[R] Decompose an irregular daily time series with missing values
Dear all, I have a daily time series for the months April to September (183 says) for a 40 years period. It contain missing values. I would like to extract a seasonal component, trend component and irregular component using an adaptive model. #Commands for making a time series: timeser-rnorm(183*40) #Add some missing values: timeser[c(5,77,98,100,105,1000,1340,2001,3277,3278,3279,4004:4009,7000)]-NA I had a look on the functions decompose and stl. But it seems unsuitable for my problem. I appreciate if someone can help me out! -- View this message in context: http://www.nabble.com/Decompose-an-irregular-daily-time-series-with-missing-values-tf4346265.html#a12382402 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Heatmap clustering
I have constructed a heatmap comparing distances between 30 elements. I get the heatmap plot with red dots when distance is close, and yellow when two elements differ. My question is...how can I get clusters of similarity from the headplot (I don´t know the total number of clusters, otherwise I´d have used Kmeans)? Lots of thanks in advance! Adrià __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] correlation structure in lmer
Hi how can I specify a correlation structure in the lmer-function as it is possible in lme(formula, ..., corr=corAR1(form=...))? Thanks Fränzi __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Simple Graph
Hi Sir We use the function mpg() for ticks on axes. If we define more number of ticks then each tick is not labeled. How to label each tick on axes? Regards -- AMINA SHAHZADI Department of Statistics GC University Lahore, Pakistan. Email: [EMAIL PROTECTED] [EMAIL PROTECTED] [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Single plot multiple levels in x?
Plotting with 2 x axis? One axis inside another, for example salary within state, 1-50 | 50 – 100 | 100+ | 1- 50 | 50 -100 | 100+ | … repeated bins for salary AL ! AR …… more states The values are all stored with a single data frame. I have tried different things with the axis function and done many searches for plotting. Can’t find a direct reference Thanks. Richard __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] rpart's loss matrix in ipred
Dear R users, I have been using the rpart procedure to predict the occurrence of depression in a large data file. Since the prevalence is very low (5%), I have been using classification trees with a loss matrix that penalized false negatives more than false positives. I have become interesested in bagging these (successful!) classification trees, and have been using the ipred package for this. However, the ipred algorithm does not seem to do anything with the loss matrix I specify in control=rpart.control(method=class,cp=0,xval=0,parms=list(loss=matrix(c(0,1,10,0))) Am I doing anyting wrong or does the ipred package simply not allow for specifying such a matrix? cheers! -- Dr. Niels Smits Research Methodology, Statistics and Data-analysis Faculty of Psychology and Education Vrije Universiteit Amsterdam Van der Boechorststraat 1 1081 BT Amsterdam The Netherlands Tel: +31 (0)20 5988713 Secr: +31 (0)20 5988757 Fax: +31 (0)20 5988758 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Converting into time series object
Hi, I have a dataframe of trading dates along with the corresponding prices. I need to convert this into a time series object. How do I do this with my price values being the time series object and the dates/time being the trading dates. BR, Shubha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Q: Mean, median and confidence intervals with functions summary boxplot.stats
Dear R ussers, My question is, How can my mean be outside the confidence intervals ?! I think i have the answer for it, but i would like to hear some other ideas on it. First my data is not continuose but categorical, it is a titre calculated on a dilution serie. It is stored as a column of values, and a column indicating the phase of the trail. Theoreticaly it is possible to have a value ranging from 0 to 4, but in practice, only sertain values will occure, and they will repeat. So it are frequencies. This is why i belief that it is better to work with a median than with a mean, because it represents the cluster of values wich occure most. Below I only give one example, but the mean being below the lowest confidence limit occures several times over different tests. does my answer seam reasonable, or should i perhapes use an other methode, any sugestion? summary_1d = summary(subset(eda_data, phase=='1' test=='test 1' ,select=lg_value), na.rm = T) conf_1d = boxplot.stats(subset(eda_data, phase=='1' test=='test 1' ,select=lg_value)) MeanMedian 95% Confidence Int. StDev. Variance 1.1981.681 1.4411.922 0.931 0.866 Kind regards, Tom W. Disclaimer: click here [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Reading lines from file
Hi. I have a text file containing a few hundred lines of numbers, each line has a different length. For example: 1 4 1 1 7 3 11 1 1 1 1 1 1 2 2 4 1 2 And so on. I need to do a simple plot function for each line, and then to save each plot into a separate file. Is there any way doing it from within R? Alternatively, is there a way to run R from another language or from the X11 (Linux) terminal, sending as input each line, then plotting and saving? Thank you very much for any idea. -- View this message in context: http://www.nabble.com/Reading-lines-from-file-tf4347936.html#a12387556 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Month end calculations
A simple example (avoiding using dates, just to show the principle) - this assumes that your data are already sorted (?order). temp - data.frame(subject = rep(1:2, each = 5), response = 1:10) print(temp) last - do.call(rbind, by(temp, temp$subject, function(x) tail(x, 1))) print(last) By changing the 2nd parameter to 'tail' you can get different numbers of observations in the tail of each subset, so it has more power than SAS's .last. If you need an equivalent of .first, then replace 'tail' with 'head'. Jim. Shubha Vishwanath Karanth wrote: Hi R users, Is there a function in R, which does some calculation only for the month end in a daily data?... In other words, is there a command in R, equivalent to last. function in SAS? BR, Shubha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Month-end-calculations-tf4347226.html#a12390874 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to signal the end of the table?
I am using a for loop to read a table row by row and I have to specify how
many records are there in the table. I need to read row by row because the
table is huge and the memory not large enough for the whole table.:
number.of.records=100
fp=file(abc.csv,r)
pos=seek(fp, rw=read)
for (i in 1:number.of.record){
current.row=scan(file=fp, sep=',', what=list(count=1, cusip6=, idate=1,
spread=1.1, vol252=1.1, vol1000=1.1, st_debt=1.1, lt_debt=1.1 , total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1, sec=1.1, cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
I need to know the number of records in the table and put it in the variable
named number.of.records. When I have a new table that I do not know how
many records it has, I use excel to open the file to figure it out and put
it in variable number.of.records. I often have many tables to try and
every one of them has thousands of recordsit takes a lot of time and
trouble to adjust the code every time I read a different table.
I am wondering if I can change the for loop to a while loop:
while (the end of the table has not been reached)
{
current.row=scan(file=fp, sep=',', what=list(count=1, cusip6=, idate=1,
spread=1.1, vol252=1.1, vol1000=1.1, st_debt=1.1, lt_debt=1.1 , total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1, sec=1.1, cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
The problem is how to articulate while (the end of the table has not been
reached), or equivalently, how to signal the end of the table?
Best Wishes
Yuchen Luo
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] customizing the color and point shape for each line drawn using lattice's xyplot
Does this help common.without.Method4 - subset(common, Method!=4) xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common.without.Method4, groups=Method.f, type=l, auto.key=T) Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gen Sent: Thursday, 30 August 2007 2:51 PM To: r-help@stat.math.ethz.ch Subject: [R] customizing the color and point shape for each line drawn using lattice's xyplot Description of what I am trying to do: I am using the xyplot code below to plot the variable MeanBxg against the variable PercentVarExplained for all 9 possible combinations of variables bdg and bdx. Within each of these 9 scenarios I am plotting a separate line for each of up to 9 different methods that I used to estimate the variable MeanBxg. These methods are identified by the numeric variable called Method. Giving me one plot with 9 figures and each of the figures contains 9 lines. My problem arises because I would like to repeat the creation of this plot 8 times, in each instance only a subset (eg 6) of the 9 methods are used (a different subset each time). What I can't figure out: I would like to learn how to specify the exact line color that corresponds to each method such that Method==1 will always be represented by the same color (in every plot that it appears in). Where two methods that I used were of the same family of methods (say method==1 and method==2 made the same assumptions about the data) I would like to, if possible, represent the two methods using the same color and distinguish them by the symbol used to represent points on the line. My code as it currently stands: xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common, groups=common$Method.f, type=l, subset= Method!=4, auto.key=T) As the code is, the default colors assigned are repeated causing different methods to be represented by the same color with no way to distinguish them (I have not succeeded in plotting lines and points simultaneously). Side question: When I subset the data to particular methods, is there a way to remove the excluded methods from the key as well? (in my code Method is a numeric variable, and Method.f corresponds to the lengthy descriptions of each method for the purpose of the key) Thank you very much for your help. Genevieve -- View this message in context: http://www.nabble.com/customizing-the-color-and-point-shape-for-each-lin e-drawn-using-lattice%27s-xyplot-tf4351934.html#a12400517 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Incomplete output with `sn' library package
Dear R users: I have a question regarding the output for two of the functions in the `sn' package, which deals with the mle fitting of skew normal curves to linear regressions. I'm using the examples and the dataset given as an example in the online documentation for this package, for the functions `msn.fit' and `msn.mle'. I'm following the example code in the documentation for these two functions exactly. Part of the data output is supposed to be se, which gives the standard errors of the estimated coefficients. This particular value comes out as being NA in the examples given, but there are three coefficients in each case and no numerical problems about why the standard errors cannot be calculated. Am I setting this program up right? Is there some other command I should use (or an option I need to use) to get the output to display standard errors of the coefficients? Thank you for your time in reading this question - Cordially, Manasi __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Reading lines from file
uv wrote:
Hi. I have a text file containing a few hundred lines of numbers, each line
has a different length. For example:
1 4 1 1 7
3 11 1 1 1 1 1 1 2
2 4 1 2
And so on. I need to do a simple plot function for each line, and then to
save each plot into a separate file. Is there any way doing it from within
R?
You can use read.table with the fill argument:
read.table(file.txt,fill=NA)
V1 V2 V3 V4 V5 V6 V7 V8 V9
1 1 4 1 1 7 NA NA NA NA
2 3 11 1 1 1 1 1 1 2
3 2 4 1 2 NA NA NA NA NA
and then loop over rows. You may have to tell read.table the maximum
number of columns in your data, since it only looks at the first five
rows to have a guess at the size.
Or you can use readLines(file.txt) to read each line into an element
of a character vector, then use strsplit() to break it up:
lapply(readLines(file.txt),
function(s){
as.numeric(unlist(strsplit(s,split= )))
})
[probably a neater way to do this but I'm not in the mood for playing R
golf today]
which gives a list:
[[1]]
[1] 1 4 1 1 7
[[2]]
[1] 3 11 1 1 1 1 1 1 2
[[3]]
[1] 2 4 1 2
You could then lapply() on this list to do whatever to the elements,
in your case the plotting.
Barry
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Re: [R] Excel
On 30/8/07 6:42 AM, Erich Neuwirth wrote: There is one feature in Excel which is extremely convenient, Pivot tables. Anybody doing any work as statistical consultant really ought to know about Pivot tables, and I am still surprised how many statisticians do not know about it. Neither Gnumeric nor OpenOffice Calc offer comparably convenient ways working with multidimensional tables. I'm not familiar with Gnumeric, but OpenOffice has provided the similar DataPilot feature for at least three years; see for example http://www.openofficetips.com/blog/archives/2004/09/datapilot_101.html James -- James Reilly Department of Statistics, University of Auckland Private Bag 92019, Auckland, New Zealand __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Incomplete output with `sn' library package
MANASI VYDYANATH wrote: Dear R users: I have a question regarding the output for two of the functions in the `sn' package, which deals with the mle fitting of skew normal curves to linear regressions. I'm using the examples and the dataset given as an example in the online documentation for this package, for the functions `msn.fit' and `msn.mle'. I'm following the example code in the documentation for these two functions exactly. Part of the data output is supposed to be se, which gives the standard errors of the estimated coefficients. This particular value comes out as being NA in the examples given, but there are three coefficients in each case and no numerical problems about why the standard errors cannot be calculated. Am I setting this program up right? Is there some other command I should use (or an option I need to use) to get the output to display standard errors of the coefficients? We cannot know if you use it right, since you have not given any details on OS, R version, sn version, and particularly a reproducible example. As each R-help message tells in the footer: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Uwe Ligges Thank you for your time in reading this question - Cordially, Manasi __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Strage result with an append/strptime combination
Hi again, Just a quick post to propose another solution (without chron) : strptime function on the whole object and not on each namefile. namefile-070707050642.dat namefile-append(namefile,namefile) namefile-append(namefile,namefile) namefile [1] 070707050642.dat 070707050642.dat 070707050642.dat 070707050642.dat jourheure-strptime(namefile,%d%m%y%H%M%S) jourheure [1] 2007-07-07 05:06:42 2007-07-07 05:06:42 2007-07-07 05:06:42 2007-07-07 05:06:42 Gabor Grothendieck wrote: Try chron: library(chron) namefile - 070707050642.dat#day-month-year-hour-minute-second.dat x - chron(substr(namefile, 1, 6), substr(namefile, 7, 12), + format = c(dmy, hms), out.format = c(m/d/y, h:m:s)) c(x, x) [1] (07/07/07 05:06:42) (07/07/07 05:06:42) See R News 4/1 Help Desk article for more. Hi, I keep on trying to write some small scripts in order to learn R but even with basic scripts I have problems ... I start with the name of a file which is in fact the time the file has been generated (I cannot change the format). Then I convert namefile with strptime. The problem occurs when I add another time from another file with append. It displays some informations I don't want. I found a post about this problem (http://www.nabble.com/Error-with-strptime-tf3607942.html#a10081942) but I don't understand the solution. I tested as.POSIXct or as.POSIX.lt but it has no effect. Do you have some ideas to solve this problem ? Thank you for your help. Ptit Bleu. --- namefile-070707050642.dat#day-month-year-hour-minute-second.dat jourheure-strptime(namefile,%d%m%y%H%M%S) jourheure [1] 2007-07-07 05:06:42 jourheure-append(jourheure,jourheure) jourheure [1] 2007-07-07 05:06:42 Paris, Madrid (heure d'été) 2007-07-07 05:06:42 Paris, Madrid (heure d'été) -- View this message in context: http://www.nabble.com/Strage-result-with-an-append-strptime-combination-tf4347401.html#a12385852 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Strange-result-with-an-append-strptime-combination-tf4347401.html#a12403499 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple Graph
amna khan wrote: Hi Sir We use the function mpg() for ticks on axes. If we define more number of ticks then each tick is not labeled. How to label each tick on axes? What is mpg()? Which version of R? Which device? Which OS? I guess R cannot print more labels because the ticks are too dense, but there is no reproducible example in your message. PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Uwe Ligges Regards __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Q: Mean, median and confidence intervals with functions summary boxplot.stats
Tom Willems wrote: Dear R ussers, My question is, How can my mean be outside the confidence intervals ?! I think i have the answer for it, but i would like to hear some other ideas on it. First my data is not continuose but categorical, it is a titre calculated on a dilution serie. It is stored as a column of values, and a column indicating the phase of the trail. Theoreticaly it is possible to have a value ranging from 0 to 4, but in practice, only sertain values will occure, and they will repeat. So it are frequencies. This is why i belief that it is better to work with a median than with a mean, because it represents the cluster of values wich occure most. Below I only give one example, but the mean being below the lowest confidence limit occures several times over different tests. does my answer seam reasonable, or should i perhapes use an other methode, any sugestion? summary_1d = summary(subset(eda_data, phase=='1' test=='test 1' ,select=lg_value), na.rm = T) conf_1d = boxplot.stats(subset(eda_data, phase=='1' test=='test 1' ,select=lg_value)) MeanMedian 95% Confidence Int. StDev. Variance 1.1981.681 1.4411.922 0.931 0.866 I do not understand which confidence has been calculated? Based on which assumptions / data? Is it pointwise or not? We need much more information - and if you think it is a problem with R or usage of R functions, then please give us a reproducible example. Uwe Ligges Kind regards, Tom W. Disclaimer: click here [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 signal the end of the table?
What about trying to use a database system and make queries?
If you are at the end of a file, you will read elements of length 0...
Uwe Ligges
Yuchen Luo wrote:
I am using a for loop to read a table row by row and I have to specify how
many records are there in the table. I need to read row by row because the
table is huge and the memory not large enough for the whole table.:
number.of.records=100
fp=file(abc.csv,r)
pos=seek(fp, rw=read)
for (i in 1:number.of.record){
current.row=scan(file=fp, sep=',', what=list(count=1, cusip6=, idate=1,
spread=1.1, vol252=1.1, vol1000=1.1, st_debt=1.1, lt_debt=1.1 , total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1, sec=1.1, cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
I need to know the number of records in the table and put it in the variable
named number.of.records. When I have a new table that I do not know how
many records it has, I use excel to open the file to figure it out and put
it in variable number.of.records. I often have many tables to try and
every one of them has thousands of recordsit takes a lot of time and
trouble to adjust the code every time I read a different table.
I am wondering if I can change the for loop to a while loop:
while (the end of the table has not been reached)
{
current.row=scan(file=fp, sep=',', what=list(count=1, cusip6=, idate=1,
spread=1.1, vol252=1.1, vol1000=1.1, st_debt=1.1, lt_debt=1.1 , total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1, sec=1.1, cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
The problem is how to articulate while (the end of the table has not been
reached), or equivalently, how to signal the end of the table?
Best Wishes
Yuchen Luo
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Single plot multiple levels in x?
Richard Yanicky wrote: Plotting with 2 x axis? One axis inside another, for example salary within state, 1-50 | 50 – 100 | 100+ | 1- 50 | 50 -100 | 100+ | … repeated bins for salary AL ! AR …… more states Sounds like the lattice package does exactly what you want, but without any reproducible example. Uwe Ligges The values are all stored with a single data frame. I have tried different things with the axis function and done many searches for plotting. Can’t find a direct reference Thanks. Richard __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Simple Graph
Hi Sir If I want 10 ticks on x-axis and 10 ticks on y-axis then I use lab=c(10,10, 20) Here first 10 is for 10 ticks on x-axis, second 10 is for 10 ticks on y-axis and 20 is the lenght of x and y labels in characters. Now my question is that how I can label all 10 ticks on x-aixs and all 10 ticks on y-axis, because R leaves some ticks unlabeled. Thank You -- AMINA SHAHZADI Department of Statistics GC University Lahore, Pakistan. Email: [EMAIL PROTECTED] [EMAIL PROTECTED] [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] (no subject)
Dear R users,I am looking for an easy (i.e., direct) way in obtaining the F and p values from the intercept in anovas with within-subject designs. Data are from a psychophysics expeirment where I am using d' (d-prime) values obtained from 3 modalities of presentation in each subject.I would like to know not only whether there is an effect of modality, but also wether the main effect is signiticant (meaning that d' is differnt from 0). I know that a t.test again the nul mean would provide me with similar stats on the main effect, but I would like to get those stats in an F form, for consistency with the stats on the other factors of interest. As far as I understand how R works, the function anova provides you with such information but it is restricted to between-group analyses. For within-subject designs, one has to use summray(aov) but stats on the intercept are not included in the the result of this function. I have pasted an exemple below. As onc can see only the Sum of Sq and Mean Sq are given from the main effect.Thank you for any advice you can provide,-Sid summary(aov(x~mod+Error(suj/(mod)), data=dp))Error: suj Df Sum Sq Mean Sq F value Pr(F)Residuals 10 19.5977 1.9598 Error: suj:mod Df Sum Sq Mean Sq F value Pr(F) mod2 8.2475 4.1237 4.2955 0.02806 *Residuals 20 19.2005 0.9600 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Error: Within Df Sum Sq Mean Sq F value Pr(F)Residuals 33 55.812 1.691 _ tiative now. Its free. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple Graph
amna khan wrote: Hi Sir If I want 10 ticks on x-axis and 10 ticks on y-axis then I use lab=c(10,10, 20) Here first 10 is for 10 ticks on x-axis, second 10 is for 10 ticks on y-axis and 20 is the lenght of x and y labels in characters. Now my question is that how I can label all 10 ticks on x-aixs and all 10 ticks on y-axis, because R leaves some ticks unlabeled. So either make the font size smaller or rotate the labels... Uwe Ligges Thank You __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to calculate weighted means by groups
Hello, I want to calculate the weighted means of a table column, aggregated by other elements of the same table. For example I want to calculate mean species numbers weighted by area and grouped by year, altitude and/or region. I think a combination of aggregate() and weighted.mean() should work, but I always get 'Errors'. this was my favorite: aggregate(tab, list(year = tab[,year], altitude= tab[,altit], region= tab[,region]), weighted.mean(tab[,spcs], tab[,area])) with a table like: spcs type altit area year region 38 a low 202 1980 W 40 a low 326 2000 W 45 b medium 207 1980 W 48 b medium 205 2000 W 50 a low 45 1980 E 53 a low 104 2000 E 43 c medium 7 1980 E 33 c medium 9 2000 E 23 c high 20 1980 E 13 c high 26 2000 E Who can help? Thanks in advance, Patrick __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 signal the end of the table?
Have you tried read.table() and the similar functions? I think they
would provide a much more simple solution for your problem.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be
Do not put your faith in what statistics say until you have carefully
considered what they do not say. ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Namens Yuchen Luo
Verzonden: woensdag 29 augustus 2007 19:41
Aan: r-help@stat.math.ethz.ch
Onderwerp: [R] How to signal the end of the table?
I am using a for loop to read a table row by row and I have
to specify how many records are there in the table. I need to
read row by row because the table is huge and the memory not
large enough for the whole table.:
number.of.records=100
fp=file(abc.csv,r)
pos=seek(fp, rw=read)
for (i in 1:number.of.record){
current.row=scan(file=fp, sep=',', what=list(count=1,
cusip6=, idate=1, spread=1.1, vol252=1.1, vol1000=1.1,
st_debt=1.1, lt_debt=1.1 , total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1,
sec=1.1, cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
I need to know the number of records in the table and put it
in the variable named number.of.records. When I have a new
table that I do not know how many records it has, I use excel
to open the file to figure it out and put it in variable
number.of.records. I often have many tables to try and every
one of them has thousands of recordsit takes a lot of
time and trouble to adjust the code every time I read a
different table.
I am wondering if I can change the for loop to a while loop:
while (the end of the table has not been reached)
{
current.row=scan(file=fp, sep=',', what=list(count=1,
cusip6=, idate=1, spread=1.1, vol252=1.1, vol1000=1.1,
st_debt=1.1, lt_debt=1.1 , total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1,
sec=1.1, cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
The problem is how to articulate while (the end of the table
has not been reached), or equivalently, how to signal the
end of the table?
Best Wishes
Yuchen Luo
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[R] Data simulation with R
Hi, I am currently on a placement here at GSK for my studies, and I'm working on Heckman Models. I have to make simulations, in order to see whether these models are efficient or not. I have to generate a dataset under the following constraints: - outcome is 0 or 1 - one control group, one treatment group, there must be no treatment effect - generate one continuous variable and one binary variable which are both perfectly correlated to treatment, but not correlated to outcome - generate one continuous variable and one binary variable which are both perfectly correlated to outcome, but not correlated to treatment - generate one continuous variable and one binary variable which are both perfectly correlated to both treatment and outcome Problem is, it's my first time generating data, so I'm having quite a hard time... I usually use SAS, but I didn't manage to generate data using it. Some people told me it's quite easy to do with R, but I'm not really used to this program. Is it really possible to simulate this kind of data with R ? If it is, can someone help me ? Thanks in advance. -- View this message in context: http://www.nabble.com/Data-simulation-with-R-tf4353531.html#a12405063 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R CMD BATCH: cat does not print
Dear All, I am trying to write my first R script. The code is simply cat(Hello!\n) However, when I run $ R CMD BATCH myscript.R I do not see Hello! on the console. I am using Fedora 7 (Linux) and R-2.5.1. Any ideas? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] bug in DEoptim package
(the same mail was sent to the author)
When I called the function DEoptim with control=list(strategy=1) or
control=list(strategy=2)
I got the error:
Error in mui[rtd + 1, i] : incorrect number of dimensions
Analysis of the source code of the DEoptim reveals the following fragment
if (con$strategy 5)
st - con$strategy - 5 ## binomial crossover
else {
st - con$strategy ## exponential crossover
mui - sort(t(mui)) ## transpose, collect 1's in each column
for (i in 1:NP) {
n - floor(runif(1) * d)
if (n 0) {
rtd - (rotd + n) %% d
mui[,i] - mui[rtd + 1,i] ## rotate column i by n
}
}
mui - t(mui) ## transpose back
}
sort returns a 1-dimensional vector, this causes the error in the
following indexing operators.
This also shows, that any strategy value from 1 through 5 will cause
this error, while values from 1 through 10 are possible.
Package's DESCRIPTION file contains the following:
Version: 1.1-8
Date: 2007-01-29
My sessionInfo():
sessionInfo()
R version 2.5.1 Patched (2007-08-19 r42614)
i386-pc-mingw32
locale:
LC_COLLATE=Russian_Russia.1251;LC_CTYPE=Russian_Russia.1251;LC_MONETARY=Russian_Russia.1251;LC_NUMERIC=C;LC_TIME=Russian_Russia.1251
attached base packages:
[1] tcltk stats graphics grDevices utils datasets
methods base
other attached packages:
debug mvbutils DEoptim
1.1.0 1.1.1 1.1-8
Unfortunately I haven't submerged deep enough to the algorithm details to
propose the bug fix.
--
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http://www.nabble.com/bug-in-DEoptim-package-tf4353533.html#a12405068
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Re: [R] R CMD BATCH: cat does not print
Paul Smith wrote: Dear All, I am trying to write my first R script. The code is simply cat(Hello!\n) However, when I run $ R CMD BATCH myscript.R I do not see Hello! on the console. I am using Fedora 7 (Linux) and R-2.5.1. Any ideas? You shouldn't see it on the console! BATCH writes its output to a file. You should find a file called myscript.Rout that does contain the 'Hello!'. Barry __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] breaking the x-axis and having two different x-axis labels
Georg Ehret wrote: Dear R community, I have two questions concerning barplots that I struggle to resolve: 1) How can I break (interrupt) the x-axis (e.g.: have it display values from -100 to -90 and 90 to 100 only)? I think you mean a horizontal barplot with a gap between -90 and 90. pyramid.plot (plotrix) doesn't quite do what you want, as the present version doesn't allow customized axis labels. The principle is the same as that used in gap.barplot, but the maintainer (me) didn't think that anyone would want to do this horizontally. If you are stuck I can add a feature or two... However, here's an unbeauteous hack. heights-c(sample(-100:-91,5),sample(91:100,5)) barplot(c(heights[1:5]+90,heights[6:10]-90), horiz=TRUE,xaxt=n,xlim=c(-10,10)) axis(1,at=seq(-9,9,by=2), labels=c(-99,-97,-95,-93,-91,91,93,95,97,99)) axis.break(1,-0.1,style=gap) 2) I overlay two horizontal barplots: one with negative values only and one with positive values only: I would wish to mark the two datasets separately on the x-axis (or some other way). If I use the legend I get a superposed result... is there another way? This sounds like a job for axis, but I can't really picture what you describe. Do you want the negative bars starting from the right and going left and the positive ones starting from the left and going right? Or a vertical line in the center with the bars going in the usual directions? Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R CMD BATCH: cat does not print
Use rscript Rscript myscript.R or Rscript -e 'cat(Hello!\n)' will show Hello! on the console. R CMD BATCH writes its output to the file myscript.Rout Paul Smith wrote: Dear All, I am trying to write my first R script. The code is simply cat(Hello!\n) However, when I run $ R CMD BATCH myscript.R I do not see Hello! on the console. I am using Fedora 7 (Linux) and R-2.5.1. Any ideas? Thanks in advance, Paul -- View this message in context: http://www.nabble.com/R-CMD-BATCH%3A-cat-does-not-print-tf4353572.html#a12405494 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R CMD BATCH: cat does not print
On 8/30/07, Barry Rowlingson [EMAIL PROTECTED] wrote: I am trying to write my first R script. The code is simply cat(Hello!\n) However, when I run $ R CMD BATCH myscript.R I do not see Hello! on the console. I am using Fedora 7 (Linux) and R-2.5.1. Any ideas? You shouldn't see it on the console! BATCH writes its output to a file. You should find a file called myscript.Rout that does contain the 'Hello!'. Thanks, Barry. Indeed, the file myscript.Rout exists and contains the output of cat. I was expecting a behavior similar to the bash scripts. And by the way, cannot a R script write only on the console and just what one tells it to write, likewise bash scripts? Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R CMD BATCH: cat does not print
On 8/30/07, Vladimir Eremeev [EMAIL PROTECTED] wrote: Use rscript Rscript myscript.R or Rscript -e 'cat(Hello!\n)' will show Hello! on the console. R CMD BATCH writes its output to the file myscript.Rout Thanks, Vladimir. Rscript is exactly what I was looking for! Paul Paul Smith wrote: Dear All, I am trying to write my first R script. The code is simply cat(Hello!\n) However, when I run $ R CMD BATCH myscript.R I do not see Hello! on the console. I am using Fedora 7 (Linux) and R-2.5.1. Any ideas? Thanks in advance, Paul -- View this message in context: http://www.nabble.com/R-CMD-BATCH%3A-cat-does-not-print-tf4353572.html#a12405494 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Simple Graph
amna khan wrote: Hi Sir We use the function mpg() for ticks on axes. If we define more number of ticks then each tick is not labeled. How to label each tick on axes? Hi Amina, The staxlab function in the plotrix package will display tick labels even if they would overlap on the standard axis display. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] piecewise linear approximation
If you want to minimize absolute error for this, then you can
try the rqss fitting in the quantreg package and tune lambda
to get one break in the fitted function.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Aug 29, 2007, at 8:05 PM, Achim Zeileis wrote:
On Wed, 29 Aug 2007, Naxerova, Kamila wrote:
Dear list,
I have a series of data points which I want to approximate with
exactly two
linear functions. I would like to choose the intervals so that the
total
deviation from my fitted lines is minimal. How do I best do this?
From the information you give it seems that you want to partition
a model
like
lm(y ~ x)
along a certain ordering of the observations. Without any further
restrictions you can do that with the function breakpoints() in
package
strucchange. If there are continuity restrictions or something like
that, you want to look at the segmented package.
hth,
Z
Thanks!
Kamila
The information transmitted in this electronic communication...
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[R] Assigning line colors in xyplot
Hi, I have a dataframe containing data from individuals 1, ..., 12 (grouping variable g in the data frame below), which belong either to A or B (grouping variable f): set.seed(1) tmp - data.frame( y=c(rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2)), x=1:10, f=c(rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10)), g=c(rep(3,10),rep(2,10),rep(1,10),rep(4,10),rep(5,10),rep(6,10),rep(8,10),rep(7,10),rep(9,10),rep(11,10),rep(12,10),rep(10,10))) I would like to draw line plots using the function xyplot: library(lattice) xyplot(y ~ x | g , groups=g, data=tmp,type=l, par.settings=list(superpose.line=list(col=c(red,blue))), auto.key=list(space=top, text=levels(tmp$f),points=FALSE,lines=TRUE)) As it is, the colors are recycled alternately in the order the individuals appear in the plot (1, 10, 11, 12, 2, ..., 9). How can I assign the red color to all individuals of group A and the blue color to all individuals of group B? Thanks in advance! Christof __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 into time series object
You can use the 'ts' function. Example df - ts(your_data_frame, start=c(1990,1), #The initial date of your data end=c(2000,12), #The final date of your data or frequency=12)#The frequency of your data, so you don't need of the 'end' argument See ?ts and ?as.ts for more. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 29/08/2007, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote: Hi, I have a dataframe of trading dates along with the corresponding prices. I need to convert this into a time series object. How do I do this with my price values being the time series object and the dates/time being the trading dates. BR, Shubha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Q: Mean, median and confidence intervals with functions summary boxplot.stats
If you look at ?boxplot.stats, you will find that the confidence interval it
reports is centred on the median and :
The notches (if requested) extend to '+/-1.58 IQR/sqrt(n)'.
If you have skewed data it is very possible (as you have found) that the mean
is outside median+/-1.58 IQR/sqrt(n).
All that is happening is that the majority of the data are around 1 or 2 and
you have a substantial number near zero. Result: mean much lower than median.
And with a high n, the boxplot notch is very narrow and excludes the mean.
But it does sound very much as if you are doing something questionable at best.
I would not trust IQR as a dispersion measure on discrete data with few
possible values even if they were on an interval scale; too much risk of
getting the same IQR for many different distributions. On an ordinal scale it
is worse; the only points that are valid at all are the valid scale values, so
a CI that uses intermediate values is formally meaningless (what is a shoe size
of 7.2, for example? Answer: Not a shoe size at all). It is of course entirely
meaningless to talk about an IQR on a categorical scale.
It sounds like boxplot.stats is an inappropriate tool for summarising your data.
Tom Willems [EMAIL PROTECTED] 30/08/2007 10:00:50
Dear R ussers,
My question is, How can my mean be outside the confidence intervals ?!
I think i have the answer for it, but i would like to hear some other
ideas on it.
First my data is not continuose but categorical, it is a titre calculated
on a dilution serie.
It is stored as a column of values, and a column indicating the phase of
the trail.
Theoreticaly it is possible to have a value ranging from 0 to 4, but in
practice, only sertain values will occure, and they will repeat.
So it are frequencies.
This is why i belief that it is better to work with a median than with a
mean, because it represents the cluster of values wich occure most.
Below I only give one example, but the mean being below the lowest
confidence limit occures several times over different tests.
does my answer seam reasonable, or should i perhapes use an other methode,
any sugestion?
summary_1d = summary(subset(eda_data, phase=='1' test=='test
1' ,select=lg_value), na.rm = T)
conf_1d = boxplot.stats(subset(eda_data, phase=='1'
test=='test 1' ,select=lg_value))
MeanMedian 95% Confidence Int. StDev.
Variance
1.1981.681 1.4411.922 0.931
0.866
Kind regards,
Tom W.
Disclaimer: click here
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Re: [R] Excel (off-topic, sort of)
Try prolog. You can do this sort of programming there.
Albicelli, Nicholas (Exchange) wrote:
Except for the ability to perform circular recalculation, I believe that
the closest programming analogy to a spreadsheet is a functional
programming language. Check out Haskell (or LISP or Erlang) to do what
you describe.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of François Pinard
Sent: Wednesday, August 29, 2007 11:36 AM
To: Alberto Monteiro
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Excel (off-topic, sort of)
[Alberto Monteiro]
Maybe I'll write a letter to Santa Claus [there are people
who write to congressman; they must have more faith than me].
:-) :-)
I wish a language where I can write
a = b + 10
and then when I write
a = 20
the language automatically assigns b = 10.
METAFONT does this (and consequently, Metapost as well). I still
remember my surprise when I found out that Donald Knuth resorts to such
sophisticated machinery for the sole purpose of designing font
characters. Knuth surely did many wonderful things :-).
--
François Pinard http://pinard.progiciels-bpi.ca
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***
Bear Stearns is not responsible for any recommendation, soli...{{dropped}}
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[R] boxplot will remember the factor levels
R-help, I'm trying to do a simple box-and-whisker plot to some data. The data are a subset of a large data frame but when running the boxplot function on the subset data all the factors are still present in the graph leaving a huge empty space until the actuals factors are shown. This produces a spurious box-and-whisker plot. If the subset data are exported to another R session the problem is gone. Why are the factors still remembered by the boxplot? Attached is a copy of the data. Thanks in advance ## the line code boxplot(mLength ~ puntar,data=test) version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 5.1 year 2007 month 06 day27 svn rev42083 language R version.string R version 2.5.1 (2007-06-27) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R CMD BATCH: cat does not print
Paul Smith wrote: Thanks, Barry. Indeed, the file myscript.Rout exists and contains the output of cat. I was expecting a behavior similar to the bash scripts. And by the way, cannot a R script write only on the console and just what one tells it to write, likewise bash scripts? Not easily, I think. The 'BATCH' command is really intended for long-running or off-line jobs that may be disconnected from a terminal. Barry __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to mask or escape =
-- View this message in context: http://www.nabble.com/How-to-mask-or-escape-%22%3D%22-tf4354174.html#a12406926 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to mask or escape = in Windows command prompt?
I have defined a function with several arguments and have it stored in the
.RData file.
The 'function head' is defined as follows
EstimALIConc
-function(sdname,SZ,W,farea,watri,biomodel,start.part=1,nparts=20,method=c(optim,DEoptim))
{
[ blah-blah-blah ]
(function body doesn't matter)
}
Then I call Rscript:
e: rscript --restore -e
EstimALIConc('17-aug',27.8,5,1,watri,biomodel,1,100,method='DEoptim');warnings();
117-aug-log.txt 217-aug.err
and get the error in the 17-aug.err:
Error in -args : invalid argument to unary operator
Execution halted
If I remove method= from the function call, then everything works fine,
and function executes.
e: rscript --restore -e
EstimALIConc('17-aug',27.8,5,1,watri,biomodel,1,100,'DEoptim');warnings();
117-aug-log.txt 217-aug.err
The problem seems to be that the equality sign acts as a separator.
How to avoid this?
This would allow skiping of the default arguments, for example start.part
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Re: [R] boxplot will remember the factor levels
Try: boxplot(mLength ~ puntar[drop=T],data=test) -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 30/08/2007, Luis Ridao Cruz [EMAIL PROTECTED] wrote: R-help, I'm trying to do a simple box-and-whisker plot to some data. The data are a subset of a large data frame but when running the boxplot function on the subset data all the factors are still present in the graph leaving a huge empty space until the actuals factors are shown. This produces a spurious box-and-whisker plot. If the subset data are exported to another R session the problem is gone. Why are the factors still remembered by the boxplot? Attached is a copy of the data. Thanks in advance ## the line code boxplot(mLength ~ puntar,data=test) version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 5.1 year 2007 month 06 day27 svn rev42083 language R version.string R version 2.5.1 (2007-06-27) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xeon processor and ATLAS
hui xie [EMAIL PROTECTED] writes: The reason why I would like to use ATLAS is that R FAQ said : The savings can be appreciable: on a 2.6GHz P4 and a 1000 x 1000 matrix svd took 16.2 sec with the standard BLAS and 7.8 sec with ATLAS. Because ATLAS is tuned to a particular chip we can't use it generally: the optimal routines for a P4 or an Athlon XP are quite different and neither will run at all on a PII. This seems to me an impressive gain to use the correct ATLAS instead of the default BLAS. I've been doing econometrics for nearly 20 years, and have not yet run across a situation that called for looking at a 1000 x 1000 matrix. I tend not to believe analyses with more than a dozen explanatory variables. But I suppose that you could run across a situation where a design matrix with hundreds of columns actually made sense. But does that really happen often enough that saving 8 seconds of computation time is worth going through the ATLAS hassle? -- Jeff __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Another issue with the Matrix package.
Hi all, I am encountering a strange issue with the Matrix package. I have just built R-devel from source on my macbook pro, and I wonder if others can reproduce this problem. I will give example code to go along: Starting a fresh R session: R version 2.6.0 Under development (unstable) (2007-08-30 r42697) Copyright (C) 2007 The R Foundation for Statistical Computing ISBN 3-900051-07-0 ... log(2) [1] 0.6931472 library(Matrix) Loading required package: lattice log(2) Error in log(2) : could not find symbol base in environment of the generic function There seems to be something wrong here and I cannot figure out what it is. Am I doing something wrong or is it an issue with Matrix (which is what I have narrowed it down to). I think that it might be a namespace collision or something, but I am certainly not sure. Here is my sessionInfo() output: sessionInfo() R version 2.6.0 Under development (unstable) (2007-08-30 r42697) i386-apple-darwin8.10.1 locale: C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] Matrix_0.99875-2 lattice_0.16-3 loaded via a namespace (and not attached): [1] grid_2.6.0 Thanks. Tony [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] categorical variable coefficients in QSAR
Dear list: I am interested in the following sort of problem, as is found frequently in the field of QSAR. I have biological activity as a function of chemical structure, with structure defined in a categorical manner in that the SUBSTITUENT is the levels of the POSITION factor. For example, data from Kubinyi (http://www.kubinyi.de/dd-12.pdf) for this type of analysis is presented as follows: factor para: H F Cl Br I Me H H H H H F F F Cl Cl Cl Br Br Br Me Me factor meta: H H H H H H F Cl Br I Me Cl Br Me Cl Br Me Cl Br Me Me Br observed biological activity: 7.46 8.16 8.68 8.89 9.25 9.30 7.52 8.16 8.30 8.40 8.46 8.19 8.57 8.82 8.89 8.92 8.96 9.00 9.35 9.22 9.30 9.52 I then think the following analysis should be appropriate meta-factor(scan(file=meta,what=character)) para-factor(scan(file=para,what=character)) ba-scan(file=ba) rslt-lm(ba~meta+para-1) What I wish to obtain is a coefficient for each substituent at each position, as does Kubinyi: H F Cl Br I Me meta 0.00 -0.30 0.21 0.43 0.58 0.45 para 0.00 0.34 0.77 1.02 1.43 1.26 However, I do not get a coefficient for the Br substituent at the para position. I would like to know if there is an error in this formulation. The technique is quite well established in the field of medicinal chemistry and it is traditional that the binary incidence matrix is formed by hand as an intermediate step in the analysis, instead of the much simpler formulation that I am considering here. Thank you for whatever insight you may give. Prof. Roy Little Dept. Chem. Universidad de los Andes Mérida, Venezuela __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] categorical variable coefficients in QSAR
Dear list: I am interested in the following sort of problem, as is found frequently in the field of QSAR. I have biological activity as a function of chemical structure, with structure defined in a categorical manner in that the SUBSTITUENT is the levels of the POSITION factor. For example, data from Kubinyi (http://www.kubinyi.de/dd-12.pdf) for this type of analysis is presented as follows: factor para: H F Cl Br I Me H H H H H F F F Cl Cl Cl Br Br Br Me Me factor meta: H H H H H H F Cl Br I Me Cl Br Me Cl Br Me Cl Br Me Me Br observed biological activity: 7.46 8.16 8.68 8.89 9.25 9.30 7.52 8.16 8.30 8.40 8.46 8.19 8.57 8.82 8.89 8.92 8.96 9.00 9.35 9.22 9.30 9.52 I then think the following analysis should be appropriate meta-factor(scan(file=meta,what=character)) para-factor(scan(file=para,what=character)) ba-scan(file=ba) rslt-lm(ba~meta+para-1) What I wish to obtain is a coefficient for each substituent at each position, as does Kubinyi: H F Cl Br I Me meta 0.00 -0.30 0.21 0.43 0.58 0.45 para 0.00 0.34 0.77 1.02 1.43 1.26 However, I do not get a coefficient for the Br substituent at the para position. I would like to know if there is an error in this formulation. The technique is quite well established in the field of medicinal chemistry and it is traditional that the binary incidence matrix is formed by hand as an intermediate step in the analysis, instead of the much simpler formulation that I am considering here. Thank you for whatever insight you may give. Prof. Roy Little Dept. Chem. Universidad de los Andes Mérida, Venezuela __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to signal the end of the table?
Well, i am surprise you have problems to read a table that is small enough to
be opened entirely in Excel.
I work with csv tables with hundreds of thousands of rows, and sometimes even
millions with no problems except that sometimes i have to wait up to 1 or
2 minutes for R to read the table. I use the command read.csv.
If your tables are always small enough to be read in Excel, the limit of number
of rows in Excel is about 64000 or something of that sort (maybe 65000 ) so
you can use this number as a limit for your number of rows although you
may have to clean-up your data.frame afterwards.
I work on a Windows machine with 4 Gb DRAM - just for comparison.
I hope this helps,
Monica
Message: 88Date: Wed, 29 Aug 2007 10:41:05 -0700From: Yuchen Luo [EMAIL
PROTECTED]Subject: [R] How to signal the end of the table?To:
r-help@stat.math.ethz.chMessage-ID:[EMAIL PROTECTED]Content-Type:
text/plainI am using a for loop to read a table row by row and I have to
specify howmany records are there in the table. I need to read row by row
because thetable is huge and the memory not large enough for the whole
table.:number.of.records=100fp=file(abc.csv,r)pos=seek(fp,
rw=read)for (i in 1:number.of.record){current.row=scan(file=fp, sep=',',
what=list(count=1, cusip6=, idate=1,spread=1.1, vol252=1.1, vol1000=1.1,
st_debt=1.1, lt_debt=1.1 , total_liab=1.1, cr=1.1, shrout=1.1, prc=1.1,
mkt_cap=1.1, rtng=1.1, sec=1.1, cr3m=1.1,cr5y=1.1, ust3m=1.1, ust5y=1.1),
flush=TRUE, nlines=1,quiet=T)...}I need to know the number of records
in the table and put it in the variablenamed nu!
mber.of.records. When I have a new table that I do not know howmany records
it has, I use excel to open the file to figure it out and putit in variable
number.of.records. I often have many tables to try andevery one of them has
thousands of recordsit takes a lot of time andtrouble to adjust the code
every time I read a different table.I am wondering if I can change the
for loop to a while loop:while (the end of the table has not been
reached){current.row=scan(file=fp, sep=',', what=list(count=1, cusip6=,
idate=1,spread=1.1, vol252=1.1, vol1000=1.1, st_debt=1.1, lt_debt=1.1 ,
total_liab=1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1, sec=1.1,
cr3m=1.1,cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE,
nlines=1,quiet=T)...}The problem is how to articulate while (the end of
the table has not beenreached), or equivalently, how to signal the end of the
table?Best WishesYuchen Luo [[alternative HTML version deleted]]
_
Discover the new Windows Vista
[[alternative HTML version deleted]]
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Re: [R] Single plot multiple levels in x?
Uwe, I have looked into lattice and can't seem to make this work. I can easily make multiple panels but this isn't what I am looking to do. Any suggestions on which functions to use? the axis function seems a natural place to start but I still can't seem to make it happen. HELP! Richard -Original Message- From: Uwe Ligges [EMAIL PROTECTED] Sent: Aug 30, 2007 5:59 AM To: Richard Yanicky [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Single plot multiple levels in x? Richard Yanicky wrote: Plotting with 2 x axis? One axis inside another, for example salary within state, 1-50 | 50 – 100 | 100+ | 1- 50 | 50 -100 | 100+ | … repeated bins for salary AL ! AR …… more states Sounds like the lattice package does exactly what you want, but without any reproducible example. Uwe Ligges The values are all stored with a single data frame. I have tried different things with the axis function and done many searches for plotting. Can’t find a direct reference Thanks. Richard __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Month end calculations
Shubha,
I apologize if this is a bit late - consequence of digest summary preference.
If I understand what you need, it is to calculate the value at month
end given a data object. For zoo objects the following should do what
you need. Actually is part of my new package on CRAN quantmod -
basically a workflow management tool for quant finance modelling.
Also visible at www.quantmod.com
If you convert your data.frame to a zoo object (designed for ordered
obs. - e.g. time-series)
# for your data - which I don't know : )
zoo.ts - zoo(youdataframe[,-1],as.Date(yourdataframe[,1]))
#^
^
# ^ ^
^ ^
# minus 'date' column the 'date'
column - in CCYY-MM-DD format
My example:
A zoo time series object consisting of 231 days:
zoo.ts - zoo(rnorm(231),as.Date(13514:13744))
start(zoo.ts)
[1] 2007-01-01
end(zoo.ts)
[1] 2007-08-19
# these are the end points of each period
breakpoints(zoo.ts,months,TRUE)
[1] 0 31 59 90 120 151 181 212 231
# get the associated values
zoo.ts[breakpoints(zoo.ts,months,TRUE)]
2007-01-31 2007-02-28 2007-03-31 2007-04-30 2007-05-31 2007-06-30
-0.008829668 -2.207802921 0.171705151 -1.820125167 1.776643162 0.884558259
2007-07-31 2007-08-19
0.655305543 0.191870144
You can also apply a function inside each of these periods (intervals)
with the function
period.apply:
e.g. the standard deviation of each period would be had with:
period.apply(zoo.ts,breakpoints(zoo.ts,months,TRUE),FUN=sd)
[1] 0.9165168 1.2483743 1.0717529 1.2002236 0.9568443 0.8112068 0.8563814
[8] 0.8671502
The functions (and many others) are in quantmod - on CRAN and most up
to date at www.quantmod.com
For those who'd rather just have the functions:
breakpoints -
function (x, by = c(weekdays, weeks, months, quarters, years), ...)
{
if (length(by) != 1)
stop(choose ONE method for \by\)
by - match.fun(by)
breaks - which(diff(as.numeric(by(x, ...))) != 0)
breaks - c(0, breaks, NROW(x))
return(breaks)
}
period.apply -
function (x, INDEX, FUN, ...)
{
FUN - match.fun(FUN)
y - NULL
for (i in 1:(length(INDEX) - 1)) {
sindex - (INDEX[i] + 1):INDEX[i + 1]
dat - x[sindex]
y - c(y, FUN(dat, ...))
}
return(y)
}
Jeff Ryan
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Re: [R] xeon processor and ATLAS
On Thu, Aug 30, 2007 at 09:33:32AM -0400, Jeffrey J. Hallman wrote: saving 8 seconds of computation time is worth going through the ATLAS hassle? For some of us, the 'hassle' is typing $ sudo apt-get install atlas3-base which gives you the low-hanging fruit of basic Atlas. Locally re-building the Debian/Ubuntu package to match _your_ cpu is also supported ... but there I share the '80/20' rule sentiment that the additional work is not always worth the additional effort. But let's remind people that they have a choice. Dirk -- Three out of two people have difficulties with fractions. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] retrieve p-value from a cox.obj
Several responders have given input on how to find and read the code for summary.coxph and print.coxph, or using str() to examine the output of a coxph object. An even better starting place would be the documentation (yes there is some). help(coxph) gives a page that contains: Value: an object of class coxph. See 'coxph.object' for details. and help(coxph.object) gives a description of all the components. Terry Therneau __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Another issue with the Matrix package *under R-devel*
I suspect you have not using a re-installed Matrix after re-building R. I can reproduce the problem using a version of Matrix I installed under 2.5.1, but not with one installed under R-devel this week. Since R-devel is 'Under development' you may need to reinstall packages when it changes. This is particularly prevalent with S4-using packages, as the methods code tends to capture the value of objects as they existed when a package was installed. In this case log() was changed quite a few weeks ago, but Matrix needs to be reinstalled after other changes last weekend. For the record, the packages I know need to reinstalled under a recent R-devel are Matrix, distr, kernlab, kinship and matlab (but there may be others). And please use the appropriately named R-devel list for questions about R-devel. On Thu, 30 Aug 2007, Tony Chiang wrote: Hi all, I am encountering a strange issue with the Matrix package. I have just built R-devel from source on my macbook pro, and I wonder if others can reproduce this problem. I will give example code to go along: Starting a fresh R session: R version 2.6.0 Under development (unstable) (2007-08-30 r42697) Copyright (C) 2007 The R Foundation for Statistical Computing ISBN 3-900051-07-0 ... log(2) [1] 0.6931472 library(Matrix) Loading required package: lattice log(2) Error in log(2) : could not find symbol base in environment of the generic function There seems to be something wrong here and I cannot figure out what it is. Am I doing something wrong or is it an issue with Matrix (which is what I have narrowed it down to). I think that it might be a namespace collision or something, but I am certainly not sure. Here is my sessionInfo() output: sessionInfo() R version 2.6.0 Under development (unstable) (2007-08-30 r42697) i386-apple-darwin8.10.1 locale: C attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] Matrix_0.99875-2 lattice_0.16-3 loaded via a namespace (and not attached): [1] grid_2.6.0 Thanks. Tony [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 signal the end of the table?
For example,
fp-file(abc.csv,r)
c.row-scan(file=fp,sep=,,nlines=1) # what argument is omitted for
bevity as it doesn't matter
rows-c.row
while(length(c.row)0) {
c.row-scan(file=fp,sep=;,nlines=1);
rows-rbind(rows,c.row)
}
close(fp)
If you want to read a file by parts, then you do something like the
following (untested)
But this looks like a C-style approach, not R-style
fp-file(abc.csv,r)
lines.skip-0
for(i in 1:nparts){
c.row-scan(file=fp,sep=,,nlines=1,skip=lines.skip)
lines.read-1
rows-c.row
while(length(c.row)0 lines.readpart.size) {
c.row-scan(file=fp,sep=;,nlines=1); # nlines can be also 1
rows-rbind(rows,c.row)
lines.read-lines.read+1 # if nlines above is 1 then +1 must be
replaced with nlines value
}
lines.skip-lines.read
# do operations with rows
}
close(fp)
Yuchen Luo wrote:
I am using a for loop to read a table row by row and I have to specify
how
many records are there in the table. I need to read row by row because the
table is huge and the memory not large enough for the whole table.:
number.of.records=100
fp=file(abc.csv,r)
pos=seek(fp, rw=read)
for (i in 1:number.of.record){
current.row=scan(file=fp, sep=',', what=list(count=1, cusip6=, idate=1,
spread=1.1, vol252=1.1, vol1000=1.1, st_debt=1.1, lt_debt=1.1 ,
total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1, sec=1.1,
cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
I need to know the number of records in the table and put it in the
variable
named number.of.records. When I have a new table that I do not know how
many records it has, I use excel to open the file to figure it out and put
it in variable number.of.records. I often have many tables to try and
every one of them has thousands of recordsit takes a lot of time and
trouble to adjust the code every time I read a different table.
I am wondering if I can change the for loop to a while loop:
while (the end of the table has not been reached)
{
current.row=scan(file=fp, sep=',', what=list(count=1, cusip6=, idate=1,
spread=1.1, vol252=1.1, vol1000=1.1, st_debt=1.1, lt_debt=1.1 ,
total_liab=
1.1, cr=1.1, shrout=1.1, prc=1.1, mkt_cap=1.1, rtng=1.1, sec=1.1,
cr3m=1.1,
cr5y=1.1, ust3m=1.1, ust5y=1.1), flush=TRUE, nlines=1,quiet=T)
...
}
The problem is how to articulate while (the end of the table has not been
reached), or equivalently, how to signal the end of the table?
Best Wishes
Yuchen Luo
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[R] Behaviour of very large numbers
Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? B [1] 47.73092 -51^B [1] -3.190824e+81 # seems fine # now this: x - seq(-51,-49,length=100) x^B [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN snip is.numeric(x^B) [1] TRUE is.real(x^B) [1] TRUE is.infinite(x^B) [1] FALSE FALSE FALSE FALSE FALSE I am lost, I checked the R mailing help, but could not find anything directly. I loaded package Brobdingnag and tried: as.brob(x^B) [1] NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) as.brob(x)^B [1] NAexp(187.67) NAexp(187.65) NAexp(187.63) NAexp(187.61) I guess I must be misunderstanding something fundamental. Any clues would be really appreciated. Willem __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R and Web Applications
Hello, I'm curious to know how people are calling R from web applications (I've been looking for Perl but I'm open to other languages). After doing a search, I came across the R package RSPerl, but I'm having difficulties getting it installed (on Mac OSX). I believe the problem probably has to do with changes in R since the package release. Below you will see where the installation process comes to an end. Does anyone have any suggestions, or perhaps a direction to point me in? Thanks in advance for your insight! Chris * Installing to library '/Library/Frameworks/R.framework/Resources/library' * Installing *source* package 'RSPerl' ... checking for perl... /usr/bin/perl No support for any of the Perl modules from calling Perl from R. * Set PERL5LIB to /Library/Frameworks/R.framework/Versions/2.5/Resources/library/RSPerl/perl * Testing: -F/Library/Frameworks/R.framework/.. -framework R Using '/usr/bin/perl' as the perl executable Perl modules (no): Adding R package to list of Perl modules to enable callbacks to R from Perl Creating the C code for dynamically loading modules with native code for Perl: R modules: R; linking: checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ISO C89... none needed Support R in Perl: yes configure: creating ./config.status config.status: creating src/Makevars config.status: creating inst/scripts/RSPerl.csh config.status: creating inst/scripts/RSPerl.bsh config.status: creating src/RinPerlMakefile config.status: creating src/Makefile.PL config.status: creating cleanup config.status: creating src/R.pm config.status: creating R/perl5lib.R making target all in RinPerlMakefile RinPerlMakefile:5: /Library/Frameworks/R.framework/Resources/etc/Makeconf: No such file or directory make: *** No rule to make target `/Library/Frameworks/R.framework/Resources/etc/Makeconf'. Stop. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] correlation structure in lmer
On 8/29/07, Fränzi Korner [EMAIL PROTECTED] wrote: how can I specify a correlation structure in the lmer-function as it is possible in lme(formula, ..., corr=corAR1(form=...))? The short answer is you can't. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Assigning line colors in xyplot
Hi Christof, You can do this in ggplot, with one exception: install.packages(ggplot2) library(ggplot2) qplot(x, y, data=tmp, facets = . ~ g, geom=line, colour=f) Unfortunately I don't yet have an implementation of facetting that works like lattice, wrapping the line of plots in to 2d dimensions. You can find out more about ggplot2 at http://had.co.nz/ggplot2 Hadley On 8/30/07, Christof Bigler [EMAIL PROTECTED] wrote: Hi, I have a dataframe containing data from individuals 1, ..., 12 (grouping variable g in the data frame below), which belong either to A or B (grouping variable f): set.seed(1) tmp - data.frame( y=c(rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2)), x=1:10, f=c(rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10)), g=c(rep(3,10),rep(2,10),rep(1,10),rep(4,10),rep(5,10),rep(6,10),rep(8,10),rep(7,10),rep(9,10),rep(11,10),rep(12,10),rep(10,10))) I would like to draw line plots using the function xyplot: library(lattice) xyplot(y ~ x | g , groups=g, data=tmp,type=l, par.settings=list(superpose.line=list(col=c(red,blue))), auto.key=list(space=top, text=levels(tmp$f),points=FALSE,lines=TRUE)) As it is, the colors are recycled alternately in the order the individuals appear in the plot (1, 10, 11, 12, 2, ..., 9). How can I assign the red color to all individuals of group A and the blue color to all individuals of group B? Thanks in advance! Christof __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- http://had.co.nz/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Excel (off-topic, sort of)
On Wed, 29 Aug 2007, Alberto Monteiro wrote: Do you know what's in my wish list? I wish spreadsheets and computer languages had gone one step further. I mean, it's nice to define Cell X to be equal to Cell Y + 10, and then when we change Cell Y, magically we see Cell X change. But why can't it be the reverse? Why can't I change Cell X _and see the change in Cell Y_? Well, most statistical calculations are data-reducing, ie, many-to-one. I don't think you are likely to find a language where you can change the confidence limits for the sample mean and have the data change in response. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] mtext, picking subscript text from colnames(df)
Hi! Could someone give advise on how to plot Greek letters where subscript text is picked from the names of a data.frame. I want what appears using this code mtext(expression(mu[Mountain]),side=2,at=max(y)-7,las=1,cex=1.5) but I would like to do it using something like this: mtext(expression(mu[colnames(PresEsts)[2]]),side=2,at=max(y)-15,las=1) BTW, colnames(PresEsts)[2] [1] Mountain Thanks! /Tord -- Tord Snäll Department of Ecology Swedish University of Agricultural Sciences (SLU) P.O. 7002, SE-750 07 Uppsala, Sweden Office/Mobile/Fax +46-18-672612/+46-730-891356/+46-18-673537 E-mail: [EMAIL PROTECTED] www.nvb.slu.se/staff_tordsnall __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Month end calculations
Shubha et al,
I forgot that the period methods may not work without the full
package. So either just download the current CRAN version 0.1-0 or
version 0.1-3 from quantmod.com, a new more complete version will be
up with a week or two.
For completeness here is my months method for zoo objects:
quantmod:::months.zoo
function (x, abbreviate = FALSE)
{
format(index(x), ifelse(abbreviate, %m, %B))
}
environment: namespace:quantmod
The primary difference is that abbreviate=TRUE will return the month
NUMBER, not the standard English abbreviation. For the other methods
it is probably wiser to install the package as opposed to my obviously
deficient copy and paste method.
Thanks,
Jeff Ryan
On 8/30/07, Jeff Ryan [EMAIL PROTECTED] wrote:
Shubha,
I apologize if this is a bit late - consequence of digest summary preference.
If I understand what you need, it is to calculate the value at month
end given a data object. For zoo objects the following should do what
you need. Actually is part of my new package on CRAN quantmod -
basically a workflow management tool for quant finance modelling.
Also visible at www.quantmod.com
If you convert your data.frame to a zoo object (designed for ordered
obs. - e.g. time-series)
# for your data - which I don't know : )
zoo.ts - zoo(youdataframe[,-1],as.Date(yourdataframe[,1]))
#^
^
# ^ ^
^ ^
# minus 'date' column the 'date'
column - in CCYY-MM-DD format
My example:
A zoo time series object consisting of 231 days:
zoo.ts - zoo(rnorm(231),as.Date(13514:13744))
start(zoo.ts)
[1] 2007-01-01
end(zoo.ts)
[1] 2007-08-19
# these are the end points of each period
breakpoints(zoo.ts,months,TRUE)
[1] 0 31 59 90 120 151 181 212 231
# get the associated values
zoo.ts[breakpoints(zoo.ts,months,TRUE)]
2007-01-31 2007-02-28 2007-03-31 2007-04-30 2007-05-31 2007-06-30
-0.008829668 -2.207802921 0.171705151 -1.820125167 1.776643162 0.884558259
2007-07-31 2007-08-19
0.655305543 0.191870144
You can also apply a function inside each of these periods (intervals)
with the function
period.apply:
e.g. the standard deviation of each period would be had with:
period.apply(zoo.ts,breakpoints(zoo.ts,months,TRUE),FUN=sd)
[1] 0.9165168 1.2483743 1.0717529 1.2002236 0.9568443 0.8112068 0.8563814
[8] 0.8671502
The functions (and many others) are in quantmod - on CRAN and most up
to date at www.quantmod.com
For those who'd rather just have the functions:
breakpoints -
function (x, by = c(weekdays, weeks, months, quarters, years), ...)
{
if (length(by) != 1)
stop(choose ONE method for \by\)
by - match.fun(by)
breaks - which(diff(as.numeric(by(x, ...))) != 0)
breaks - c(0, breaks, NROW(x))
return(breaks)
}
period.apply -
function (x, INDEX, FUN, ...)
{
FUN - match.fun(FUN)
y - NULL
for (i in 1:(length(INDEX) - 1)) {
sindex - (INDEX[i] + 1):INDEX[i + 1]
dat - x[sindex]
y - c(y, FUN(dat, ...))
}
return(y)
}
Jeff Ryan
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Re: [R] xeon processor and ATLAS
On 8/29/07, hui xie [EMAIL PROTECTED] wrote: Thanks very much for all your advice. To be clear, my OS is window XP. I bought this server last year. It's Dell Precision PWS690. THe processor is Xeon(TM) CPU 3GHZ, 2G RAM. I am not sure how to check more details of processor on my computer. But I went to Dell website and from what I can recall, it seemed that I ordered a Dual-core Intel Xeon 5160 3GHz, 1333FSB, 4MB L2 Cache, 80watts. I think from time to time, I will need linear algebra to do things such as Choleskey Factorization, and matrix inverse etc... At the same time, I considered myself quite unskilled on building R by myself. It seems there were lots of details that I can get things wrong. So if there is an existing ATLAS one on R website that I can use, I would be very happy to use it to replace the default one. The reason why I would like to use ATLAS is that R FAQ said : Rather than worrying about versions of ATLAS you may want to spend your time considering exactly how you are going about the calculations. A general result in numerical linear algebra is that if your algorithm involves computing the inverse of a matrix you have a sub-optimal algorithm. The savings can be appreciable: on a 2.6GHz P4 and a 1000 x 1000 matrix svd took 16.2 sec with the standard BLAS and 7.8 sec with ATLAS. Because ATLAS is tuned to a particular chip we can't use it generally: the optimal routines for a P4 or an Athlon XP are quite different and neither will run at all on a PII. This seems to me an impressive gain to use the correct ATLAS instead of the default BLAS. I guess my Xeon processor is either a P4 or Core2Duo, but I am really not sure which one to use. Could you please offer me some suggestions? Again, many thanks for all your advice! Best, Hui Uwe Ligges [EMAIL PROTECTED] wrote: Prof Brian Ripley wrote: On Tue, 28 Aug 2007, hui xie wrote: hi everyone: I have a Dell Server that has a Xeon processor, and I would like to use the best ATLAS posted in the R website. I find that R has ATLAS for core2duo and P4. I am not sure which one of these two is best suited for Xeon processor, or is that neither of these two is good and I should stick with the default one that was installed originally? And your OS is? There are many different 'Xeon' processors with very different capabilities. ... the earlier similar to P4 and some similar to Core2Duo. You won't make use of bigger L2/L3 caches in Xeon processors. You really ought to build ATLAS for yourself if numerical linear algebra performance matters to you (and it makes little difference to most people: I think Uwe Ligges quoted 10% for testing all CRAN packages). Right, it depends on what you are really doing. If most time is spend in certain numerical matrix operations, ATLAS is your friend. In all other cases, it does not matter so much, as Brian cited correctly. Uwe Ligges Your advice is very much appreciated! Best, Hui - Park yourself in front of a world of choices in alternative vehicles. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Please do! - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] t-test within tapply
Hello,
My data are as following:
Data - data.frame(Ind=rep(1:3,c(10,10,10)),
Replicate=rep(c(rep(a,5),rep(b,5)),3),
EggSize=rep(rnorm(5,mean=10),6)
)
attach(Data)
Using a t-test, I want to check if the mean egg sizes are significantly
different between replicates a and b for each individuals (ie. In that case,
3 outputs with 3 p-values).
I tried the following, but doesn't work:
Fun - function(x,y) {
print(t.test(x ~ y))
}
tapply(EggSize,Ind,Fun(EggSize,Replicate))
What should I do?
Many thanks.
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Re: [R] Single plot multiple levels in x?
Richard Yanicky wrote: Uwe, I have looked into lattice and can't seem to make this work. I can easily make multiple panels but this isn't what I am looking to do. Any suggestions on which functions to use? the axis function seems a natural place to start but I still can't seem to make it happen. If lattice is not what you want, I do not understand what you mean. Can you give a more elaborated example, please? Uwe HELP! Richard -Original Message- From: Uwe Ligges [EMAIL PROTECTED] Sent: Aug 30, 2007 5:59 AM To: Richard Yanicky [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Single plot multiple levels in x? Richard Yanicky wrote: Plotting with 2 x axis? One axis inside another, for example salary within state, 1-50 | 50 – 100 | 100+ | 1- 50 | 50 -100 | 100+ | … repeated bins for salary AL ! AR …… more states Sounds like the lattice package does exactly what you want, but without any reproducible example. Uwe Ligges The values are all stored with a single data frame. I have tried different things with the axis function and done many searches for plotting. Can’t find a direct reference Thanks. Richard __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Behaviour of very large numbers
Hello, it seems to be an R bug. It gives strange errors for non-integer exponents: version platform i686-redhat-linux-gnu version.string R version 2.4.1 (2006-12-18) x^47.0 [1] -1.802180e+80 -1.768932e+80 -1.736284e+80 -1.704227e+80 -1.672748e+80 [...] x^47.10 [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN [...] x[2]^b [1] NaN But: x[2] [1] -50.9798 -50.9798^b [1] -3.131003e+81 So it is not a merely numerical problem. Maybe a bug in memory allocation for vectors containing such big numbers? Scion __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to obtain intercept statistics in anova with within-subject factors?
Dear R users, I am looking for an easy (i.e., direct) way of obtaining the F and p values from the intercept in anovas with within-subject designs. My data are from a psychophysics experiment where I am using d' (d-prime) values obtained from 3 modalities of presentation in each subject. I would like to know not only whether there is an effect of modality, but also whether the main effect is significant (meaning that d' 0). I know that a t.test again the null mean would provide me with similar stats on the main effect, but I would like to get those stats in an F form, for consistency with the stats on the other factors of interest. As far as I understand how R works, the function anova provides you with such information but it is restricted to between-group analyses. For within-subject designs, one has to use summary(aov) but stats on the intercept are not included in the result of this function. I have pasted an example below. As one can see, only the Sum of Sq and Mean Sq are given for the main effect. Thank you for any advice you can provide, -Sid summary(aov(x~mod+Error(suj/(mod)), data=dp)) Error: suj Df Sum Sq Mean Sq F value Pr(F) Residuals 10 19.5977 1.9598 Error: suj:mod Df Sum Sq Mean Sq F value Pr(F) mod 2 8.2475 4.1237 4.2955 0.02806 * Residuals 20 19.2005 0.9600 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Error: Within Df Sum Sq Mean Sq F value Pr(F)Residuals 33 55.812 1.691 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Excel
On 8/28/2007 3:16 AM, J Dougherty wrote: On Monday 27 August 2007 22:21, David Scott wrote: On Tue, 28 Aug 2007, Robert A LaBudde wrote: If you format the column as Text, you won't have this problem. By leaving the cells as General, you leave it up to Excel to guess at the correct interpretation. Not true actually. I had converted the column to Text because I saw the interpretation as a date in the .xls file. I saved the .csv file *after* the column had been converted to Text. Looking at the .csv file in a text editor, the entry is correct. I have just rechecked this. On reopening the .csv using Excel, the entry AUG2699 had been interpreted as a date, and was showing as Aug-99. Most bizarre is that the NHI value of AUG1838 has *not* been interpreted as a date. Actually, in Excel 2000, he's right. What you have to is be sure of is that the ' that denotes a text entry precedes EVERY entry that can be confused with a date. Selecting the entire column and setting the format to text *before* data is entered does this. It will also create an appropriate *.csv file. Excel is notable too because it will automatically convert date-like entries as you type. In a column of IDs or similar critical data, that behaviour is really bad. I have never tried the MS site, but I haven't been able to find any entry about how to turn that particular automatic behaviour off. However, while I have not experimented extensively, as far as I have experimented, OpenOffice spreadsheet does not behave this way. I don't use Excel, but in OpenOffice 2.2.1 the ' is lost when a file is saved as .csv and reloaded. So if I take care and enter 'November 15 in a cell, then save it, OO will change it to 11/15/2007 when I reload. I can override this change by manually changing Standard format to Text *every time* I load the file. There's a help index entry date formats;avoiding conversion to, but it offers no more help than add an apostrophe at the beginning of the entry. This is brain-dead behaviour. Duncan Murdoch JWDougherty PS, I quit using Excel for most important work after it returned a negative variance on some data I was collecting descriptive statistics on. JWD __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Incomplete output with `sn' library package
You have my sincere apologies for the incompleteness of my message. I have given the details below, including my dataset and my code. I'm using R, version 2.5.0. My OS is a Mac, (version Tiger). The sn package is Version 0.4-1 My code was as follows: mydata - read.table(url(http://www.statsci.org/data/oz/ ais.txt), header = T) attach(mydata) a - msn.fit(X = cbind(1,Ht,Wt), y = BMI, control = list(x.tol=1e-6)) b - msn.mle(X=cbind(1,Ht,Wt), y=SSF) a b My problem is that neither the a nor the b output gives me any standard errors - those should appear under $se. In both the regressions, this field is left blank with NA under it. I would appreciate some help on this matter - are the standard errors not supposed to appear here, or is there something else I should put into the inputs? Thank you once again for your time, Manasi On Aug 30, 2007, at 5:47 AM, Uwe Ligges wrote: MANASI VYDYANATH wrote: Dear R users: I have a question regarding the output for two of the functions in the `sn' package, which deals with the mle fitting of skew normal curves to linear regressions. I'm using the examples and the dataset given as an example in the online documentation for this package, for the functions `msn.fit' and `msn.mle'. I'm following the example code in the documentation for these two functions exactly. Part of the data output is supposed to be se, which gives the standard errors of the estimated coefficients. This particular value comes out as being NA in the examples given, but there are three coefficients in each case and no numerical problems about why the standard errors cannot be calculated. Am I setting this program up right? Is there some other command I should use (or an option I need to use) to get the output to display standard errors of the coefficients? We cannot know if you use it right, since you have not given any details on OS, R version, sn version, and particularly a reproducible example. As each R-help message tells in the footer: PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. Uwe Ligges Thank you for your time in reading this question - Cordially, Manasi __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R and Web Applications
The R packages and projects for the web and R are listed here: http://www.lmbe.seu.edu.cn/CRAN/doc/FAQ/R-FAQ.html#R-Web-Interfaces On 8/30/07, Chris Parkin [EMAIL PROTECTED] wrote: Hello, I'm curious to know how people are calling R from web applications (I've been looking for Perl but I'm open to other languages). After doing a search, I came across the R package RSPerl, but I'm having difficulties getting it installed (on Mac OSX). I believe the problem probably has to do with changes in R since the package release. Below you will see where the installation process comes to an end. Does anyone have any suggestions, or perhaps a direction to point me in? Thanks in advance for your insight! Chris * Installing to library '/Library/Frameworks/R.framework/Resources/library' * Installing *source* package 'RSPerl' ... checking for perl... /usr/bin/perl No support for any of the Perl modules from calling Perl from R. * Set PERL5LIB to /Library/Frameworks/R.framework/Versions/2.5/Resources/library/RSPerl/perl * Testing: -F/Library/Frameworks/R.framework/.. -framework R Using '/usr/bin/perl' as the perl executable Perl modules (no): Adding R package to list of Perl modules to enable callbacks to R from Perl Creating the C code for dynamically loading modules with native code for Perl: R modules: R; linking: checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ISO C89... none needed Support R in Perl: yes configure: creating ./config.status config.status: creating src/Makevars config.status: creating inst/scripts/RSPerl.csh config.status: creating inst/scripts/RSPerl.bsh config.status: creating src/RinPerlMakefile config.status: creating src/Makefile.PL config.status: creating cleanup config.status: creating src/R.pm config.status: creating R/perl5lib.R making target all in RinPerlMakefile RinPerlMakefile:5: /Library/Frameworks/R.framework/Resources/etc/Makeconf: No such file or directory make: *** No rule to make target `/Library/Frameworks/R.framework/Resources/etc/Makeconf'. Stop. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Behaviour of very large numbers
On 8/30/2007 11:08 AM, willem vervoort wrote: Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? B [1] 47.73092 -51^B [1] -3.190824e+81 You should be using parentheses. You evaluated -(51^B), not (-51)^B. The latter gives NaN. # seems fine # now this: x - seq(-51,-49,length=100) x^B [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN snip is.numeric(x^B) [1] TRUE is.real(x^B) [1] TRUE is.infinite(x^B) [1] FALSE FALSE FALSE FALSE FALSE I am lost, I checked the R mailing help, but could not find anything directly. I loaded package Brobdingnag and tried: as.brob(x^B) [1] NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) as.brob(x)^B [1] NAexp(187.67) NAexp(187.65) NAexp(187.63) NAexp(187.61) I guess I must be misunderstanding something fundamental. Two things: operator precedence (the ^ has higher precedence than the unary minus), and the mathematical definition of raising something to a fractional power. The approach R takes to the latter is to define x^B to be exp(B * ln(x)), and ln(x) is undefined for negative x. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] customizing the color and point shape for each line drawn using lattice's xyplot
Thank you for your suggestion. Unfortunately it does not appear to matter if the data is subsetted before the xyplot command or within it. The problem remains. I was able to remove method 4 entirely from the key by using your suggestion to subset the dataset “common,” and then creating a new variable to assign labels for just this subset “common.without.Method4$method.not4.f” The resulting code was: xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common.without.Method4, groups=method.not4.f, type=l, auto.key=T) This however leads to lack of continuity between plots, as when I exclude methods 4 and 6 in my second plot, using the same approach, the color that had corresponds to methods 7, 8 and 9 in plot one is automatically changed in plot two. The subsetting may work, so long as I can dictate line color and symbol type in each plot. Ross Darnell wrote: Does this help common.without.Method4 - subset(common, Method!=4) xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common.without.Method4, groups=Method.f, type=l, auto.key=T) Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gen Sent: Thursday, 30 August 2007 2:51 PM To: r-help@stat.math.ethz.ch Subject: [R] customizing the color and point shape for each line drawn using lattice's xyplot Description of what I am trying to do: I am using the xyplot code below to plot the variable MeanBxg against the variable PercentVarExplained for all 9 possible combinations of variables bdg and bdx. Within each of these 9 scenarios I am plotting a separate line for each of up to 9 different methods that I used to estimate the variable MeanBxg. These methods are identified by the numeric variable called Method. Giving me one plot with 9 figures and each of the figures contains 9 lines. My problem arises because I would like to repeat the creation of this plot 8 times, in each instance only a subset (eg 6) of the 9 methods are used (a different subset each time). What I can't figure out: I would like to learn how to specify the exact line color that corresponds to each method such that Method==1 will always be represented by the same color (in every plot that it appears in). Where two methods that I used were of the same family of methods (say method==1 and method==2 made the same assumptions about the data) I would like to, if possible, represent the two methods using the same color and distinguish them by the symbol used to represent points on the line. My code as it currently stands: xyplot(MeanBxg ~ PercentVarExplained | bdg.f * bdx.f, data=common, groups=common$Method.f, type=l, subset= Method!=4, auto.key=T) As the code is, the default colors assigned are repeated causing different methods to be represented by the same color with no way to distinguish them (I have not succeeded in plotting lines and points simultaneously). Side question: When I subset the data to particular methods, is there a way to remove the excluded methods from the key as well? (in my code Method is a numeric variable, and Method.f corresponds to the lengthy descriptions of each method for the purpose of the key) Thank you very much for your help. Genevieve -- View this message in context: http://www.nabble.com/customizing-the-color-and-point-shape-for-each-lin e-drawn-using-lattice%27s-xyplot-tf4351934.html#a12400517 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/customizing-the-color-and-point-shape-for-each-line-drawn-using-lattice%27s-xyplot-tf4351934.html#a12409939 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] R and Web Applications
Hi Chris -- RWebServices provides a way to expose R functionality as (SOAP-based) web services. http://bioconductor.org/packages/2.0/bioc/html/RWebServices.html This might be more than you are looking for, and has not been tested on MacOS (which seems to be your platform). Martin Chris Parkin [EMAIL PROTECTED] writes: Hello, I'm curious to know how people are calling R from web applications (I've been looking for Perl but I'm open to other languages). After doing a search, I came across the R package RSPerl, but I'm having difficulties getting it installed (on Mac OSX). I believe the problem probably has to do with changes in R since the package release. Below you will see where the installation process comes to an end. Does anyone have any suggestions, or perhaps a direction to point me in? Thanks in advance for your insight! Chris * Installing to library '/Library/Frameworks/R.framework/Resources/library' * Installing *source* package 'RSPerl' ... checking for perl... /usr/bin/perl No support for any of the Perl modules from calling Perl from R. * Set PERL5LIB to /Library/Frameworks/R.framework/Versions/2.5/Resources/library/RSPerl/perl * Testing: -F/Library/Frameworks/R.framework/.. -framework R Using '/usr/bin/perl' as the perl executable Perl modules (no): Adding R package to list of Perl modules to enable callbacks to R from Perl Creating the C code for dynamically loading modules with native code for Perl: R modules: R; linking: checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ISO C89... none needed Support R in Perl: yes configure: creating ./config.status config.status: creating src/Makevars config.status: creating inst/scripts/RSPerl.csh config.status: creating inst/scripts/RSPerl.bsh config.status: creating src/RinPerlMakefile config.status: creating src/Makefile.PL config.status: creating cleanup config.status: creating src/R.pm config.status: creating R/perl5lib.R making target all in RinPerlMakefile RinPerlMakefile:5: /Library/Frameworks/R.framework/Resources/etc/Makeconf: No such file or directory make: *** No rule to make target `/Library/Frameworks/R.framework/Resources/etc/Makeconf'. Stop. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Martin Morgan Bioconductor / Computational Biology http://bioconductor.org __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Q: how to interrupt long calculation?
Paul Smith said the following at 08/29/2007 04:32 PM : The instance of R running will be immediately killed and then you can start R again. But then I would lose all the work. There must be some way to merely interrupt the current calculation. Mustn't there? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] correlation structure in lmer
On 8/30/07, Douglas Bates [EMAIL PROTECTED] wrote: On 8/29/07, Fränzi Korner [EMAIL PROTECTED] wrote: how can I specify a correlation structure in the lmer-function as it is possible in lme(formula, ..., corr=corAR1(form=...))? The short answer is you can't. My response was incorrect. I should have remembered the words of Simon Yoda Bloomberg. Of course you can do this but you will need to write the code to implement it. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Behaviour of very large numbers
On 8/30/07, Gustaf Rydevik [EMAIL PROTECTED] wrote: On 8/30/07, willem vervoort [EMAIL PROTECTED] wrote: Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? B [1] 47.73092 -51^B [1] -3.190824e+81 # seems fine # now this: x - seq(-51,-49,length=100) x^B [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN snip is.numeric(x^B) [1] TRUE is.real(x^B) [1] TRUE is.infinite(x^B) [1] FALSE FALSE FALSE FALSE FALSE I am lost, I checked the R mailing help, but could not find anything directly. I loaded package Brobdingnag and tried: as.brob(x^B) [1] NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) as.brob(x)^B [1] NAexp(187.67) NAexp(187.65) NAexp(187.63) NAexp(187.61) I guess I must be misunderstanding something fundamental. Any clues would be really appreciated. Willem non-integer exponents of non-positive values are not defined, thus the NaN. -1^2.5 [1] -1 (-1)^2.5 [1] NaN Best, Gustaf To modify, that is if you are working with reals. x [1] -1 x^(1/2) [1] NaN class(x)-complex x^(1/2) [1] 0+1i x--51 x^47.7 [1] NaN class(x)-complex x^47.7 [1] 1.660795e+81-2.285889e+81i Best, Gustaf -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Excel (off-topic, sort of)
On 8/30/2007 10:43 AM, Thomas Lumley wrote: On Wed, 29 Aug 2007, Alberto Monteiro wrote: Do you know what's in my wish list? I wish spreadsheets and computer languages had gone one step further. I mean, it's nice to define Cell X to be equal to Cell Y + 10, and then when we change Cell Y, magically we see Cell X change. But why can't it be the reverse? Why can't I change Cell X _and see the change in Cell Y_? Well, most statistical calculations are data-reducing, ie, many-to-one. I don't think you are likely to find a language where you can change the confidence limits for the sample mean and have the data change in response. But just think how much money you could make as a consultant if you could change the p-value from 0.08 to 0.01. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Behaviour of very large numbers
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of willem vervoort Sent: Thursday, August 30, 2007 8:09 AM To: r-help@stat.math.ethz.ch Subject: [R] Behaviour of very large numbers Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? B [1] 47.73092 -51^B [1] -3.190824e+81 # seems fine It seems fine because the precedence of the '^' operator is higher than the unary negation operator '-'. Your example is actually evaluated as -(51^B). # now this: x - seq(-51,-49,length=100) x^B [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN snip is.numeric(x^B) [1] TRUE is.real(x^B) [1] TRUE is.infinite(x^B) [1] FALSE FALSE FALSE FALSE FALSE I am lost, I checked the R mailing help, but could not find anything directly. I loaded package Brobdingnag and tried: as.brob(x^B) [1] NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) as.brob(x)^B [1] NAexp(187.67) NAexp(187.65) NAexp(187.63) NAexp(187.61) I guess I must be misunderstanding something fundamental. Yes, you can't raise a negative number to a fractional power. Hope this is helpful, Dan Daniel J. Nordlund Research and Data Analysis Washington State Department of Social and Health Services Olympia, WA 98504-5204 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Nested functions.
nalluri pratap schreef: Hi All, I have two variables X, Y. The question is if the value of X is equal to one, then the values in Y have to be reversed other wise it should not perfom any action. I think this should be done using lapply function? Example Y values : 1 2 3 NA X Y (ORIGINAL) Y (REVERSED) 1 NA 1 0 --- 1 2 3 1 1 4 1 3 2 ... Can anyone provide solution to this? Thanks, Pratap - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dear Pratap, You could try something like this: x = c(1,0,1,1) y = c(1,2,3,NA) y_rev = rev(y) ifelse(x == 1, y_rev, y) hope this helps, Paul -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +31302535773 Fax:+31302531145 http://intamap.geo.uu.nl/~paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Behaviour of very large numbers
willem vervoort wrote: Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? B [1] 47.73092 -51^B [1] -3.190824e+81 # seems fine Well, this seems not to be what you intended to do, you didn't raise a negative value to a power, but you got the negative of a positive number raised to that power (operator precedence, -51^B is the same as -(51^B) and not the same as (-51)^B...). If you really want to raise a negative value to a fractional power, you may want to tell R to use complex numbers: B - 47.73092 x - complex(real=seq(-51,-49,length=10)) x^B [1] 2.117003e+81-2.387323e+81i 1.718701e+81-1.938163e+81i [3] 1.394063e+81-1.572071e+81i 1.129702e+81-1.273954e+81i [5] 9.146212e+80-1.031409e+81i 7.397943e+80-8.342587e+80i [7] 5.978186e+80-6.741541e+80i 4.826284e+80-5.442553e+80i [9] 3.892581e+80-4.389625e+80i 3.136461e+80-3.536955e+80i Regards, Martin # now this: x - seq(-51,-49,length=100) x^B [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN snip is.numeric(x^B) [1] TRUE is.real(x^B) [1] TRUE is.infinite(x^B) [1] FALSE FALSE FALSE FALSE FALSE I am lost, I checked the R mailing help, but could not find anything directly. I loaded package Brobdingnag and tried: as.brob(x^B) [1] NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) as.brob(x)^B [1] NAexp(187.67) NAexp(187.65) NAexp(187.63) NAexp(187.61) I guess I must be misunderstanding something fundamental. Any clues would be really appreciated. Willem __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot() groups scope issue
Hi all,
Tinkering with a wrapper for xyplot that will help me plot a bunch of
plots in a data analysis I'm doing and I ran into an odd error that
I'm guessing is a scope issue. Here's a very simple version of the code:
###
#load lattice
library(lattice)
#create the wrapper function
do.xyplot = function( plot.formula, data.frame, plot.groups){
print(plot.groups) #show me what the wrapper function thinks
'plot.groups' is
xyplot(
plot.formula
,data = data.frame
,groups = eval(plot.groups)
)
}
#create some data
mydata = as.data.frame(cbind( x = c(1:4), y = rep(1:2), z = rep(1:2,
each = 2) ))
#try to plot the data (fails)
do.xyplot(
plot.formula=formula(x~y)
,data.frame = mydata
,plot.groups = expression(z)
)
#after error message try again, this time declaring plot.groups as a
global variable (succeeds)
plot.groups = expression(z)
do.xyplot(
plot.formula=formula(x~y)
,data.frame = mydata
,plot.groups = expression(z)
)
#
I'm fine with declaring plot.groups before each call to do.xyplot,
but I'm curious if there's a simpler solution.
Mike
--
Mike Lawrence
Graduate Student, Department of Psychology, Dalhousie University
Website: http://memetic.ca
Public calendar: http://icalx.com/public/informavore/Public
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/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] R and Web Applications
Dear Chris, I use Python (http://www.python.org) in combination with Rpy (http://rpy.sourceforge.net/). Rpy enables you to use R commands inside Python, not the other way around. Works quite well, also for different R versions (I currently run R 2.5.1 under Linux). cheers, Paul Chris Parkin schreef: Hello, I'm curious to know how people are calling R from web applications (I've been looking for Perl but I'm open to other languages). After doing a search, I came across the R package RSPerl, but I'm having difficulties getting it installed (on Mac OSX). I believe the problem probably has to do with changes in R since the package release. Below you will see where the installation process comes to an end. Does anyone have any suggestions, or perhaps a direction to point me in? Thanks in advance for your insight! Chris * Installing to library '/Library/Frameworks/R.framework/Resources/library' * Installing *source* package 'RSPerl' ... checking for perl... /usr/bin/perl No support for any of the Perl modules from calling Perl from R. * Set PERL5LIB to /Library/Frameworks/R.framework/Versions/2.5/Resources/library/RSPerl/perl * Testing: -F/Library/Frameworks/R.framework/.. -framework R Using '/usr/bin/perl' as the perl executable Perl modules (no): Adding R package to list of Perl modules to enable callbacks to R from Perl Creating the C code for dynamically loading modules with native code for Perl: R modules: R; linking: checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ISO C89... none needed Support R in Perl: yes configure: creating ./config.status config.status: creating src/Makevars config.status: creating inst/scripts/RSPerl.csh config.status: creating inst/scripts/RSPerl.bsh config.status: creating src/RinPerlMakefile config.status: creating src/Makefile.PL config.status: creating cleanup config.status: creating src/R.pm config.status: creating R/perl5lib.R making target all in RinPerlMakefile RinPerlMakefile:5: /Library/Frameworks/R.framework/Resources/etc/Makeconf: No such file or directory make: *** No rule to make target `/Library/Frameworks/R.framework/Resources/etc/Makeconf'. Stop. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +31302535773 Fax:+31302531145 http://intamap.geo.uu.nl/~paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Behaviour of very large numbers
On Thu, 30 Aug 2007, willem vervoort wrote: Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? Where are the 'very large numbers' here? R can cope with much larger numbers (over 10^300). B [1] 47.73092 -51^B [1] -3.190824e+81 Yes, that is -(51^B). # seems fine # now this: x - seq(-51,-49,length=100) x^B [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN snip is.numeric(x^B) [1] TRUE is.real(x^B) [1] TRUE is.infinite(x^B) [1] FALSE FALSE FALSE FALSE FALSE I am lost, I checked the R mailing help, but could not find anything directly. I loaded package Brobdingnag and tried: as.brob(x^B) [1] NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) NAexp(NaN) as.brob(x)^B I guess I must be misunderstanding something fundamental. You are. A negative number to a non-integer power is undefined in the real number system. Look at (x+0i)^B. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] 7 Courses: Upcoming September-October 2007 R/S+ schedule by XLSolutions Corp
Our upcoming September-October 2007 R/S+ course schedule is now available. Please check out this link for additional information and direct enquiries to Sue Turner [EMAIL PROTECTED] Phone: 206 686 1578 http://www.xlsolutions-corp.com/courselist.htm -- (1) An Introduction to S and R *** Seattle / October 29-30, 2007 *** (2) R/S Fundamentals and Programming Techniques *** Seattle / September 24-25, 2007 *** *** Boston / September 27-28, 2007 *** (3) R/S System: Complementing and Extending Statistical Computing forSAS Users *** Raleigh / October 22-23, 2007 *** (4) R/S System: Advanced Programming *** Princeton / October 29-30, 2007 *** *** San Francisco / October 25-26, 2007 -- http://www.xlsolutions-corp.com/courselist.htm (5) Introduction to R/S+ programming: Microarrays Analysis andBioconductor. *** Los Angeles 9/24/2007 9/25/2007 *** (6) Microarrays Data Analysis with R/S+ and GGobi *** New York City 10/18/2007 10/19/2007 *** (7) Data Mining: Practical Tools and Techniques in R/Splus *** San Francisco 9/27/2007 9/28/2007 *** --- Payment due AFTER the class Email us for group discounts. Email Sue Turner: [EMAIL PROTECTED] Phone: 206-686-1578 Visit us: www.xlsolutions-corp.com/courselist.htm Please let us know if you and your colleagues are interested inthis class to take advantage of group discount. Register now to secureyour seat! Cheers, Elvis Miller, PhD Manager Training. XLSolutions Corporation 206 686 1578 www.xlsolutions-corp.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] mtext, picking subscript text from colnames(df)
Try: mtext(eval(parse(tex=paste(expression(mu[, colnames(PresEsts)[2],]), sep=))),side=2,at=max(y)-15,las=1) -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Behaviour of very large numbers
On 8/30/2007 12:11 PM, Martin Becker wrote: willem vervoort wrote: Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? B [1] 47.73092 -51^B [1] -3.190824e+81 # seems fine Well, this seems not to be what you intended to do, you didn't raise a negative value to a power, but you got the negative of a positive number raised to that power (operator precedence, -51^B is the same as -(51^B) and not the same as (-51)^B...). If you really want to raise a negative value to a fractional power, you may want to tell R to use complex numbers: B - 47.73092 x - complex(real=seq(-51,-49,length=10)) x^B [1] 2.117003e+81-2.387323e+81i 1.718701e+81-1.938163e+81i [3] 1.394063e+81-1.572071e+81i 1.129702e+81-1.273954e+81i [5] 9.146212e+80-1.031409e+81i 7.397943e+80-8.342587e+80i [7] 5.978186e+80-6.741541e+80i 4.826284e+80-5.442553e+80i [9] 3.892581e+80-4.389625e+80i 3.136461e+80-3.536955e+80i But watch out if you do this, because of the arbitrary choice of a root. You get oddities like this: x - complex(real = -1) x [1] -1+0i 1/x [1] -1+0i x^(1/3) [1] 0.5+0.8660254i (1/x)^(1/3) [1] 0.5-0.8660254i i.e. even though x and 1/x are equal, the 1/3 powers of them are not. Duncan Murdoch P.S. I'm tempted to say, But don't worry about it, the difference is only imaginary, but I'll refrain. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] xyplot() groups scope issue
On 8/30/07, Mike Lawrence [EMAIL PROTECTED] wrote:
Hi all,
Tinkering with a wrapper for xyplot that will help me plot a bunch of
plots in a data analysis I'm doing and I ran into an odd error that
I'm guessing is a scope issue. Here's a very simple version of the code:
It's indeed a scoping issue. For writing wrappers, the best solution
I've been able to come up with involves match.call() and
eval.parent(). See lattice:::dotplot.formula for a template.
If you really want to do it your way, try changing the environment of
the formula:
do.xyplot = function( plot.formula, data.frame, plot.groups){
print(plot.groups)
environment(plot.formula) - environment() # new
xyplot(
plot.formula
,data = data.frame
,groups = eval(plot.groups)
)
}
-Deepayan
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Re: [R] Assigning line colors in xyplot
On 8/30/07, Christof Bigler [EMAIL PROTECTED] wrote: Hi, I have a dataframe containing data from individuals 1, ..., 12 (grouping variable g in the data frame below), which belong either to A or B (grouping variable f): set.seed(1) tmp - data.frame( y=c(rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2),rnorm(10,0,1),rnorm(10,4,2)), x=1:10, f=c(rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10),rep(A,10),rep(B,10)), g=c(rep(3,10),rep(2,10),rep(1,10),rep(4,10),rep(5,10),rep(6,10),rep(8,10),rep(7,10),rep(9,10),rep(11,10),rep(12,10),rep(10,10))) I would like to draw line plots using the function xyplot: library(lattice) xyplot(y ~ x | g , groups=g, data=tmp,type=l, par.settings=list(superpose.line=list(col=c(red,blue))), auto.key=list(space=top, text=levels(tmp$f),points=FALSE,lines=TRUE)) As it is, the colors are recycled alternately in the order the individuals appear in the plot (1, 10, 11, 12, 2, ..., 9). How can I assign the red color to all individuals of group A and the blue color to all individuals of group B? Why not simply use f as the grouping variable rather than g: xyplot(y ~ x | g , groups=f, data=tmp,type=l, par.settings=list(superpose.line=list(col=c(red,blue))), auto.key=list(space=top, points=FALSE,lines=TRUE)) ? Or am I missing something? -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] percentage explained by fixed effects in random model
Hi, I realise this has come up before in various reincarnations but I couldnt find the answer... I wish to quote the percentage variance explained by each of three components in my mixed model. If I didnt have a random effect I would just use r squared. I can work out the percentage explained by the random effect using summary() but this doesnt give variance for the fixed effects. Linear mixed-effects model fit by REML Formula: yell ~ carot.code + weight16 + (1 | fosterbrood) Data: colour AIC BIClogLik MLdeviance REMLdeviance 1555.455 1576.282 -772.7276 1536.585 1545.455 Random effects: Groups NameVariance Std.Dev. fosterbrood (Intercept) 1.0747 1.0367 Residual1.1363 1.0660 # of obs: 476, groups: fosterbrood, 61 Fixed effects: Estimate Std. Error DF t value Pr(|t|) (Intercept) -3.187028 1.011828 473 -3.1498 0.001737 ** carot.code1 0.201951 0.098442 473 2.0515 0.040771 * weight16 0.168861 0.054273 473 3.1114 0.001975 ** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Correlation of Fixed Effects: (Intr) crt.c1 carot.code1 0.035 weight16-0.989 -0.084 I was thinking of (cheating by) taking the residuals from a regression with the random effect (fosterbrood) as a fixed effect, then correlating these with my two x variables? Any better ideas? Thanks in advance, Simon. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Month end calculations
The zoo package includes the yearmon class to facilitate such manipulations. Here are a few solutions assuming you store you series in a zoo variable: # test data library(zoo) z - zoo(1001:1100, as.Date(101:200))[-(45:55)] # Solution 1. tapply produces indexes of last of month tt - time(z) z[ c(tapply(seq_along(tt), as.yearmon(tt), tail, 1)) ] # If we want to create a last variable which corresponds # to last in sas then do it this slightly longer way: # Solution 2 tt - time(z) last - seq_along(tt) %in% tapply(seq_along(tt), as.yearmon(tt), tail, 1) z[last] # Solution 3. another solution with a last variable. f(x) is # vector same length as x with all 0's except last element is 1. tt - time(z) f - function(x) replace(0*x, length(x), 1) last - ave(seq_along(tt), as.yearmon(tt), FUN = f) z[last] In all these solutions the last point in the series is always included. We have not assumed that every day is necessarily included in your series but if every day is included then even simpler solutions are possible. On 8/29/07, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote: Hi R users, Is there a function in R, which does some calculation only for the month end in a daily data?... In other words, is there a command in R, equivalent to last. function in SAS? BR, Shubha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Month end calculations
The last line is wrong (see below for correction): On 8/30/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: The zoo package includes the yearmon class to facilitate such manipulations. Here are a few solutions assuming you store you series in a zoo variable: # test data library(zoo) z - zoo(1001:1100, as.Date(101:200))[-(45:55)] # Solution 1. tapply produces indexes of last of month tt - time(z) z[ c(tapply(seq_along(tt), as.yearmon(tt), tail, 1)) ] # If we want to create a last variable which corresponds # to last in sas then do it this slightly longer way: # Solution 2 tt - time(z) last - seq_along(tt) %in% tapply(seq_along(tt), as.yearmon(tt), tail, 1) z[last] # Solution 3. another solution with a last variable. f(x) is # vector same length as x with all 0's except last element is 1. tt - time(z) f - function(x) replace(0*x, length(x), 1) last - ave(seq_along(tt), as.yearmon(tt), FUN = f) z[last] This last line should be: z[last == 1] In all these solutions the last point in the series is always included. We have not assumed that every day is necessarily included in your series but if every day is included then even simpler solutions are possible. On 8/29/07, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote: Hi R users, Is there a function in R, which does some calculation only for the month end in a daily data?... In other words, is there a command in R, equivalent to last. function in SAS? BR, Shubha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Q: how to interrupt long calculation?
On Thu, 30 Aug 2007, D. R. Evans wrote: Paul Smith said the following at 08/29/2007 04:32 PM : The instance of R running will be immediately killed and then you can start R again. But then I would lose all the work. There must be some way to merely interrupt the current calculation. Mustn't there? Only if it is long-running in R code, when ctrl-C or equivalent (Esc in Rgui) works. If it is long-running in C or Fortran code, there is not. Assuming a Unix-alike, sending SIGUSR1 will save the current workspace and quit. Even that is a little dangerous as the workspace need not be in a consistent state. People who have been bitten will learn safer programming practices, for example to call save.image() at suitable checkpoint times. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Barplot2 using for loop, how to adjust margins?
Hi R-users,
I inted to make multiple plots using for loop. The question is how can
I adjust the left hand side margin of the plot according to the
names.arg argument in barplot2. In every plot I have different
annotations in the y axis and they vary in length. Now when I have
fixed margins
opar - par(mar=c(3,15,0,2)...
I get the same margins in all of the plots. That leaves lots of white
space in plots where the annotations are short.
Here is the code I'm using:
library(gplots)
opar - par(mar=c(3,15,0,2), bg=white, cex=1, oma = c(0, 0, 2, 0),
mgp=c(3,0.5,0))
for (i in names(spl)) {
.order - order(spl[[i]]$x)
barplot2(spl[[i]]$x[.order],
names.arg=as.character(spl[[i]]$os[.order]),
horiz=TRUE,
las=1,
cex.names=0.7,
cex.main=0.9,
cex.axis=0.7,
xlim=c(0, max(spl[[i]]$x)+10),
col=as.character(spl[[i]][1,7]),
plot.grid = TRUE,
)
box()
mtext(paste(paste(as.character(spl[[i]][1,2]), :, sep=),
texthere, (texthere), as.character(spl[[i]][1,3])), outer=T, line
= 0.5, cex=1.1)
mtext(texthere, side=1, line=1.5, adj=0.5)
dev.copy(png, filename=paste(i, .png, sep=), height=400,
width=480)
dev.off()
}
Thanks in advance
Lauri
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[R] Additions to xyplot (lattice)? - legend, ticks, axis label size, text
I have created an xyplot of a time series with the following code...
win.graph(width = 10, height = 7)
panel1 = function(x, y) {
panel.loess(x, y, lwd=2.5, span=0.5, col=gray)
panel.xyplot(x, y, pch=19, col=blue, cex=1.25)
}
xyplot(oneplusdensity ~ year, data=figdata, aspect=fill, cex=1.5,
xlab=NULL, ylab=expression(Crabs per 1,000 m^2),
xlim=c(1964, 2007), ylim=c(-2, 62),
scales=list (x=list(tick.number=41, cex=0.8, rot=90),
y= list(tick.number=7, cex=1.2)), panel=panel1)
I want to do the following things:
(1) Remove the ticks on the top and right boundaries
(2) Increase the size of the y-axis label to be larger than the tick labels
(3) Add a legend with the points and lowess line
(4) Add some annotation (text) in the lower left corner ('text' for plot()
did not work)
I'm relatively new to lattice and spent a few hours with the manual and
other help pages. I've begun to wonder if a regular old plot() for this
would be better. Appreciate any redirection or suggestions.
Thanks,
Dave Hewitt
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Re: [R] Q: how to interrupt long calculation?
On linux and solaris, I have had some luck doing a CTRL-Z (STOP signal) and the resuming later... some libraries have issues (I don't recall which) but...usually pretty good. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of D. R. Evans Sent: Thursday, August 30, 2007 11:56 AM Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Q: how to interrupt long calculation? Paul Smith said the following at 08/29/2007 04:32 PM : The instance of R running will be immediately killed and then you can start R again. But then I would lose all the work. There must be some way to merely interrupt the current calculation. Mustn't there? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - This email transmission and any accompanying attachments may contain CSX privileged and confidential information intended only for the use of the intended addressee. Any dissemination, distribution, copying or action taken in reliance on the contents of this email by anyone other than the intended recipient is strictly prohibited. If you have received this email in error please immediately delete it and notify sender at the above CSX email address. Sender and CSX accept no liability for any damage caused directly or indirectly by receipt of this email. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Q: how to interrupt long calculation?
Prof Brian Ripley said the following at 08/30/2007 11:00 AM : On Thu, 30 Aug 2007, D. R. Evans wrote: Paul Smith said the following at 08/29/2007 04:32 PM : The instance of R running will be immediately killed and then you can start R again. But then I would lose all the work. There must be some way to merely interrupt the current calculation. Mustn't there? Only if it is long-running in R code, when ctrl-C or equivalent (Esc in Rgui) works. If it is long-running in C or Fortran code, there is not. It's inside loess()... so isn't that R code? I can sit hitting ctrl-C all day (well, it seems like it), but the code does not get interrupted :-( Assuming a Unix-alike, sending SIGUSR1 will save the current workspace and quit. Even that is a little dangerous as the workspace need not be in a consistent state. That's helpful, thank you; at least it means I stand a chance of being able to interrupt the code and recover. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Additions to xyplot (lattice)? - legend, ticks, axis label size, text
On 8/30/07, Dave Hewitt [EMAIL PROTECTED] wrote:
I have created an xyplot of a time series with the following code...
win.graph(width = 10, height = 7)
panel1 = function(x, y) {
panel.loess(x, y, lwd=2.5, span=0.5, col=gray)
panel.xyplot(x, y, pch=19, col=blue, cex=1.25)
}
xyplot(oneplusdensity ~ year, data=figdata, aspect=fill, cex=1.5,
xlab=NULL, ylab=expression(Crabs per 1,000 m^2),
xlim=c(1964, 2007), ylim=c(-2, 62),
scales=list (x=list(tick.number=41, cex=0.8, rot=90),
y= list(tick.number=7, cex=1.2)), panel=panel1)
I want to do the following things:
(1) Remove the ticks on the top and right boundaries
scales = list(tck = c(1, 0), x = ..., y = ...)
(2) Increase the size of the y-axis label to be larger than the tick labels
ylab = list( expression(...), cex = 2)
(3) Add a legend with the points and lowess line
See the entry for 'key' in ?xyplot. As a starting point,
key = list(text = list(loess), lines = list(lwd=2.5, span=0.5, col=gray),
text = list(points), points = list(pch=19, col=blue, cex=1.25))
(4) Add some annotation (text) in the lower left corner ('text' for plot()
did not work)
Use panel.text() instead inside your panel function.
-Deepayan
I'm relatively new to lattice and spent a few hours with the manual and
other help pages. I've begun to wonder if a regular old plot() for this
would be better. Appreciate any redirection or suggestions.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Month end calculations
And one more yearmon solution. Here z is a zoo series as before: tt - time(z) aggregate(z, ave(tt, as.yearmon(tt), FUN = max), tail, 1) On 8/30/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: The last line is wrong (see below for correction): On 8/30/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: The zoo package includes the yearmon class to facilitate such manipulations. Here are a few solutions assuming you store you series in a zoo variable: # test data library(zoo) z - zoo(1001:1100, as.Date(101:200))[-(45:55)] # Solution 1. tapply produces indexes of last of month tt - time(z) z[ c(tapply(seq_along(tt), as.yearmon(tt), tail, 1)) ] # If we want to create a last variable which corresponds # to last in sas then do it this slightly longer way: # Solution 2 tt - time(z) last - seq_along(tt) %in% tapply(seq_along(tt), as.yearmon(tt), tail, 1) z[last] # Solution 3. another solution with a last variable. f(x) is # vector same length as x with all 0's except last element is 1. tt - time(z) f - function(x) replace(0*x, length(x), 1) last - ave(seq_along(tt), as.yearmon(tt), FUN = f) z[last] This last line should be: z[last == 1] In all these solutions the last point in the series is always included. We have not assumed that every day is necessarily included in your series but if every day is included then even simpler solutions are possible. On 8/29/07, Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote: Hi R users, Is there a function in R, which does some calculation only for the month end in a daily data?... In other words, is there a command in R, equivalent to last. function in SAS? BR, Shubha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Q: how to interrupt long calculation?
On Thu, 30 Aug 2007, D. R. Evans wrote: Prof Brian Ripley said the following at 08/30/2007 11:00 AM : On Thu, 30 Aug 2007, D. R. Evans wrote: Paul Smith said the following at 08/29/2007 04:32 PM : The instance of R running will be immediately killed and then you can start R again. But then I would lose all the work. There must be some way to merely interrupt the current calculation. Mustn't there? Only if it is long-running in R code, when ctrl-C or equivalent (Esc in Rgui) works. If it is long-running in C or Fortran code, there is not. It's inside loess()... so isn't that R code? No, it is mainly Fortran, called from C called from R. I can sit hitting ctrl-C all day (well, it seems like it), but the code does not get interrupted :-( Assuming a Unix-alike, sending SIGUSR1 will save the current workspace and quit. Even that is a little dangerous as the workspace need not be in a consistent state. That's helpful, thank you; at least it means I stand a chance of being able to interrupt the code and recover. -- 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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
