similar to: about vector to the power of a decimal

I had a question about the basic power functions in R. For example from the R console I enter: -1 ^ 2 [1] -1 but also -1^3 [1] -1 -0.1^2 [1] -0.01 Normally pow(-1, 2) return either -Infinity or NaN. Has R taken over the math functions? If so I would think that -1^2 is 1 not -1 and -0.1^2 is 0.01 not -0.01. Thank you. Kevin

Why does the expression "(-8)^(1/3)" return NaN, instead of -2? This is not answered by http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-powers-of-negative-numbers-wrong_003f Thanks, Dave [[alternative HTML version deleted]]

Dear R developers, is that right? > -27^(1/3) [1] -3 > c(-27,27)^(1/3) [1] NaN 3 i'm using sign( c(-27,27) ) * abs( c(-27,27)) ^(1/3) , thanks [[alternative HTML version deleted]]

Can someone please enlighten me as to why the following happens? > -2.7^8.6 [1] -5125.407 > p<- -2.7 > q<- 8.6 > p^q [1] NaN R seems perfectly able to calculate -2.7^8.6, but fails when the exact same values are assigned to variables and then the computation is repeated. Thanks in advance for any suggetsions. Kris.

Hi Second question in a day, i'm beginnning to feel incompetent... This time i'm having a weird problem, i'm importing the next data base: >car<-read.csv2("Historicos.csv") 'data.frame': 1818 obs. of 6 variables: $ Dpto : Factor w/ 11 levels "ANTIOQUIA","ATL?NTICO",..: 2 2 2 2 2 1 1 1 1 5 ... $ Rio : Factor w/ 43 levels

XCP Users: I received an email from Alberto Gonzalez Rodriguez who created OpenXenCenter/OpenXenManager and is now working on OpenXenWebManager. Info to access the tool is: steps to test are: download from http://bit.ly/bres2U uncompress: tar xvfz xenwebmanager_rev24_full.tar.gz cd xenwebmanager python frontend.py and with a browser open http://localhost:8080 (or http://ip:8080) you need

XCP Users: As many of you are aware, there are several open source products available for everyone to use when working with XCP. I would like to get some feedback on the various tools to get a feeling on how people are using XCP and these tools. Here are the tools that I am aware of (please respond if you are using another tool): * OpenXenManager -

my list al is as below: al=list(c(2,3),5,7) > al [[1]] [1] 2 3 [[2]] [1] 5 [[3]] [1] 7 and I check the second component, its element is 5, then I remove this, now my al is: al[[2]][al[[2]]!=5]->al[[2]] > al [[1]] [1] 2 3 [[2]] numeric(0) [[3]] [1] 7 The Question is, how I can get the new list without the second component, that is : > alwanted [[1]] [1] 2 3 [[2]] [1] 7 Thank

Xen Community: As you are aware, the Xen Cloud Platform has been available for download since last November and is becoming a very successful release for Xen.org. In fact, we are seeing substantial network traffic for this solution as we had 26 TB of data from XCP downloads in the past 4 weeks. Of course, this amount of traffic is forcing a re-examination of our co-location agreements as we are

Xen Community: As you are aware, the Xen Cloud Platform has been available for download since last November and is becoming a very successful release for Xen.org. In fact, we are seeing substantial network traffic for this solution as we had 26 TB of data from XCP downloads in the past 4 weeks. Of course, this amount of traffic is forcing a re-examination of our co-location agreements as we are

Team: It is time to start thinking about our 2nd XCP meeting for the Xen.org XCP team. The last meeting was held on May 25, 2010 at 14:00 UTC. Notes are at http://wiki.xensource.com/xenwiki/XCP_Monthly_Meetings. For the next meeting I hope to schedule a time that is better for our Asian participants so please contact me with a time that would be best if you live in Japan, Korea, Australia, etc.

Team: It is time to start thinking about our 2nd XCP meeting for the Xen.org XCP team. The last meeting was held on May 25, 2010 at 14:00 UTC. Notes are at http://wiki.xensource.com/xenwiki/XCP_Monthly_Meetings. For the next meeting I hope to schedule a time that is better for our Asian participants so please contact me with a time that would be best if you live in Japan, Korea, Australia, etc.

Hi R Users, I'm trying to re-order the "site names" ("Waseca", "Morris", ...). I'm using following code: libarry(lattice) barchart(yield ~ variety | site, data = barley, groups = year, layout = c(6,1), aspect=.7, ylab = "Barley Yield (bushels/acre)", scales = list(x = list(abbreviate = TRUE, rot=45,

Citrix Project Satori is the result of a collaborative agreement between XenSource and Microsoft, and was carried forward after XenSource was acquired by Citrix Systems. The base Satori components are released by Microsoft as the Linux Integration Components for Hyper-V, and provide support for paravirtualized XenLinux guests running on Hyper-V. The Linux Integration Components can be downloaded

Citrix Project Satori is the result of a collaborative agreement between XenSource and Microsoft, and was carried forward after XenSource was acquired by Citrix Systems. The base Satori components are released by Microsoft as the Linux Integration Components for Hyper-V, and provide support for paravirtualized XenLinux guests running on Hyper-V. The Linux Integration Components can be downloaded

Citrix Project Satori is the result of a collaborative agreement between XenSource and Microsoft, and was carried forward after XenSource was acquired by Citrix Systems. The base Satori components are released by Microsoft as the Linux Integration Components for Hyper-V, and provide support for paravirtualized XenLinux guests running on Hyper-V. The Linux Integration Components can be downloaded

Hello, All, How can one sort on column and row names. For example: How can this X1 X3 X2 X1 1 0 0 X3 0 1 0 X2 0 0 1 become this? X1 X2 X3 X1 1 0 0 X2 0 1 0 X3 0 0 1 Thank you for your time! Jim [[alternative HTML version deleted]]

Hello! I wrote a function that returns a data frame. Nowhere in the function do I say print(my.data.frame), but when I run the function - the data frame is printed on the console. Is there any way to suppress it? Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hi, I have a (correlation) matrix and I want to select a subset of its cells depending on the combination of their column and row names. This illustrates my problem: mtrx <- matrix(c(1,2,3,4,5,6,7,8,9), nrow=3, ncol=3, dimnames = list(c('c132','c432', 'c233'), c('r132','r233', 'r432')))> mtrx r132 r233 r432c132 1 4 7c432 2

Hello. More than 1 year ago, someone said that HXEN would have an installer (*.dmg) for Mac OS X. Is such an installer now available? leopard _________________________________________________________________ Your E-mail and More On-the-Go. Get Windows Live Hotmail Free. https://signup.live.com/signup.aspx?id=60969 _______________________________________________ Xen-users mailing