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Hello, I've had this same model CyberPower connected to a linux server and a freebsd server. [root at sensor003 menon]# lsusb Bus 005 Device 001: ID 1d6b:0001 Linux Foundation 1.1 root hub Bus 004 Device 001: ID 1d6b:0001 Linux Foundation 1.1 root hub Bus 003 Device 004: ID 0764:0501 Cyber Power System, Inc. CP1500 AVR UPS Bus 003 Device 001: ID 1d6b:0001 Linux Foundation 1.1 root hub Bus 002 Device 001: ID 1d6b:0...

...W)) but this doesn't work (it returns 1, but I have 2800 rows in each of bar_signal and cif_signal, so I should get 2800) Is there a smart way to do this? And is there a way to tell R that the entries in the Signal column of tradingSignal are the names of objects? Thanks, Murali Murali Menon/GB/ABNAMRO/NL 29/03/2007 11:13 To r-help at stat.math.ethz.ch cc Subject creating conditional list of elements Folks, I have a matrix as follows (first column is the rownames, first row is the columnnames) Rule Enabled Signal Foo False foo_signal Bar True...

folks, is there a clever way to compute the sum of the product of two vectors such that the common indices are not multiplied together? i.e. if i have vectors X, Y, how can i compute Sum (X[i] * Y[j]) i != j where i != j also, what if i wanted Sum (X[i] * Y[j] * R[i, j]) i != j where R is a matrix? thanks, murali

...ber of rows in the smallest matrix, chopping off the top rows of the other matrices to reduce them this size, then doing a for-loop across each currency pair, row-wise, to create a 3-vector which I then apply median() on. Surely there's a better way to do this? Please advise. Thanks, Murali Menon _________________________________________________________________ Valentine?s Day -- Shop for gifts that spell L-O-V-E at MSN Shopping

...single code of R ? For example, I need to run the same code for different datasets (here list of companies) and since individual files are quite large, appending the files into one file is not a desirable option. Can this be done through a macro or sql kind of command? Thanks and Regards, Rahul S Menon Research Associate ISB, Hyderabad Ph-040-2318 7288 ________________________________ DISCLAIMER : This e-mail (including any attachments) is ...{{dropped:11}}

Folks, I'm trying to get stats from a matrix for each transition from one state to another. I have a matrix x as below. structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0, 0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19, -0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.81, 0.39, -1.38, 0.06, 0.84, 0.51, -1, -1.29, 2.15, 0.39, 0.78, 0.85, 1.18, 1.66, 0.9, -0.94,

Apologies if you received duplicate copies of this Call for Papers. Sincerely, Aravind Menon MBDS 2012 ----------------------------------------------------------------- CALL FOR PAPERS International Workshop on Management of Big Data Systems (MBDS 2012) http://www.cercs.gatech.edu/mbds12 In conjunction with ICAC 2012 http://icac2012.cs.fiu.edu September 21, 2012, San Jose, CA -------...

Apologies if you received duplicate copies of this Call for Papers. Sincerely, Aravind Menon MBDS 2012 ----------------------------------------------------------------- CALL FOR PAPERS International Workshop on Management of Big Data Systems (MBDS 2012) http://www.cercs.gatech.edu/mbds12 In conjunction with ICAC 2012 http://icac2012.cs.fiu.edu September 21, 2012, San Jose, CA -------...

Folks, I have a 3000 x 4 matrix (y), which I need to regress row-by-row against a 4-vector (x) to create a matrix lm.y of intercepts and slopes. To illustrate: y <- matrix(rnorm(12000), ncol = 4) x <- c(1/12, 3/12, 6/12, 1) system.time(lm.y <- t(apply(y, 1, function(z) lm(z ~ x)$coefficient))) [1] 44.72 18.00 69.52 NA NA Takes more than a minute to do (and I need to do many

Folks, If I have a vector such as the following: x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0) and I want to replace the zeroes by the nearest non-zero number to the left, is there a more elegant way to do this than the following loop? y <- x for (i in 2 : length(x)) { if (y[i] == 0) { y[i] <- y[i - 1] } } > y [1] 0 -1 -1 -1 -1 -1 1 -1 1 1 You can see the

Hi folks, Not sure what this sort of estimation is called. I have a 2-column time-series x(i,t) [with (i=1,2; t=1,...T)], and I want to do the following 'simultaneous' regressions: x(1,t) = (d - 1)(x(1, t-1) - mu(1)) x(2,t) = (d - 1)(x(2, t-1) - mu(2)) And I want to determine the coefficients d, mu(1), mu(2). Note that the d should be the same for both estimations, whereas the

Folks, I'm a bit puzzled by the fact that if I generate 100,000 standard normal variates using rnorm() and perform the Jarque-Bera on the resulting vector, I get p-values that vary drastically from run to run. Is this expected? Surely the p-val should be close to 1 for each test? Are 100,000 variates sufficient for this test? Or is it that rnorm() is not a robust random number generator?

Folks, I have a matrix: set.seed(123) a <- matrix(rnorm(100), 10) And a vector: b <- rnorm(10) Now, I want to switch the signs of those rows of a corresponding to indices in b whose values exceed the 75 %-ile of b which(b > quantile(b)[4]) [1] 2 6 10 so I want, in effect: a[2, ] <- -a[2, ] a[6, ] <- -a[6, ] a[10, ] <- -a[10, ] I thought I could do a[which(b >

Folks, If I have a series mm of, say, monthly observations, and a series dd of daily dates, what's a good way of expanding mm such that corresponding to each day in dd within the corresponding month in mm, the values of mm are repeated? So e.g., if I have mm: mm <- c(15, 10, 12, 13, 11) names(mm)<-c("Nov 2008", "Dec 2008", "Jan 2009", "Feb

...ader. Is there any way i can drop the number of observation counts being added into the matrix. Thanks in advance for your help. Sincerely Anup "However bad life may seem, there is always something you can do and succeed at. While there is life, there is hope." Stephen Hawking Anup Menon Nandialath ********************************************************* http://www.soundclick.com/bands/7/tailgunner_music.htm * ********************************************************* ____________________________________________________________________________________ Looking for earth-friendl...

Folks, I have a 'matching' matrix between variables A, X, L, O: > a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58, 0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list( c("A", "X", "L", "O"), c("A", "X", "L", "O"))) > a A X L O A 1.00 0.41

folks, if i have three matrices, a, b, cc with some colnames in common, and i want to create a matrix which consists of the common columns added up, and the other columns tacked on, what's a good way to do it? i've got the following roundabout code for two matrices, but if the number of matrices increases, then i'm a bit stymied. > a <- matrix(1:20,ncol=4); colnames(a) <-

Hi. I''m wondering if there''s a good way to configure a Linux firewall box to failover to a single backup server, while preserving connection state. This question has been asked before, but the latest reference I can find is from 2004, at which time Linux had no equivalent of OpenBSD''s pfsync, though Harald was said to be working on one. Did anything come of those

...ot; "1" [3,] "09/02/2007" "0" [4,] "12/02/2007" "1" [5,] "15/02/2007" "0" Is that what you want or did I miss it ?! Hope this helps. Olivier > essage: 92 > Date: Fri, 16 Feb 2007 10:14:30 +0000 > From: "Murali Menon" <feanor0 at hotmail.com> > Subject: [R] array searches > To: r-help at stat.math.ethz.ch > Message-ID: <BAY113-F191B663BED1069C6C1EDE9EE950 at phx.gbl> > Content-Type: text/plain; format=flowed > > Folks, > > I have a dataframe comprising a column of dates...

Hi I have a data frame with 40,000 rows and 4 columns, one of which is "class". For each row, the "class" column can be one of 10 possible NUMERIC values. I wish to substitute these numeric values with words/characters. For example, I wish to substitute all occurences of "5467" in the column "class" with "alpha", "7867" with