similar to: Developing the UPS side of the UPS-NUT equation (via usbhid)

On 5/30/2017 10:47 PM, Charles Lepple wrote: > On May 22, 2017, at 11:13 AM, Mike <the.lists at mgm51.com> wrote: >> >> On 5/21/2017 10:39 AM, Charles Lepple wrote: [snip] > Even though the CPS sub-driver has been written already, I would like to take a look at the raw "explore" mode output: > >

## the following is a dream: add some sugar syntax to allow for contracts with teeth (in and out checking) > is.positive <- function(x) (all(x>0)) > exponentiate <- function( x ::is.data.frame , exponent ::is.numeric is.positive) :: is.vector is.numeric { x$base :: is.positive ## error also if base does not exist in x; may need some special IQ x$base^exponent }

Dear all, I would like to (i) produce boxplot graphs with axis in logarithm in base 10 and (ii) showing the values on the axis in 10^exponent format rather than 10E+exponent. To illustrate with an example, I have some widely spread data that I chart plot using boxplot() [figure on the left]; the log="y" option of boxplot() I obtained the natural logarithm conversion of the data and

> > Hello All, > Please help me with this problem.I have been having this problem for over a month now and I could not find any information.I later realised that error is with MATRIX package. I am working on graph object using IGRAPH version 0.5.2-2 package & wanted to do Bonacich Power. What I have is undirected GRAPH. This is my graph object. The file 'Bonacich

Hi all, I need to convert a floating point value from Microsoft Basic format to IEEE format. There's a simple way to achieve this in R or I have to write my own function? (e.g. convert the C code below) thanks t #include <string.h> /* for strncpy */ int _fmsbintoieee(float *src4, float *dest4) { unsigned char *msbin = (unsigned char *)src4; unsigned char *ieee

Hello, I'm using fArma package to estimate the value of Hurst exponent using R/S method. However, for a certain set of data I get H ~ 1.8. How do I interpret this? Following are the output that I get for this set: > mean(data[,2]) [1] 400.5433 > sd(data[,2]) [1] 1139.786 > > rsFit(data[,2], levels = 64) Title: Hurst Exponent from R/S Method Call: rsFit(x = data[, 2], levels

I am having difficulty with the calculation of the power law exponent for set of nodes within a graph. Specifically, I am interested in the distribution of in-degree and out-degree among communities of web pages where the web pages are the nodes of the graph and the hyperlinks the edges. According to the literature, the distribution of incoming and outgoing links obeys a power law distribution

Hello, I am trying to collect several global measures or statistics for time-series as well as packages of R that can compute them. I have found several of them in papers and books, but the literature is so big i am sure i am missing several of them. skewness kurtosis min max mean SD trend seasonality periodicity chaos (Lyapunov Exponent) / Largest Lyapunov Exponent (i think is the same

Hi all, I need to solve this following equation which the unknown value (here is "a") is on the exponent place, could anybody please help me to figure it out? I tried to use lm.sol() or solve() function but it doesn't work...... Thanks for help, Peiling (x, y) = (2.21, 1.01) y = 1 + x - ( 1 + (x)^a)^(1/a)

On Mar 23, 2014, at 11:12 PM, Bill S wrote: > The output of the driver is below. The incorrect voltage of 136 is > shown at 0.265256. It should match input voltage of 119 as shown at > 0.265219 but it doesn't. The voltages displayed on the UPS LCD panel > information however are equal and match the input voltage provided by > the driver. There might be other incorrect

I would like to have an expression f^(-1)(x) in a legend of plot. For this I used expression(f^{-1}(x)), but variable is always in exponent. How could I change it in order to (x) be in a line, not in exponent? Thank you for your responses! -- View this message in context: http://r.789695.n4.nabble.com/Expression-in-legend-of-plot-tp4647393.html Sent from the R help mailing list archive at

Hi Folks; I just bought a Polycom SoundPoint 500 off of ebay after having spent way too much time trying to get updated sip images for our cisco phones. The phone I bought didn't have an AC power adapter; Could someone please tell me the volts & amps that the dc plug that comes with the phone puts out? Thanks! Mike

On Jun 3, 2017, at 2:21 PM, Mike <the.lists at mgm51.com> wrote: > > On 5/30/2017 10:47 PM, Charles Lepple wrote: >> On May 22, 2017, at 11:13 AM, Mike <the.lists at mgm51.com> wrote: >>> >>> On 5/21/2017 10:39 AM, Charles Lepple wrote: > [snip] > >> Even though the CPS sub-driver has been written already, I would like to take a look at the

This patch supplies software IEEE floating point support. The comment from the patch reproduced below says all there is to say. This patch contains the prior "cleanup" patch; please don't apply that one. Please let me know of any bugs. It is tested reasonably well, but until I put together random tests it's hard to have 100% confidence. Neil. /* A self-contained host- and

Hi, I'm using the power.law.fit function from the igraph package to fit a power law distribution to some data. This function returns the power law exponent as it's only result. I would like to have some sort of goodness-of-fit and/or error estimate of the exponent returned. This paper: http://www.edpsciences.org/articles/epjb/pdf/2004/18/b04111.pdf suggests using the

Dear R, I have two small questions confused me recently. Now assume I have a matrix "a", like this, > a <- matrix(1:6, 2, 3) > a [,1] [,2] [,3] [1,] 1 3 5 [2,] 2 4 6 I sometimes need each row of "a" raised to a different exponent. So I do a trick like this, > a^c(2, 3) [,1] [,2] [,3] [1,] 1 9 25 [2,] 8 64 216 My first

Dear All, I am a novice when it comes to time-series analysis and at the moment I am actually interested in calculating the Hurst exponent of a time series. This question has already been asked quite some time ago http://bit.ly/98dZsi and I trust some progress has been made ever since. I was able to find some functions in the packages http://cran.r-project.org/web/packages/Rwave/index.html

I'm breaking this out from LLVM asserts thread. Here are all the assertions in APInt I consider should be recoverable: APInt::APInt: bitwidth too small Null pointer detected! APInt::getBitsNeeded: Invalid string length string is only a minus! Invalid radix APInt::fromString: Radix should be 2, 8, 10, or 16! Invalid string length string is only a minus Insufficient bit width Invalid digit in

I am curious if there is code developed to plot confidence regions in 3D. The scatterplot3d function generates the plot I want, but would like an 3D equivalent to the data.ellipse function. Any help in this direction would be appreciated, be it theoretical, graphical, or otherwise. Melanie Edwards Senior Statistician Exponent 15375 SE 30th PL, Suite 250 Bellevue, WA 98007 Tel: (425) 519-8714

So I have this key information that I got by generating a private and public key in VB, but it gave it to me in the form of a modulus and exponent in an XML file. Now, I need to encrypt some stuff in ruby using that info. On this site: http://blog.internautdesign.com/2007/11/2/ruby-openssl-rsa-public-key-from-modulus-and-exponent/comments/165, they explain how to do that, but clearly this info is