Re: [R] question about the axis label
Unfortunately, this fine control is only hardly possible: 1. [x,y].ticklabs labels the ticks that are in the plot. If there are less ticks than ticklabs, the ticklabs are recycled and hence overlap. 2. xlim should do the trick to restrict to a certain range of values. It does not precisely and this is a known unfixed (and hard to fix by design) bug. 3. What you can do is enhance the number of ticks by telling R how many ticks to use approximately as in: scatterplot3d(dat, type = h, lab = c(12, 10,0), lab.z = 8, scale.y = 0.7) where dat is your data.frame. This tells R to - use approx. 12 ticks on x-axis - use approx. 10 ticks on y-axis - use approx. 8 ticks on z axis - scale y length by factor of 0.7 Best wishes, uwe On 25.01.2011 08:53, 孟欣 wrote: Hello sir: I have a question about the axis label of scatterplot3d function. The data is in the attachment. If I use the command: scatterplot3d(x,y,z,type=h) I want the plot's x-axis lab to be 1,2,3,...,13, y-axis lab to be 1,2,3,...11 But if I use the command: scatterplot3d(x,y,z,type=h,x.ticklabs=1:13,y.ticklabs=1:11) The axis labs are overlap, still can't get what I want. I'm in difficulty in how to set the axis labels,which need your suggestion very much. Thanks for your help. My best. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Extract NA data rows
hi i have the following dataframe x y 1 345 6 NA 8 123 32 123 12 NA 6 124 7 NA and i want to extract the data rows which contains NA data, I tried subset(dataframe,y==NA) but fail. if you know the answers, please let me know thanks. typhoong -- View this message in context: http://r.789695.n4.nabble.com/Extract-NA-data-rows-tp3235568p3235568.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] setting headers in POST
I need to use an API which requires me set certain attribute=values pairs in the header. The postToHost() method allows only to send data in the body. Please help me. -- View this message in context: http://r.789695.n4.nabble.com/setting-headers-in-POST-tp3235542p3235542.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Using open calais in R
I am using calais api in R for text analysis. But im facing a some problem when fetching the rdf from the server. I'm using the getToHost() method for the api call but i get just a null string. The same url in browser returns an RDF document. getToHost(www.api.opencalais.com,/enlighten/rest/?licenseID=dkzdggsre232ur97c6be269gcontent=HomeparamsXML=) [1] http://api.opencalais.com/enlighten/rest/?licenseID=dkzdggsre232ur97c6be269gcontent=HomeparamsXML=; -- View this message in context: http://r.789695.n4.nabble.com/Using-open-calais-in-R-tp3235597p3235597.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question about the axis label
Hello sir: I have a question about the axis label of scatterplot3d function. The data is in the attachment. If I use the command: scatterplot3d(x,y,z,type=h) I want the plot's x-axis lab to be 1,2,3,...,13, y-axis lab to be 1,2,3,...11 But if I use the command: scatterplot3d(x,y,z,type=h,x.ticklabs=1:13,y.ticklabs=1:11) The axis labs are overlap, still can't get what I want. I'm in difficulty in how to set the axis labels,which need your suggestion very much. Thanks for your help. My best. way nation fre 1 1 0 1 2 1 1 3 1317 1 4 0 1 5 0 1 6 0 1 7 1 1 8 1 1 9 1 1 10 16 1 11 2 2 1 0 2 2 0 2 3 35 2 4 0 2 5 0 2 6 0 2 7 0 2 8 0 2 9 3 2 10 0 2 11 0 3 1 12 3 2 8 3 3 539 3 4 38 3 5 7 3 6 0 3 7 6 3 8 7 3 9 325 3 10 36 3 11 15 4 1 0 4 2 0 4 3 2 4 4 0 4 5 0 4 6 0 4 7 0 4 8 0 4 9 1 4 10 0 4 11 0 5 1 0 5 2 0 5 3 4 5 4 1 5 5 0 5 6 0 5 7 0 5 8 0 5 9 4 5 10 0 5 11 2 6 1 0 6 2 0 6 3 0 6 4 0 6 5 0 6 6 0 6 7 0 6 8 0 6 9 1 6 10 0 6 11 0 7 1 0 7 2 0 7 3 1 7 4 0 7 5 0 7 6 0 7 7 0 7 8 0 7 9 0 7 10 0 7 11 0 8 1 0 8 2 0 8 3 128 8 4 2 8 5 0 8 6 0 8 7 0 8 8 0 8 9 0 8 10 0 8 11 1 9 1 0 9 2 0 9 3 22 9 4 0 9 5 0 9 6 0 9 7 0 9 8 0 9 9 3 9 10 0 9 11 0 10 1 0 10 2 0 10 3 8 10 4 0 10 5 0 10 6 0 10 7 0 10 8 0 10 9 2 10 10 0 10 11 0 11 1 0 11 2 0 11 3 6 11 4 0 11 5 0 11 6 0 11 7 0 11 8 1 11 9 0 11 10 0 11 11 0 12 1 0 12 2 0 12 3 124 12 4 1 12 5 0 12 6 10 12 7 0 12 8 0 12 9 5 12 10 1 12 11 0 13 1 5 13 2 3 13 3 561 13 4 7 13 5 3 13 6 0 13 7 7 13 8 9 13 9 50 13 10 7 13 11 45 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] crazy loop error.
On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. The error resulting in the NA was pretty easy to fix, and my loop works, but the results are still wrong (new script below). Ideally it should print single hello for the single letters and grouped '3 hellos' for the fives, grouped '2 hellos' for the sixes etc. Based on the run results, if the value of n is being tracked, it changes quite unpredictably. Can someone explain how the value of n changes from end of the loop to the top without anything being done to it? Hi. A for-loop in R is different from a for-loop in C. It is similar to foreach loop in Perl. If v is a vector, then for (n in v) first creates the vector v and then always performs length(v) iterations. Before iteration i, n is assigned v[i] even if n is changed in the previous iteration. If you want to control the loop variable during execution, it is possible to use a while loop, where you have full control. While loop may be better also if v has a very large length, since, for example for (n in 1:100) creates a vector of length 100 in memory. It should also be noted that the for-loop for (n in 1:k) performs 2 iterations, if k is 0, since 1:0 is a vector of length 2. If k may be 0, then it is better to use for (n in seq(length=k)) since seq(length=0) has length 0. Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] crazy loop error.
Hi
r-help-boun...@r-project.org napsal dne 24.01.2011 23:18:35:
Thanks for the reply Erik, As you mentioned, grouping consecutive
elements
of 'a' was my idea.
I am unaware of any R'ish way to do it. It would be nice if someone in
the
community knows this.
The error resulting in the NA was pretty easy to fix, and my loop works,
but
the results are still wrong (new script below).
Ideally it should print single hello for the single letters and
grouped '3
hellos' for the fives, grouped '2 hellos' for the sixes etc.
Based on the run results, if the value of n is being tracked, it changes
quite unpredictably.
Can someone explain how the value of n changes from end of the loop to
the
top without anything being done to it?
Put it into a foo function and make use debug(fun())
Regards
Petr
BTW mapply solution is shorter and probably quicker and easier to
maintain.
I cannot figure out what I am doing wrong.
a-c(2,3,5,5,5,6,6,7)
for(n in 1:length(a))
{
print(paste(n: ,n))
z1-a[n]
print(paste(z1:,z1))
#make a list container
ldata-list()
t=1
while(z1==a[n])
{
#add dataframes to list
ldata[[t]]-paste(hello)
n=n+1
t=t+1
if(nlength(a))
{
break;
}
}
print(--End of while loop---)
for(y in 1:length(ldata))
{
print(ldata[[y]])
}
print(paste(n: ,n))
print(**End of for loop)
}
Thanks,
Roy
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Re: [R] Extract NA data rows
On 25/01/2011 8:07 p.m., typhoong wrote: hi i have the following dataframe x y 1 345 6 NA 8 123 32 123 12 NA 6 124 7 NA and i want to extract the data rows which contains NA data, I tried subset(dataframe,y==NA) but fail. if you know the answers, please let me know thanks. typhoong Your test is wrong for NA. You must use is.na: df x y 1 1 345 2 6 NA 3 8 123 4 32 123 5 12 NA 6 6 124 7 7 NA subset(df, is.na(y)) x y 2 6 NA 5 12 NA 7 7 NA -- _ David Scott Department of Statistics The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055 Email: d.sc...@auckland.ac.nz, Fax: +64 9 373 7018 Director of Consulting, Department of Statistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Odp: Extract NA data rows
Hi r-help-boun...@r-project.org napsal dne 25.01.2011 08:07:10: hi i have the following dataframe x y 1 345 6 NA 8 123 32 123 12 NA 6 124 7 NA and i want to extract the data rows which contains NA data, I tried subset(dataframe,y==NA) See ?is.na however ?complete.cases is my preferred choice when working with data frames and several columns. dataframe[complete.cases(dataframe),] Regards Petr but fail. if you know the answers, please let me know thanks. typhoong -- View this message in context: http://r.789695.n4.nabble.com/Extract-NA-data- rows-tp3235568p3235568.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extract NA data rows
Hi! Try subset(dataframe, is.na(y)) or df[is.na(df$y),] HTH, Ivan Le 1/25/2011 08:07, typhoong a écrit : hi i have the following dataframe x y 1 345 6 NA 8 123 32 123 12 NA 6 124 7 NA and i want to extract the data rows which contains NA data, I tried subset(dataframe,y==NA) but fail. if you know the answers, please let me know thanks. typhoong -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Problems plotting the efficient frontier with fPortfolio
Hello, I have some simulations of financial data, I have 17 variables simulated 1000 times to three horizons. I am tring to plot the efficient frontier which I already obtained using th fPortfolio package. I am using the following commands: Data=timeSeries(X[1,,]) lppSpec - portfolioSpec() longFrontier - portfolioFrontier(Data, lppSpec) plot(longFrontier) Selección: 1 Error en dimnames(x) - dn : la longitud de 'dimnames' [1] no es igual a la extensión del arreglo tailoredFrontierPlot(object = longFrontier, mText = MV Portfolio - LongOnlyConstraints,risk = Cov) Error en dimnames(x) - dn : la longitud de 'dimnames' [1] no es igual a la extensión del arreglo and getting the error that appears. I also tried to do the same with the same data changing the solver to solveRshortExact and using the Short constraints and got the same error. Does anybody know what might be going on? Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] crazy loop error.
On Tue, 25 Jan 2011, Petr Savicky wrote: On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. The error resulting in the NA was pretty easy to fix, and my loop works, but the results are still wrong (new script below). Ideally it should print single hello for the single letters and grouped '3 hellos' for the fives, grouped '2 hellos' for the sixes etc. Based on the run results, if the value of n is being tracked, it changes quite unpredictably. Can someone explain how the value of n changes from end of the loop to the top without anything being done to it? Hi. A for-loop in R is different from a for-loop in C. It is similar to foreach loop in Perl. If v is a vector, then for (n in v) first creates the vector v and then always performs length(v) iterations. Before iteration i, n is assigned v[i] even if n is changed in the previous iteration. And also if v is changed during the loop. If you want to control the loop variable during execution, it is possible to use a while loop, where you have full control. While loop may be better also if v has a very large length, since, for example for (n in 1:100) creates a vector of length 100 in memory. It should also be noted that the for-loop for (n in 1:k) performs 2 iterations, if k is 0, since 1:0 is a vector of length 2. If k may be 0, then it is better to use for (n in seq(length=k)) since seq(length=0) has length 0. Since you keep mentioning that, it is actually much better to use seq_len(k) (and seq_along(x) instead of your earlier recommendation of seq(along=x)). And if you are using seq() in other cases in programs, consider seq.int() instead. Hope this helps. Petr Savicky. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Subtracting elements of data.frame
Dear R helpers I have a dataframe as df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) df x y 1 1 102 2 14 500 3 3 40 4 21 101 5 11 189 # Actually I am having dataframe having multiple columns. I am just giving an example. I need to subtract all the rows of df by the first row of df i.e. I need to subtract each element of 'x' column by 1. Likewise I need to subtract all elements of column 'y' by 11. Thus I need an output like df_new x y 1 0 0 2 13 398 3 2 -62 4 20 -1 5 10 87 As I had mentioned above, I have number of columns in reality and thus I can't use the command say df_new = data.frame(x = df$x-df$x[1], y = df$y-df$y[1]) Kindly guide Thanking you all in advance Regards Vincy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot geom_boxplot and stat_smooth
Hi Petr: I had to do a little bit of finagling, but this seems to work. I basically did the following: (i) coordinated the dodging of the points and boxplots by typ within konc.f; (ii) summarized the group medians in a separate data frame and added an additional column to compensate for the offsets in the boxplots due to dodging (iii) used geom_line() with group typ to get the connecting lines. # library(ggplot2) # Summarize group medians ads - ddply(ad, .(konc.f, typ), summarise, m = median(bel)) # Offset x-positions ads$pos - rep(c(1:4, 6, 7), each = 4) + rep(c(-0.3, -0.1, 0.1, 0.3), 6) # Note use of position_dodge() with same width in both the boxplot and point geoms # Add geom_line with summarized data frame and offset x-positions # group = typ produces a separate line per typ p + geom_boxplot(position = position_dodge(width = 0.8)) + geom_point(position = position_dodge(width = 0.8)) + geom_line(data = ads, aes(x = pos, y = m, colour = typ, group = typ)) I chose to create a separate data frame for the summaries rather than use stat_summary(), for example, because it is a bit easier to plot individual layers that way. HTH, Dennis On Mon, Jan 24, 2011 at 11:05 PM, Petr PIKAL petr.pi...@precheza.cz wrote: Dear all I would like to superpose some smoothing line through boxplot in ggplot dput(ad) structure(list(konc.f = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 6L, 7L, 7L, 7L), .Label = c((189,196], (196,202], (202,208], (208,215], (215,221], (221,227], (227,234]), class = factor), bel = c(99.28, 99.29, 99.25, 98.13, 99.51, 99.21, 99.09, 97.84, 98.97, 98.48, 98.64, 98.09, 98.44, 98.19, 98.25, 97.54, 99.11, 98.23, 97.62, 97.01, 97.62, 97.58, 97.42, 97.75, 97.16, 96.79, 96.82, 98.8, 99.02, 98.86, 98.85, 99.25, 98.46, 98.68, 98.3, 98.67, 98.54, 98.39, 98.18, 98.54, 99.13, 98.92, 98.29, 98.78, 98.58, 98.78, 98.9, 98.18, 97.5, 98.63, 97.53, 96.55, 96.52, 96.23, 95.54, 95.54, 96.33, 95.91, 95.41, 94.98, 94.69, 93.95, 94.05, 95.22, 94.6, 94.27, 93.44, 95.15, 94.62, 93.86, 94.51, 94.83, 93.66, 92.95, 94.4, 93.17, 92.77, 95.79, 95.03, 94.96, 95.94, 95.71, 95.19, 95.11, 94.91, 94.42, 94.68, 94.91, 94.66, 94.05, 93.57, 93.43, 94.77, 93.58, 93.84, 93.24, 94.45, 93.57, 93.46, 92.38, 92.39, 94.07 ), typ = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(alrut, nealrut, stan, vlakan), class = factor)), .Names = c(konc.f, bel, typ), row.names = c(NA, -102L), class = data.frame) Here is what I did p-ggplot(ad, aes(x=konc.f, y=bel, colour=typ)) p+geom_boxplot()+geom_jitter(position=position_jitter(w=0.1))+stat_smooth() geom_smooth: Only one unique x value each group.Maybe you want aes(group = 1)? I get nice picture with boxes but I expected to get something like smoothing line through box centres. Is it possible without some hack to stat_smooth code? Regards Petr __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extract NA data rows
Hi, May be try this : data[which(is.na(data[,2])),] 2011/1/25 typhoong graham...@eurus-energy.com hi i have the following dataframe x y 1 345 6 NA 8 123 32 123 12 NA 6 124 7 NA and i want to extract the data rows which contains NA data, I tried subset(dataframe,y==NA) but fail. if you know the answers, please let me know thanks. typhoong -- View this message in context: http://r.789695.n4.nabble.com/Extract-NA-data-rows-tp3235568p3235568.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Kamel Gaanoun (+33) (0)6.76.04.65.77 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] crazy loop error.
On Tue, Jan 25, 2011 at 09:05:03AM +0100, Petr Savicky wrote: [...] to foreach loop in Perl. If v is a vector, then for (n in v) first creates the vector v and then always performs length(v) iterations. I forgot that ‘break’ may stop the loop. See ?for for further information. In particular, it says You can assign to ‘var’ within the body of the loop, but this will not affect the next iteration. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] crazy loop error.
Dear Erik, Thanks for the mapply idea. I never got around to understand all those apply functions. I am still curious as to why the other loop didnt work. I even tried the debug but doesnt help. Anyway I will leave that for now. Thanks a lot for your help. Regards, Roy On Mon, Jan 24, 2011 at 11:43 PM, Erik Iverson er...@ccbr.umn.edu wrote: Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. Is this the idea you're trying to execute? It uses ?rle and ?mapply. a - c(2,3,5,5,5,6,6,7) mapply(rep, hello, rle(a)$lengths, USE.NAMES = FALSE) [[1]] [1] hello [[2]] [1] hello [[3]] [1] hello hello hello [[4]] [1] hello hello [[5]] [1] hello -- Best Regards, Roy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subtracting elements of data.frame
Hi: df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) apply(df, 2, function(x) x - x[1]) x y [1,] 0 0 [2,] 13 398 [3,] 2 -62 [4,] 20 -1 [5,] 10 87 HTH, Dennis On Tue, Jan 25, 2011 at 1:20 AM, Vincy Pyne vincy_p...@yahoo.ca wrote: Dear R helpers I have a dataframe as df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) df x y 1 1 102 2 14 500 3 3 40 4 21 101 5 11 189 # Actually I am having dataframe having multiple columns. I am just giving an example. I need to subtract all the rows of df by the first row of df i.e. I need to subtract each element of 'x' column by 1. Likewise I need to subtract all elements of column 'y' by 11. Thus I need an output like df_new x y 1 0 0 2 13 398 3 2 -62 4 20 -1 5 10 87 As I had mentioned above, I have number of columns in reality and thus I can't use the command say df_new = data.frame(x = df$x-df$x[1], y = df$y-df$y[1]) Kindly guide Thanking you all in advance Regards Vincy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subtracting elements of data.frame
Hi, Try this: df_new - as.data.frame(lapply(df, FUN=function(x) x-x[1])) I hope it works! Ivan Le 1/25/2011 10:20, Vincy Pyne a écrit : Dear R helpers I have a dataframe as df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) df x y 1 1 102 2 14 500 3 3 40 4 21 101 5 11 189 # Actually I am having dataframe having multiple columns. I am just giving an example. I need to subtract all the rows of df by the first row of df i.e. I need to subtract each element of 'x' column by 1. Likewise I need to subtract all elements of column 'y' by 11. Thus I need an output like df_new x y 1 0 0 2 13 398 3 2 -62 4 20 -1 5 10 87 As I had mentioned above, I have number of columns in reality and thus I can't use the command say df_new = data.frame(x = df$x-df$x[1], y = df$y-df$y[1]) Kindly guide Thanking you all in advance Regards Vincy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subtracting elements of data.frame
I always forget about sapply(): df_new - sapply(df, FUN=function(x) x-x[1]) Ivan Le 1/25/2011 10:33, Ivan Calandra a écrit : Hi, Try this: df_new - as.data.frame(lapply(df, FUN=function(x) x-x[1])) I hope it works! Ivan Le 1/25/2011 10:20, Vincy Pyne a écrit : Dear R helpers I have a dataframe as df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) df x y 1 1 102 2 14 500 3 3 40 4 21 101 5 11 189 # Actually I am having dataframe having multiple columns. I am just giving an example. I need to subtract all the rows of df by the first row of df i.e. I need to subtract each element of 'x' column by 1. Likewise I need to subtract all elements of column 'y' by 11. Thus I need an output like df_new x y 1 0 0 2 13 398 3 2 -62 4 20 -1 5 10 87 As I had mentioned above, I have number of columns in reality and thus I can't use the command say df_new = data.frame(x = df$x-df$x[1], y = df$y-df$y[1]) Kindly guide Thanking you all in advance Regards Vincy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] review of R Graphs Cookbook
If you are foolish enough not to be following R Bloggers via RSS or twitter, you might miss: http://www.portfolioprobe.com/2011/01/24/review-of-r-graphs-cookbook-by-hrishi-mittal/ Executive summary: Extremely useful for new users, informative to even quite seasoned users. -- Patrick Burns pbu...@pburns.seanet.com twitter: @portfolioprobe http://www.portfolioprobe.com/blog http://www.burns-stat.com (home of 'Some hints for the R beginner' and 'The R Inferno') __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] crazy loop error.
Mr Ripley, May I ask why seq_len() and seq_along() are better than seq()? Thanks, Ivan Le 1/25/2011 09:58, Prof Brian Ripley a écrit : On Tue, 25 Jan 2011, Petr Savicky wrote: On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. The error resulting in the NA was pretty easy to fix, and my loop works, but the results are still wrong (new script below). Ideally it should print single hello for the single letters and grouped '3 hellos' for the fives, grouped '2 hellos' for the sixes etc. Based on the run results, if the value of n is being tracked, it changes quite unpredictably. Can someone explain how the value of n changes from end of the loop to the top without anything being done to it? Hi. A for-loop in R is different from a for-loop in C. It is similar to foreach loop in Perl. If v is a vector, then for (n in v) first creates the vector v and then always performs length(v) iterations. Before iteration i, n is assigned v[i] even if n is changed in the previous iteration. And also if v is changed during the loop. If you want to control the loop variable during execution, it is possible to use a while loop, where you have full control. While loop may be better also if v has a very large length, since, for example for (n in 1:100) creates a vector of length 100 in memory. It should also be noted that the for-loop for (n in 1:k) performs 2 iterations, if k is 0, since 1:0 is a vector of length 2. If k may be 0, then it is better to use for (n in seq(length=k)) since seq(length=0) has length 0. Since you keep mentioning that, it is actually much better to use seq_len(k) (and seq_along(x) instead of your earlier recommendation of seq(along=x)). And if you are using seq() in other cases in programs, consider seq.int() instead. Hope this helps. Petr Savicky. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplot geom_boxplot and stat_smooth
Thank you I did not realise I can simply add data from different source data frame to constructed graph. It even works with stat_smooth() quite straightforward. One is always learning new tricks. Best regards Petr r-help-boun...@r-project.org napsal dne 25.01.2011 10:24:42: Hi Petr: I had to do a little bit of finagling, but this seems to work. I basically did the following: (i) coordinated the dodging of the points and boxplots by typ within konc.f; (ii) summarized the group medians in a separate data frame and added an additional column to compensate for the offsets in the boxplots due to dodging (iii) used geom_line() with group typ to get the connecting lines. # library(ggplot2) # Summarize group medians ads - ddply(ad, .(konc.f, typ), summarise, m = median(bel)) # Offset x-positions ads$pos - rep(c(1:4, 6, 7), each = 4) + rep(c(-0.3, -0.1, 0.1, 0.3), 6) # Note use of position_dodge() with same width in both the boxplot and point geoms # Add geom_line with summarized data frame and offset x-positions # group = typ produces a separate line per typ p + geom_boxplot(position = position_dodge(width = 0.8)) + geom_point(position = position_dodge(width = 0.8)) + geom_line(data = ads, aes(x = pos, y = m, colour = typ, group = typ)) I chose to create a separate data frame for the summaries rather than use stat_summary(), for example, because it is a bit easier to plot individual layers that way. HTH, Dennis On Mon, Jan 24, 2011 at 11:05 PM, Petr PIKAL petr.pi...@precheza.cz wrote: Dear all I would like to superpose some smoothing line through boxplot in ggplot dput(ad) structure(list(konc.f = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 7L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 6L, 7L, 7L, 7L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 6L, 7L, 7L, 7L), .Label = c((189,196], (196,202], (202,208], (208,215], (215,221], (221,227], (227,234]), class = factor), bel = c(99.28, 99.29, 99.25, 98.13, 99.51, 99.21, 99.09, 97.84, 98.97, 98.48, 98.64, 98.09, 98.44, 98.19, 98.25, 97.54, 99.11, 98.23, 97.62, 97.01, 97.62, 97.58, 97.42, 97.75, 97.16, 96.79, 96.82, 98.8, 99.02, 98.86, 98.85, 99.25, 98.46, 98.68, 98.3, 98.67, 98.54, 98.39, 98.18, 98.54, 99.13, 98.92, 98.29, 98.78, 98.58, 98.78, 98.9, 98.18, 97.5, 98.63, 97.53, 96.55, 96.52, 96.23, 95.54, 95.54, 96.33, 95.91, 95.41, 94.98, 94.69, 93.95, 94.05, 95.22, 94.6, 94.27, 93.44, 95.15, 94.62, 93.86, 94.51, 94.83, 93.66, 92.95, 94.4, 93.17, 92.77, 95.79, 95.03, 94.96, 95.94, 95.71, 95.19, 95.11, 94.91, 94.42, 94.68, 94.91, 94.66, 94.05, 93.57, 93.43, 94.77, 93.58, 93.84, 93.24, 94.45, 93.57, 93.46, 92.38, 92.39, 94.07 ), typ = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(alrut, nealrut, stan, vlakan), class = factor)), .Names = c(konc.f, bel, typ), row.names = c(NA, -102L), class = data.frame) Here is what I did p-ggplot(ad, aes(x=konc.f, y=bel, colour=typ)) p+geom_boxplot()+geom_jitter(position=position_jitter(w=0.1))+stat_smooth() geom_smooth: Only one unique x value each group.Maybe you want aes(group = 1)? I get nice picture with boxes but I expected to get something like smoothing line through box centres. Is it possible without some hack to stat_smooth code? Regards Petr __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Masking commands - Permutation in gregmisc and e1071
On 24.01.2011 23:53, Peter Langfelder wrote: On Mon, Jan 24, 2011 at 2:47 PM, Yanika Borgakina...@gmail.com wrote: I am using the function permutations from the package *gregmisc*. However, I am also making use of the package *e1071*, which also contains a function called permutations. I want to use the function permutations from the * gregmisc* package, however, the other package is masking this function. This happens both when I load the *e1071* package before *gregmisc* and when I load *e1071* after I load *gregmisc*. Is there any specific command to use the permutation from one package and not the other please? To specify package when you call a function, use package::function(...), for example gregmisc::permutations. You may also want to read about the NAMESPACE mechanism in R. Best, Uwe Ligges Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to get loglik parameter from splm package?
Dear useR,
although I admit that getting the log likelihood is important, you must
concede that obtaining the parameter estimates is not bad either.
Regarding craze, well there are crazier things in the world than this,
just look at the political situation in Italy.
Anyway, the loglik has always been there, although it wasn't exported
(hence the NULL value). In the most recent versions of 'splm' we have
made it available, at least for RE models, through the standard way: a
logLik() method. Usage:
logLik(yourmodel)
You can download it from R-forge, as usual.
Best,
Giovanni
original message --
Message: 42
Date: Mon, 24 Jan 2011 06:59:39 -0800 (PST)
From: zhaowei zao_...@msn.com
To: r-help@r-project.org
Subject: [R] how to get loglik parameter from splm package?
Message-ID: 1295881179014-3234185.p...@n4.nabble.com
Content-Type: text/plain; charset=us-ascii
splm package is a r implemention of spatial panel data models.
and the loglik paremeter is most important infomation for splm methods.
but i found the loglik always been null ,it's craze to get right
estimation
in splm with null loglik.
Any one knows the splm package and can get the right loglik ? please
help
me.
thanks
--
View this message in context:
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age-tp3234185p3234185.html
Sent from the R help mailing list archive at Nabble.com.
--- end original message -
Giovanni Millo
Research Dept.,
Assicurazioni Generali SpA
Via Machiavelli 4,
34132 Trieste (Italy)
tel. +39 040 671184
fax +39 040 671160
Ai sensi del D.Lgs. 196/2003 si precisa che le informazi...{{dropped:13}}
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[R] Integration of two lines
Hello, I need to integrate the absolute difference between two lines measured on different points. # For example : x - seq(0, 1, 1/100) f_x - runif(101) + x y - seq(0, 1, 1/23) f_y - runif(24) + (1 - y) plot(x, f_x, type=l) lines(y, f_y) Then I would like to compute Integral( | f_x - f_y | )dx. (This is not the same as | Integral(f_x)dx - Integral(f_y)dx |.) Computing this integral looks non trivial. I guess I should interpolate the points of f_y over x and integrate both lines on these intervals. Even then I would miss points where the lines cross. There are functions to integrate below *one* line (I'm thinking about the trapz function in caTools). Do you know if there is a function to do this integration properly with two lines (and especially their absolute difference)? Regards, Xavier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Map an Area to another
Dear All,
I would like to ask you help with the following:
Assume that I have an area of 36 cells (or sub-areas)
sr-matrix(data=seq(from=1,to=36),nrow=6,ncol=6,byrow=TRUE)
sr
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123456
[2,]789 10 11 12
[3,] 13 14 15 16 17 18
[4,] 19 20 21 22 23 24
[5,] 25 26 27 28 29 30
[6,] 31 32 33 34 35 36
This 6*6 matrix denotes an area and the contents of the cell denote the values
the area gets inside that cell-zone (or sub-area).
I would like to map this AREA into x,y function for x e[-1,1] ye[-1,1]
I really didn't know how to implement that, so with help I got from a list
member I tried to use FindInterval for the x,y plane. I am sure that it might
be an easier way to do that so please do suggest it. If there is not then
continue reading.
In the picture below:
http://img521.imageshack.us/f/maparea.gif/
You can see the matrix presented above placed inside the x e[-1,1] ye[-1,1]
which from now on these are the results of the f(x,y) function.
I.e f(-0.5,0.3) will return the appropriate cell (x,y) index.
As I told you I tried using the findInterval to get for the f(x,y) as an output
the correct cell number.
This is what I have implemented so far (you can copy paste the code there is a
small call of my function at the end):
sr.map - function(sr){
# This function converts the s(x,y) matrix into a function x spans from -1 #to
1 and y spans from -1 to 1.
# Input: sr a x,y matrix containing the shadowing values of a Region
breaksX - seq(from=-1, to = 1, length = nrow(sr) +1L )
breaksY - seq(from=-1, to = 1, length = ncol(sr) + 1L)
function(x,y){ # SPAGGETI CODE FOR EVER
indx - findInterval(x, breaksX,rightmost.closed=TRUE)
indy - findInterval(y, breaksY,rightmost.closed=TRUE)
if ( (x0) (y0) ) # x0,y0
c(indx,ncol(sr)-indy+1)
else if ( (x0) (y0) ) # x0,y0
c(nrow(sr)-indx+1,indy)
else if ( (x0) (y0) ) # x0,y0
c(nrow(sr)-indx+1,indy)
else if ( (x0) (y0) ) # x0,y0
c(indx,ncol(sr)-indy+1)
else
c(indx,indy)
}
}
sr-matrix(data=seq(from=1,to=36),nrow=6,ncol=6,byrow=TRUE)
f.sr.map-sr.map((sr))
f.sr.map(-0.1,-0.1)
If you are worried about my so many if inside the main body of my function you
can find why in the picture above. In the right part of the images you can see
some of my calculations
Top right part: What I wanted to have and what was the output was when there
was no if statement but only the last line c(indx,indy) (what two findIntervals
returned)
Based on the what I wanted (column on the top right part) I wrote the changes
in the logic by introducing if statements. You can find them at the bottom
right part.
This works fine for values close to the 'borders'
f.sr.map(-1,-1)
f.sr.map(-1,1)
f.sr.map(1,-1)
f.sr.map(1,1)
but not for values for example close to zero. This is a wrong output
f.sr.map(0.1,1)
[1] 3 6
I am sure that it might be an easier way to do all that so please inform me .
I would like to thank all of you for your kind efforts to help me
Best Regards
Alex
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Re: [R] Subtracting elements of data.frame
On 2011-01-25 01:20, Vincy Pyne wrote:
Dear R helpers
I have a dataframe as
df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189))
df
x y
1 1 102
2 14 500
3 3 40
4 21 101
5 11 189
# Actually I am having dataframe having multiple columns. I am just giving an
example.
I need to subtract all the rows of df by the first row of df i.e. I need to
subtract each element of 'x' column by 1. Likewise I need to subtract all
elements of column 'y' by 11. Thus I need an output like
df_new
x y
1 0 0
2 13 398
3 2 -62
4 20 -1
5 10 87
As I had mentioned above, I have number of columns in reality and thus I can't
use the command
say
df_new = data.frame(x = df$x-df$x[1], y = df$y-df$y[1])
Kindly guide
You've already had good simple solutions. Here's one more,
using the plyr package:
require(plyr)
df_new - colwise(.fun = function(x) {x - x[1]})(df)
plyr is very handy for all sorts of data manipulations.
Peter Ehlers
Thanking you all in advance
Regards
Vincy
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subtracting elements of data.frame
Try this: sweep(as.matrix(df), 2, as.matrix(df[1,])) On Tue, Jan 25, 2011 at 7:20 AM, Vincy Pyne vincy_p...@yahoo.ca wrote: Dear R helpers I have a dataframe as df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189)) df x y 1 1 102 2 14 500 3 3 40 4 21 101 5 11 189 # Actually I am having dataframe having multiple columns. I am just giving an example. I need to subtract all the rows of df by the first row of df i.e. I need to subtract each element of 'x' column by 1. Likewise I need to subtract all elements of column 'y' by 11. Thus I need an output like df_new x y 1 0 0 2 13 398 3 2 -62 4 20 -1 5 10 87 As I had mentioned above, I have number of columns in reality and thus I can't use the command say df_new = data.frame(x = df$x-df$x[1], y = df$y-df$y[1]) Kindly guide Thanking you all in advance Regards Vincy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Train error:: subscript out of bonds
What version of caret and R? We'll also need a reproducible example.
On Mon, Jan 24, 2011 at 12:44 PM, Neeti nikkiha...@gmail.com wrote:
Hi,
I am trying to construct a svmpoly model using the caret package (please
see code below). Using the same data, without changing any setting, I am
just changing the seed value. Sometimes it constructs the model
successfully, and sometimes I get an “Error in indexes[[j]] : subscript out
of bounds”.
For example when I set seed to 357 following code produced result only for 8
iterations and for 9th iteration it reaches to an error that “subscript out
of bonds” error. I don’t understand why
Any help would be great
thanks
###
for (i in 1:10)
{
fit1-NULL;
x-NULL;
x-which(number==i)
trainset-d[-x,]
testset-d[x,]
train1-trainset[,-ncol(trainset)]
train1-train1[,-(1)]
test_t-testset[,-ncol(testset)]
species_test-as.factor(testset[,ncol(testset)])
test_t-test_t[,-(1)]
#CARET::TRAIN
fit1-train(train1,as.factor(trainset[,ncol(trainset)]),svmpoly,trControl
= trainControl((method = cv),10,verboseIter = F),tuneLength=3)
pred-predict(fit1,test_t)
t_train[[i]]-table(predicted=pred,observed=testset[,ncol(testset)])
tune_result[[i]]-fit1$results;
tune_best-fit1$bestTune;
scale1[i]-tune_best[[3]]
degree[i]-tune_best[[2]]
c1[i]-tune_best[[1]]
}
--
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--
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Re: [R] Integration of two lines
g - function(x) abs(f1(x)-f2(x)) now you have one function and you can integrate it. Rich Sent from my iPhone On Jan 25, 2011, at 7:32, Xavier Robin xavier.ro...@unige.ch wrote: Hello, I need to integrate the absolute difference between two lines measured on different points. # For example : x - seq(0, 1, 1/100) f_x - runif(101) + x y - seq(0, 1, 1/23) f_y - runif(24) + (1 - y) plot(x, f_x, type=l) lines(y, f_y) Then I would like to compute Integral( | f_x - f_y | )dx. (This is not the same as | Integral(f_x)dx - Integral(f_y)dx |.) Computing this integral looks non trivial. I guess I should interpolate the points of f_y over x and integrate both lines on these intervals. Even then I would miss points where the lines cross. There are functions to integrate below *one* line (I'm thinking about the trapz function in caTools). Do you know if there is a function to do this integration properly with two lines (and especially their absolute difference)? Regards, Xavier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to simulate a variable Xt=Wit+0.5Wit-1 with
Dear Carlos,
please refrain from posting the same question umpteen times. Please
consider that code is hard to read and people might not have the time to
run your simulation etc. etc..
As I told you privately in response to your message on 18/1,
Re: generating correlated effects, I tried this only once, but I
didn't get it right. Simulations using this are, e.g., Hansen (2007)
http://faculty.chicagobooth.edu/christian.hansen/research/panel_cov_t.
pdf
and Drukker (2003)
http://ideas.repec.org/a/tsj/stataj/v3y2003i2p168-177.html
I suggest you take inspiration from what they did.
so the references are here for the possible benefit (?) of the list too.
In the meantime I looked at your code and it looks more or less OK to
me. I would have generated the correlated effects simply as
x_i=something random + gamma*u_i, but your Choleski approach is much
better. Sorry but a thorough check is beyond my time availability now.
arima.sim() may be a good way to go, I can't tell. The simplest way is
just to use loops: generating (in pseudo-R)
w - runif(n, parameters)
x[1] - random
and then
for(i in 2:n) x[i] - 0.5*x[i-1]+w[i]
PS remember to generate a sufficiently long sequence of x's *before*
those in your sample, in order to make the initial conditions
irrelevant.
Let me suggest Ch. 7.1 of Kleiber and Zeileis, Applied Econometrics
with R for a nice and organized approach to (econometric) simulations
in R.
Other than this, it's just basic R, trying it out and seeing if it
works.
Best wishes,
Giovanni
--- original message ---
Message: 18
Date: Mon, 24 Jan 2011 14:38:43 +
From: car...@sapo.pt
To: r-help r-help@r-project.org
Subject: [R] How to simulate a variable Xt=Wit+0.5Wit-1 with
Wit~U(0,2)
Message-ID: 20110124143843.10742vvdq8rvq...@mail.sapo.pt
Content-Type: text/plain; charset=ISO-8859-15; DelSp=Yes;
format=flowed
Dear all
I simulate a panel data:
n - 10
t - 5
nt - n*t
pData - data.frame(id = rep(paste(JohnDoe, 1:n, sep = .), each =
t),time = rep(1981:1985, n))
rho -0.99#simulate alphai corelated with the xi
print(rho)
alphai - rnorm(n,mean=0,sd=1)#alphai simulation
x- as.matrix(rnorm(nt,1))#xi simulation
akro - kronecker(alphai ,matrix(1,t,1))#kronecker of alphai
cormat-matrix(c(1,rho,rho,1),nrow=2,ncol=2)#correlation matrix
cormat.chold - chol(cormat)#choleski transformation of correlation
matrix
akrox - cbind(akro,x)
ax - akrox%*%cormat.chold
ai - as.matrix(ax[,1])
pData$alphai-as.vector(ai)
xcorr - as.matrix(ax[,2:(1+ncol(x))])
pData$xcorrei-as.vector(xcorr)
library(plm)
pData-plm.data(pData, index = c(id, time))
pData
But now i need a variable Xt=Wit+0.5Wit-1 with Wit~U(0,2), the code i
Try to use is:
for (i in 1:n) {
p - i*t
m - (i-1)*t+1
for (j in m:p){
xt-arima.sim(n=nt, list(ar=c(0.5))) }
}
Is this the correct way to simulate the AR(1), without the assumption
Wit~U(0,2)? How i simulate the variable with the assumption
Wit~U(0,2), and put it in my dataframe correctly?
Comments and tips are welcome,
regards
Carlos Br?s
Statistics Portugal-INE
-- end original message -
Giovanni Millo
Research Dept.,
Assicurazioni Generali SpA
Via Machiavelli 4,
34132 Trieste (Italy)
tel. +39 040 671184
fax +39 040 671160
Ai sensi del D.Lgs. 196/2003 si precisa che le informazi...{{dropped:13}}
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[R] xlsReadWrite 1.5.4 and xlsReadWritePro 1.6.4 released
The xlsReadWrite[Pro] package allows to natively read and write Excel files (.xls) on the Win 32-bit platform. Changes: o fix bug with integer conversion (http://dev.swissr.org/issues/113) PROBLEM: values outside the integer range (i.e. 12345678901) didn't give an NA (and a warning message) but have been converted to a wrong value WHERE: - free/pro: reading an Excel sheet and explicitely defining: 'colClasses = integer' or 'type = integer. It's assumed that only very few people do this. - pro: when (auto) detecting a data.frame column type for numeric values the type becomes an 'integer' in some situations. This probably affects more people than above SOLUTION: - an error will be raised for too large/small values (same as in read.table) - numeric data.frame column types default to 'double' (as in the free version) o adapt RUnit tests to new behaviour o some small cosmetic changes Credits to Gyorgy Ottucsak who found the problem in xlsReadWritePro and send us a report (incl. testfile). Thanks a lot! The new xlsReadWrite version will be submitted to CRAN in a moment. Both updated packages are available now from www.swissr.org/download or - full listing - here: http://dl.dropbox.com/u/2602516/swissrpkg/index.html. Cheers, Hans-Peter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem reading PostgreSQL data with RODBC
Am 24.01.2011 14:08, schrieb Bart Joosen: I think this is a problem with quotes. If you look good, you see: seiz.df- sqlFetch(chnl, 'source.MAIN') ... 'source.main': table not found on channel You asked MAIN, but your db can't find main. If you use seiz.df- sqlFetch(chnl, '\source\.\MAIN\') , you problem should be gone. I think the deeper problem lies with upper/lower case in the table name - is it MAIN, Main or main? If a tablename in PostgreSQL has at least one upper case, PostgreSQL uses quotes, and therefore the solution proposed above applies... HTH, Albin -- | Albin Blaschka, Mag.rer.nat. | Etrichstrasse 26, A-5020 Salzburg | * www.albinblaschka.info * www.thinkanimal.info * | - It's hard to live in the mountains, hard but not hopeless! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Integration of two lines
Xavier Robin Xavier.Robin at unige.ch writes: Hello, I need to integrate the absolute difference between two lines measured on different points. # For example : x - seq(0, 1, 1/100) f_x - runif(101) + x y - seq(0, 1, 1/23) f_y - runif(24) + (1 - y) plot(x, f_x, type=l) lines(y, f_y) Then I would like to compute Integral( | f_x - f_y | )dx. (This is not the same as | Integral(f_x)dx - Integral(f_y)dx |.) First define a function from those points: fx - approxfun(x, f_x) fy - approxfun(y, f_y) f - function(x) abs(fx(x)-fy(x)) and now you can apply integrate() or trapz(): xx - sort(c(x, y)) yy - f(xx) trapz(xx, yy) trapz() should return the more accurate (i.e. exact) result. --Hans Werner Computing this integral looks non trivial. I guess I should interpolate the points of f_y over x and integrate both lines on these intervals. Even then I would miss points where the lines cross. There are functions to integrate below *one* line (I'm thinking about the trapz function in caTools). Do you know if there is a function to do this integration properly with two lines (and especially their absolute difference)? Regards, Xavier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Integration of two lines
Le 25.01.2011 15:23, Rmh a écrit : g - function(x) abs(f1(x)-f2(x)) now you have one function and you can integrate it. Thank you Rich. Unfortunately I have no f1 and f2 functions, only a set of observed points on two lines - and no idea about the underlying distribution to create a function. Other ideas? Xavier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using open calais in R
fayazvf wrote: I am using calais api in R for text analysis. But im facing a some problem when fetching the rdf from the server. I'm using the getToHost() method for the api call but i get just a null string. You haven't told us nearly enough for us to be able to reproduce what you are doing. Where and how is the R function getToHost() ? is it in an R package? The same url in browser returns an RDF document. getToHost(www.api.opencalais.com,/enlighten/rest/?licenseID=dkzdggsre232ur97c6be269gcontent=HomeparamsXML=) [1] http://api.opencalais.com/enlighten/rest/?licenseID=dkzdggsre232ur97c6be269gcontent=HomeparamsXML=; Yes, and library(RCurl) getURLContent(http://api.opencalais.com/enlighten/rest/?licenseID=dkzdggsre232ur97c6be269gcontent=HomeparamsXML=;) returns RDF content as does download.file(http://api.opencalais.com/enlighten/rest/?licenseID=dkzdggsre232ur97c6be269gcontent=HomeparamsXML=;, eg.txt) But since we have no way of knowing what getToHost() does (or the postToHost() in your earlier mail), we cannot figure out what is happening for you. Please do read the posting guidelines, specifically telling us about your session and what packages you are using. D. -- View this message in context: http://r.789695.n4.nabble.com/Using-open-calais-in-R-tp3235597p3235597.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- There are men who can think no deeper than a fact - Voltaire Duncan Temple Langdun...@wald.ucdavis.edu Department of Statistics work: (530) 752-4782 4210 Mathematical Sciences Bldg. fax: (530) 752-7099 One Shields Ave. University of California at Davis Davis, CA 95616, USA pgpV6jduW3uDW.pgp Description: PGP signature __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Learn Vectorization (Vectorize)
Greetings Friends,
I would be grateful if you can help me undestand how to make my R code more
efficiently.
I have read in R intoductory tutorial that a for loop is not used so ofter (and
is not maybe not that efficient) compared to other languages.
So I am trying to build understanding how to get the equivalent of a for loop
using more R-oriented thinking.
If I got it right one way to do that in R is Vectorize. So I have writen a
small snippet of code with two nested for loops. I would be grateful if you can
help me find the equivalent of this code using Vectorize (or any other R
thinking)
My code takes as input a n*m matrix and prints the x,y coordinates where a
cells starts and ends.
remap - function (sr){
# Input this funcion takes as arguments
# sr: map
startpos- -1 #
endpos- +1 #
stepx- (endpos - (startpos)) / nrow(sr)
stepy- (endpos - (startpos)) / ncol(sr)
for (i in seq(from=-1,to=1,by=stepx) ) {
for (j in seq(from=-1,to=1,by=stepx) ){
cat(' \n',i,j)
}
}
}
sr-matrix(data=seq(from=1,to=9),nrow=3,ncol=3,byrow=TRUE)
remap(sr)
Regards
Alex
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Re: [R] Train error:: subscript out of bonds
Version: R = 2.11.1 CARET = 4.68 -- View this message in context: http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3236251.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Downloading data from internet
From: ggrothendi...@gmail.com Date: Mon, 24 Jan 2011 22:43:55 -0500 To: megh700...@gmail.com CC: r-help@r-project.org Subject: Re: [R] Downloading data from internet On Mon, Jan 24, 2011 at 8:48 PM, Megh Dal wrote: Dear all, I need to download an excel file from net, on which I have address like http://www.2shared.com/file/MMSMWv4B/MyData.html. Can I somehow directly download this file into my R workbook? If you have a direct URL to the file (not the URL in your post which just points a web page) then see the gdata example on this page: http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows Also, try to get the problem fixed properly by contacting site webmaster and admin ( the webmaster is unlikely to have authority himself to do something useful and indeed may think that complicated download procedures are cool). Tell them you are trying to load data into a computer program and that is more complicated than it needs to be. This comes up a lot. Scraping html is often difficult and unstable. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] crazy loop error.
ooh.. I have another question. What if I want to add the value in the vector a to the hello each time it prints. Here is your output a - c(2,3,5,5,5,6,6,7) mapply(rep, hello, rle(a)$lengths, USE.NAMES = FALSE) [[1]] [1] hello [[2]] [1] hello [[3]] [1] hello hello hello [[4]] [1] hello hello [[5]] [1] hello If I want something like this, based on values in vector a. [[1]] [1] hello 2 [[2]] [1] hello 3 [[3]] [1] hello 5 hello 5 hello 5 [[4]] [1] hello 6 hello 6 [[5]] [1] hello 7 What i am actually doing is hmm.. I have a bunch of text files which is output from another program. I want to extract some specific information from these files and write to a new file and save it. All these files have a certain variable k which maybe 2, or 3 or 5 etc. The vector a shows the k values of 8 of such files. I want the contents of all files with k value 5 to be written into one file. Thanks, Roy On Mon, Jan 24, 2011 at 11:43 PM, Erik Iverson er...@ccbr.umn.edu wrote: Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. Is this the idea you're trying to execute? It uses ?rle and ?mapply. a - c(2,3,5,5,5,6,6,7) mapply(rep, hello, rle(a)$lengths, USE.NAMES = FALSE) [[1]] [1] hello [[2]] [1] hello [[3]] [1] hello hello hello [[4]] [1] hello hello [[5]] [1] hello -- Best Regards, Roy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with matchit() and zelig()
Dear all, Does anybody know why the following code returns an error message? library(MatchIt) library(Zelig) data(lalonde) m.out1-matchit(treat~age+educ+black+hispan+nodegree+married +re74+re75, method=full, data=lalonde) z.out1-zelig(re78~age+educ+black+hispan+nodegree+married+re74+re75, data=match.data(m.out1, control), model=ls , weights=weights ) x.out1-setx(z.out1, data=match.data(m.out1, treat), cond=T ) s.out1-sim(z.out1, x=x.out1) Error in model.frame.default(formula = re78 ~ age + educ + black + hispan + : variable lengths differ (found for '(weights)') I noticed that the problem disappears if either the option weights=weights in the zelig() function or the option cond=T in the setx() function is removed. Is there a conflict between the options weights=weights in zelig() and cond=T in setx()/sim()? Or is there an inconsistency in my syntax? Any suggestion would be very much appreciated. Best wishes, Flavia Giammarino London School of Economics, UK Please access the attached hyperlink for an important electronic communications disclaimer: http://lse.ac.uk/emailDisclaimer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 3D Binning
I am trying to do binning on three variables (3d binning). The bin boundaries are specified by the user separately for each variable. I used the bin2 function in the 'ash' package for 2d binning that involves only two variables but didn't any package for similar binning with three variables. Are there any packages or codes available for 3d binning?? Thank you. -- View this message in context: http://r.789695.n4.nabble.com/3D-Binning-tp3236223p3236223.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Paired data survival analysis
--- begin included message --- Im an honours student at Monash University. I'm trying to analyse some data for my project, which involved 2 treatments. My subjects were exposed to both treatments, and i gave them 60 minutes to perform a certain behaviour. 3 of my subjects performed the behaviour in one treatment but not the other. Therefore, i need to do a survival analysis using paired data. Im little confused about how to go about this in R. Im able to perfrom a normal surival analyses not taking the paired data into account, but im just wondering if there is some way to take the pairing into account. I know there are 3 different ways to deal with grouping in the survival package, strata, cluster and frailty but i struggle to understand the meaning of these arguments and therefore do not know which one to use (if any). --- end inclusion --- All 3 methods can be defended. Adding cluster(id) to the model is equivalent to a generalized estimating equations approach (if this were a glm) or to the variance estimates commonly used in survey sampling (if this were a linear model). Adding frailty(id) is equivalent to fitting a linear mixed model. Using strata corresponds to a matched-pair analysis, and will essentially reduce to a sign test: for each subject treatment A was better, B was better, or tied. It's overkill in this case (lower power). If this were a linear model, you could find strong advocates for either the GEE and mixed approach being better. I somewhat prefer the GEE method myself. Terry T. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Learn Vectorization (Vectorize)
Try this:
expand.grid(seq(startpos, endpos, by = diff(c(startpos, endpos)) /
nrow(sr)),
seq(startpos, endpos, by = diff(c(startpos, endpos)) / nrow(sr)))
On Tue, Jan 25, 2011 at 1:29 PM, Alaios ala...@yahoo.com wrote:
Greetings Friends,
I would be grateful if you can help me undestand how to make my R code more
efficiently.
I have read in R intoductory tutorial that a for loop is not used so ofter
(and is not maybe not that efficient) compared to other languages.
So I am trying to build understanding how to get the equivalent of a for
loop using more R-oriented thinking.
If I got it right one way to do that in R is Vectorize. So I have writen a
small snippet of code with two nested for loops. I would be grateful if you
can help me find the equivalent of this code using Vectorize (or any other R
thinking)
My code takes as input a n*m matrix and prints the x,y coordinates where a
cells starts and ends.
remap - function (sr){
# Input this funcion takes as arguments
# sr: map
startpos- -1 #
endpos- +1 #
stepx- (endpos - (startpos)) / nrow(sr)
stepy- (endpos - (startpos)) / ncol(sr)
for (i in seq(from=-1,to=1,by=stepx) ) {
for (j in seq(from=-1,to=1,by=stepx) ){
cat(' \n',i,j)
}
}
}
sr-matrix(data=seq(from=1,to=9),nrow=3,ncol=3,byrow=TRUE)
remap(sr)
Regards
Alex
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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[R] Extracting SSE from lm
Apologies for this simple question - Given the number of comparisons I need to do it has become somewhat laborious to compute the SSE manually. I first have to extract the coefficients, build the model and run the model on the data. So far I haven't found any method in R that will do this for me. Is there a method that I haven't seen, or is there a small function I could write that would do this, and how might I go about that? Thanks, Brian [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] deSolve: Problem solving ODE including modulo-operator
I have a problem integrating the 'standard map' (
http://en.wikipedia.org/wiki/Standard_map
http://en.wikipedia.org/wiki/Standard_map ) with deSolve:
By using the modulo-operator '%%' with 2*pi in the ODEs (standardmap1), the
resulting values of P and Theta, should not be greater than 2pi. Because
this was not the case, i was thinking that the function 'ode' has some
internal problems with the '%%' or integrating periodical ODEs, so I wrote a
modulo-function by myself (modulo and standardmap2). But still I get values
much higher than 2pi and I cannot find the error... Any guess?
Thanks
Full code:
library(deSolve)
iterations - 100
Parameter - c(k = 0.6)
State - c(Theta = 1 , P = 1)
Time - 0:iterations/10
standardmap1 - function(Time, State, Parameter){
with(as.list(c(State, Parameter)), {
dP - (P + k * sin(Theta)) %% (2 * pi)
dTheta - (P + Theta) %% (2 * pi)
return(list(c(dP, dTheta)))
})
}
#solve ode using standardmap1
out1 - as.data.frame(ode(func = standardmap1, y = State,parms = Parameter,
times = Time))
# x mod y, end: maximal iterations
modulo - function(x,y,end=1000){
for (n in 0:end)
if (x (n-1)*y)
z = x - (n-1)*y
if (z 0)
return(z)
else break
}
standardmap2 - function(Time, State, Parameter){
with(as.list(c(State, Parameter)), {
dTheta - modulo((P + Theta),(2*pi))
dP - modulo(P + k *sin(Theta),(2*pi))
return(list(c(dP, dTheta)))
})
}
#solve ode using standardmap2
out2 - as.data.frame(ode(func = standardmap2, y = State,parms = Parameter,
times = Time))
#plot the results
matplot(out1[2],out1[3], type = p, pch = .)
matplot(out2[2],out2[3], type = p, pch = .)
--
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[R] barplot with varaible-width bars
I would like to produce a bar plot with varying-width bars. Here is an example
to illustrate:
ww - c(417,153,0.0216,0.0065,556,256,0.0162,0.0117,
+ 726,379,0.0358,0.0501,786,502,0.0496,0.0837,
+ 892,591,0.0785,0.0795)
yy-t(t(array(ww,c(2,10
barplot(yy[,2*1:5],las=1,space=c(.1,.5),beside=T)
produces a barplot of 5 pairs of bars that are of equal width
barplot(yy[,2*1:5],las=1,width=c(yy[,(2*1:5)-1]),space=c(.1,.5),beside=T)
makes the bars in each pair of unequal width, but the two widths do not vary
from pair to pair
I would like the width of each bar to be proportional to its corresponding
value in the width statement of this last call of barplot, like what I think
could be done with the mulbar function of SPlus. Can I do this with barplot
itself, or is this something for which lattice or ggplot 2 is needed? And, if
so, what would typical code look like?
Thanks for your help.
Larry Gould
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Re: [R] crazy loop error.
Hi r-help-boun...@r-project.org napsal dne 25.01.2011 10:58:36: ooh.. I have another question. What if I want to add the value in the vector a to the hello each time it prints. Here is your output a - c(2,3,5,5,5,6,6,7) mapply(rep, hello, rle(a)$lengths, USE.NAMES = FALSE) [[1]] [1] hello [[2]] [1] hello [[3]] [1] hello hello hello [[4]] [1] hello hello [[5]] [1] hello If I want something like this, based on values in vector a. Not sure how to use mapply test-vector(list, 5) lll-rle(a) for (i in seq_along(lll$lengths)) test[[i]] - rep(paste(hello, lll$values[i]), lll$lengths[i]) [[1]] [1] hello 2 [[2]] [1] hello 3 [[3]] [1] hello 5 hello 5 hello 5 [[4]] [1] hello 6 hello 6 [[5]] [1] hello 7 What i am actually doing is hmm.. I have a bunch of text files which is output from another program. I want to extract some specific information from these files and write to a new file and save it. All these files have a certain variable k which maybe 2, or 3 or 5 etc. The vector a shows the k values of 8 of such files. I want the contents of all files with k value 5 to be written into one file. That is rather vague description. Does those files have some structure? How do you know the variable k? Loops are not so ineffective if you use them for what they are good and if you do not expand the object within loop. See R-Inferno from P.Burns. Regards Petr Thanks, Roy On Mon, Jan 24, 2011 at 11:43 PM, Erik Iverson er...@ccbr.umn.edu wrote: Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. Is this the idea you're trying to execute? It uses ?rle and ?mapply. a - c(2,3,5,5,5,6,6,7) mapply(rep, hello, rle(a)$lengths, USE.NAMES = FALSE) [[1]] [1] hello [[2]] [1] hello [[3]] [1] hello hello hello [[4]] [1] hello hello [[5]] [1] hello -- Best Regards, Roy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] applying a function to output a matrix
Dear List,
I am using function distCosine from package geosphere to a list of lat/lon
coordinates, and I want to calculate the great circle distance between a
pair of coordinates in the list and all other pairs --- essentially, the
output should be a matrix. I have been able to achieve this with two nested
loops, as in the example below, but this is rather slow. Can someone please
suggest how to do this with apply or similar?
Many thanks,
Lara
install.packages(geosphere)
library(geosphere)
##generate sets of random points
n - 100
lon - runif(n, -180, 180)
lat - runif(n, -90, 90)
#package geosphere
##spherical law of cosines method
dCos - matrix( , nrow = length(lon), ncol = length(lat))
for (i in 1:length(lon)) {
for (j in 1:length(lat)) {
dCos[[i,j]] - distCosine(matrix(c(lon[i], lat[i]), ncol=2), matrix(c(
lon[j], lat[j]), ncol=2))
}}
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Re: [R] crazy loop error.
Well, I'm not Prof. Ripley, but the answer is: Look at the code. seq_len, seq.int, and seq_along call Primitives, which are implemented in C, and therefore MUCH faster than seq(), which is implemented as pure R code (and is also a generic, so requires method dispatch). Though for small n (up to a few thousand, say), it probably doesn't make much difference.(Here, to be corrected by Prof. Ripley is needed). -- Bert On Tue, Jan 25, 2011 at 2:22 AM, Ivan Calandra ivan.calan...@uni-hamburg.de wrote: Mr Ripley, May I ask why seq_len() and seq_along() are better than seq()? Thanks, Ivan Le 1/25/2011 09:58, Prof Brian Ripley a écrit : On Tue, 25 Jan 2011, Petr Savicky wrote: On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. The error resulting in the NA was pretty easy to fix, and my loop works, but the results are still wrong (new script below). Ideally it should print single hello for the single letters and grouped '3 hellos' for the fives, grouped '2 hellos' for the sixes etc. Based on the run results, if the value of n is being tracked, it changes quite unpredictably. Can someone explain how the value of n changes from end of the loop to the top without anything being done to it? Hi. A for-loop in R is different from a for-loop in C. It is similar to foreach loop in Perl. If v is a vector, then for (n in v) first creates the vector v and then always performs length(v) iterations. Before iteration i, n is assigned v[i] even if n is changed in the previous iteration. And also if v is changed during the loop. If you want to control the loop variable during execution, it is possible to use a while loop, where you have full control. While loop may be better also if v has a very large length, since, for example for (n in 1:100) creates a vector of length 100 in memory. It should also be noted that the for-loop for (n in 1:k) performs 2 iterations, if k is 0, since 1:0 is a vector of length 2. If k may be 0, then it is better to use for (n in seq(length=k)) since seq(length=0) has length 0. Since you keep mentioning that, it is actually much better to use seq_len(k) (and seq_along(x) instead of your earlier recommendation of seq(along=x)). And if you are using seq() in other cases in programs, consider seq.int() instead. Hope this helps. Petr Savicky. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Multivariate polynomials Howto
Good Evening, I would like to work with multivariate polynomials (x and y variables). I know that there is a package called multipol but I am not sure that supports my needs. I use a function (in reality legendre.polynomials) which creates me the polynomials I want. For example the following returns legendre.polynomials(2)[[2]] x (first order polynomial) I would like to calculate the polynomials with variable x with the polynomials with variable y. I do not know how I can do that in R as my function always returns output with x. Is it possible to change variable x with y? Regards Alex __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Learn Vectorization (Vectorize)
Inline below.
-- Bert
On Tue, Jan 25, 2011 at 7:48 AM, Henrique Dallazuanna www...@gmail.com wrote:
Try this:
expand.grid(seq(startpos, endpos, by = diff(c(startpos, endpos)) /
nrow(sr)),
seq(startpos, endpos, by = diff(c(startpos, endpos)) / nrow(sr)))
On Tue, Jan 25, 2011 at 1:29 PM, Alaios ala...@yahoo.com wrote:
Greetings Friends,
I would be grateful if you can help me undestand how to make my R code more
efficiently.
I have read in R intoductory tutorial that a for loop is not used so ofter
(and is not maybe not that efficient) compared to other languages.
So I am trying to build understanding how to get the equivalent of a for
loop using more R-oriented thinking.
If I got it right one way to do that in R is Vectorize.
-- You got it wrong. Vectorize() is just another (disguised) way of
doing R level looping, via apply functions.It offers no efficiency
advantage over explicit for loops, though it may make for cleaner
programming, which IS a great advantage in the bigger scheme of
things, I admit. But thatwasn't your question.
The basic idea of vectorization in R is to use **built-in** commands
that act on whole objects through their internal (usually C) code.
These **will** be much faster than looping in R.
So for example to generate 10 random numbers one can do:
system.time(x -rnorm(1e5))
user system elapsed
0.020.000.02
## OR
system.time({
+ y - numeric(1e5)
+ for(i in 1:(1e5))y[i] - rnorm(1)
+ })
user system elapsed
0.860.010.87
The former is vectorized, producing all 10 randoms at a go and
runs 40 times faster than the latter, which is not, instead producing
one random number at a time (and requiring the overhead of a separate
call for each, I think).
Please read an Intro to R for more. VR's MASS also has some helpful
remarks on vectorization. I'm sure that other texts (e.g. Dalgaard's)
are worth consulting on this, too, but I haven't read them.
-
So I have writen a
small snippet of code with two nested for loops. I would be grateful if you
can help me find the equivalent of this code using Vectorize (or any other R
thinking)
My code takes as input a n*m matrix and prints the x,y coordinates where a
cells starts and ends.
remap - function (sr){
# Input this funcion takes as arguments
# sr: map
startpos- -1 #
endpos- +1 #
stepx- (endpos - (startpos)) / nrow(sr)
stepy- (endpos - (startpos)) / ncol(sr)
for (i in seq(from=-1,to=1,by=stepx) ) {
for (j in seq(from=-1,to=1,by=stepx) ){
cat(' \n',i,j)
}
}
}
sr-matrix(data=seq(from=1,to=9),nrow=3,ncol=3,byrow=TRUE)
remap(sr)
Regards
Alex
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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--
Bert Gunter
Genentech Nonclinical Biostatistics
467-7374
http://devo.gene.com/groups/devo/depts/ncb/home.shtml
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[R] ggplot - controlling point size
Can anyone illuminate the following for me? How can I get rid of the blue line in the key in the second plot? ## Create a simple data frame df=data.frame(x=1:1000, y=2*1:1000+rnorm(1000,sd=1000), type=sample(letters[1:2],1000, replace=TRUE)) ## Very nice! Almost what I want qplot(x, y, data=df, colour=factor(type)) + geom_smooth() ## Make a nicer plot, with smaller points ## but why does that add the little blue line with a 1? qplot(x, y, data=df, colour=factor(type), size=1) + geom_smooth() [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Failing to install {rggobi} on win-7 R 2.12.0
Greetings all, I am failing to install the package rggobi on windows 7 with R 2.12.0. On R 2.11.1, the package was installed fine. I asked for help on the rggobi google group 4 days ago, and didn't receive any help, so I was wondering if someone here might have a suggestion. Here are the details: --- I am having a similar (bot not exact) problem as Tom had here: http://groups.google.com/group/ggobi/browse_thread/thread/67b7260d074d710c I downloaded and installed all components. (GTK was placed in d:\\GTK) When I try to load the rggobi library (library(rggobi) it offers me to install GTK+ because it can't find the dll (even that all of the dll it is looking for are present in the path environment Sys.getenv(PATH) Sys.getenv(GTK_BASEPATH) ) When answering no to the installation of GTK, I get the following error in a popping window: the procedure entry point g_malloc_n could not be located in the dynamic link library libglic-2.0-0.dll I then get the following errors in the R console: Loading required package: rggobi Error in library.dynam(RGtk2, pkgname, libname) : DLL 'RGtk2' not found: maybe not installed for this architecture? Failed to load RGtk2 dynamic library, attempting to install it. Learn more about GTK+ at http://www.gtk.org If the package still does not load, please ensure that GTK+ is installed and that it is on your PATH environment variable IN ANY CASE, RESTART R BEFORE TRYING TO LOAD THE PACKAGE AGAIN Error in inDL(x, as.logical(local), as.logical(now), ...) : unable to load shared object 'd:/Program Files/R/library/rggobi/libs/ i386/rggobi.dll': LoadLibrary failure: The specified procedure could not be found. I also tried running R as an administrator, but it didn't fix the problem (only added another error popup window with the massage: the procedure entry point cairo_glyph_allocate could not be located in the dynamic link library libcairo-2.dll) Here is my sessionInfo: sessionInfo() R version 2.12.0 (2010-10-15) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=Hebrew_Israel.1255 LC_CTYPE=Hebrew_Israel.1255 [3] LC_MONETARY=Hebrew_Israel.1255 LC_NUMERIC=C [5] LC_TIME=Hebrew_Israel.1255 attached base packages: [1] stats graphics grDevices utils datasets methods base Thank you for any advice. Best, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NA replacing
Hello R user, I have following data frame: df=data.frame(id=c(1:10),strata=rep(c(1,2),c(5,5)),y=c( 10,12,10,NA,15,70,NA,NA,55,100),x=c(3,4,5,7,4,10,12,8,3,15)) and I would like to replace NA's with: instead of first NA tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[1]* *7 */tapply(na.exclude(df)$x,na.exclude(df)$strata,sum)[1] where 7 is the value of x (id=4) in strata 1 where y=NA instead of second NA tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[2]* *12 */tapply(na.exclude(df)$x,na.exclude(df)$strata,sum)[2] where 12 is the value of x (id=7) in strata 2 where y=NA instead of third NA tapply(na.exclude(df)$y,na.exclude(df)$strata,sum)[2]* * 8 */tapply(na.exclude(df)$x,na.exclude(df)$strata,sum)[2] where 8 is the value of x(id=8) in strata 2 where y=NA. So, I would like to replace NA inside the stratas on above explained way. Does anyone know how to do this? thanks in advance Andrija [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot - controlling point size
try this: qplot(x, y, data=df, colour=factor(type), size=I(1)) + geom_smooth() Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA http://www.fws.gov/redbluff/rbdd_jsmp.aspx - Original Message From: Gene Leynes gleyne...@gmail.com To: r-help@r-project.org Sent: Tue, January 25, 2011 9:28:20 AM Subject: [R] ggplot - controlling point size Can anyone illuminate the following for me? How can I get rid of the blue line in the key in the second plot? ## Create a simple data frame df=data.frame(x=1:1000, y=2*1:1000+rnorm(1000,sd=1000), type=sample(letters[1:2],1000, replace=TRUE)) ## Very nice! Almost what I want qplot(x, y, data=df, colour=factor(type)) + geom_smooth() ## Make a nicer plot, with smaller points ## but why does that add the little blue line with a 1? qplot(x, y, data=df, colour=factor(type), size=1) + geom_smooth() [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Train error:: subscript out of bonds
after using options(error=utils::recover) option, following is the output. if i am correct this means that in ksvm there is some problem, but really could not understand. could anyone please tell me what is wrong... any help will be great thank you so much.. Enter a frame number, or 0 to exit 1: train(train1, as.factor(trainset[, ncol(trainset)]), svmpoly, trControl = trainControl((method = cv), 10, 2: train.default(train1, as.factor(trainset[, ncol(trainset)]), svmpoly, trControl = trainControl((method = c 3: do.call(trControl$computeFunction, argList) 4: function (X, FUN, ...) 5: FUN(X[[25]], ...) 6: do.call(createModel, args) 7: function (data, method, tuneValue, obsLevels, pp = NULL, ...) 8: ksvm(as.matrix(trainX), trainY, kernel = polydot(degree = tuneValue$.degree, scale = tuneValue$.scale, offset 9: ksvm(as.matrix(trainX), trainY, kernel = polydot(degree = tuneValue$.degree, scale = tuneValue$.scale, offset 10: .local(x, ...) 11: ksvm(x[c(indexes[[i]], indexes[[j]]), , drop = FALSE][cind, ], yd[cind], type = type(ret), kernel = kernel, k 12: ksvm(x[c(indexes[[i]], indexes[[j]]), , drop = FALSE][cind, ], yd[cind], type = type(ret), kernel = kernel, k 13: .local(x, ...) -- View this message in context: http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3236494.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] determining the order in which points are plotted
With large numbers of points you might want to consider hexagonal binning instead of scatter plots. I don't know of any tools that both do the binning and take groups into account, but you could think it through and work something out. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Mike Miller Sent: Monday, January 24, 2011 5:39 PM To: David Winsemius Cc: R-Help List Subject: Re: [R] determining the order in which points are plotted On Mon, 24 Jan 2011, David Winsemius wrote: On Jan 24, 2011, at 6:49 PM, Mike Miller wrote: I make plenty of scatterplots, especially using scatterplot.matrix from library(car). One thing I don't know how to do is determine which points are plotted last. Sometimes I plot a large number of points for multiple groups represented by different colors. ?points Thanks for the tip. I guess I would make vectors for x, y and col in the desired order and the first elements would be plotted first: Graphical parameters ‘pch’, ‘col’, ‘bg’, ‘cex’ and ‘lwd’ can be vectors (which will be recycled as needed) giving a value for each point plotted. If lines are to be plotted (e.g. for ‘type = b’/ the first element of ‘lwd’ is used. Suppose I'm plotting 10,000 points in a 10 x 10 scatterplot matrix (roughly what I'm actually doing). That's a total of 1 million points. It might take a while, but I can wait. However, I'm not sure how to get the coordinates right for additional points in a scatterplot matrix. Maybe I need to study that source code. I did figure out recently how to use transparent points to get the axes right. Color #ff00 does that trick for me -- that's white color with zero opaqueness, full transparency. Mike -- Michael B. Miller, Ph.D. Minnesota Center for Twin and Family Research Department of Psychology University of Minnesota __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] coxme and random factors
Hi I would really appreciate some help with my code for coxme... My data set I'm interested in survival of animals after an experiment with 4 treatments, which was performed on males and females. I also have two random factors: Response variable: survival (death) Factor 1: treatment (4 levels) Factor 2: sex (male / female) Random effects 1: person nested within day (2 people did the experiment over 2 days) Random effects 2: box nested within treatment (animals were kept in boxes in groups of 6, and there were multiple boxes per treatment) I've read the introductions to coxme by Terry Therneau, and something like the following is what I think I should use: model1-coxme(Surv(death,censor)~treatment*sex+(1|day/person)+(1| treatment/box)) Which gives me the following: Cox mixed-effects model fit by maximum likelihood events, n = 154, 291 Iterations= 57 305 NULL Integrated Penalized Log-likelihood -823.276 -795.2354 -784.4807 Chisq df p AIC BIC Integrated loglik 56.08 11.00 4.9096e-08 34.08 0.67 Penalized loglik 77.59 17.91 2.0958e-09 41.78 -12.60 Model: Surv(death, censor) ~ treatment * sex + (1 | day/person) + (1 | treatment/box) Fixed coefficients coef exp(coef) se(coef) zp teratmentb -0.0838877 0.9195345 0.3744511 -0.22 0.8200 treatmentb2 -0.4731922 0.6230103 0.3136199 -1.51 0.1300 treatmentn -1.0154149 0.3622521 0.4097754 -2.48 0.0130 sexmale -0.1838885 0.8320286 0.2602169 -0.71 0.4800 treatmentb:sexmale -0.3905856 0.6766605 0.2132936 -1.83 0.0670 treatmentb2:sexmale 0.6742202 1.9625020 0.3836907 1.76 0.0790 treatmentn:sexmale 1.2628977 3.5356520 0.4603589 2.74 0.0061 Random effects Group VariableStd Dev Variance day/person(Intercept) 0.32690104 0.10686429 day (Intercept) 0.49516113 0.24518455 treatment/box (Intercept) 0.26837158 0.07202330 treatment (Intercept) 0.29263637 0.08563604 My questions (1) Does anyone know how I can test the significance of each of the random effects in turn? For example, to find the significance of (1| treatment/box) can I compare the results of the above model to one without this term? (i.e. by subtracting the integrated loglikelihood values of the model without (1|treatment/box) from the model with that term). (2) Can I include 'treatment' as a factor as well as including it as part of a nested term? (incidentally I did wonder if I should include it as (treatment|box), but an error message comes back that factors cannot be used as a covariate within a random effect) (3) Is it possible to test the proportionality assumption within coxme. Previously I used cox.zph(model1) with coxph, but that does not work with coxme. Very many thanks to anyone who can offer me some advice! Sophie [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] crazy loop error.
Now I understand what the difference between a primitive and a non-primitive! Thanks for the clarification! Ivan Le 1/25/2011 18:03, Bert Gunter a écrit : Well, I'm not Prof. Ripley, but the answer is: Look at the code. seq_len, seq.int, and seq_along call Primitives, which are implemented in C, and therefore MUCH faster than seq(), which is implemented as pure R code (and is also a generic, so requires method dispatch). Though for small n (up to a few thousand, say), it probably doesn't make much difference.(Here, to be corrected by Prof. Ripley is needed). -- Bert On Tue, Jan 25, 2011 at 2:22 AM, Ivan Calandra ivan.calan...@uni-hamburg.de wrote: Mr Ripley, May I ask why seq_len() and seq_along() are better than seq()? Thanks, Ivan Le 1/25/2011 09:58, Prof Brian Ripley a écrit : On Tue, 25 Jan 2011, Petr Savicky wrote: On Mon, Jan 24, 2011 at 11:18:35PM +0100, Roy Mathew wrote: Thanks for the reply Erik, As you mentioned, grouping consecutive elements of 'a' was my idea. I am unaware of any R'ish way to do it. It would be nice if someone in the community knows this. The error resulting in the NA was pretty easy to fix, and my loop works, but the results are still wrong (new script below). Ideally it should print single hello for the single letters and grouped '3 hellos' for the fives, grouped '2 hellos' for the sixes etc. Based on the run results, if the value of n is being tracked, it changes quite unpredictably. Can someone explain how the value of n changes from end of the loop to the top without anything being done to it? Hi. A for-loop in R is different from a for-loop in C. It is similar to foreach loop in Perl. If v is a vector, then for (n in v) first creates the vector v and then always performs length(v) iterations. Before iteration i, n is assigned v[i] even if n is changed in the previous iteration. And also if v is changed during the loop. If you want to control the loop variable during execution, it is possible to use a while loop, where you have full control. While loop may be better also if v has a very large length, since, for example for (n in 1:100) creates a vector of length 100 in memory. It should also be noted that the for-loop for (n in 1:k) performs 2 iterations, if k is 0, since 1:0 is a vector of length 2. If k may be 0, then it is better to use for (n in seq(length=k)) since seq(length=0) has length 0. Since you keep mentioning that, it is actually much better to use seq_len(k) (and seq_along(x) instead of your earlier recommendation of seq(along=x)). And if you are using seq() in other cases in programs, consider seq.int() instead. Hope this helps. Petr Savicky. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/1525_8_1.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplot - controlling point size
On 1/25/2011 9:44 AM, Felipe Carrillo wrote: try this: qplot(x, y, data=df, colour=factor(type), size=I(1)) + geom_smooth() Felipe very nicely answered the how of your question. I thought I'd followup with the why. Using qplot, it assumes that you are giving a set of aesthetic mappings. As such, size is being mapped to a variable that is 1 for all entries. Then, in the usual way, a mapping is created between values of the variable in the data space and in the aesthetic space (actual sizes) and a legend is created to display this mapping. Enclosing the 1 in an I() indicates that you are giving a literal value to set the aesthetic to, in which case no mapping between aesthetic and data space (nor legend) is needed. Alternatively, you can use an identity scale to state that the data and aesthetic spaces are the same, and tell it to not plot the legend. qplot(x, y, data=df, colour=factor(type), size=1) + geom_smooth() + scale_size_identity(legend=FALSE) The distinction between aesthetic mapping and setting is even more evident in ggplot notation: ggplot(df, aes(x=x, y=y, colour=factor(type))) + geom_point(size=1) + geom_smooth() Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA http://www.fws.gov/redbluff/rbdd_jsmp.aspx - Original Message From: Gene Leynesgleyne...@gmail.com To: r-help@r-project.org Sent: Tue, January 25, 2011 9:28:20 AM Subject: [R] ggplot - controlling point size Can anyone illuminate the following for me? How can I get rid of the blue line in the key in the second plot? ## Create a simple data frame df=data.frame(x=1:1000, y=2*1:1000+rnorm(1000,sd=1000), type=sample(letters[1:2],1000, replace=TRUE)) ## Very nice! Almost what I want qplot(x, y, data=df, colour=factor(type)) + geom_smooth() ## Make a nicer plot, with smaller points ## but why does that add the little blue line with a 1? qplot(x, y, data=df, colour=factor(type), size=1) + geom_smooth() -- Brian S. Diggs, PhD Senior Research Associate, Department of Surgery Oregon Health Science University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ggplot - controlling point size
Thank you both, very much. Using the identity function I() is a very nice trick, but it still feels like a trick. Using ggplot makes the most sense to me. ggplot(df, aes(x=x, y=y, colour=factor(type))) + geom_point(size=1) + geom_smooth() Thank you very much for taking the time to explain the syntax. I had tried using the ggplot function, but I couldn't figure out how to handle thecolour=factor(type) part. (my erroneous formulas are omitted for clarity) On Tue, Jan 25, 2011 at 12:06 PM, Brian Diggs dig...@ohsu.edu wrote: On 1/25/2011 9:44 AM, Felipe Carrillo wrote: try this: qplot(x, y, data=df, colour=factor(type), size=I(1)) + geom_smooth() Felipe very nicely answered the how of your question. I thought I'd followup with the why. Using qplot, it assumes that you are giving a set of aesthetic mappings. As such, size is being mapped to a variable that is 1 for all entries. Then, in the usual way, a mapping is created between values of the variable in the data space and in the aesthetic space (actual sizes) and a legend is created to display this mapping. Enclosing the 1 in an I() indicates that you are giving a literal value to set the aesthetic to, in which case no mapping between aesthetic and data space (nor legend) is needed. Alternatively, you can use an identity scale to state that the data and aesthetic spaces are the same, and tell it to not plot the legend. qplot(x, y, data=df, colour=factor(type), size=1) + geom_smooth() + scale_size_identity(legend=FALSE) The distinction between aesthetic mapping and setting is even more evident in ggplot notation: ggplot(df, aes(x=x, y=y, colour=factor(type))) + geom_point(size=1) + geom_smooth() Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA http://www.fws.gov/redbluff/rbdd_jsmp.aspx - Original Message From: Gene Leynesgleyne...@gmail.com gleynes%...@gmail.com To: r-help@r-project.org Sent: Tue, January 25, 2011 9:28:20 AM Subject: [R] ggplot - controlling point size Can anyone illuminate the following for me? How can I get rid of the blue line in the key in the second plot? ## Create a simple data frame df=data.frame(x=1:1000, y=2*1:1000+rnorm(1000,sd=1000), type=sample(letters[1:2],1000, replace=TRUE)) ## Very nice! Almost what I want qplot(x, y, data=df, colour=factor(type)) + geom_smooth() ## Make a nicer plot, with smaller points ## but why does that add the little blue line with a 1? qplot(x, y, data=df, colour=factor(type), size=1) + geom_smooth() -- Brian S. Diggs, PhD Senior Research Associate, Department of Surgery Oregon Health Science University [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting SSE from lm
It is not clear what you are doing or why you are doing it. If you tell us your ultimate goal we may be able to help you find a way that does not require all the computing that you are doing. How do you get your coefficients? Are you using lm? Have you looked at the resid function? -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Brian J Mingus Sent: Tuesday, January 25, 2011 9:08 AM To: R-help@r-project.org Subject: [R] Extracting SSE from lm Apologies for this simple question - Given the number of comparisons I need to do it has become somewhat laborious to compute the SSE manually. I first have to extract the coefficients, build the model and run the model on the data. So far I haven't found any method in R that will do this for me. Is there a method that I haven't seen, or is there a small function I could write that would do this, and how might I go about that? Thanks, Brian [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting SSE from lm
On 2011-01-25 08:08, Brian J Mingus wrote: Apologies for this simple question - Given the number of comparisons I need to do it has become somewhat laborious to compute the SSE manually. I first have to extract the coefficients, build the model and run the model on the data. So far I haven't found any method in R that will do this for me. Is there a method that I haven't seen, or is there a small function I could write that would do this, and how might I go about that? You can always write a function to automate whatever you're doing now. But if I understand what you need, try: tail( anova( yourModel )[, 2], 1) Peter Ehlers Thanks, Brian [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Warning with predict.glm method
Dear list, When I use the predict.glm method on a glm fitted object, I get the following warning message: In addition: Warning message: In predict.lm(object, x) : prediction from a rank-deficient fit may be misleading As the documentation says this happens if the fit is rank-deficient, some of the columns of the design matrix will have been dropped. It is my understanding that this means that at least one predictor variable in the design matrix can be replicated by a linear combination of the other predictors. But this doesn't seem the case in my data set. I noticed that the problem goes away after removing higher order interaction terms from the model. Some of the coefficients from those terms were NA as shown in the summary(glm). I'm wondering if my interpretation of the message is correct, and/or in what type of situations would I get this warning message. Thanks in advance for your help. Lars. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Counting number of rows with two criteria in dataframe
Hi R-users, I'm trying to find an elegant way to count the number of rows in a dataframe with a unique combination of 2 values in the dataframe. My data is specifically one column with a year, one with a month, and one with a day. I'm trying to count the number of days in each year/month combination. But for simplicity's sake, the following dataset will do: x-c(1,1,1,1,2,2,2,2,3,3,3,3) y-c(1,1,2,2,3,3,4,4,5,5,6,6) z-c(1,2,3,4,5,6,7,8,9,10,11,12) X-data.frame(x y z) So with dataset X, how would I count the number of z values (3rd column in X) with unique combinations of the first two columns (x and y)? (for instance, in the above example, there are 2 instances per unique combination of the first two columns). I can do this in Matlab and it's easy, but since I'm new to R this is royally stumping me. Thanks, Ryan -- Ryan Utz Postdoctoral research scholar University of California, Santa Barbara (724) 272 7769 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Counting number of rows with two criteria in dataframe
If you want count: xtabs( ~ x + y, X) or sum: xtabs(z ~ x + y, X) On Tue, Jan 25, 2011 at 5:25 PM, Ryan Utz utz.r...@gmail.com wrote: Hi R-users, I'm trying to find an elegant way to count the number of rows in a dataframe with a unique combination of 2 values in the dataframe. My data is specifically one column with a year, one with a month, and one with a day. I'm trying to count the number of days in each year/month combination. But for simplicity's sake, the following dataset will do: x-c(1,1,1,1,2,2,2,2,3,3,3,3) y-c(1,1,2,2,3,3,4,4,5,5,6,6) z-c(1,2,3,4,5,6,7,8,9,10,11,12) X-data.frame(x y z) So with dataset X, how would I count the number of z values (3rd column in X) with unique combinations of the first two columns (x and y)? (for instance, in the above example, there are 2 instances per unique combination of the first two columns). I can do this in Matlab and it's easy, but since I'm new to R this is royally stumping me. Thanks, Ryan -- Ryan Utz Postdoctoral research scholar University of California, Santa Barbara (724) 272 7769 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] subsetting based on joint values of critera
Dear colleagues, I have a dataset that looks as below. I would like to make a new dataset that excludes the cases which are joint conjunctions of particular state names and years, so Connecticut and 2010, Maryland and 2010 and Vermont and 2010. I'm trying the following subset code: newdata- subset(bpa, (!State==Connecticut year2010)) It appears that it's only evaluating both criteria independently and not jointly, so this is returning all cases in 2008 and 2009, leaving out connecticut for those years as well. How do I get subset to return a dataset based on the joint occurrence of values of two variables? Yours, Simon Kiss str(bpa) 'data.frame': 150 obs. of 5 variables: $ State : Factor w/ 50 levels Alabama,Alaska,..: 1 2 3 4 5 6 7 8 9 10 ... $ year: num 2008 2008 2008 2008 2008 ... $ ban : num 0 0 0 0 0 0 0 0 0 0 ... $ partisan: Factor w/ 3 levels democrat,mixed,..: 1 1 1 1 1 1 1 2 3 2 ... $ news: num 1.67 2 0 0 2.38 ... * Simon J. Kiss, PhD Assistant Professor, Wilfrid Laurier University 73 George Street Brantford, Ontario, Canada N3T 2C9 Cell: +1 519 761 7606 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Importing xls from a http://
I know a lot of people asked similar questions like this. I have tried using read.xls () Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Incorrect number of arguments (11), expecting 10 for ReadXls read.table or read.csv (Wrong table format) odbcConnectExcel have problem with the URL I want to import this excel file (http://www.econ.yale.edu/~shiller/data/ie_data.xls) from row 8 in Data sheet. Thanks Cameron -- View this message in context: http://r.789695.n4.nabble.com/Importing-xls-from-a-http-tp3236800p3236800.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help Derivate for Nonlinear Growth Models
Hi!! Im doing my graduated work in Onion Curves Growth with Nonlinear Models, I'm amateur in R so i have doubt how i put or program next models, http://r.789695.n4.nabble.com/file/n3236748/96629508.png Also, i cant derivate for Gauss Model, and Richard Model dont have funtion, If someone could help me, i was so grate, -- View this message in context: http://r.789695.n4.nabble.com/Help-Derivate-for-Nonlinear-Growth-Models-tp3236748p3236748.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Does anybody knows the default value of starting value in glm?
Hello ; Do you know what is the default value of starting value in glm ? glm(..., start=c(),... ) I know that it is NULL by default but it need a value to start iteration . what is this value? Thanks; [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with calculating correlation coefficient - creating a network
Hi, I am creating a correlation based graph for the data 27 X 3040. I am not sure if I am allocating enough space for the data. I am using the following .R file (which is an example for geneData)! http://r.789695.n4.nabble.com/file/n3236801/sampleClimate.R sampleClimate.R/nabble_a and the data is in the file http://r.789695.n4.nabble.com/file/n3236801/data3y3w_landmask.txt data3y3w_landmask.txt I am getting the following error in the line, net - network(corm, directed = F) Error in which.matrix.type(x) : subscript out of bounds But if I use the geneData 500 X 26, it doesnt throw any error! http://r.789695.n4.nabble.com/file/n3236801/geneData.txt geneData.txt Please help to fix the issue. (Note since I was having issue I took the transpose in the cor function.) Thanks, Kumaraguru -- View this message in context: http://r.789695.n4.nabble.com/Help-with-calculating-correlation-coefficient-creating-a-network-tp3236801p3236801.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Counting number of rows with two criteria in dataframe
Hi Ryan, One option would be X$a - paste(X$x, X$y, sep=.) table(X$a) Best, Ista On Tue, Jan 25, 2011 at 2:25 PM, Ryan Utz utz.r...@gmail.com wrote: Hi R-users, I'm trying to find an elegant way to count the number of rows in a dataframe with a unique combination of 2 values in the dataframe. My data is specifically one column with a year, one with a month, and one with a day. I'm trying to count the number of days in each year/month combination. But for simplicity's sake, the following dataset will do: x-c(1,1,1,1,2,2,2,2,3,3,3,3) y-c(1,1,2,2,3,3,4,4,5,5,6,6) z-c(1,2,3,4,5,6,7,8,9,10,11,12) X-data.frame(x y z) So with dataset X, how would I count the number of z values (3rd column in X) with unique combinations of the first two columns (x and y)? (for instance, in the above example, there are 2 instances per unique combination of the first two columns). I can do this in Matlab and it's easy, but since I'm new to R this is royally stumping me. Thanks, Ryan -- Ryan Utz Postdoctoral research scholar University of California, Santa Barbara (724) 272 7769 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Importing xls from a http://
On Tue, Jan 25, 2011 at 2:06 PM, cameron raymond...@invesco.com wrote: I know a lot of people asked similar questions like this. I have tried using read.xls () Error in .Call(ReadXls, file, colNames, sheet, type, from, rowNames, : Incorrect number of arguments (11), expecting 10 for ReadXls read.table or read.csv (Wrong table format) odbcConnectExcel have problem with the URL I want to import this excel file (http://www.econ.yale.edu/~shiller/data/ie_data.xls) from row 8 in Data sheet. How to import a spreadsheet from the internet was asked less than 24 hours ago on this list. See: http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows e.g. library(gdata) URL - http://www.econ.yale.edu/~shiller/data/ie_data.xls; DF - read.xls(URL, pattern = Fraction) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Predictions with 'missing' variables
Dear List, I think I'm going crazy here...can anyone explain why do I get the same predictions in train and test data sets below when the second has a missing input? y - rnorm(1000) x1 - rnorm(1000) x2 - rnorm(1000) train - data.frame(y,x1,x2) test - data.frame(x1) myfit - glm(y ~ x1 + x2, data=train) summary(myfit) all(predict(myfit, test) == predict(myfit, train)) [1] TRUE Thanks, Axel. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] subsetting based on joint values of critera
Hi Simon, You almost had it! Just need to move the negation outside the rest of the logic, and remove the quotes from year. Not actually tested (no data), but I think newdata- subset(bpa, !(State==Connecticut year2010)) should do it. Best, Ista On Tue, Jan 25, 2011 at 1:34 PM, Simon Kiss simonjk...@yahoo.ca wrote: Dear colleagues, I have a dataset that looks as below. I would like to make a new dataset that excludes the cases which are joint conjunctions of particular state names and years, so Connecticut and 2010, Maryland and 2010 and Vermont and 2010. I'm trying the following subset code: newdata- subset(bpa, (!State==Connecticut year2010)) It appears that it's only evaluating both criteria independently and not jointly, so this is returning all cases in 2008 and 2009, leaving out connecticut for those years as well. How do I get subset to return a dataset based on the joint occurrence of values of two variables? Yours, Simon Kiss str(bpa) 'data.frame': 150 obs. of 5 variables: $ State : Factor w/ 50 levels Alabama,Alaska,..: 1 2 3 4 5 6 7 8 9 10 ... $ year : num 2008 2008 2008 2008 2008 ... $ ban : num 0 0 0 0 0 0 0 0 0 0 ... $ partisan: Factor w/ 3 levels democrat,mixed,..: 1 1 1 1 1 1 1 2 3 2 ... $ news : num 1.67 2 0 0 2.38 ... * Simon J. Kiss, PhD Assistant Professor, Wilfrid Laurier University 73 George Street Brantford, Ontario, Canada N3T 2C9 Cell: +1 519 761 7606 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Predictions with 'missing' variables
See the note in the help page for ?predict.glm Best, Ista On Tue, Jan 25, 2011 at 2:59 PM, Axel Urbiz axel.ur...@gmail.com wrote: Dear List, I think I'm going crazy here...can anyone explain why do I get the same predictions in train and test data sets below when the second has a missing input? y - rnorm(1000) x1 - rnorm(1000) x2 - rnorm(1000) train - data.frame(y,x1,x2) test - data.frame(x1) myfit - glm(y ~ x1 + x2, data=train) summary(myfit) all(predict(myfit, test) == predict(myfit, train)) [1] TRUE Thanks, Axel. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Manual two-stage least squares in R
Hi, I am trying to manipulate a gls regression model output to adjust for use of two-stage least squares. Basically, I want to estimate a model, then feed in a new set of residuals, then re-calculate all of the model output (i.e. the standard errors of the estimators, etc.). I have found some documentation on doing this in stata, which is below: http://www.stata.com/help.cgi?ereturn I am wondering whether there is a function like this ereturn() (see http://www.stata.com/help.cgi?ereturn) in R, and whether this might allow me to achieve something similar. Thanks so much! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lattice draw.key(): position of key in panels
Good afternoon,
I am working on a plot that requires custom legends to be placed in some
panels of the plot; other panels do not contain legends. The problem
that I run into is positioning of the legend in individual panels. In
particular, the 'x' and 'y' elements of the key-list are ignored by
draw.key() when it is called from inside a panel function. As a result,
the legend is placed in the middle of the panel. The example below
illustrates this problem.
df - data.frame(x=c(1,1),y=c(1,2),type=c(A,B))
panel.xyplot.x - function(...) {
# draw data
panel.xyplot(...)
# create key-list
pnlid - panel.number()
lbl - ifelse(pnlid==1,AA,BB)
pts - Rows(trellis.par.get(superpose.symbol),pnlid)
key - list(points=pts,text=list(lbl),x=0.1,y=0.9,corner=c(0,1))
# draw key
draw.key(key,draw=TRUE)
}
oltc - xyplot(y~x|type,data=df,panel=panel.xyplot.x)
print(oltc)
I tried using 'vp=current.viewport()' in the call to draw.key() but it
did not help. Can anybody suggest the proper way to specify position of
the key-list so that it is respected by draw.key() when called within a
panel function?
Sincerely,
Boris Vasiliev.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Manual two-stage least squares in R
Your question has some similarities this paper: Alison Smith, Brian Cullis, and Arthur Gilmour. The analysis of crop variety evaluation data in Australia. Aust. N. Z. J. Stat., 43:129--145, 2001. In that paper, the authors fit a mixed model with several random effects. The variances are then held fixed while one of the model terms is changed from a random effect to a fixed effect and the model is re-fit using the constrained variances. They refer to this as unshrinking the BLUPs. This is accomplished with ASREML or the R version asreml-r, a commercial package (does have a 30-day free trial). Not sure if this would help you at all. Good luck, Kevin Wright On Tue, Jan 25, 2011 at 2:47 PM, Katharina Ley kat...@umich.edu wrote: Hi, I am trying to manipulate a gls regression model output to adjust for use of two-stage least squares. Basically, I want to estimate a model, then feed in a new set of residuals, then re-calculate all of the model output (i.e. the standard errors of the estimators, etc.). I have found some documentation on doing this in stata, which is below: http://www.stata.com/help.cgi?ereturn I am wondering whether there is a function like this ereturn() (see http://www.stata.com/help.cgi?ereturn) in R, and whether this might allow me to achieve something similar. Thanks so much! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Kevin Wright __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] MAtrix addressing
Hello I would like to ask you if it is possible In R Cran to change the default way of addressing a matrix. for example matrix(data=seq(from=1,to=4,nrow=2,ncol=2, by row numbering) # not having R at this pc will create something like the following 1 2 3 4 the way R address this matrix is from top left corner moving to bottom right. The cell numbers in that way are 1 2 3 4 IS it possible to change this default addresing number to something that goes bottom left to top right? In this simple case I want to have 3 4 1 2 Would that be possible? I would like to thank y for your help Regards Alex __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] FW: question about the pt() calculation
From: Leitch, Matthew C.
Sent: Monday, January 24, 2011 6:53 PM
To: 'i...@network-theory.co.uk'
Subject: question about the pt() calculation
Hello
Thank you for your time. I am a graduate student at the University of Texas
Medical Branch, and I was wondering if you could help me with a R program I am
writing. I have some data that is stored a file that has 1733 rows and 4
columns. Each row is independent, so I have a loop system so I don't have to
manually perform 1733 t-tests. However I got most of it to work, thanks to the
An Introduction to R book and a helpful colleague. But I am having some
difficulty with getting my t-test scores converted to p-values. The pt() test
won't take the t object, but I am unsure how to make it create a p-value for
each row. If you have any suggestions they would be greatly appreciated.
Best Wishes
Matt
cancer-read.table(group2.txt, header=TRUE)
noncancer-read.table(group1.txt, header=TRUE)
#meanc-mean(cancer)
#meannc-mean(noncancer)
#varc-var(cancer)
#varnc-var(noncancer)
#Sx1x2-sqrt((varc+varnc)/2)
#n-4
#t-(noncancer-cancer)/(Sx1x2*(sqrt(2/n)))
#df-(2*n)-2
#pt(t,df)*2
answer = matrix(,1733,1);
alpha = 0.05;
#for(i=1:1733)
#{
# meanc = mean(cancer[i,]);
# meannc = mean(noncancer[i,]);
#
# varc = var(cancer[i,]);
# varnc = var(noncancer[i,]);
#
# Sx1x2 = sqrt((varc+varnc)/2);
# n = 4;
# t = (noncancer-cancer)/(Sx1x2*(sqrt(2/n)));
# df = 2*n - 2;
#
#
# answer[i,1] = pt(t,df)*2;
#
# if (answer[i,1] = alpha)
# {
#sing.
# }
# else
# {
#nonsing.
# }
#}
for(i in 1:1733)
{
meanc = mean(cancer[i,]);
meannc = mean(noncancer[i,]);
varc = var(cancer[i,]);
varnc = var(noncancer[i,]);
Sx1x2 = sqrt((varc+varnc)/2);
n = 4;
t = (noncancer-cancer)/(Sx1x2*(sqrt(2/n)));
df = 2*n - 2;
answer[i,1] = pt(t,df)*2 ;
if (answer[i,1] alpha)
{
banswer[i,1] = 1;
}
else
{
banswer[i,1] = 0;
}
#ifelse(answer[i,1] alpha, banswer[i,1] = 1, banswer[i,1] = 0);
}__
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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] 3D Binning
On Tue, Jan 25, 2011 at 06:00:36AM -0800, vioravis wrote: I am trying to do binning on three variables (3d binning). The bin boundaries are specified by the user separately for each variable. I used the bin2 function in the 'ash' package for 2d binning that involves only two variables but didn't any package for similar binning with three variables. Are there any packages or codes available for 3d binning?? Thank you. Try the following, possibly with different breaks for each dimension. x - matrix(rnorm(3), ncol=3) breaks - seq(-1, 1, length=5) xints - data.frame( x1=cut(x[, 1], breaks=breaks), x2=cut(x[, 2], breaks=breaks), x3=cut(x[, 3], breaks=breaks)) table(complete.cases(xints)) xtabs(~ ., xints) Hope this helps. Petr Savicky. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] function application
Suppose I have a function, like list, that takes a variable number of arguments, and I have those arguments in some vector x. How can execute the function with the *contents* of x as its arguments? I.e., I don't want list(x), but rather list(x[[1]], x[[2]], ..., x[[n]]), but I don't want to spell out the individual elements of x (either because I want to do this programmatically, so I cannot code an expression like list(x[[1]],...,x[[n]]) because the value of n is not know until run time, or else, simply to avoid the tedium of typing out all the elements of x individually). Thanks! Roy P.S. In Python, if x is some sequence-like object (e.g. a list or a tuple), and f is some function, the expression f(*x) causes f to be called with the *contents* of x as its arguments. (This is to be distinguished from f(x), which calls f with x as its sole argument.) In Mathematica, one can achieve a similar effect using the Apply function: Apply[f, x]. I'm looking for the equivalent of this in R. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] FW: question about the pt() calculation
From: Leitch, Matthew C.
Sent: Monday, January 24, 2011 6:53 PM
To: 'i...@network-theory.co.uk'
Subject: question about the pt() calculation
Hello
Thank you for your time. I am a graduate student at the University of Texas
Medical Branch, and I was wondering if you could help me with a R program I am
writing. I have some data that is stored a file that has 1733 rows and 4
columns. Each row is independent, so I have a loop system so I don't have to
manually perform 1733 t-tests. However I got most of it to work, thanks to the
An Introduction to R book and a helpful colleague. But I am having some
difficulty with getting my t-test scores converted to p-values. The pt() test
won't take the t object, but I am unsure how to make it create a p-value for
each row. If you have any suggestions they would be greatly appreciated.
Best Wishes
Matt
cancer-read.table(group2.txt, header=TRUE)
noncancer-read.table(group1.txt, header=TRUE)
#meanc-mean(cancer)
#meannc-mean(noncancer)
#varc-var(cancer)
#varnc-var(noncancer)
#Sx1x2-sqrt((varc+varnc)/2)
#n-4
#t-(noncancer-cancer)/(Sx1x2*(sqrt(2/n)))
#df-(2*n)-2
#pt(t,df)*2
answer = matrix(,1733,1);
alpha = 0.05;
#for(i=1:1733)
#{
# meanc = mean(cancer[i,]);
# meannc = mean(noncancer[i,]);
#
# varc = var(cancer[i,]);
# varnc = var(noncancer[i,]);
#
# Sx1x2 = sqrt((varc+varnc)/2);
# n = 4;
# t = (noncancer-cancer)/(Sx1x2*(sqrt(2/n)));
# df = 2*n - 2;
#
#
# answer[i,1] = pt(t,df)*2;
#
# if (answer[i,1] = alpha)
# {
#sing.
# }
# else
# {
#nonsing.
# }
#}
for(i in 1:1733)
{
meanc = mean(cancer[i,]);
meannc = mean(noncancer[i,]);
varc = var(cancer[i,]);
varnc = var(noncancer[i,]);
Sx1x2 = sqrt((varc+varnc)/2);
n = 4;
t = (noncancer-cancer)/(Sx1x2*(sqrt(2/n)));
df = 2*n - 2;
answer[i,1] = pt(t,df)*2 ;
if (answer[i,1] alpha)
{
banswer[i,1] = 1;
}
else
{
banswer[i,1] = 0;
}
#ifelse(answer[i,1] alpha, banswer[i,1] = 1, banswer[i,1] = 0);
}__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] function application
Suppose I have a function, like list, that takes a variable number of arguments, and I have those arguments in some vector x. How can execute the function with the *contents* of x as its arguments? I.e., I don't want list(x), but rather list(x[[1]], x[[2]], ..., x[[n]]), but I don't want to spell out the individual elements of x (either because I want to do this programmatically, so I cannot code an expression like list(x[[1]],...,x[[n]]) because the value of n is not know until run time, or else, simply to avoid the tedium of typing out all the elements of x individually). In this particular case, because you want to create a list, x - 1:10 as.list(x) will do. P.S. In Python, if x is some sequence-like object (e.g. a list or a tuple), and f is some function, the expression f(*x) causes f to be called with the *contents* of x as its arguments. (This is to be distinguished from f(x), which calls f with x as its sole argument.) In Mathematica, one can achieve a similar effect using the Apply function: Apply[f, x]. I'm looking for the equivalent of this in R. In general, if you already have a *list* and want to call a function with the contents of that list as the arguments, then ?do.call is what you need. a - list(example, of, do.call) #compare the two following expressions paste(a) do.call(paste, a) --Erik __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] function application
?do.call Please provide a reproducible example if the help file is not sufficient. - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Tue, 25 Jan 2011, Roy Shimizu wrote: Suppose I have a function, like list, that takes a variable number of arguments, and I have those arguments in some vector x. How can execute the function with the *contents* of x as its arguments? I.e., I don't want list(x), but rather list(x[[1]], x[[2]], ..., x[[n]]), but I don't want to spell out the individual elements of x (either because I want to do this programmatically, so I cannot code an expression like list(x[[1]],...,x[[n]]) because the value of n is not know until run time, or else, simply to avoid the tedium of typing out all the elements of x individually). Thanks! Roy P.S. In Python, if x is some sequence-like object (e.g. a list or a tuple), and f is some function, the expression f(*x) causes f to be called with the *contents* of x as its arguments. (This is to be distinguished from f(x), which calls f with x as its sole argument.) In Mathematica, one can achieve a similar effect using the Apply function: Apply[f, x]. I'm looking for the equivalent of this in R. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Train error:: subscript out of bonds
You should try different tuning parameters; the defaults are not likely to work for many datasets. I don't use the polynomial kernel too much but scale parameter values that are really of could cause this. Unlike the rbf, I don't know of any good techniques for estimating this. Max On Jan 25, 2011, at 11:47 AM, Neeti nikkiha...@gmail.com wrote: after using options(error=utils::recover) option, following is the output. if i am correct this means that in ksvm there is some problem, but really could not understand. could anyone please tell me what is wrong... any help will be great thank you so much.. Enter a frame number, or 0 to exit 1: train(train1, as.factor(trainset[, ncol(trainset)]), svmpoly, trControl = trainControl((method = cv), 10, 2: train.default(train1, as.factor(trainset[, ncol(trainset)]), svmpoly, trControl = trainControl((method = c 3: do.call(trControl$computeFunction, argList) 4: function (X, FUN, ...) 5: FUN(X[[25]], ...) 6: do.call(createModel, args) 7: function (data, method, tuneValue, obsLevels, pp = NULL, ...) 8: ksvm(as.matrix(trainX), trainY, kernel = polydot(degree = tuneValue$.degree, scale = tuneValue$.scale, offset 9: ksvm(as.matrix(trainX), trainY, kernel = polydot(degree = tuneValue$.degree, scale = tuneValue$.scale, offset 10: .local(x, ...) 11: ksvm(x[c(indexes[[i]], indexes[[j]]), , drop = FALSE][cind, ], yd[cind], type = type(ret), kernel = kernel, k 12: ksvm(x[c(indexes[[i]], indexes[[j]]), , drop = FALSE][cind, ], yd[cind], type = type(ret), kernel = kernel, k 13: .local(x, ...) -- View this message in context: http://r.789695.n4.nabble.com/Train-error-subscript-out-of-bonds-tp3234510p3236494.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] FW: question about the pt() calculation
Matthew -
Others will probably tell you about the folly of performing
1733 t-tests on groups with 4 observations each, but an alternative
to your approach would be to use R to solve your problem. (I'm
using var.equal=TRUE because that's what you're calculating, but
you might consider using the default behaviour assuming unequal
variances.)
onerow = function(i){
thetest = t.test(noncancer[i,],cancer[i,],var.equal=TRUE)
c(thetest$statistic,thetest$p.value,as.numeric(thetest$p.value 0.05))
}
answer = t(sapply(1:nrow(cancer),onerow))
Then answer should have the information that you want.
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Tue, 25 Jan 2011, Leitch, Matthew C. wrote:
From: Leitch, Matthew C.
Sent: Monday, January 24, 2011 6:53 PM
To: 'i...@network-theory.co.uk'
Subject: question about the pt() calculation
Hello
Thank you for your time. I am a graduate student at the University of Texas Medical
Branch, and I was wondering if you could help me with a R program I am writing. I have
some data that is stored a file that has 1733 rows and 4 columns. Each row is
independent, so I have a loop system so I don't have to manually perform 1733 t-tests.
However I got most of it to work, thanks to the An Introduction to R book and
a helpful colleague. But I am having some difficulty with getting my t-test scores
converted to p-values. The pt() test won't take the t object, but I am unsure how to make
it create a p-value for each row. If you have any suggestions they would be greatly
appreciated.
Best Wishes
Matt
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] ANOVA table look
Y'all, I need to get the look of a standard fixed effect ANOVA table: anova(aov(meas~op*part,data=fs)) Analysis of Variance Table Response: meas Df Sum Sq Mean Sq F value Pr(F) op 22.62 1.308 1.3193 0.2750 part 19 1185.43 62.391 62.9151 2e-16 *** op:part 38 27.05 0.712 0.7178 0.8614 Residuals 60 59.50 0.992 when I perform a two-factor factorial with random factors analysis using: fs.anova=lmer(meas~(1|op)+(1|part),data=fs) print(fs.anova) Linear mixed model fit by REML Formula: meas ~ (1 | op) + (1 | part) Data: fs AIC BIC logLik deviance REMLdev 417.6 428.7 -204.8410.7 409.6 Random effects: Groups NameVariance Std.Dev. part (Intercept) 1.0250e+01 3.2015e+00 op (Intercept) 1.8497e-16 1.3600e-08 Residual 8.9167e-01 9.4428e-01 Number of obs: 120, groups: part, 20; op, 3 Fixed effects: Estimate Std. Error t value (Intercept) 22.3917 0.7211 31.05 The data is twenty parts, three operators with two replications of measurements on the twenty parts. Any help would be appreciated. Thanks, Keith Jones __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Does anybody knows the default value of starting value in glm?
On Tue, Jan 25, 2011 at 1:14 PM, Akram Khaleghei Ghosheh balagh a.khaleg...@gmail.com wrote: Hello ; Do you know what is the default value of starting value in glm ? glm(..., start=c(),... ) I know that it is NULL by default but it need a value to start iteration . what is this value? Actually the typical call to glm ends up setting a value of mu and eta from which the parameter values are calculated on the first iteration. See the initialize component for the particular family in which you are interested to see how it creates mustart. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] write.table -- maintain decimal places
Hello, All, How can I maintain the decimal places when using write.table()? Jim e.g. df: EFFECT2 PVALUE 1 0.0230.88080 2 -0.260 0.08641 3 -0.114 0.45200 write.table(df,file='df.txt',quote=F,sep='\t',row.names=F) df.txt: EFFECT2PVALUE 0.023 0.8808 -0.26 0.08641 -0.114 0.452 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] post-hoc comparisons in GAMs (mgcv) with parametric terms
Dear list, I´m wondering if there is something analogous to the TukeyHSD function that could be used for parametric terms in a GAM. I´m using the mgcv package to fit models that have some continuous predictors (modeled as smooth terms) and a single categorical predictor. I would like to do post hoc test on the categorical predictor in the models where it is significant. Any suggestions? Thanks, Julian -- Julian Mariano Burgos Hafrannsóknastofnunin/Marine Research Institute Skúlagata 4, 121 Reykjavík, Iceland Sími/Telephone : +354-5752037 Bréfsími/Telefax: +354-5752001 Netfang/Email: jul...@hafro.is, jmbur...@uw.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.table -- maintain decimal places
On Tue, 25 Jan 2011 16:16:37 -0800, Jim Moon moo...@ohsu.edu wrote: Hello, All, How can I maintain the decimal places when using write.table()? Have a look at ?format.data.frame -- Seb __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] crash when using hazard.ratio.plot from rms package
Dear all, I always encounter a crash when running hazard.ratio.plot from rms package with my predictor as a factor. It works fine when the predictor is a continous score. Anyone encounters this too? Is this a bug or something? Thanks, Lilian [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.table -- maintain decimal places
On 2011-01-25 16:16, Jim Moon wrote: Hello, All, How can I maintain the decimal places when using write.table()? Jim e.g. df: EFFECT2 PVALUE 1 0.0230.88080 2 -0.260 0.08641 3 -0.114 0.45200 write.table(df,file='df.txt',quote=F,sep='\t',row.names=F) write.table(format(df, drop0trailing=FALSE), ) Peter Ehlers df.txt: EFFECT2PVALUE 0.023 0.8808 -0.26 0.08641 -0.114 0.452 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.table -- maintain decimal places
Thank you for the response, Peter. The approach: write.table(format(df, drop0trailing=FALSE),file='df.txt',quote=F,sep='\t',row.names=F) surprisingly still results in some loss of trailing 0's. df: EFFECT2 PVALUE 1 0.0230.88080 2 -0.260 0.08641 3 -0.114 0.45200 df.txt: EFFECT2PVALUE 0.023 8.808e-01 -0.26 8.641e-02 -0.114 4.520e-01 -Original Message- From: Peter Ehlers [mailto:ehl...@ucalgary.ca] Sent: Tuesday, January 25, 2011 5:09 PM To: Jim Moon Cc: r-help@r-project.org Subject: Re: [R] write.table -- maintain decimal places On 2011-01-25 16:16, Jim Moon wrote: Hello, All, How can I maintain the decimal places when using write.table()? Jim e.g. df: EFFECT2 PVALUE 1 0.0230.88080 2 -0.260 0.08641 3 -0.114 0.45200 write.table(df,file='df.txt',quote=F,sep='\t',row.names=F) write.table(format(df, drop0trailing=FALSE), ) Peter Ehlers df.txt: EFFECT2PVALUE 0.023 0.8808 -0.26 0.08641 -0.114 0.452 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.table -- maintain decimal places
On 2011-01-25 17:22, Jim Moon wrote: Thank you for the response, Peter. The approach: write.table(format(df, drop0trailing=FALSE),file='df.txt',quote=F,sep='\t',row.names=F) surprisingly still results in some loss of trailing 0's. What version of R? I'm using R version 2.12.1 Patched (2010-12-27 r53883) and it works for me. Peter Ehlers df: EFFECT2 PVALUE 1 0.0230.88080 2 -0.260 0.08641 3 -0.114 0.45200 df.txt: EFFECT2PVALUE 0.023 8.808e-01 -0.26 8.641e-02 -0.114 4.520e-01 -Original Message- From: Peter Ehlers [mailto:ehl...@ucalgary.ca] Sent: Tuesday, January 25, 2011 5:09 PM To: Jim Moon Cc: r-help@r-project.org Subject: Re: [R] write.table -- maintain decimal places On 2011-01-25 16:16, Jim Moon wrote: Hello, All, How can I maintain the decimal places when using write.table()? Jim e.g. df: EFFECT2 PVALUE 1 0.0230.88080 2 -0.260 0.08641 3 -0.114 0.45200 write.table(df,file='df.txt',quote=F,sep='\t',row.names=F) write.table(format(df, drop0trailing=FALSE), ) Peter Ehlers df.txt: EFFECT2PVALUE 0.023 0.8808 -0.26 0.08641 -0.114 0.452 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] return object from loop inside a function
Hi All,
I have a for loop inside the function and I cannot get UUU to give me an
updated grid.dens object when I run the function (it does update when I
run just the for loop). Here's a simplified version of my function:
UUU=function(pop, grid.dens) {
for (i in 1:10){
Food=grid.dens[pop$yloc[i],pop$xloc[i]] #use initial grid.dens values
Consumed=(pop$weight[i]*0.25)
Left=Food-Consumed
grid.dens[pop$yloc[i],pop$xloc[i]]=Left #update grid.dens values on i
pop$birth[i]=pop$birth[i]+1
}
return(pop)
return(grid.dens)
}
I get an updated pop, but not an updated grid.dens. What am I doing wrong?
Thanks!
Nico
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