search for: 0.0625

Dear R help, I have some readings in three dimensions (x, y, z) and an amplitude for each. I'd like to visualize the data using ggplot, using tile plots, as I have some additional point data I would like to eventually overlay on the tile plots. I would like to subset the data by sections, slices if you will, in the z dimension, and plot the data for that slice. I can do all of this, but am

#Here is a vector of IDs > cwaves [1] 86 90 185 196 197 209 210 215 216 217 218 #Here is a matrix. The rows and columns correspond to the IDs in cwaves, and the matrix is populated with a coefficient > mat 86 90 185 196 209 210 215 216 217 218 86 0 0 0 0 0 0.00000 0 0.000000 0.000000 0.000000 90 0 0 0 0 0 0.00000 0

Hello all, This is something that I am sure has a really suave solution in R, but I can't quite figure out the best (or even a basic) way to do it. I have a simple linear regression that is fit with lm for which I would like to estimate the x intercept with some measure of error around it (confidence interval). In biology, there is the concept of a developmental zero - a temperature under

I am tangled with a syntax question. I want to calculate basic statistics for a large dataset provided in weights and values and I can't figure out an elegant way to expand the data. For example here are the counts: > counts n4 n3 n2 n1 p0 p1 p2 p3 p4 1 0 0 0 1 1 3 16 55 24 2 0 0 0 0 2 8 28 47 15 3 1 17 17 13 4 5 12 24 8 ... and the values: > values

Optim's Nelder-Mead works correctly for this example. > optim(par=10, fn=fn, method="Nelder-Mead") x=10, ret=100.02 (memory) x=11, ret=121 (calculate) x=9, ret=81 (calculate) x=8, ret=64 (calculate) x=6, ret=36 (calculate) x=4, ret=16 (calculate) x=0, ret=0 (calculate) x=-4, ret=16 (calculate) x=-4, ret=16 (memory) x=2, ret=4 (calculate) x=-2, ret=4 (calculate) x=1, ret=1

Hi I have the feeling, that the argument stringsAsFactors has no impact in the function expand.grid: a <- c("PR", "NC", "A2", "BS") b <- c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125) class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]]) [1] "factor" class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]]) [1] "factor" Also, when

Dear Dr.Goslee and anyone may intrested in matrix manipulate, I am using your ecodist to do mantel and partial mantel test, I have locality data and shape variation data, and the two distance matrixs are given as belowings. When I run the analysis, it is always report that the matrix is not square, but I didn't know what's wrong with my data. Would you please help me on this. I am quite

Right now, all profile information is funneled through two analysis passes prior to any part of the optimizer using it. First, we have BranchProbabilityInfo, which provides a simple interface to the simplest form of profile information: local and relative branch probabilities. These merely express the likelihood of taking one of a mutually exclusive set of exit paths from a basic block. They are

Hello, the more general thing I'd like to learn here is how to compute Function of Data on the basis of grouping determiend by n variables. In terms of the reason why I am interested in this, I need to compute the average of my data based on the value of the month and day across years. I have come up withy the code below which, as far as I can see, does what I need but getting either a more

That's consistent/not surprising if the problem lies in the numerical gradient calculation step ... On 2019-05-06 10:06 a.m., Ravi Varadhan wrote: > Optim's Nelder-Mead works correctly for this example. > > >> optim(par=10, fn=fn, method="Nelder-Mead") > x=10, ret=100.02 (memory) > x=11, ret=121 (calculate) > x=9, ret=81 (calculate) > x=8, ret=64

Hi May you help me correct my loop function. I want optim to estimates al_j; au_j; sigma_j; b_j by looking at 0 to 20, 21 to 40, 41 to 60 data points. The final result should have 4 columns of each of the estimates AND 4 rows of each of 0 to 20, 21 to 40, 41 to 60. ###MY code is n=20 runs=4 out=matrix(0,nrow=runs) llik = function(x) { al_j=x[1]; au_j=x[2]; sigma_j=x[3]; b_j=x[4]

Hi, I'm trying to plot some data (z) that is linked to lat&long coordinates (x&y). These coordinates are not however on a regular grid. I also have some shapefiles on which I would like to overlay the data. I can plot the shapefiles (country/city outlines) and overplot the data, but only using quilt.plot because I otherwise always get the error message that 'Error in

It seems that it's an old bug that was found in some other packages, but at that time not optim: https://bugs.r-project.org/bugzilla/show_bug.cgi?id=15958 and that Duncan Murdoch posted a patch already last Friday :) Thomas Am 06.05.2019 um 16:40 schrieb Ben Bolker: > That's consistent/not surprising if the problem lies in the numerical > gradient calculation step ... >

Hi there, I checked google for aov. usually one uses summary to see whether the p-value is small. but I want to put aov in my script. how can I get the p-value, (0.1115, 0.6665, 0.6665 in the following example)? thanks YU > datafilename="http://personality-project.org/r/datasets/R.appendix2.data" > data.example2=read.table(datafilename,header=T) > aov.ex2 =

On Sun, Feb 2, 2014 at 6:18 PM, Andrew Trick <atrick at apple.com> wrote: > > On Feb 2, 2014, at 2:13 AM, Chandler Carruth <chandlerc at gmail.com> wrote: > > > Right now, all profile information is funneled through two analysis > passes prior to any part of the optimizer using it. > > > > First, we have BranchProbabilityInfo, which provides a simple

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

Dear List members, I would like to ask for advise on quantile regression in R. I am trying to perform an analysis of a relationship between species abundance and its habitat requirements - the habitat requirements are, however, codes - 0,1,2,3... where 0<1<2<3 and the scale is linear - so I would be happy to treat them as continuos The analysis of the data somehow does not work, I am

Dear R people, I notice that the confidence intervals of a very small sample (e.g. n=6) derived from the one-sample wilcox.test are just the maximum and minimum values of the sample. This only occurs when the required confidence level is higher than 0.93. Example: > sample <- c(1.22, 0.89, 1.14, 0.98, 1.37, 1.06) > summary(sample) Min. 1st Qu. Median Mean 3rd Qu. Max.

On Feb 3, 2014, at 12:13 AM, Andrew Trick <atrick at apple.com> wrote: > On Feb 2, 2014, at 6:55 PM, Chandler Carruth <chandlerc at gmail.com> wrote: > >> On Sun, Feb 2, 2014 at 6:18 PM, Andrew Trick <atrick at apple.com> wrote: >> >> On Feb 2, 2014, at 2:13 AM, Chandler Carruth <chandlerc at gmail.com> wrote: >> >> > Right now,

Yes, I think you are right. I was at first confused by the fact that after the optim() call, > environment(fn)$xx [1] 10 > environment(fn)$ret [1] 100.02 so not 9.999, but this could come from x being assigned the final value without calling fn. -pd > On 3 May 2019, at 11:58 , Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > > Your results below make it look like a