search for: 0.8

I just started to write tiny functions and therefore I appologise in advance if I am asking stupid question. I wrote a tiny function to give me back from the original matrix, a matrix showing only the values smaller -0.8 and bigger 0.8. y<-c(0.1,0.2,0.3,-0.8,-0.4,0.9) x<-c(0.5,0.3,0.9,-0.9,-0.7,0.3) XY<-rbind(x,y) extract.values<-function (x) { if(x>=0.8|x<=-0.8)x

CentOS Errata and Security Advisory 2013:0897 Important Upstream details at : https://rhn.redhat.com/errata/RHSA-2013-0897.html The following updated files have been uploaded and are currently syncing to the mirrors: ( sha256sum Filename ) i386: 593fa4ebcae095c02f486fd0fdb9449fd4ac1e03bf26082f015177a34f146d72 glx-utils-9.0-0.8.el6_4.3.i686.rpm

Hi all, I try to generate sets of strategies that contain probability distributions for a defined number of elements, e.g. imagine an animal that can produce 5 different types of offspring and I want to figure out which percentage of each type it should produce in order to maximize its fitness. In order to do so, I need to calculate the fitness for all potential strategies. As an example, if I

I have a FreeBSD 4.8-Release machine that I upgraded from a 4.5 or 4.4 machine. It works just great for about 9 days, then processes will not die. It starts to become a problem when I can't kill processes, restart services, connections hang, and I can't reboot or shutdown, I have to do a hard reset. I can't really find anything in the logs to indicate what is giving me

hi i try to do a graph of a time series which shows in red the values > -0.05, in blue the values >0.05 and in white the values between -0.05 and 0.05 for un exemple :http://www.appinsys.com/globalwarming/enso.htm thanks !!!!! denisse -- View this message in context: http://r.789695.n4.nabble.com/how-to-do-a-graph-with-tree-different-colors-tp4635206.html Sent from the R help mailing

Hi, I'm looking the best way to fill irregular polygons with patterns, Something like the function grid.pattern do, but my case is with irregular polygons. Whit this script I can get it, but I'm looking for an "elegant" solution.. library(grid) grid.polygon(x=c(0.2, 0.8, 0.6, 0.6, 0.8, 0.2), y=c(0.2, 0.2, 0.3, 0.5, 0.7,0.7), gp=gpar(fill="grey",

Hello all. I have two data frames. Group Start End G1 200 700 G2 500 1000 G3 2000 3000 G4 4000 6000 G5 7000 8000 and Pos C0 C1 200 0.9 0.6 500 0.8 0.8 800 0.9 0.7 1000 0.7 0.6 2000 0.6 0.4 2500 1.2 0.8

Dear R programmers, is there a sensible explanation for the following behaviour? The second command seems not to be interpreted correctly. > seq(0.6, 0.9, by=0.1) == 0.8 [1] FALSE FALSE TRUE FALSE > seq(0.7, 0.9, by=0.1) == 0.8 [1] FALSE FALSE FALSE > c(0.7, 0.8, 0.9) == 0.8 [1] FALSE TRUE FALSE > seq(0.9, 0.7, by=-0.1) == 0.8 [1] FALSE TRUE FALSE I am running R version 1.7.1 on

I am receiving the following error while executing KLdiv method: kl<-KLdiv(y) Error in function (classes, fdef, mtable) : unable to find an inherited method for function "KLdiv", for signature "data.frame" the code is as follows: require(flexmix) KLdiv(y) the object y (snippet, actual object contains 188 entries) is as follows: s1p5 s2p5 1 0.6 0.8 2 0.8

Bonjour I have this data frame and I am newbie in R. I want to ask if it is possible to draw 10 lines in a plot such that: a line for every colomn, the x - axis is the second column and the y-axis is the third one. Thank you for any input 1 0 1094442 1 0.2 1163576.2 1 0.4 1238539.6 1 0.6 1303510.4 1 0.8 1376573.2 1 1 1454175 2 0 1076068 2 0.2 1139246 2 0.4 1212344 2

Folks, The contour() function wants x and y to be in increasing order. I have a situation where I have a grid in x and y, and associated z values, which looks like this: x y z [1,] 0.00 20 1.000 [2,] 0.00 30 1.000 [3,] 0.00 40 1.000 [4,] 0.00 50 1.000 [5,] 0.00 60 1.000 [6,] 0.00 70 1.000 [7,] 0.00 80 0.000 [8,] 0.00 90

Hi, We have a serious issue running on our POP/IMAP servers these days. The load average of a servers spikes up to 400-500 as a uptime command result, for a particular time period , to be specific mostly in noon time and evening, but it last for few minutes only. We have 2 servers running dovecot 1.1.20 , in loadbanlancer, We have used KEEPLIVE (1.1.13) for loadbalacing. Server

hello! Today I tried to compile the new FLAC 0.8 sourcecode on my SuSE Linux 7.1 machine and failed. Below is the output of "make" and "configure". I hope the information is enough that somebody can help me to compile the source. Thanks a lot. PS: A lot of thanks to all developers of FLAC. Its very useful to me!! ------------------------------------------------------ This

All, I'm witnessing some strange behavior when checking the values of one of my variables. My guess is that it has something to do with the number of significant digits being represented, but perhaps not as my variable was created from other variables that only had one decimal place. See below. I'm sure this is something basic, any suggestions much appreciated. Cheers, David >

Hi, Use ?split() #dat1 is the dataset: lst1<- split(dat1,dat1$BaseProd) lst1 #$`2231` ?# BaseProd? CF OSA #1???? 2231 0.5 0.7 #2???? 2231 0.8 0.6 #3???? 2231 0.4 0.8 # #$`2232` ?# BaseProd CF OSA #4???? 2232? 1?? 2 #5???? 2232? 3?? 1 # #$`2233` ?# BaseProd? CF OSA #6???? 2233 0.9 0.5 #7???? 2233 0.7 0.5 #8???? 2233 4.0 5.0 #9???? 2233 5.0 7.0 lst1[[1]] #? BaseProd? CF OSA #1???? 2231 0.5 0.7

Hi Group, Consider a data frame like this: mylabel1 <- rep(c("A","B","C"),each=3) mylabel2 <- rep(c("A","B","C"),3) corrs <- c(1,.8,.7,.8,1,.7,.7,.7,1) myData <- data.frame(mylabel1,mylabel2,corrs) myData mylabel1 mylabel2 corrs 1 A A 1.0 2 A B 0.8 3 A C 0.7 4 B

Hi I am a newbie to R but have tried a number of ways in R to do this and can't find a good solution. (I could do it out of R in perl or awk but would like to know how to do this in R). I have a large data frame 49 variables and 7000 observations however for simplicity I can express it in the following data frame Base, Image, LVEF, ES_Time A, 1, 4.32, 0.89 A, 2, 4.98, 0.67 A, 3, 3.7, 0.5

Hi, I would like to generate a series in the following form (0.8^1, 0.8^2, ..., 0.8^600) Could anyone tell me how can I achieve that? I am really new to R. -- View this message in context: http://www.nabble.com/sequences-tf4019146.html#a11414836 Sent from the R help mailing list archive at Nabble.com.

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Hello > orderulcount Group.1 Group.2 Group.3 x V5 7 C L 0.0 30 C / L 19 C L 0.2 27 C / L 31 C L 0.4 15 C / L 43 C L 0.6 7 C / L 54 C L 0.8 2 C / L 10 C S 0.0 27 C / S 22 C S 0.2 10 C / S 34 C S 0.4 6 C / S 46 C S 0.6 1 C / S 1 D