search for: 0.625

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 all, I have a question about basic statistics. Given a PDF value of 0.328161, how can I find out the value of -0.625 in R? It is like reversing the dnorm function but I do not know how to do it in R. > pdf.xb <- dnorm(-0.625) > pdf.xb [1] 0.328161 > qnorm(pdf.xb) [1] -0.444997 > pnorm(pdf.xb) [1] 0.628605 Many thanks, Edwin -- View this message in context:

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

Hi All, simple question: do you know how to graph the following object/matrix in a 'surface manner': [,1] [,2] [,3] [,4] [,5] [,6] [1,] -0.154 -0.065 0.129 0.637 0.780 0.221 [2,] 0.236 0.580 0.448 0.729 0.859 0.475 [3,] 0.401 0.506 0.310 0.650 0.822 0.448 [4,] 0.548 0.625 0.883 0.825 0.945 0.637 [5,] 0.544 0.746 0.823 0.877 0.861 0.642 [6,] 0.262 0.399

http://r.789695.n4.nabble.com/file/n3216865/Inverse_Gamma.png Hello, I am seeking help in estimating the parameters of an inverse gamma distribution (from the 'actuar' package) using a function like 'fitdistr'. Unfortunately I haven't found such a package using findFn('fit Inverse Gamma') from the 'sos' package and was therefore hoping someone might be aware

THANKS so very much for your help (previous and future!). I have a two follow-up questions. 1) You say that dispersion = 1 by definition ....dispersion changes from 1 to 13.5 when I go from binomial to quasibinomial....does this suggest that I should use the binomial? i.e., is the dispersion factor more important that the 2) Is there a cutoff for too much overdispersion - mine seems to be

Hi Ricky / AJ Progress of sorts. I got passed the last problem by looking at the makefiles but run in to the next one, see below. It has created files in /contrib/R-2.4.0. there is an " R " under /contrib/R-2.4.0/bin/R. Running it gives :- /contrib/R-2.4.0/bin/R R version 2.4.0 (2006-10-03) Copyright (C) 2006 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is

Hello, I did a cross-validation using cvlm from DAAG package but wasn't sure how to assess the result. Does this result means my model is a good model? I understand that the overall ms is the mean of sum of squares. But is 0.0987 a good number? The response (i.e. gailRel5yr) has min,1st Quantile, median, mean and 3rd Quantile, and max as follows: (0.462, 0.628, 0.806, 0.896, 1.000, 2.400) ?

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?

In order to see how llvm-gcc-4.2 svn performs in code generation compared to the upcoming gcc 4.4, I ran the Polyhedron 2005 benchmarks on a MacPro with the -ffast-math -funroll-loops -msse3 -O3 optimization flags for both compilers. The results are summarized below. Ave Run (secs) Ave Run (secs) llvm-gcc-4.2/ Benchmark llvm-gcc-4.2 svn gcc trunk gcc trunk

Hello, my test.R file contains two huge arrays (>3000 entries), from which R needs to calculate the Pearson Correlation, if I look at the file the numbers look correct. if I run R R < test.R --no-save I see things like this: 0.723, 0.838, 1.002, 0.364, 0.357, 0.227, 0.982+ , 0.963, 0.535, 1.214, 1.270, 0.832, 1.033, 0.632, 2.482, 1.239, 0.743, 1.077, 0.962, 1.052, 1.075, 1.427, 1.395,

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

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

hi, knowing there are a lot of statistics guru's in the R-mailing list, i throw in a statistics question. i have data that i do not know how to statistically test: subjects are repeatedly asked to make a decision (e.g. left-right -> coded as 0 or 1). i have 20 subjects, each subject made 8 decisions. i now want to analyse whether my experimental manipulation induced a systematic bias

Dear R - Users I have some problems fitting a linear mixed effects model using the lme function (nlme library). A sample data is as shown at the bottom of this mail. I fit my linear mixed model using the following R code: bmr <-lme (outcome~ -1 + as.factor(endpoint)+ as.factor(endpoint):trt, data=datt, random=~-1 + as.factor(endpoint) + as.factor(endpoint):trt|as.factor(Trial),

Dear R-Helpers, I am trying to bootstrap the coefficient of variation on a suite of vectors, here I provide an example using one of the vectors in my study. When I ran this script with the vector x <-c(0.625, 0.071428571, 0.133333333, 0.125, 0), it returned CV(boot) [the second one], and stderr(boot) [the second one] without problem. However, when I ran it with the vector in the

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

Hi, I got two data point vectors. Now I want to make a ks.test(). I you print both vectors you will see, that they fit pretty fine. Here is a picture: http://www.jochen-bauer.net/downloads/kstest-r-help-list-plot.png As you can see there is one histogram and moreover there is the gumbel density function plotted. Now I took to bin-mids and the bin-height for vector1 and computed the

I have a data set called datastep4 with 211484 rows and 95 columns > dim(datastep4) [1] 211484 95 The first few column names are given below, note the first one is "RESPONDED" > names(datastep4)[1:5] [1] "RESPONDED" "VAR_30" "VAR_31" "VAR_32" "VAR_33" A table of RESPONDED shows mostly zeros >

I am starting to use the sample profiler to analyze new performance opportunities. The loop unroller has popped up in several of the benchmarks I'm running. In particular, libquantum. There is a ~12% opportunity when the runtime unroller is triggered. This helps functions like quantum_sigma_x (http://sourcecodebrowser.com/libquantum/0.2.4/gates_8c_source.html#l00149). The function accounts