similar to: search and replace in a list of strings

Dear All. I have some problem with combined two data frame. .... I have first data frame .. GPAX THAI MATH SCINCE SOCIAL HEALT ART CAREER LANGUAGE 1227 2.99 3.32 2.50 2.64 3.05 3.60 3.72 3.57 2.62 1704 2.81 2.56 2.48 2.86 3.22 3.19 3.55 3.20 2.51 617 2.18 1.90 1.97 2.06 2.38 3.50 3.54 2.33 1.70 876 2.82 3.14 2.73 2.46 2.71 3.11 3.04 3.24 2.90

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

Dear list members, I am new member and fairly new into R world! I hope what I have is not beyond the purpose of this list. I did first search for similar experimental designs without success. I want to perform an ANOVA analysis using the aov() function. I am not 100% sure that I have it right. If anyone can help me, that will be greatly appreciated. My design is not balanced for any of the

Hi all, I'm plotting to get the intersection value of three curves. Defining the x-axis as dsm, the following code works; dsm = c(800,600,NA,525,NA,450,400,NA,NA,NA,0) s3 = seq(0.05,1.05,0.1) plot(dsm,s3,col="blue",las=1,ylab="fraction",xlab="distance (km)") fc <- function(x,a,b){a*exp(-b*x)} fm <- nls(s3~fc(dsm,a,b),start=c(a=1,b=0)) co <- coef(fm)

Dear List, I ran into some trouble by calculating differences. For me it is important that differences are either 0 or not. So I don't understand the outcome of this calculation 865.56-(782.86+0+63.85+18.85+0) [1] -1.136868e-13 I run R version 2.71 on WinXP I could solve my problem by using round() but I would like to know the reason. Maybe someone can help me? Thanx

Hi R users, I am trying to use the optim function to maximize a likelihood funciton, and I got the following warning messages. Could anyone explain to me what messege 31 means exactly? Is it a cause for concern? Since the value of convergence turns out to be zero, it means that the converging is successful, right? So can I assume that the parameter estimates generated thereafter are reliable MLE

Hi R list, I imported values from Excel, there is a column with numbers like 45%, 65%, 12%. I want to find its mean. What should I use? strisplit() split() parse() Data from dput(), structure(c(78L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "-0.15%", "-0.34%", "-1.3%", "-10.77%", "-100.00%", "-11.45%",

I've found the following strange behaviour R (RStudio) which has been confirmed by another user in RGui. Inside a script I want to set two variables: default.wd = getwd() tmp.wd = setwd(choose.dir()) After choosing tmp.wd the value of default.wd is shown in Workspace, but getwd() is giving back the correct string of tmp.wd. Is there a workaround for the problem? I'm working on

Hi, I'm getting horrible performance on my samba server, and I am unsure of the cause after reading, benchmarking, and tuning. My server is a K6-500 with 43MB of RAM, standard x86 hardware. The OS is Slackware 10.0 w/ 2.6.7 kernel I've had similar problems with the 2.4.26 kernel. I use samba version 3.0.5. I've listed my partitions below, as well as the drive models. I have a

Dear R Expert   allow me to ask a quick qestion: I have a mean value of 6 and a SD of 3 describing my distribution. I would like to "convert" this distribution into a log normal distribution that would best describe it when resimulated using log normal distribution. Currently I am using another software to estimate the respective mean and SD on the log scale and the results are: 1.6667

Should there be a [[<-.factor() that either throws an error or acts like [<-.factor() to avoid making an illegal object of class factor? > z <- factor(c("Two","Two","Three"), levels=c("One","Two","Three")) > z [1] Two Two Three Levels: One Two Three > str(z) Factor w/ 3 levels

Hi >i have this data > X [1] 5.79 1579.52 2323.70 68.85 426.07 110.29 108.29 1067.60 17.05 22.66 [11] 21.02 175.88 139.07 144.12 20.46 43.40 194.90 47.30 7.74 0.40 [21] 82.85 9.88 89.29 215.10 1.75 0.79 15.93 3.91 0.27 0.69 [31] 100.58 27.80 13.95 53.24 0.96 4.15 0.19 0.78 8.01 31.75 [41] 7.35 6.50

Currently sub(perl=TRUE) allows you to specify \U and \L in the replacement argument so that the rest of the subpatterns in the line (the \\<digit> things) will be converted to upper or lower case, respectively. perl also also has a \E operator to end these case conversions for the rest of the subpatterns (so they retain whatever case they had in the original text). For symmetry's sake

Hello, I have recently installed Ubuntu (8.10) Intrepid Ibex and I am trying to get R version 2.8. I have changed my /etc/apt/sources.list file to include a mirror. I have also added the Vincent Goulet key. However, whenever I run sudo apt-get update sudo apt-get install r-base it installs version 2.71. Given that I am new to Ubuntu, is there anything that I need to do to make ubuntu get

The current llvm/llvm-gcc-4.2 2.6 branch passes all of the Polyhedron 2005 benchmarks built with its gfortran. The results compare as follows... Compile Command : gfortran -ffast-math -funroll-loops -msse3 -O3 %n.f90 -o %n benchmark gcc-4.2.4 llvm-gcc-svn llvm-gcc-2.6 llvm-gcc-2.6 at -m32 20081031 -m32 at -m32 at -m64 ac 18.30

Richard Zimmerman wrote: > hw wrote: >> Next question: you want RAID, how much storage do you need? Will 4 or 8 3.5" drives be enough (DO NOT GET crappy 2.5" drives - they're *much* more expensive than the 3.5" drives, and >smaller disk space. For the price of a 1TB 2.5", I can get at least a 4TB WD Red. > > I will second Marks comments here. Yes,

Hi, I am trying to assemble a three-way contingency table examining the presence/absence of mussels, water depth (Depth1 and Depth 2) and water velocity (Flow vs. No Flow). I have written the following code listed below; however, when run the glm I get the following message, "Error in model.frame.default(formula = Count ~ MP + wd + wv, drop.unused.levels = TRUE) : variable lengths differ

A little while ago I asked a question on CFE-Dev about a change in the behaviour of programs using the ISO C math functions, although that question should have been put to LLVM-Dev. But I got excellent clarification of the problem anyway. However, since then I have been trying to adapt our out-of-tree implementation to get the previous behaviour. The problem is that something like: #include

I have 3 4TB WD drives I want to put in a RAID1 array. Two WD4000FYYZ and One WD4000F9YZ All enterprise class but two are WD Re and one is WD Se. I ordered the first two thinking 2 drives in the raid array would be sufficient but later decided its a long drive to the server so I would rather have 3 drives and ordered a third in accidentally did not get EXACT same thing. Would there be ANY

Hello R-community, I am trying to populate a column (doy) in a large dataset with the first column number that exceeds the value in another column (thold) using the 'apply' function. Sample data: pt D1 D17 D33 D49 D65 D81 D97 D113 D129 D145 D161 D177 D193 D209 D225 D241 D257 1 39177 0 0 0 0 0 0 0 0 0.4336 0.4754 0.5340667 0.5927334 0.6514 0.6966