isdst = -1L), .Names = c(sec, min, hour, mday, mon, year, wday, yday, isdst ), class = c(POSIXt, POSIXlt)) dput(time4) structure(list(sec = 0, min = 57L, hour = 0L, mday = 1L, mon = 5L, year = 109L, wday = 1L, yday = 151L, isdst = -1L), .Names = c(sec, min, hour, mday, mon, year, wday, yday, isdst ), class = c(POSIXt, POSIXlt)) sessionInfo() R version 2.11.0 (2010-04-22) i386-pc-mingw32 locale: [1] LC_COLLATE=Icelandic_Iceland.1252 LC_CTYPE=Icelandic_Iceland.1252 [3] LC_MONETARY=Icelandic_Iceland.1252 LC_NUMERIC=C [5] LC_TIME=Icelandic_Iceland.1252 attached base packages: [1] grDevices datasets splines graphics stats tcltk utils [8] methods base other attached packages: [1] svSocket_0.9-48 TinnR_1.0.3 R2HTML_2.0.0Hmisc_3.7-0.1 [5] survival_2.35-8 loaded via a namespace (and not attached): [1] cluster_1.12.3 grid_2.11.0lattice_0.18-5 svMisc_0.9-57 tools_2.11.0 Thanks, Julian [[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] Strange behaviour when using diff with POSIXt and POSIXlt objects
Dear list, I´m calculating time differences between series of time stamps and I noticed something odd: If I do this... time1=strptime(2009 05 31 22 57 00,format=%Y %m %d %H %M) time2=strptime(2009 05 31 23 07 00,format=%Y %m %d %H %M) diff(c(time1,time2),units=mins) Time difference of 10 mins .. I get the correct response in minutes. But if I try the same thing with different values, say.. time3=strptime(2009 06 01 00 47 00,format=%Y %m %d %H %M) time4=strptime(2009 06 01 00 57 00,format=%Y %m %d %H %M) diff(c(time3,time4)) Time difference of NA secs ...which is not what I´m looking for. The difference should also be 10 minutes. I burned a few neurons (and searched the documentation) and I cannot figure why this happens. Any ideas? All the best, Julian Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.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] Strange behaviour when using diff with POSIXt and POSIXlt objects
Hi Jim I´m using R 2.11.0, on Windows XP. On Thu, May 20, 2010 at 4:55 PM, jim holtman jholt...@gmail.com wrote: Please provide information as to what version you are using; works fine for me: time3=strptime(2009 06 01 00 47 00,format=%Y %m %d %H %M) time4=strptime(2009 06 01 00 57 00,format=%Y %m %d %H %M) diff(c(time3,time4)) Time difference of 10 mins I have version 2.10.1 On Thu, May 20, 2010 at 12:36 PM, Julian Burgos jmbur...@uw.edu wrote: Dear list, I´m calculating time differences between series of time stamps and I noticed something odd: If I do this... time1=strptime(2009 05 31 22 57 00,format=%Y %m %d %H %M) time2=strptime(2009 05 31 23 07 00,format=%Y %m %d %H %M) diff(c(time1,time2),units=mins) Time difference of 10 mins .. I get the correct response in minutes. But if I try the same thing with different values, say.. time3=strptime(2009 06 01 00 47 00,format=%Y %m %d %H %M) time4=strptime(2009 06 01 00 57 00,format=%Y %m %d %H %M) diff(c(time3,time4)) Time difference of NA secs ...which is not what I´m looking for. The difference should also be 10 minutes. I burned a few neurons (and searched the documentation) and I cannot figure why this happens. Any ideas? All the best, Julian Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.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] GAM for non-integer proportions
Dear list, I´m using the mgcv package to model the proportion by weight of certain prey on the stomach content of a predator. This proportion is the ratio of two weights (prey weight over stomach weight), and ranges between 0 and 1. The variance is low when proportion is close to 0 and 1, and higher at intermediate values. It seems that the best way to go is to model this using the quasi family with a logit link and a mu(1-mu) variance. Or I am missing something obvious? I will be thankful for any input. All the best, Julian -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.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] linear models with variance in dependent and independent variables
Dear list, I need to fit a multiple regression in which individual data point in the independent and the dependent variables have an associated variance or standard deviation. To make things clear, instead of having just a table of dependent (X) and independent (Y1, Y2) variable values like this: X Y1Y2 2.2 3.4 4.5 4.1 2.6 5.2 7.4 1.4 4.6 I have a table in which each value has an associated standard deviation: X Y1 Y2 2.2 (0.31) 3.4 (0.45)4.5 (0.21) 4.1 (0.28) 2.6 (0.23)5.2 (0.54) 7.4 (0.45) 1.4 (0.63)4.6 (0.37) What would be the best way to fit a multiple regression taking into account the variance of the individual values of the independent and dependent variables? Any references to methods or R packages will be greatly welcomed. Julian -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.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] SOM library - where do I find it
You have to download it from CRAN and install it. From the GUI, do Packages-Install package(s). Pretty basic stuff...you should check the documentation before posting. Julian R version 2.9.2 (2009-08-24) - for windows library(SOM) Error in library(SOM) : there is no package called 'SOM' Where can I get the SOM library from? Thanks in advance -- View this message in context: http://old.nabble.com/SOM-library---where-do-I-find-it-tp26415633p26415633.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] compute differences
Here is an approach...probably there are more elegant ways mydata=data.frame(ID=c(A,B,C,D,E),val=c(.3,1.2,3.4,2.2,2.0)) index=expand.grid(1:nrow(mydata),1:nrow(mydata)) dif=data.frame(difference=mydata$val[index$Var2]-mydata$val[index$Var1]) rownames(dif)=paste(mydata$ID[index$Var2],-,mydata$ID[index$Var1]) alessandro carletti wrote: Hi, I have a problem. I have a data frame looking like: ID val A .3 B 1.2 C 3.4 D 2.2 E 2.0 I need to CREATE the following TABLE: CASE DIFF A-A0 A-B-0.9 A-C-3.1 A-D-1.9 A-E-1.7 B-A... B-B... B-C B-D B-E C-A C-B C-C C-D C-E D-A D-B D-C D-D D-E E-A E-B E-C E-D E-E WHERE CASE IS THE COUPLE OF ELEMENTS CONSIDEREDM AND DIFF IS THE computed DIFFERENCE between their values. Could you give me suggestions? Alessandro Carletti __ 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. -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@u.washington.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ggplot2 legend text....a basic question
Hello fellow R's, I´ve been learning to use the ggplot2 library, and after a full day of work I still have a couple of basic questions. Here is an example: mydata=data.frame(x=runif(20),y=runif(20),n=runif(20)) mydata2=data.frame(x=c(0.4,0.6,0.5),y=c(0.4,0.4,0.6)) ggplot(mydata, aes(x, y)) + geom_point(aes(size = n)) + geom_polygon(data=mydata2,aes(x,y,alpha=0.5)) In this plot, the points are labeled as n (the name of the variable) and the polygon is labeled as 0.5 (the alpha value used). My questions are: a) How to change the text in the legend (for example, number instead of n). b) How to avoid having a legend for the polygon? Many thanks, Julian -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@u.washington.edu __ 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] Out of memory issue
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Otherwise...the only other possible suggestion is to use a smaller file. Julian Neotropical bat risk assessments wrote: Hi all, I am trying to run some plots on data, but when loading he CSV data file R is stopping and I am getting an out of memory error. Anyway to tweak this somehow to get it to run? Using WinXP with 4 GB RAM Tnx Bruce __ 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] Forcing the extrapolation of loess through the origin
Hi Torsten, If you are fitting a line, why are you using loess? Why not simply use lm to fit a regression line that goes through the origin? (i.e. with no intercept). Julian jimm-pa...@gmx.de wrote: Hi all, I'm fitting a line to my dataset. Later I want to predict missing values that exceed the [min,max] interval of my empirical data, therefore I choose surface=direct for extrapolation. l1-loess(y1~x1,span=0.1,data.frame(x=x1,y=y1),control=loess.control(surface=direct)) In my application it is highly important that the fitted line intercepts at the point of origin. Is it possible to do this in R? Thanks in advance. Cheers, Torsten -- __ 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] coalesce columns within a data frame
You could do something like this:
Name.x=c('nx1','nx2',NA,NA)
Name.y=c('ny1','NA','ny3',NA)
Name=Name.x
Name[is.na(Name.x)]=Name.y[is.na(Name.x)]
Name
[1] nx1 nx2 ny3 NA
Julian
Ivan Alves wrote:
Dear all,
I searched the mail archives and the R site and found no guidance (tried
merge, cbind and terms like coalesce with no success). There
surely is a way to coalesce (like in SQL) columns in a dataframe,
right? For example, I would like to go from a dataframe with two
columns to one with only one as follows:
From
Name.x Name.y
nx1 ny1
nx2 NA
NA ny3
NA NA
...
To
Name
nx1
nx2
ny3
NA
...
where column Name.x is taken if there is a value, and if not then column
Name.y
Any help would be appreciated
Kind regards,
Ivan
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Re: [R] 3 curves / 1 plot
You can use the points() and lines() functions to add points and lines to an existing plot. Julian Michel PETITJEAN wrote: I am a new user of R. Please does somebody knows how to plot 3 datasets (x1,a1),...,(xn,an), (x1,b1),...,(xn,bn), and (x1,c1),...,(xn,cn) on a single x,y plot, each of the three datasets being plotted with its own character pch() ? (three calls to plot() erase the two first datasets). Thank you very much. Michel Petitjean, DSV/iBiTec-S/SB2SM (CNRS URA 2096), CEA Saclay, bat. 528, 91191 Gif-sur-Yvette Cedex, France. Phone: +331 6908 4006 / Fax: +331 6908 4007 E-mail: [EMAIL PROTECTED] http://petitjeanmichel.free.fr/itoweb.petitjean.html __ 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] Inverting data frame...row wise
How about something like my.data=my.data[,4:1] Julian milicic.marko wrote: Hi, I have the data.frame with 4 columns. I simply want to invert dataset so that last row becomes first... I tried with rev(my_data-frame) but I got my columns inverted... not my rows 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. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about col.names in write.csv
Hi Luz, No entiendo bien tu pregunta. Querés grabar una tabla con nombres en las columnas y tambien en la primera fila? Si es asi, tenés que asignarle los nombres a la tabla antes de grabar. Por ejemplo: mi.tabla=matrix(runif(30),ncol=3) colnames(mi.tabla)=c(A,B,C) rownames(mi.tabla)=c(D,rep(,9)) write.csv(mi.tabla,file=mi.archivo.csv) Cuando asignás nombres a las filas o columnas, el vector tiene que tener la misma longitud que el numero de filas o columnas. Entonces, para darle un nombre solamente a la primera fila, hice un vector con el nombre y nueve espacios en blanco (para completar las diez filas que tiene mi tabla). Al grabar la tabla usando write.csv, el comportamiento default es guardar los nombres de las columnas y filas. Saludos, Julian Luz Milena Zea Fernandez wrote: Dear support, I don't ignore col.names in write.csv. I want to write names for the firts row. How can I do? Thanks in advance __ 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] Mclust - which cluster is each observation in?
Well, you are dealing with probability based clustering, so for each bird you will get a probability of belonging to each cluster. If your clusters are well defined, then each bird should have a very high probability of belonging to one of the clusters. You can get this probability matrix from your mclust object. For the iris dataset example, my.clusters=Mclust(iris[,-5]) This will give you the probability matrix my.clusters$z You can assign membership based on these probabilities (i.e. each bird belongs to the cluster with highest probability). You can obtain this membership by doing my.clusters$membership Hope this helps, Julian cnagy wrote: I'm trying to test a method of identifying individuals (birds) based on measured data (their calls). I have test data from known individual birds, and I am using the Mclust package to see if the program can correctly identify which calls come from different birds. So far, mclust has correctly ID'd the number of birds in the test data set (i.e., the correct # of clusters). However I also need to correctly assign each call to the right bird (i.e., data rows (calls) 1 - 10 are in cluster (bird) 1; rows 2 - 20 are in cluster 2, etc.). Is there a way to get mclust to show the cluster assignments of each observation? Thank you __ 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 which to identify a range of values
Do simply which(a100 a=200) Julian sj wrote: Hello, I am trying to identify values that fall within a certain range. I thought that I might be able to use the which function to do this but I have been unable to figure out a way to do it. Perhaps a little code will illustrate what i am trying to do. a - rnorm(1000, 100, 50) which( 100 a = 200) of course this doesn't work but illustrates what I ma trying to do, If anyone has suggestions I would greatly appreciate it Best, Spencer [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problemas usando jri0.4-1 y R 2.7.0
Hola Borja, Creo que vas a tener muy pocas respuestas a menos que escribas a la lista en inglés. Lamentablemente yo conozco poco de Java y no puedo responder tu pregunta. Saludos, Julian Borja Soto Varela wrote: Hola, es la primera vez que mando un correo a cualquiera de las listas de correo de R y no se si esta consulta se ajusta al próposito de la r-help list o debería haberlo mandado a otra de las listas que hay. Mi problema es el siguiente: Estoy desarrollando un programa en java con llamadas a R y no puedo usar jri 0.4-1 con la version 2.7 o 2.6 de R. Curiosamente si me funciona si uso jri 0.4 o la propia 0.4-1 con la version 2.4. El problema que me da al usar la ultima versión de R es el siguiente: Cannot find JRI native library! Please make sure that the JRI native library is in a directory listed in java.library.path. java.lang.UnsatisfiedLinkError: C:\JRI_0.4-1\JRI\jri.dll: No se encontró el proceso especificado at java.lang.ClassLoader$NativeLibrary.load(Native Method) at java.lang.ClassLoader.loadLibrary0(Unknown Source) at java.lang.ClassLoader.loadLibrary(Unknown Source) at java.lang.Runtime.loadLibrary0(Unknown Source) at java.lang.System.loadLibrary(Unknown Source) at org.rosuda.JRI.Rengine.clinit(Rengine.java:9) at vista.main.Main.main(Main.java:12) El directorio donde tengo jri.dll lo incluí en el path del sistema en esa ruta (de hecho con 2.4 funciona sin hacer nada más). Me preguntaba si a alguien también le ha pasado esto y como se puede solucionar. Gracias [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] more columns that column names
Hi Paul, The easiest thing to to is to open the file using a text editor (Notepad will do) and examine the first few lines. You can add add a column name if needed. Julian Paul Adams wrote: Hello to everyone, I have gotten my file to print to screen but when I use read.table I am getting an error message that says there are more columns than column names.This is a file that was not created by me so I am not sure how to investigate and solve this problem.I looked in the help file and it suggested an auxilliary function called count.fieldsThe code that was used was: read.table(file=C:\\Documents and Settings\\Owner\\My Documents\\colon cancer1.txt,header=T,row.names=1) Any help would be appreciated paul [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plot colors
You could do something like this: mydata=c(1,2,1,1,6,7,-1,-1,5,-1) color= as.numeric(mydata== -1) +1 plot(mydata,col=color) This will give you a plot where the -1's are in red (color = 2) and the other numbers in black (color=1). Julian uv wrote: Hi. I am plotting graphs for values ranging between -1 and 10, for example: (1,2,1,1,6,7,-1,-1,5,-1) I am trying to plot the graphs so that the points with value of -1 will be in one specific color, and the rest of the points will be in one different specific color. I would be grateful for any idea of how to do that in two colors. Thanks for any hint. -- Julian M. Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Phone: 206-221-6864 __ 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] Excluding/removing row and column names on text output files
Hi Steve, You can use write.table: write.table(x, file=/Users/Desktop/Data.txt, sep=,row.names=F,col.names=F) Cheers, Julian Stropharia wrote: Dear R users, I've had no joy finding a solution to this online or in any of my R books. Many thanks in advance for any help you can give. I'm seeking to output a data frame (or matrix - it doesn't matter which for my purposes) to a .txt file, but omit any row or column names. The data frame that I'm using doesn't actually have column or row names to start with as it has been coerced into its present form from a matrix. When I use: capture.output(x, file=/Users/Desktop/Data.txt, append=TRUE) I get the following (this is a small fraction of the actual data frame): 1 2 3 4 5 X1 0 0 0 0 2 X2 0 0 0 2 0 X3 1 1 2 0 0 If the data frame is transformed into a matrix, I still get row and column 'names' (or at least numbers like v1, etc.). Using the sink function also produces the exact same result. I've tried using row.names=FALSE (as you would when writing to a .csv file), in the capture.output function, but it doesn't work. I would also like the remove the horizontal spaces between numbers on the same row, to produce: 2 00020 11200 But, I want each row to remain a separate entity (not be concatenated with the others). I know I can remove the blank spaces by doing Find and Replace in a text editor, but is it possible to remove the row and column names, and the row spaces, directly in R so that it outputs like the above example? Thanks, Steve ~~ Steven Worthington Ph.D. Candidate New York Consortium in Evolutionary Primatology Department of Anthropology New York University 25 Waverly Place New York, NY 10003 U.S.A. ~~ -- Julian M. Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Phone: 206-221-6864 __ 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 on nested FOR loops
Hey Philip, I'm not sure if I understand what your x11, x12, etc. are. You can combine the values of your two vectors using the expand.grid function. There is no need to do nester FOR loops: i=c(1,2,3,4,5) j=c(1,2,3) x=expand.grid(i,j) print (x) Var1 Var2 1 11 2 21 3 31 4 41 5 51 6 12 7 22 8 32 9 42 1052 1113 1223 1333 1443 1553 Hope this helps, Julian Philip Twumasi-Ankrah wrote: I am new to more radical programming in R. I am trying to write a nested 'for' loop to produce output that takes subscripts like: for i taking values 1,2,3,4,5 and j taking values 1,2,3 I want to output for a computation using the combination values of i and j a value x like this; i jx 1 1 x11 1 2 x12 1 3 x13 2 1 x21 2 2 x22 2 3 x23 3 1 x31 3 2 x32 3 3 x33 ... Need help urgently. Thanks. Philip A Smile costs Nothing But Rewards Everything Happiness is not perfected until it is shared -Jane Porter [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] wavlet analysis
Hi Stephen, Your link doesn't work. In any case, check out the wavCWT function in the wmtsa package. Julian stephen sefick wrote: http://ion.researchsystems.com/cgi-bin/ion-p I would like a continuous wavelet transform. I have downloaded wavethresh, Rwave, and waveslim. I would like an output very similar to the above website. any suggestions? __ 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] A very simple question
Try this:
k=c(1,1,1,2,2,1,1,1)
k[(k!=1)]
[1] 2 2
k[(k!=2)]
[1] 1 1 1 1 1 1
k[(k!=3)]
[1] 1 1 1 2 2 1 1 1
Julian
Shubha Vishwanath Karanth wrote:
Hi R,
Suppose
l=c(1,1,1,2,2,1,1,1)
k[-which(k==1)]
[1] 2 2
k[-which(k==2)]
[1] 1 1 1 1 1 1
But,
k[-which(k==3)]
numeric(0)
I do not want this numeric(0), instead the whole k itself should be my
result... How do I do this?
Thanks,
Shubha
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
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Re: [R] Can R handle large dataset?
Depends on the RAM in your machine. And in your definition of 'handle'. You may be able to load a very large dataset into R, but won't be able to use some functions that require additional memory. A vague answer to a vague question... :) Julian Mingjun Huang wrote: Hello, I am new to R, can anyone give me an idea of how R handle a large dataset (e.g. couple of Gbytes)? Thanks a lot! Best, Mingjun __ 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] use list elements to subtract values from the dataframe
Try get(paste(wf$,fl[[1]],sep=)) See ?get Julian Dirkheld wrote: Hi, I have a dataframe wf existing of a header with different labels and beneath the values of those labels : wf: label1 label2 ... 0,450,21 0,100,45 I have a list fl - c(label2,label3,..) Isn't possible to use the list elements in the list in order to subtract values from the dataframe? like : wf$fl[[1]] When I do in R I get :NULL fl[[1]] gives label2 so no problem here... While wf$label1 works fine. __ 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 count the overlapped in two vectors
See ?match ss wrote: Dear list, If I have two vector, t1 and t2, of different lengths. Is there an easy way to count the number of the overlapped in two vectors and show the result in the graph? Thanks much, Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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 do you test for consecutivity?
Hey Anthony,
There must be many ways to do this. This is one of them:
#First, define a function to calculate the proportion of consecutive
numbers in a vector.
prop.diff=function(x){
d=diff(sort(x))
prop=(sum(d==1)+1)/length(x)
return(prop)}
#Note that I am counting both numbers in a consecutive pair. For
example, the vector c(1,2,6,9,10) will contain 4 consecutive numbers. I
think this is what you wanted do do, right?
#Next, generate a matrix with 1000 columns (one for each experiment) and
5 rows (the five numbers in each experiment). Note the use of the
'replicate' function to generate multiple sets of random numbers
selection=replicate(1000,sort(sample(1:30,5)))
#Third, use the apply function to apply the function we defined above to
each column of the matrix
diffs=apply(selection,2,prop.diff)
# This will give you a vector with the 1000 proportions of consecutive
numbers
Julian
Anthony28 wrote:
I need to use R to model a large number of experiments (say, 1000). Each
experiment involves the random selection of 5 numbers (without replacement)
from a pool of numbers ranging between 1 and 30.
What I need to know is what *proportion* of those experiments contains two
or more numbers that are consecutive. So, for instance, an experiment that
yielded the numbers 2, 28, 31, 4, 27 would be considered a consecutive =
true experiment since 28 and 27 are two consecutive numbers, even though
they are not side-by-side.
I am quite new to R, so really am puzzled as to how to go about this. I've
tried sorting each experiment, and then subtracting adjacent pairs of
numbers to see if the difference is plus or minus 1. I'm also unsure about
whether to use an array to store all the data first.
Any assistance would be much appreciated.
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Re: [R] How do you test for consecutivity?
Hey Anthony,
My previous function may not work in all cases. Say one of the
experiments yields these numbers:
1,2,3,6,7
Would you say that the proportion of consecutive numbers is 100%? If
so, this will work:
prop.diff=function(x){
d=diff(sort(x))
prop=sum((c(0,d==1)+c(d==1,0))0)
prop=prop/length(x)
return(prop)}
This function first identifies which numbers in your original vector are
part of a sequence of consecutive numbers.
Julian
Julian Burgos wrote:
Hey Anthony,
There must be many ways to do this. This is one of them:
#First, define a function to calculate the proportion of consecutive
numbers in a vector.
prop.diff=function(x){
d=diff(sort(x))
prop=(sum(d==1)+1)/length(x)
return(prop)}
#Note that I am counting both numbers in a consecutive pair. For
example, the vector c(1,2,6,9,10) will contain 4 consecutive numbers. I
think this is what you wanted do do, right?
#Next, generate a matrix with 1000 columns (one for each experiment) and
5 rows (the five numbers in each experiment). Note the use of the
'replicate' function to generate multiple sets of random numbers
selection=replicate(1000,sort(sample(1:30,5)))
#Third, use the apply function to apply the function we defined above to
each column of the matrix
diffs=apply(selection,2,prop.diff)
# This will give you a vector with the 1000 proportions of consecutive
numbers
Julian
Anthony28 wrote:
I need to use R to model a large number of experiments (say, 1000). Each
experiment involves the random selection of 5 numbers (without
replacement)
from a pool of numbers ranging between 1 and 30.
What I need to know is what *proportion* of those experiments contains
two
or more numbers that are consecutive. So, for instance, an experiment
that
yielded the numbers 2, 28, 31, 4, 27 would be considered a consecutive =
true experiment since 28 and 27 are two consecutive numbers, even though
they are not side-by-side.
I am quite new to R, so really am puzzled as to how to go about this.
I've
tried sorting each experiment, and then subtracting adjacent pairs of
numbers to see if the difference is plus or minus 1. I'm also unsure
about
whether to use an array to store all the data first.
Any assistance would be much appreciated.
__
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http://www.R-project.org/posting-guide.html
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Re: [R] means and variances of several groups in the matrix
Hy Katie, There are many ways to do this. A simple one is to create a vector of the same length than your 'x' vector, containing a group label. group=rep(c(1,2,3),times=nr[1,]) Then you can use tapply to apply a function (in this case mean and variance) of the values of x within each group. means=tapply(x,group,mean) vars=tapply(x,group,var) means 1 2 3 -0.14711206 0.28314274 -0.07861427 vars 1 2 3 1.4584971 0.3611996 0.6300624 Julian kathie wrote: Dear R users, I have 32 observations in data x. After sorting this, I want to compute means and variances of 3 groups divided by nr. Actually, the number of groups is flexible. Any suggestion will be greatly appreciated. Kathryn Lord --- x=rnorm(32) y=sort(x) nr=matrix(c(12,11,10,10,10,11),2,3) nr [,1] [,2] [,3] [1,] 12 10 10- sum=32 [2,] 11 10 11- sum=32 For the 1st row in nr, index of y = (1,..,12, 13,...,23, 24,...32) I want to compute means and variances for 3 groups (1st group is 1 through 12; 2nd group is 13-23; 3rd group is 24-32) For the 2nd row in nr, index of y = (1,..,11, 12,...,22, 23,...32) also, I want to compute means and variances for 3 groups (1st group is 1 through 11; 2nd group is 12-22; 3rd group is 23-32) __ 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] cumsum list..
In this case you can simply do cumsum(a[x,]+a[y,]) Julian yoo wrote: Hi all, i have the following.. a - data.frame(data = seq(1,10)) i have indices: x - c(1, 5, 3, 9) y - c(2, 7, 4, 10) I want the cumsum of a[1:2], a[5:7], a[3:4]... is there an elegant way to do it without any loop? 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] I need to buy a book in R
Hello Kayj, There are very good tutorials at the R website. See here: http://cran.r-project.org/other-docs.html Julian kayj wrote: Hi All, I am a new user in R and I would like to buy a book that teaches me how to use R. In addition, I may nees to do some advanced statistical analysis. Does anyone recommend some books or websites where I can learn R. 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] how to go to a line in R
Tinn-R also has this option. I suspect most editors will also do. Julian -Halcyon- wrote: RWinEdt has line indication. You might want to try that. Uwe Ligges-3 wrote: This depends on the editor you use for writing R code rather than on R. Uwe Ligges Jack Luo wrote: Hi, List When I was writing R code, I notice that there is no number indicating how many lines of codes you are writing. Is there a way to go to a line with defined number? say I want to go to the 20th line. Thanks, Jack [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] bootstrap sampling
See rnorm(). If you are sampling from a continuous normal distribution, it makes no sense to define a sample with replacement, because the probability of sampling twice the same number is zero. Julian sigalit mangut-leiba wrote: Hello, How do I sample observations with replacement from a normal distribution with a specific mean and s.d? (I want to see the sample, not only the statistic.) Thank you, Sigalit. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Great looking plot - but what does it mean?
Hi Mika03, It would be useful to know what function you used to create your plot. Assuming you used boxplot, do this: ?boxplot ?boxplot.stats Julian mika03 wrote: http://www.nabble.com/file/p14668788/paragraphs.png Hi, R is is world full of wonders... I created the attached plot, and I think it's exactly what I need! Well, actually I think it is more that wht I need... I wanted R to show the mean values of the categories on the x-axis and maybe the standard derivation as well. I am pretty confident that the bold horrizontal lines in the plot show the mean values. But what are the white boxes and the dashed lines? And what's that one small circle on the Section column supposed to mean. And if I would like to get rid of that small circle, how can I? Thanks a lot! __ 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] [OT] vernacular names for circular diagrams
I should say that the name of this chart varies even among
Spanish-speaking countries. In Argentina is diagrama de torta which
is something like cake-chart.
Julian
ahimsa campos-arceiz wrote:
Two non-eatable examples from Spain and Japan:
in Spanish we call them diagrama de sectores or gráfico de sectores. As
you can imagine it means sectors diagram (or graph).
in Japanese it is called 円グラフ (en gurafu), which means circular
graph
a link with its name in other languages:
http://isi.cbs.nl/glossary/term550.htm
Cheers,
Ahimsa
On Dec 13, 2007 3:01 AM, R Heberto Ghezzo, Dr [EMAIL PROTECTED]
wrote:
From Montreal,
Some people here call it the 'pizza diagram'
?some not eatable names?
salut
-Original Message-
From: [EMAIL PROTECTED] on behalf of Peter Dalgaard
Sent: Wed 12/12/2007 9:33 AM
To: Jean lobry
Cc: r-help@r-project.org
Subject: Re: [R] [OT] vernacular names for circular diagrams
Jean lobry wrote:
Dear useRs,
by a circular diagram representation I mean what you will get by
entering
this at your R promt:
pie(1:5)
Nice to have R as a lingua franca :-)
The folowing quote is from page 360 in this very interesting paper:
@article{SpenceI2005,
title = {No Humble Pie: The Origins and Usage of a Statistical
Chart},
author = {Spence, I.},
journal = {Journal of Educational and Behavioral Statistics},
volume = {30},
pages = {353-368},
year = {2005}
}
QUOTE
Like us, the French employ a gastronomical metaphor when
they refer to Playfair's pie chart, but they have preferred
instead to invoke the name of the wonderful round soft
cheese from Normandy - the camembert. When I spent 4 months
in Paris a few years ago, a friend invited my wife and me to
lunch with her elderly father who lives in Rouen, Normandy,
about an hour North of Paris. Her father inquired -
coincidentally during the cheese course - what work I was
doing in Paris; I replied that I was researching the
activities of a Scot, William Playfair, during the
revolutionary period. I told him that Playfair had invented
several statistical graphs, including the pie chart, which I
referred to, in French, as le camembert. After a stunned
silence of perhaps a couple of seconds, the distinguished
elderly gentleman looked me in the eye and exclaimed, Mon
Dieu ! Notre camembert?
UNQUOTE
So, I'm just curious: how do you refer in your own language to
this kind of graphic? How do you call it?
Best,
Jean
Grin
In Danish it is Lagkagediagram as in the layer cakes that are
traditional at birthday parties (and thrown at eachother's faces in
slapstick comedy).
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45)
35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45)
35327907
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Re: [R] Importing Large Dataset into Excel
Hi Wayne, I'm assuming that you file is really a comma-separated file (*.csv) and not an Excel workbook (*.xls) saved with a .csv extension, right? That (in my experience) is a common mistake. You should open your file with a simple text editor (notepad will do if the file is not too large) and review line 528, instead of reviewing the spreadsheet in Excel. You should be able to spot the problem right away. Julian Wayne Aldo Gavioli wrote: Hello all, I seem to be having a problem importing a data set from Excel into R. I'm using the read.table command to import the data with the following line of code: newborn-read.table(newborn edit.csv, header=T, sep=,) where newborn edit.csv is the name of the file. Unfortunately, I'm getting back the following error message: Error in scan(file,, what, nmax, sep, dc, quote, skip, nlines, na.string, : line 528 did not have 44 elements As far as I can tell, line 528 of the spreadsheet table does have the same number of elements as the other rows - by chance can this error message mean anything else? Also, is there an easier way to import data from R into Excel using a single line of R code? Thanks, Wayne __ 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] combine variables to matrix
Hi Andre, I don't quite understand what you are trying to do. Why you are using cbind to join columns of a dataset that it is already in table form? It is true that read.table will give you a data.frame instead of a matrix, but if for some reason you need a matrix you can do simply data.matrix=as.matrix(data) Julian Andre Jung wrote: I just got stuck with a quite simple question. I've just read in an ASCII table from a plain text file with read.table(). It's a 1200x1200 table. R has assigned variables for each column: V1,V2,V3,V4,... For small data sets data - read.table(data.txt); data.matrix - cbind(V1,V2,V3); works. But how could I put together 1200 columns? I've searched the R mailing help and stumbled upon this entry: https://stat.ethz.ch/pipermail/r-help/2007-July/137121.html which doesn't help me. thanks for your help. andre __ 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] matrix graph
The basic functions you need are image() contour() although I like better the plot.surface() function in the 'fields' package. Julian threshold wrote: Hi All, simple question: do you know how to graph the following object/matrix in a 'surface manner': [,1] [,2] [,3][,4] [,5][,6] [1,] -0.154 -0.065 0.129 0.637 0.780 0.221 [2,] 0.236 0.580 0.448 0.729 0.859 0.475 [3,] 0.401 0.506 0.310 0.650 0.822 0.448 [4,] 0.548 0.625 0.883 0.825 0.945 0.637 [5,] 0.544 0.746 0.823 0.877 0.861 0.642 [6,] 0.262 0.399 0.432 0.620 0.711 0.404 will be very grateful for hints. rob __ 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 rewriting looping structure?
Hi TLowe,
I'm not quite sure if I understand what you are trying to do. If you
are trying to get the cumulative sum of your data frame along each
column you can simply do
rcumsum=function(x){cumsum(x)/sum(x)}
apply(tdat,2,rcumsum)
Yet that is not what your code is doing. With a bit of clarification I
may help you some more.
Julian
TLowe wrote:
Hey Folks,
Could somebody help me rewrite the following code?
I am looping through all records across 5 fields to calculate the cumulative
percentage of each record (relative to each individual field).
Is there a way to rewrite it so I don't have to loop through each individual
record?
# tdat is my data frame
# j is my field index
# k is my record index
# tsum is the sum of all values in field j
# tmp is a vector containing the values in field j
# tdat[k,paste(cpct,j,sep=)] creates new fields cpct1,...,cpct5
for(j in 1:5) {
tsum- sum(tdat[,j]);
for(k in 1:nrow(tdat)) {
td- tdat[k,j];
tmp-tdat[,j];
# sum values = to current value and divide by the total sum
tdat[k,paste(cpct,j,sep=)]- sum(tmp[tmp = td]) / tsum;
}
}
Thanks,
TLowe
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Re: [R] if/else for plot/lines?
The simplest way would be to have a flag, an indicator variable that
stores a value that indicates if a plot has been done before. Something
like this
plot (do my first plot here...)
is.plot=T
later in the code...
if (is.plot) {plot (do new plot here)} else {lines(add lines to the
previous plot)}
Julian
Roger Levy wrote:
I'm interested in writing a function that constructs a new plot on the
current graphics device if no plot exists there yet, but adds lines to
the existing plot if a plot is already there. How can I do this? It
seems to me that the exists() function might be co-opted to do this, but
it's not obvious how.
Many thanks,
Roger
--
Julian M. Burgos
Fisheries Acoustics Research Lab
School of Aquatic and Fishery Science
University of Washington
1122 NE Boat Street
Seattle, WA 98105
Phone: 206-221-6864
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Re: [R] image() plot with z not in matrix format
Hello Marc, Well, image() requires data values in a regular grid. So you need to interpolate your data to a regular grid before you do your plot. There are many interpolation methods, but a good place to start is to do linear interpolation. You should first use expand.grid() to create the regular grid where to interpolate your data. Then use the interp() function (in the akima package) to do the interpolation. Finally, to do the plot you can use image() or better yet, the plot.surface() function in the 'fields' package, which is an improved version of the image() function. I hope that helps, Julian [EMAIL PROTECTED] wrote: Hello R cracks The image() function requires strictly increasing x and y values and z as a matrix. Actually, I don't have equally spaced variables, but anyway want to plot an colored image() (with z-information). An example of my problem is here: a-data.frame(rnorm(100), rnorm(100), runif(100)*100) How can I plot this data as an image (x=a[,1], y=a[,2] and z=a[,3])- according to ?image. It has to be possible to adapt the grid size so that every grid cell in the plot is coloured consequently. Thanks for your help Marc -- __ 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] Filling in a Zero Matrix
Hi Amy,
Many (perhaps most) of the people on the list do not receive emails with
html...so we can't see colored text. Also it would be helpfully to have
a bit of your data, so we can run your code (see the posting guide in
this regard,http://www.R-project.org/posting-guide.html).
Please explain what are the 'xlabel' and the 'ylabel' that you
mentioned. I don't see those variables in your code.
Julian
Amit Patel wrote:
Hi
I am very new to R and statistical programming in general. I am trying to
reorder data from a .csv file. I have managed to import the data and create a
zero matrix. I am now trying to fill the matrix. There seems to be some
problem with this section of my code. I have highlighted the dodgy code in
red. Please help if possible.
##
### Create matrix ###
##
#Open the csv file
OGSdata - read.table(MG3199.csv,sep=,,header=TRUE)
#creates 3 separate vectors
sample - OGSdata[,1]
mci - OGSdata[,2]
pct - OGSdata[,3]
#change mci range
offset - min(mci)-1
mci - (mci - offset)
#matrix sizes
mci_count - max(mci)
sample_count - max(sample)
#creates a zero matrix
OGS - mat.or.vec(mci_count,sample_count)
#Create labels
sample_lab - (A-9,B-9, C-9, D-9, E-9,A-12,B-12, C-12,
D-12, E-12)
#add data
for (i in 1:length(pct)) {
OGS(mci(i),sample(i))- pct(i);
}
What I want is to have colum1 from original data to be the xlabel, column 2
to be the ylabel and the 3rd colum to be the values in the matrix
Any help is appreciated.
Kind Regards
Amit Patel
___
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Re: [R] Filling in a Zero Matrix
My mistake, by xlabel and ylabel you are refering to the row and column
names of your matrix, right? To label a matrix you can simply use
colnames() and rownames().
Still its not clear to me what you are trying to do with your code.
Perhaps if you give us a sample of your data and explain clearly what
you are trying to achieve, we can help you to simplify your code.
Julian
Julian Burgos wrote:
Hi Amy,
Many (perhaps most) of the people on the list do not receive emails with
html...so we can't see colored text. Also it would be helpfully to have
a bit of your data, so we can run your code (see the posting guide in
this regard,http://www.R-project.org/posting-guide.html).
Please explain what are the 'xlabel' and the 'ylabel' that you
mentioned. I don't see those variables in your code.
Julian
Amit Patel wrote:
Hi
I am very new to R and statistical programming in general. I am trying to
reorder data from a .csv file. I have managed to import the data and create
a zero matrix. I am now trying to fill the matrix. There seems to be some
problem with this section of my code. I have highlighted the dodgy code in
red. Please help if possible.
##
### Create matrix ###
##
#Open the csv file
OGSdata - read.table(MG3199.csv,sep=,,header=TRUE)
#creates 3 separate vectors
sample - OGSdata[,1]
mci - OGSdata[,2]
pct - OGSdata[,3]
#change mci range
offset - min(mci)-1
mci - (mci - offset)
#matrix sizes
mci_count - max(mci)
sample_count - max(sample)
#creates a zero matrix
OGS - mat.or.vec(mci_count,sample_count)
#Create labels
sample_lab - (A-9,B-9, C-9, D-9, E-9,A-12,B-12, C-12,
D-12, E-12)
#add data
for (i in 1:length(pct)) {
OGS(mci(i),sample(i))- pct(i);
}
What I want is to have colum1 from original data to be the xlabel, column 2
to be the ylabel and the 3rd colum to be the values in the matrix
Any help is appreciated.
Kind Regards
Amit Patel
___
now.
[[alternative HTML version deleted]]
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Re: [R] better curve
Hello Mysimbaa, If you want to fit a smooth line to your data, there are many ways to do it. One option is to use splines. See the smooth.spline() function. If you only want to add a line to highlight the trend in your data, that should be enough. But if you want to do more serious analytical work, it is probably a good idea to learn about the different methods, its assumptions and limitations. Julian mysimbaa wrote: http://www.nabble.com/file/p13880048/Fluctuation.jpeg Hi R users, I have collected data which I plot(x,y).The problem it has oscillations. Now i'm trying to make better this curve with a smooth line. And then collect my new datas. But I don't know how doing this. Perhaps a linear regression ?? See .jpeg foto. Thanks for any help it will be given. __ 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] Reconstruct array dataset
Hey Marc, You can use the function scan() directly to read your file as a single vector. Then, as Rob suggested, use the matrix function to give it the dimensions you want. Other option (perhaps less elegant) is to do something like this. x=read.table (myfile,...etc.etc.) x=as.vector(as.matrix(x)) x=matrix(x,ncol=11,byrow=T) Julian marcg wrote: This was my intention, but I'm not able to read it in as a vector (because i don't know the function, neither I can convert the read in table to a vector and then to matrix or directly. What did I miss or where do I have to look up? thanks alot marc Original-Nachricht Datum: Wed, 21 Nov 2007 14:17:41 - Von: Rob Robinson [EMAIL PROTECTED] An: \'marcg\' [EMAIL PROTECTED] CC: [EMAIL PROTECTED] Betreff: RE: [R] Reconstruct array dataset Can you not read it into a single vector and then use as.matrix to shape it into a an appropriate sized matrix? Cheers rob *** Wan==t to know about Britain's birds? Try www.bto.org/birdfacts *** Dr Rob Robinson, Senior Population Biologist British Trust for Ornithology, The Nunnery, Thetford, Norfolk, IP24 2PU Ph: +44 (0)1842 750050 E: [EMAIL PROTECTED] Fx: +44 (0)1842 750030 W: http://www.bto.org How can anyone be enlightened, when truth is so poorly lit = -Original Message- From: [EMAIL PROTECTED] [mailto:] On Behalf Of marcg Sent: 21 November 2007 14:08 To: [EMAIL PROTECTED] Subject: [R] Reconstruct array dataset Hi there I have an interesting problem: My csv file is of array dimensions [12,50], but it was saved the wrong way: there should be only 11 colums. What happens now if I read it into R is that the whole data set is shifted ( in the first row, the last column contains already the first value of the supposed second row and so on...) how can I tell R to switch after 11 read values to the next row, taking the value from column 12 as first in the new row (for row 3 the two second last of the upper row etc...) Thanks for suggestions marc -- Ist Ihr Browser Vista-kompatibel? Jetzt die neuesten __ 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-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] Trying to get around R
Hi Loren,
It wasn't my intention to sound arrogant, cruel or off putting.
English is not my first language, and perhaps my message had a tone I
did not intended. The person posting the message I responded was
literally asking for somebody on the list to do his/hers homework (there
are even references to hints from the textbook). This is not only
academically non-kosher, is also not the purpose of the R-help list. I
pointed this to this person and invited him/her to come back with
specific questions on R coding.
The posting guidelines are easily accessible. The first line in the
Mailing Lists section of the R website
(http://www.r-project.org/mail.html) states Please read the
instructions below and the posting guide before sending anything to any
mailing list!, and displays a link to the guide. There also a link to
the guide at the bottom of every message send through this list. The
posting guide is not hard to find, and (in my opinion) it isn't long or
difficult to understand (in particular for anyone taking college level
statistics).
Julian
Loren Engrav wrote:
I am a newbie to R and Bio emails and
It is clear that newbies make mistakes, I made several which were pointed
out and I am trying to fix them, and as I fix one I make another, in time
perhaps I will know it all, but if it is like surgery, I will make
mistakes until I retire
But the response of the old-timers to these mistakes seems arrogant and
cruel and off putting and does NOT encourage more participation. In fact
it takes real stuff to continue after this putdown and that putdown.
There are 3,783 links to posting guidelines, which took 1.5 hours to find
and read and understand.
Why not a link on how the mistakes of the newbies will be dealt with?
Or a kindly response from the moderator personal to the newbie rather than
to the entire world?
Or a kindly general response as from Ben Bolker to my last infraction which
was You might have better luck with this on the Bioconductor mailing list
...
Rather than to the universe...
Using the wrong list: this is for R-sig-mac, and the topic occcurred
there recently.
All in an effort to encourage promote useful and increasing exchange
participation
Or not
Loren Engrav, MD
Univ Washington
From: Julian Burgos [EMAIL PROTECTED]
Date: Mon, 19 Nov 2007 10:44:49 -0800
To: Epselon [EMAIL PROTECTED]
Cc: r-help@r-project.org
Subject: Re: [R] Trying to get around R
Hello Epselon (if that is your name),
This sounds like homework questions.
From the R-help posting guide:
Basic statistics and classroom homework:
R-help is not intended for
these.
If you have a specific question on R
coding, do ask it (and provide
reproducible code). But you should not expect
for people on the list to
do your homework for you. That is a big
no-no.
Cheers,
Julian
Epselon wrote:
I have three problems I am trying to
simulate, that I am having difficulty
getting around with.
Problem 1.
I want to determine the 85 percentile (the x value for which the sum of
probabilities becomes 0.85) of the following distributions (two binomials
and a Poisson with rate Lmbda= np of the two binomials): X ~B(10, 0.3),
Y~P(3) ,
Z~B(30, 0.1). I want to show that that Y is a good approximation
for Z but
not for X...(by examining these distributions for few
different
percentiles)
Problem 2:
For a binomial distribution X ~ B(20, 0.4), I
want to use R to calculate
P{|X - μ| 2} and verify that it is near or
larger than 0.95. (Hint from
the text book: Since μ = 8 and 2.3 then
you
may want to read the
weights, or probabilities, of the values 6:10, into a
vector v and then use
the command sum(v) to
calculate the sum.) Repeat
this for another set of parameters of your
choice.
Problem 3:
Draw a
sample of size 10, from a Poisson with Lambda= 5, and calculate the
mean
and
the standard deviation of this sample, Repeat this calculation with
size 20
and 30 and demonstrate
that ¯X gets closer to μ as the sample size
increases.
Thanks.
I would appreciate it if someone accompanied the
codes with a brief
explanation so I can be able to replicate it
myself.
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and provide commented
Re: [R] Logarithmic axis
Hey John, You can do simply plot(reizstaerke, kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)),log=x) but if you want to fine tune where the tick marks are, you can do it by hand. kennlinie1 - c(8.0746909, 3.9916973, 9.9789444, 19.962869); kennlinie2 - c(6.0994206, 8.9661081, 19.924883, 31.879496); reizstaerke - c(76, 92, 108, 124); plot(log10(reizstaerke), kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)),axes=F) box() axis(1,at=log10(seq(75,125,by=5)),label=seq(75,125,by=5)) axis(2) The obvious reason that you are getting a plot with a single 100 and nothing else is that the range of the values in the reizstaerke vector is away from 10 and from 1000. If you really want to have the log scale in the way you described, you could do something like this: plot(log10(reizstaerke), kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)),axes=F,xlim=log10(c(1,1000))) box() axis(1,at=log10(c(1,10,100,1000)),label=c(1,10,100,1000)) axis(2) Julian John Wiedenhoeft wrote: Hi there, I guess this must be a standard issue, but I'm starting to go crazy with it. I simply want a plot with the x axis being logarithmic, having labels 1, 10, 100..., and ten unlabelled ticks between each of them - just as they introduce logarithmic axis at school. I've played around a bit with log=x, xlog=T (where exactly is the difference here?), xaxp, and xaxt (unfortunately xaxt=l isn't implemented). The best I get is a plot with an axis having a single 100 and nothing else... here is what I've tried: pdf(file=kennlinien.pdf); par(log=x, xlog=TRUE); kennlinie1 - c(8.0746909, 3.9916973, 9.9789444, 19.962869); kennlinie2 - c(6.0994206, 8.9661081, 19.924883, 31.879496); reizstaerke - c(76, 92, 108, 124); #plot(reizstaerke, kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)), xlim=c(0, max(reizstaerke)), log=x, xlog=TRUE, xaxp=c(1, 2, 1), type=b); #plot(reizstaerke, kennlinie1, type=b, log=x, xlog=TRUE, xaxp=c(1, 2, 3)); plot(reizstaerke, kennlinie1, type=b,usr=c(min(reizstaerke), max(reizstaerke), min(kennlinie1, kennlinie2), max(kennlinie1, kennlinie2)), log=x, xlog=TRUE, xaxp=c(1, 2, 3)); #points(reizstaerke, kennlinie2, xlog=TRUE, xaxp=c(1, 3, 3), type=b); dev.off(); Certainly I've missed something, but I can't figure it out. Any help appreciated, Cheers, John platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) __ 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] Logarithmic axis
Like always, there is much to be learned from the R-help list! Another message had a much simpler approach. plot(xy.coords(reizstaerke, kennlinie1, log=x), log=x) Julian Julian Burgos wrote: Hey John, You can do simply plot(reizstaerke, kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)),log=x) but if you want to fine tune where the tick marks are, you can do it by hand. kennlinie1 - c(8.0746909, 3.9916973, 9.9789444, 19.962869); kennlinie2 - c(6.0994206, 8.9661081, 19.924883, 31.879496); reizstaerke - c(76, 92, 108, 124); plot(log10(reizstaerke), kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)),axes=F) box() axis(1,at=log10(seq(75,125,by=5)),label=seq(75,125,by=5)) axis(2) The obvious reason that you are getting a plot with a single 100 and nothing else is that the range of the values in the reizstaerke vector is away from 10 and from 1000. If you really want to have the log scale in the way you described, you could do something like this: plot(log10(reizstaerke), kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)),axes=F,xlim=log10(c(1,1000))) box() axis(1,at=log10(c(1,10,100,1000)),label=c(1,10,100,1000)) axis(2) Julian John Wiedenhoeft wrote: Hi there, I guess this must be a standard issue, but I'm starting to go crazy with it. I simply want a plot with the x axis being logarithmic, having labels 1, 10, 100..., and ten unlabelled ticks between each of them - just as they introduce logarithmic axis at school. I've played around a bit with log=x, xlog=T (where exactly is the difference here?), xaxp, and xaxt (unfortunately xaxt=l isn't implemented). The best I get is a plot with an axis having a single 100 and nothing else... here is what I've tried: pdf(file=kennlinien.pdf); par(log=x, xlog=TRUE); kennlinie1 - c(8.0746909, 3.9916973, 9.9789444, 19.962869); kennlinie2 - c(6.0994206, 8.9661081, 19.924883, 31.879496); reizstaerke - c(76, 92, 108, 124); #plot(reizstaerke, kennlinie1, ylim=c(0, max(kennlinie1, kennlinie2)), xlim=c(0, max(reizstaerke)), log=x, xlog=TRUE, xaxp=c(1, 2, 1), type=b); #plot(reizstaerke, kennlinie1, type=b, log=x, xlog=TRUE, xaxp=c(1, 2, 3)); plot(reizstaerke, kennlinie1, type=b,usr=c(min(reizstaerke), max(reizstaerke), min(kennlinie1, kennlinie2), max(kennlinie1, kennlinie2)), log=x, xlog=TRUE, xaxp=c(1, 2, 3)); #points(reizstaerke, kennlinie2, xlog=TRUE, xaxp=c(1, 3, 3), type=b); dev.off(); Certainly I've missed something, but I can't figure it out. Any help appreciated, Cheers, John platform i486-pc-linux-gnu arch i486 os linux-gnu system i486, linux-gnu status major 2 minor 4.1 year 2006 month 12 day18 svn rev40228 language R version.string R version 2.4.1 (2006-12-18) __ 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] Trying to get around R
Hello Epselon (if that is your name),
This sounds like homework questions. From the R-help posting guide:
Basic statistics and classroom homework: R-help is not intended for
these.
If you have a specific question on R coding, do ask it (and provide
reproducible code). But you should not expect for people on the list to
do your homework for you. That is a big no-no.
Cheers,
Julian
Epselon wrote:
I have three problems I am trying to simulate, that I am having difficulty
getting around with.
Problem 1.
I want to determine the 85 percentile (the x value for which the sum of
probabilities becomes 0.85) of the following distributions (two binomials
and a Poisson with rate Lmbda= np of the two binomials): X ~B(10, 0.3),
Y~P(3) ,
Z~B(30, 0.1). I want to show that that Y is a good approximation for Z but
not for X...(by examining these distributions for few
different percentiles)
Problem 2:
For a binomial distribution X ~ B(20, 0.4), I want to use R to calculate
P{|X − μ| 2} and verify that it is near or larger than 0.95. (Hint from
the text book: Since μ = 8 and 2.3 then you may want to read the
weights, or probabilities, of the values 6:10, into a vector v and then use
the command sum(v) to
calculate the sum.) Repeat this for another set of parameters of your
choice.
Problem 3:
Draw a sample of size 10, from a Poisson with Lambda= 5, and calculate the
mean and the standard deviation of this sample, Repeat this calculation with
size 20 and 30 and demonstrate
that ¯X gets closer to μ as the sample size increases.
Thanks.
I would appreciate it if someone accompanied the codes with a brief
explanation so I can be able to replicate it myself.
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Re: [R] delete a row in a matrix
Something like this? a=matrix(1:9, ncol=3) print(a) [,1] [,2] [,3] [1,]147 [2,]258 [3,]369 a=a[-2,] print(a) [,1] [,2] [,3] [1,]147 [2,]369 You can use negative numbers to reference rows and/or columns, and you'll get everything except the rows/columns referenced. Julian Barb, Jennifer (NIH/CIT) [E] wrote: Can anyone tell me how to delete a row in a matrix? I have searched around and couldn't find a straightforward way to do this. Thanks, any help will be greatly appreciated. Jennifer [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] About print a label in plot
affy snp wrote: Dear list, Hello! I have a question about how to print a label in the plot. I am using the following code: pdf(mel4_chr3_11cancer_cghFLasso.pdf, height=6, width=5);plot(Disease.FL, index=i, type=Single,main=Plot of Labels);dev.off(); But Plot of Labels has not been printed. Any suggestions? Thanks a lot! Allen __ 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. -- Julian M. Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Phone: 206-221-6864 __ 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] a repetition of simulation
summary(log_v)
Julian
sigalit mangut-leiba wrote:
Hello,
In addition to my question a few days ago,
Now I have a matrix of the coefficients,
how can I see all the P.Values (Pr(|z|)) of the covariates from the 1000
iterations?
I tried names(log_v) and couldn'n find it.
Thank you,
Sigalit.
On 11/13/07, Julian Burgos [EMAIL PROTECTED] wrote:
Well, the obvious (but perhaps not the most elegant) solution is put
everything in a loop and run it 600 times.
coefficients=matrix(NA,ncol=3,nrow=600)
for (loop in 1:600){
[all your code here]
coefficients[loop,]=coef(log_v)
}
That will give you a matrix with the coefficients of each model run in
each row.
Julian
sigalit mangut-leiba wrote:
I want to repeat the simulation 600 times and to get a vector of 600
coefficients for every covariate: aps and tiss.
Sigalit.
On 11/13/07, Julian Burgos [EMAIL PROTECTED] wrote:
And what is your question?
Julian
sigalit mangut-leiba wrote:
Hello,
I have a simple (?) simulation problem.
I'm doing a simulation with logistic model and I want to reapet it 600
times.
The simulation looks like this:
z - 0
x - 0
y - 0
aps - 0
tiss - 0
for (i in 1:500){
z[i] - rbinom(1, 1, .6)
x[i] - rbinom(1, 1, .95)
y[i] - z[i]*x[i]
if (y[i]==1) aps[i] - rnorm(1,mean=13.4, sd=7.09) else aps[i] -
rnorm(1,mean=12.67, sd=6.82)
if (y[i]==1) tiss[i] - rnorm(1,mean=20.731,sd=9.751) else tiss[i] -
rnorm(1,mean=18.531,sd=9.499)
}
v - data.frame(y, aps, tiss)
log_v - glm(y~., family=binomial, data=v)
summary(log_v)
I want to do a repetition of this 600 times (I want to have 600
logistic
models), and see all the coefficients of the covariates aps tiss.
Thanks in advance,
Sigalit.
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Re: [R] Plot problem
Hi Allen, Its difficult to know what is the problem without knowing what type of object is 'Disease.FL'. plot() is a generic function and it will act differently depending on the type of object you are passing to it. As always, you should provide 'provide commented, minimal, self-contained, reproducible code' so we can find the problem. As a general comment, you can do any number of plots on a device (a window or a pdf file). The limit is only given by the number and size of the plots and the size of the device. Julian affy snp wrote: Dear list, I have a question about using plot(). I tried the code: pdf(mel_chr_all_13cancer_cghFLasso_all.pdf, height=6, width=11);plot( Disease.FL, index=1:4, type=All);dev.off(); and it went through well which outputed 4 plots for 4 samples in one page. But if I increase the numbers of plots(samples) which I want, saying to 11, pdf(mel_chr_all_13cancer_cghFLasso_all.pdf, height=6, width=11);plot( Disease.FL, index=1:11, type=All);dev.off(); then I got an error message as: Error in segments((1:n)[y 0], jp, (1:n)[y 0], jp + y[y 0], col = downcol) :invalid first argument I suspect that it has sth to do with the maxium plots which can be outputed on one page, which means less or equal to 4 will be fine but beyond that there will be a problem. I have tried the number 5 yet. Is there a way that I could specify that the plots can be put on multiple pages with 4 plots per one. Thank you very much for your help! Best, Allen [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get row numbers of a subset of rows
One way to do this is range(which(B[,2]==1)) Julian affy snp wrote: Hello list, I read in a txt file using B-read.table(file=data.snp,header=TRUE,row.names=NULL) by specifying the row.names=NULL so that the rows are numbered. Below is an example after how the table looks like using B[1:10,1:3] SNPChromosome PhysicalPosition 1 SNP_A-1909444 1 7924293 2 SNP_A-2237149 1 8173763 3 SNP_A-4303947 1 8191853 4 SNP_A-2236359 1 8323433 5 SNP_A-2205441 1 8393263 6 SNP_A-1909445 1 7924293 7 SNP_A-2237146 2 8173763 8 SNP_A-4303946 2 8191853 9 SNP_A-2236357 2 8323433 10 SNP_A-2205442 2 8393263 I am wondering if there is a way to return the start and end row numbers for a subset of rows. For example, If I specify B[,2]=1, I would like to get start=1 and end=6 if B[,2]=2, then start=7 and end=10 Is there any way in R to quickly do this? Thanks a bunch! Allen __ 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] Log random number
Hi Tobias, You'll have to explain what you mean with log transformed values. If your question is if it is possible to generate random numbers from a normal distribution (using rnorm()) and then getting their logarithm, then you can do that with the obvious caveat that the logarithm is not defined for negative numbers. Julian Wensui Liu wrote: dont think so, unless i miss something here. please do check the range of normal random number and the domain of log function. On 11/14/07, Tobias Schlottmann [EMAIL PROTECTED] wrote: Dear R users, Simply my question is that how it is possible to generate some random numbers using rnorm( ) but in log transformed values. Thank you, Tobias - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] factors levels ?
What you are looking for is the findInterval() function. Julian W Eryk Wolski wrote: Hi, It's just some example code.. The application is uninteresting. I am searching for some functionality. X - rnorm(100) //my data Y - seq(-3,3,by=0.1) // bin boundaries. Now I would like to generate a - list of factors, length as X... i.e.: all values in the range [-3,-2.9) have the same factor... [-3,-2.9) etc. I would assume R has such a function but I cant recall which one it is. __ 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] a repetition of simulation
Well, the obvious (but perhaps not the most elegant) solution is put
everything in a loop and run it 600 times.
coefficients=matrix(NA,ncol=3,nrow=600)
for (loop in 1:600){
[all your code here]
coefficients[loop,]=coef(log_v)
}
That will give you a matrix with the coefficients of each model run in
each row.
Julian
sigalit mangut-leiba wrote:
I want to repeat the simulation 600 times and to get a vector of 600
coefficients for every covariate: aps and tiss.
Sigalit.
On 11/13/07, Julian Burgos [EMAIL PROTECTED] wrote:
And what is your question?
Julian
sigalit mangut-leiba wrote:
Hello,
I have a simple (?) simulation problem.
I'm doing a simulation with logistic model and I want to reapet it 600
times.
The simulation looks like this:
z - 0
x - 0
y - 0
aps - 0
tiss - 0
for (i in 1:500){
z[i] - rbinom(1, 1, .6)
x[i] - rbinom(1, 1, .95)
y[i] - z[i]*x[i]
if (y[i]==1) aps[i] - rnorm(1,mean=13.4, sd=7.09) else aps[i] -
rnorm(1,mean=12.67, sd=6.82)
if (y[i]==1) tiss[i] - rnorm(1,mean=20.731,sd=9.751) else tiss[i] -
rnorm(1,mean=18.531,sd=9.499)
}
v - data.frame(y, aps, tiss)
log_v - glm(y~., family=binomial, data=v)
summary(log_v)
I want to do a repetition of this 600 times (I want to have 600 logistic
models), and see all the coefficients of the covariates aps tiss.
Thanks in advance,
Sigalit.
[[alternative HTML version deleted]]
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[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] TRUNCATED error with data frame
Hi Amit, Please read carefully the Mailing List Posting Guide (available at http://www.r-project.org/posting-guide.html). In particular, this section: For new subjects, compose a new message and include the '[EMAIL PROTECTED]' (or '[EMAIL PROTECTED]') address specifically. (Replying to an existing post and then changing the subject messes up the threading in the archives and in many people's mail readers.) About your question. You have a mistake in your model formula. The basic way to code a formula in R is by doing y ~ 'model', where y is your dependent variable and 'model'. Notice that '~' is not the same as '-'. For details, see ?formula, or read the manual titled An introduction to R that can be found here http://cran.r-project.org/manuals.html. Julian apsawant wrote: Hi , I am trying to modify the same example script to calculate AOV. Below is the script file (aov.R) I am trying to execute: aov.R -- data1-c(49,47,46,47,48,47,41,46,43,47,46,45,48,46,47,45,49,44,44,45,42,45,45,40 ,49,46,47,45,49,45,41,43,44,46,45,40,45,43,44,45,48,46,40,45,40,45,47,40) matrix(data1, ncol= 4, dimnames = list(paste(subj, 1:12), c(Shape1.Color1, Shape2.Color1, Shape1.Color2, Shape2.Color2))) Hays.df-data.frame(rt = data1, subj=factor(rep(paste(subj, 1:12, sep=), 4)), shape=factor(rep(rep(c(shape1,shape2), c(12, 12)), 2)), color=factor(rep(c(color1,color2), c(24, 24 aov(rt - shape * color + Error(subj/(shape * color)), data=Hays.df) summary(aov(rt - shape * color + Error(subj/(shape * color)), data=Hays.df)) Here is the error I get: source(aov.R) Error in terms(formula, Error, data = data) : Object shape not found I would appreciate your help as to why I am getting this error message. Thanks, Amit. Prof Brian Ripley wrote: Try ?source: it is not an 'error'. max.deparse.length: integer; is used only if 'echo' is 'TRUE' and gives the maximal length of the printout of a single expression. So the 'echo' has been truncated at 150 characters, not the expression used. On Mon, 12 Nov 2007, apsawant wrote: Hi , I am new to R. I am trying to run a simple R script as shown below: aov.R -- data1-c(49,47,46,47,48,47,41,46,43,47,46,45,48,46,47,45,49,44,44,45,42,45,45,40 ,49,46,47,45,49,45,41,43,44,46,45,40,45,43,44,45,48,46,40,45,40,45,47,40) matrix(data1, ncol= 4, dimnames = list(paste(subj, 1:12), c(Shape1.Color1, Shape2.Color1, Shape1.Color2, Shape2.Color2))) Hays.df-data.frame(rt = data1, subj=factor(rep(paste(subj, 1:12, sep=), 4)), shape=factor(rep(rep(c(shape1,shape2), c(12, 12)), 2)), color=factor(rep(c(color1,color2), c(24, 24 I am getting the following error message: Output at the R prompt -- source(aov.R,echo=T) data1 - c(49, 47, 46, 47, 48, 47, 41, 46, 43, 47, 46, 45, 48, 46, 47, 45, 49, 44, 44, 45, 42, 45, 45, 40, 49, 46, 47, 45, 49, 45, 41, 43, 4 [TRUNCATED] matrix(data1, ncol = 4, dimnames = list(paste(subj, 1:12), c(Shape1.Color1, Shape2.Color1, Shape1.Color2, Shape2.Color2))) Shape1.Color1 Shape2.Color1 Shape1.Color2 Shape2.Color2 subj 1 49484945 subj 2 47464643 subj 3 46474744 subj 4 47454545 subj 5 48494948 subj 6 47444546 subj 7 41444140 subj 8 46454345 subj 9 43424440 subj 1047454645 subj 1146454547 subj 1245404040 Hays.df - data.frame(rt = data1, subj = factor(rep(paste(subj, 1:12, sep = ), 4)), shape = factor(rep(rep(c(shape1, shape2), c(12, [TRUNCATED] I would appreciate if anyone could point out as to why I am getting this TRUNCATED message. Eventually, I want to run the following command: aov(rt - shape * color + Error(subj/(shape * color)), data=Hays.df) summary(aov(rt - shape * color + Error(subj/(shape * color)), data=Hays.df)) Thanks, Amit. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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
Re: [R] How to subset a portion of columns from a matrix
There are many ways. For example, you can do something like A[seq(1,dim(A)[2],2)] Julian Julian M. Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Phone: 206-221-6864 affy snp wrote: Dear List, Hi! I am wondering what is the simplest way to subset a portion of columns from a matrix. For example, there is a matrix A (238,304*243). What is the simplest way to get a sub-matrix B which comprises of column 1, 3, 5, 7,9,...(odd column number) from matrix A? Thank you very much for your help! Best, Allen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to subset a portion of columns from a matrix
No problem. Actually, I missed a comma. You should do A[,seq(1,dim(A)[2],2)] Julian affy snp wrote: Hi Julian, Thanks for the help! Best, Allen On Nov 9, 2007 11:08 PM, Julian Burgos [EMAIL PROTECTED] wrote: There are many ways. For example, you can do something like A[seq(1,dim(A)[2],2)] Julian Julian M. Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Phone: 206-221-6864 affy snp wrote: Dear List, Hi! I am wondering what is the simplest way to subset a portion of columns from a matrix. For example, there is a matrix A (238,304*243). What is the simplest way to get a sub-matrix B which comprises of column 1, 3, 5, 7,9,...(odd column number) from matrix A? Thank you very much for your help! Best, Allen __ 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. -- Julian M. Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Phone: 206-221-6864 __ 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] please help me
Hi Azadeh, As the warning message is telling you, it seems that your initial parameters for the covariance functions are not very good. Something that you can do is to use the eyefit() function (package geoR) to fit your variogram by eye and get a first approximation for your covariance parameter values (or to test if the values you are using do generate a variogram curve that is close to your data). Then you can use these parameters values as initial values in the variofit() function. Julian azadeh sadeghian wrote: I have thise problem in work with the function variofit and nls and dont know how to solve it. var1-variog(data,option=bin) var2-variog(data,option=cloud) v1-var1$v u1-var1^u v2-var2$v u2-var2$u variofit(var1,ini.cov.pars=c(0.005,1.5),cov.model=power,fix.nugget=F,weight=equal) variofit: weights used: equal variofit: minimisation function used: optim Error in if (loss (.Machine$double.xmax^0.5) | loss == Inf | loss == : missing value where TRUE/FALSE needed In addition: Warning message: unreasonable initial value for sigmasq + nugget (too low) in: variofit(var1, ini.cov.pars = c(0.005, 1.5), cov.model = power, ... variofit(var2,ini.cov.pars=c(0.005,1.5),cov.model=power,fix.nugget=F,weight=equal) variofit: weights used: equal warning: minimisation function nls can not be used with given cov.model. changing for optim. variofit: minimisation function used: optim Error in if (loss (.Machine$double.xmax^0.5) | loss == Inf | loss == : missing value where TRUE/FALSE needed In addition: Warning message: In variofit(var2, ini.cov.pars = c(0.005, 1.5), cov.model = power, : unreasonable initial value for sigmasq + nugget (too low) ... nls(v2~c0+ce*(1-exp(-(u2^2)/(ae^2))),start=list(c0=0,ae=3,ce=2)) Error in nls(v2 ~ c0 + ce * (1 - exp(-(u2^2)/(ae^2))), start = list(c0 = 0, : number of iterations exceeded maximum of 50 best regards. Sadeghian. __ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extract correlations from a matrix
Hey Christoph, It is not clear what do you want to extract. w[w0.6] does give you the correlation values above 0.6. What is your question? Julian Christoph Scherber wrote: Dear R users, suppose I have a matrix of observations for which I calculate all pair-wise correlations: m=matrix(sample(1:100,replace=T),10,10) w=cor(m,use=pairwise.complete.obs) How do I extract only those correlations that are 0.6? w[w0.6] #obviously doesn´t work, and I can´t find a way around it. I would very much appreciate any help! Best wishes Christoph (using R 2.5.1 on Windows XP) __ 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] partially sum variable of a dataframe
I'm assuming that you want to add b if 3a5.25. If so, there are many ways. One of them is sum (b[a3 a5.25]) This is very simple R coding. I recommend you spend some time learning the basics. There are very good tutorials at the R website. Julian [EMAIL PROTECTED] wrote: Hello, A stupid question: I have an array with two columns, the first a acting as my index in 0.25 steps, the second one b the column of interest. How can i sum up b only for a specified window in a (as the window command for time series) a=seq(0,10,0.25) b=runif(41) c=data.frame(a,b) Sum up c if 3a5.25 How to do that? thanks marc __ 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] Can I replace NA by 0 (if yes, how) ?
Do this: pfit$coefficients[is.na(pfit$coefficients)]=0 Julian Ptit_Bleu wrote: Hello, I'm trying to fit some points with a 8-degrees polynom (result of lm is stored in pfit). In most of the case, it is ok but for some others, some coefficients are NA. I don't really understand the meaning of these NA. And the problem is that I can't perform a derivation (pderiv-as.function((deriv(polynomial(pfit$coefficients) on pfit due to the presence of these NA. I tried the functions na.omit and na.exclude but NA are still there. I tried to replace manually NA by 0. The fit seems ok and then I can derive the polynom. But can I do this and, if yes, how can I detect and automatically replace these NA ? To conclude, I must say that I read some posts on NA but I must confess that it is still not clear to me (maybe because I'm french and R-newbie ...) Thanks in advance for your help, Ptit Bleu. pfit Call: lm(formula = pfitmax ~ poly(vfitmax, 8, raw = T)) Coefficients: (Intercept) poly(vfitmax, 8, raw = T)1 poly(vfitmax, 8, raw = T)2 poly(vfitmax, 8, raw = T)3 poly(vfitmax, 8, raw = T)4 -2.790e+04 6.276e+03 -5.645e+02 2.591e+01 -6.241e-01 poly(vfitmax, 8, raw = T)5 poly(vfitmax, 8, raw = T)6 poly(vfitmax, 8, raw = T)7 poly(vfitmax, 8, raw = T)8 6.681e-03 NA -3.942e-07 NA pderiv-as.function((deriv(polynomial(pfit$coefficients Erreur dans while ((la - length(a)) 1 a[la] == 0) a - a[-la] : valeur manquante là où TRUE / FALSE est requis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] loops sampling
Hi Garth,
Your code is really confusing! You should start by reading the help file
on the for() function and understanding what it does:
?for
Your line
for(i in 1:nboot){
}
is simply starting a loop around the variable 'i', which will change
values following the sequence 1:nboot.
It seems that the problem (or part of it) is that your are calling the
sample() function using a 'n' variable that is not defined anywhere.
Also, what nboot is supposed to be? The numbers of samples to be taken
(10, 20, etc.) or the number of iterations (1000). In your example, you
are calling your function as
bt.cor - npboot.function(nboot=10)
so in this case your function will loop around 10 times.
Here is a function that will do what you want:
npboot.function - function(data,nboot){
boot.cor - vector(length=1000)
for (i in 1:1000){
abc2=data[-(1:nboot),] #Remove the first 'nboot' rows
my.sample=sample(1:(250-nboot),nboot,replace=T) # Sample rows
abc2=rbind(abc2,abc2[my.sample,]) # Add the sampled rows to the
truncated dataset
model - lm(asin(sqrt(abc2$y/100)) ~ abc2$x1 + abc2$x2) #Fit the model
boot.cor[i]=cor(abc2$y,model$fit) #Get correlation
}
return (boot.cor)}
bt.cor - npboot.function(abc,nboot=120)
bootmean - mean(bt.cor)
[EMAIL PROTECTED] wrote:
Hi,
I'm new to R (and statistics) and my boss has thrown me in the deep-end with
the following task:
We want to evaluate the impact that sampling size has on our ability to
create a robust model, or evaluate how robust the model is to sample size for
the purpose of cross-validation i.e. in our current project we have collected
a series of independent data at 250 locations, from which we have built a
predictive model, we want to know whether we could get away with collecting
fewer samples and still build a decent model; for the obvious operational
reasons of cost, time spent in the field etc..
Our thinking was that we could apply a bootstrap type procedure:
We would remove 10 records or samples from the total n=250 and then replace
those 10 removed with replacements (or copies) from the remaining 240. With
this new data-frame we would apply our model and calculate an r², we would
then repeat through looping 1000 times before generating the mean r² from
those 1000 r² values generated. After which we would start the process again
by remove 20 samples from our data with replacements from the remaining 230
records and so on...
Below is a simplified version of the real code which contains most of the
basic elements. My main problem is I'm not sure what the 'for(i in 1:nboot)'
line is doing, originally I though what this meant was that it removed 1
sample or record from the data which was replaced by a copy of one of the
records from the remaining n, such that 'for(i in 10:nboot)' when used in the
context of the below code removed 10 samples with replacements as I have said
above. I'm almost positive that this isn't happening and if not how can I
make the code below for example do what we want it to?
library(utils)
#data
a - c(5.5, 2.3, 8.5, 9.1, 8.6, 5.1)
b - c(5.2, 2.2, 8.6, 9.1, 8.8, 5.7)
c - c(5.0,14.6, 8.9, 9.0, 9.1, 5.5)
#join
abc - data.frame(a,b,c)
#set column names
names(abc)[1]-y
names(abc)[2]-x1
names(abc)[3]-x2
abc2 - abc
#sample
abc3 - as.data.frame(t(as.matrix(data.frame(abc2
n - length(abc2)
npboot.function - function(nboot)
{
boot.cor - vector(length=nboot)
for(i in 1:nboot){
rdata - sample(abc3,n,replace=T)
abc4 - as.data.frame(t(as.matrix(data.frame(rdata
model - lm(asin(sqrt(abc4$y/100)) ~ I(abc4$x1^2) + abc4$x2)
boot.cor[i] - cor(abc4$y, model$fit)}
boot.cor
}
bt.cor - npboot.function(nboot=10)
bootmean - mean(bt.cor)
Any assistance would be greatly appreciated, also the sooner the better as we
are under pressure to reach a conclusion.
Cheers,
Garth
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Re: [R] Some problem in opening connection with .dat extention file in matrix(scan) function of R 2.5
The error message is telling you that R cannot find your file. Is your 'motives_pc.dat' file in your R working directory? If not, you have to give a complete path to the scan() function. Julian [EMAIL PROTECTED] wrote: Dear helpers please provide me some helpful answer to my problem while I m trying to run a program .I m attaching both the program and the data to which I have to obtain my estimation results. Motives.dat is the data file, and OBTfile4.3 is the complete code of program. by Running this // rawdata-matrix(scan(inputFile, n = nsubj*ncomp), nsubj, ncomp, byrow = TRUE) \\ The error appears to be // Error in file(file, r) : unable to open connection In addition: Warning message: cannot open file 'motives_pc.dat', reason 'No such file or directory' in: file(file, r) \\ where # name of preferences data file is assigned as, inputFile - motives_pc.dat Thanking Regards SYED ADIL HUSSAIN (+923455205402) QUAID-E-AZAM UNIVERSITY ISLAMABAD, PAKISTAN. This is Virus Free Email Scanned by QAU's McAfee Virus Scanner THIS COMMUNICATION MAY CONTAIN CONFIDENTIAL AND/OR OTHERWISE PROPRIETARY MATERIAL and is thus for use only by the intended recipient. If you received this in error, please contact the sender and delete the e-mail and its attachments from all computers. www.qau.edu.pk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to switch off accepting the shortcut of column names
You cannot call a column on a dataframe using the first letter (or first few letters) if the letters match more than one name. Extraction methods for data frames allow partially matching row names, but if the result is undefined you get NULL in return. Try this. first_item - seq(1,10) second_item - seq(11,20) dat - data.frame(first_item, second_item) dat$s [1] 11 12 13 14 15 16 17 18 19 20 #Now add an extra column with matching name ssecond_item - seq(21,30) dat - data.frame(dat,ssecond_item) dat$s NULL Julian P. Olsson wrote: Dear R-users, currently I am working with the R version 2.4.1. I realized it has a feature, which might be wonderful (as so many things in R), but in this case might be a bit dangerous as well. It seems that columns of a data frame can be called just by indicating the first letter of the name of the column. For example: first_item - seq(1,10) second_item - seq(11,20) dat - data.frame(first_item, second_item) names(dat) # [1] first_item second_item dat$f # [1] 1 2 3 4 5 6 7 8 9 10 dat$s # [1] 11 12 13 14 15 16 17 18 19 20 The good thing is, that if there is more than one column starting with the same letter(s), more than one letter has to be given to call the column. However, I would appreciate if I could choose an option in my workspace, whether this type of shortcut is allowed or not. Is there such an option? Thanks for any potential hints. Best wishes, P. Olsson [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data format
There are many ways. A simple one is to use split() to divide your
'Value' column using your 'Label' column as index. For example,
# Create dataset
mydata=data.frame(Label=c('Good','Bad','Good','Good','Good','Bad','Bad'),
Value=c(10,12,15,18,12,15,10))
# Split the data
mydata=split(mydata$Value,mydata$Label)
# Do a ks test
ks.test(mydata[[1]],mydata[[2]])
Julian
Emre Unal wrote:
Hi,
How can I analyze the data collected in database formatting (with labels)
rather than splitted by individual columns (almost in excel)?
For example (comma separated data);
Label,Value
Good,10
Bad,12
Good,15
Good,18
Good,12
Bad,15
Bad,10
etc...
ks.test or chisq.test can be done.
Splitting the data into new columns is not applicable cos' I'll use
R-integration in another software.
Thanks for your concern
Emre
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Re: [R] gam for longitudinal data?
You can use the gamm function (in the mgcv package) to fit generalized additive mixed models and specify your covariance structure. Julian gallon li wrote: I used gam for data analysis a lot. Is it possible to use gam to analyze longitudinal data? I mean, besides the working independence assumption, can i specify other more reasonable covariance structure in gam? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding pagebreaks on files???
I do this all the time. Simply, a) Open your pdf file using the pdf function. b) Do a bunch of plots. Because now the pdf file is your active device, every time you call a new plot you should get a new page. You can also use par(mfrow=...) to split the page (the same way you do it on a window), etc. c) Close the connection to the pdf using dev.off(). Julian Tom.O wrote: Hi Does anyone know if its possible to add pagebreaks to an pdf through an R command? I'm running a loop where each session exports an pdf graph. Each image becomes a separate file, but I would like to have them in the same document but on separate pages. Does anyone know a solution or workaround? Regards, Tom __ 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] Converting a string
I'm not sure what you mean. You should provide an example (i.e. some code). Julian Gang Chen wrote: This must be very simple, but I'm stuck. I have a command line in R defined as a variable of a string of characters. How can I convert the variable so that I can execute it in R? Really appreciate any help, Gang __ 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] data frame usage
Check out the tapply function. ?tapply Julian Bernd Jagla wrote: Hi, I am new to R and couldn't find any information on how to handle my table data that I just read in the way I want to use it.. I read in a table from a file: x - read.delim(filenam, header=TRUE) one column (x$label) hold the class labels. Another holds some values (x$val). I want to calculate summary statistics for different classes. How would I do this? Thanks, Bernd __ 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] Spatial autocorrelation
Hi Geertje, You should look into linear mixed-effects models. In these you can incorporate spatial correlation explicitly. The basic function to use is lme(), but you should do some reading about this type of models before jumping into it. An excellent resource is the book Mixed Effects Models in S and S-Plus by Jose Pinheiro and Douglas Bates. Good Luck! Julian Geertje Van der Heijden wrote: Hi, I have collected data on trees from 5 forest plots located within the same landscape. Data within the plots are spatially autocorrelated (calculated using Moran's I). I would like to do a ANCOVA type of analysis combining these five plots, but the assumption that there is no autocorrelation in the residuals is obviously violated. Does anyone have any ideas how to incorporate these spatial effects in my analysis? I have been reading up on autoregressive techniques, but I am not sure if it works with more than one plot. All help is greatly appreciated! Many thanks, Geertje van der Heijden [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looped t.test results according to a subset variable
See by()
Matthew Dubins wrote:
Hi all,
I wrote a simple function that gives me multiple t.test results
according to a subset variable and am wondering whether or not I
reinvented the wheel. Observe:
t.test.sub - function (formula, data, sub, ...)
{
for(i in 1:max(sub))
{
print(t.test(formula, data = subset(data, sub == i), ...))
}
}
Is there already a similar function in some package?
Thanks,
Matthew Dubins
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looped t.test results according to a subset variable
Could you post some of your data and your initial test, and explain why
it didn't worked? It is difficult to figure out what is the problem
with your call to by().
Julian
Matthew Dubins wrote:
I've tried to use by(), but the closest i got to it doing what I wanted
was using the following:
by(percent, quiz, function(percent) {t.test(percent~group,
data=marks.long)})
But the results it gave me weren't t.tests of percent by group according
to quiz number.
Julian Burgos wrote:
See by()
Matthew Dubins wrote:
Hi all,
I wrote a simple function that gives me multiple t.test results
according to a subset variable and am wondering whether or not I
reinvented the wheel. Observe:
t.test.sub - function (formula, data, sub, ...)
{
for(i in 1:max(sub))
{
print(t.test(formula, data = subset(data, sub == i),
...))
}
}
Is there already a similar function in some package?
Thanks,
Matthew Dubins
__
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] variance explained by each term in a GAM
Dear Prof. Wood, Just another quick question. I am doing model selection following Wood and Augustin (2002). One of the criteria for retaining a term is to see if removing it causes an increase in the GCV score. When doing this, do I also need to fix the smooth parameters? Thanks, Julian Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Simon Wood wrote: I think that your approach is reasonable, except that you should use the same smoothing parameters throughout. i.e the reduced models should use the same smoothing parameters as the full model. Otherwise you get in trouble if x1 and x2 are correlated, since the smoothing parameters will then tend to change alot when terms are dropped as one smooth tries to `do the work' of the other. Here's an example, (which is modifiable to illustrate the problem with not fixing the sp's) ## simulate some data set.seed(0) n-400 x1 - runif(n, 0, 1) ## to see problem with not fixing smoothing parameters ## remove the `##' from the next line, and the `sp' ## arguments from the `gam' calls generating b1 and b2. x2 - runif(n, 0, 1) ## *.1 + x1 f1 - function(x) exp(2 * x) f2 - function(x) 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10 f - f1(x1) + f2(x2) e - rnorm(n, 0, 2) y - f + e ## fit full and reduced models... b - gam(y~s(x1)+s(x2)) b1 - gam(y~s(x1),sp=b$sp[1]) b2 - gam(y~s(x2),sp=b$sp[2]) b0 - gam(y~1) ## calculate proportions deviance explained... (deviance(b1)-deviance(b))/deviance(b0) ## prop explained by s(x2) (deviance(b2)-deviance(b))/deviance(b0) ## prop explained by s(x1) On Monday 08 October 2007 20:19, Julian M Burgos wrote: Hello fellow R's, I do apologize if this is a basic question. I'm doing some GAMs using the mgcv package, and I am wondering what is the most appropriate way to determine how much of the variability in the dependent variable is explained by each term in the model. The information provided by summary.gam() relates to the significance of each term (F, p-value) and to the wiggliness of the fitted smooth (edf), but (as far as I understand) there is no information on the proportion of variance explained. One alternative may be to fit alternative models without each term, and calculate the reduction in deviance. For example: m1=gam(y~s(x1) + s(x2)) # Full model m2=gam(y~s(x2)) m3=gam(y~s(x1)) ddev1=deviance(m1)-deviance(m2) ddev2=deviance(m1)-deviance(m3) Here, ddev1 would measure the relative proportion of the variability in y explained by x1, and ddev2 would do the same for x2. Does this sound like an appropriate approach? Julian Julian Burgos FAR lab University of Washington __ 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] combining vectors on unequal length
Well, if you bind two vectors you form an array with dimensions 2 x length of the longest vector. So you need to decide how to fill up the 'empty' spacies corresponding to the shorter vector. Recycling the shorter vector is the default action. If you just want to save the data, you could create a list and save it as a R object. my.list=list(X,Y) save(my.list) Julian Nair, Murlidharan T wrote: If I have two vectors X-1:10 Y-1:5 When I combine them using cbind, the shorter one is repeated and both are made of the same length. Is there a methods that does this without duplicating the shorter one. I want to use this to store the data back to a file. Thanks ../Murli [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
