search for: 0.375

I have a file: x y z 0.025 0.025 1.65775 0.025 0.050 1.62602 0.025 0.075 1.63683 0.025 0.100 1.91847 0.025 0.125 2.00913 0.025 0.150 1.82222 0.025 0.175 1.70901 0.025 0.200 1.39759 0.025 0.225 1.39089 0.025 0.250 1.04762 If I read the file like this: data<-read.table("file.dat") How do I access the vectors x,y,z that are inside the dataframe data? I studied Venables and

Hello. I asked before and it was great, cause as a beginner I learned a lot. But, if I have this in R (1 and 2 are codes for sex): > sex<-c(1,2,2,1,1,2,2,2) > sex [1] 1 2 2 1 1 2 2 2 I´d like to obtain the proportion according to sex.So I type: > prop.table(sex) [1] 0.07692308 0.15384615 0.15384615 0.07692308 0.07692308 0.15384615 0.15384615 [8] 0.15384615 The result is OK, but I

the following the the lower.tri matrix in a file named luxry.car and i want to read it in R as a lower.tri matrix.how can i do? i have try to use help.search("read"),but no result what i want. 1.000 0.591 1.000 0.356 0.350 1.000

Hi, I would like to get from a plot the size of the symbols plotted. Imagine I have the following plot function : plot(1:2,1:2, pch=15, cex=4) I would like the get the values SIZE1 and SIZE2 so that if I plot the following rectangle : rect(1.5,1.5, 1.5+SIZE1, 1.5+SIZE2) then the size of this square is exactely the same as the one of the symbols that have been plotted. Thanks for any idea. --

I'm trying to convert an S-Plus program to R.  Since I'm a SAS programmer I'm not facile is either S-Plus or R, so I need some help.  All I did was convert the underscores in S-Plus to the assignment operator <-.  Here are the first few lines of the S-Plus file:   sshc _ function(rc, nc, d, method, alpha=0.05, power=0.8,              tol=0.01, tol1=.0001, tol2=.005, cc=c(.1,2),

Hi All, I need to evaluate a series expansion using Legendre polynomials. Using the 'orthopolinom' package I can get a list of the first n Legendre polynomials as character strings. > library(orthopolynom) > l<-legendre.polynomials(4) > l [[1]] 1 [[2]] x [[3]] -0.5 + 1.5*x^2 [[4]] -1.5*x + 2.5*x^3 [[5]] 0.375 - 3.75*x^2 + 4.375*x^4 But I can't figure out how to

When I look at ppoints I see: ppoints<-function (x) { n <- length(x) if (n == 1) n <- x (1:n - 0.5)/n } However Venables & Ripley (2nd ed, p 165) say ppoints() should return (i-1/2)/n for n>=11; (i-3/8)/(n+1/4) for n<=10. The version below should work as described: ppoints<-function (x) { n <- length(x) if (n <= 10) (1:n - 0.375)/(n + 0.25) else (1:n - 0.5)/n

Dear Sirs: I've been working with several variables in a dataframe that serve as part of a calculation that I need to perform in a different way depending on its value. Let me explain: The main dataframe is called llmcc llmcc : 'data.frame': 283 obs. of 11 variables: $ Area : num 308.8 105.6 51.4 51.4 52.9 ... $ mFondo : num 30.1 10 10.2 10.2 40.4 ... $ mFachada :

Dear all, I have a question concerning the ks.test() function. I tryed to calculate the example given on the German wikipedia page. xi <- c(9.41,9.92,11.55,11.6,11.73,12,12.06,13.3) I get the right results when I calculate: ks.test(xi,pnorm,11,1) Now the question: shouldn't I obtain the same or a very similar result if I commpare the sample and a calculated sample from the distribution?

Do you know of a reference that discusses alternative choices for plotting positions for a normal probability plot? The documentation for qqnorm says it calls ppoints, which returns qnorm((1:m-a)/(m+1-2*a)) with "a" = ifelse(n<=10, 3/8, 1/2)? The help pages for qqnorm and ppoints just refer to Becker, Chambers and Wilks (1988) The New S Language (Wadsworth & Brooks/Cole),

Dear R People: I have used the command round(x,3) to produce values with 3 places to the right of the decimal. Is there any command to remove the leading zero before the decimal point, please: that is, if I have 0.375, how do I produce just .375, please? Thanks in advance R 2.0.1 for Windows Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com

Hi Folks, Ran into something I'd really like to do in R simply/elegantly, but my R - coding skills seem surpassed. This is the thing. Imagine the following data: labs<-c("abcdef","abcgg","tgthefdk","tgtijuel","tgtnjmoi","gbnt","dlift") dat<-c(0.5,0.25,1,2,16,0.250,4) dframe<-data.frame(labs,dat) I would like to

Hi, I hope some one can help. I need to compute Thurston's case 5 on a large set of data. I have gotten as far as computing the proportional preference matrix but the next math is beyond me. Here us my matrix 0.500 0.472 0.486 0.587 0.366 0.483 0.496 0.434 0.528 0.500 0.708 0.578 0.633 0.554 0.395 0.620 0.514 0.292 0.500 0.370 0.557 0.580 0.615 0.329 0.413 0.422 0.630 0.500 0.783 0.641 0.731

Hello, I''m trying to define my own analyzer by doing something like: #----------------------------------------------------- require ''ferret'' include Ferret class MyAnalyzer < Analysis::Analyzer def token_stream(field, str) # Display results of analysis puts ''Analyzing: field:%s str:%s'' % [field, str] t =

In the first example you are performing a one-sample test against a continuous cumulative distribution (in this case a normal distribution). In the second case you are performing a two-sample test. You drew your values for x non-randomly by specifying fixed intervals along a normal distribution, but ks.test() just sees that you have provided two samples, not one sample and values along a

Dear forum members, What is the formula to calculate denominator degrees of freedom (den df) for nonlinear mixed-effect models with covariates? My model is similar to a CO2 uptake example from Pinheiro and Bates (2000, page 376). In this CO2 dataset, there are two treatments and two types (84 observations in total), but den df for each parameter of the model is 64. Isn’t it too high? Your

Dear r-helpers, After successfully running require(nlme) vfr.lmL <- lmList( estimate ~ (slant + respType + visField + hand)^2 | subject, vfr ) pairs(vfr.lmL, id = 0.01, adj = -0.5) # Pinheiro & Bates (p. 141) produces the following error: Error in sprintf(gettext(fmt, domain = domain), ...) : object "form" not found Any guesses as to what I may have done wrong?

I have the manual for S+ 6 and I'm trying to use R for the Principle Component Analysis example and I'm getting a few interesting answers... The log is as follows: R : Copyright 2001, The R Development Core Team Version 1.3.0 (2001-06-22) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type `license()' or

On 16/01/2014, 23:47 , Andrew Trick wrote: > > On Jan 15, 2014, at 4:13 PM, Diego Novillo <dnovillo at google.com > <mailto:dnovillo at google.com>> wrote: > >> Chandler also pointed me at the vectorizer, which has its own >> unroller. However, the vectorizer only unrolls enough to serve the >> target, it's not as general as the runtime-triggered

Dear R-ophiles, I've found something very odd when I apply convolve to ever larger vectors. Here is an example below with vectors ranging from 2^11 to 2^17. There is a funny bump up at 2^12. Then it gets very slow at 2^16. > for( i in 11:20 )print( system.time(convolve(1:2^i,1:2^i,type="o"))) user system elapsed 0.002 0.000 0.002 user system elapsed 0.373