similar to: model frame problem

I am experiencing some problems at working with lists at high level. In the following "coef" contains the original DWT coefficients organized in a list. Thorugh applying the following two commands: coef.abs <- lapply(unlist(coef,recursive=FALSE,use.names =TRUE),abs) coef.abs.sorted <- sort(unlist(coef.abs),decreasing=TRUE) I get vector "coef.abs.sorted" containing

Dear R-list members, I've had a hard time trying to solve a non-linear system (nls) of equations which structure for the equation i, i=1,...,4, is as follows: f_i(d_1,d_2,d_3,d_4)-k_i(l,m,s) = 0 (1) In the expression above, both f_i and k_i are known functions and l, m and s are known constants. I would like to estimate the vector d=(d_1,d_2,d_3,d_4) which is solution

System: Cenots Linux 6.9 Application: nut-2.7.5-0.20170613gitb1314c6 [with usb 0.1 from distro] Device: Mustek PowerMust 1060 LCD Comunication log file: dump.txt We are looking at the possibility of successful communicating with this device UPS Mustek PowerMust 1060 LCD. PS: wolfy on the list gives me assistance and i can install any new compiled nut version from sources. Thanks, Catalin.

Otras variantes con y sin paquetes adicionales... > sapply(split(datIn$Gain, as.factor(datIn$Diet)), mean) d1 d2 d3 280 278 312 > by(datIn$Gain, datIn$Diet, mean) datIn$Diet: d1 [1] 280 -------------------------------------------------------------- datIn$Diet: d2 [1] 278 -------------------------------------------------------------- datIn$Diet: d3 [1] 312 > > library(dplyr) >

Estimados Cuando existia epicalc, había una manera muy fácil de determinar la media de una variable (en esta caso Gain) por grupos, en este caso (Diet). ?Como se puede hacer ahora? Diet Gain 1 d1 270 2 d1 300 3 d1 280 4 d1 280 5 d1 270 6 d2 290 7 d2 250 8 d2 280 9 d2 290 10 d2 280 11 d3 290 12 d3 340 13 d3 330 14 d3 300 15 d3 300

Me gusta la respuesta uqe has dado, pero si por ejemplo, alguno de los datos tiene datos faltantes, entonces devuelve NA. He probado con: sapply(split(datos$uno, as.factor(datos$dos)), mean(na.rm=TRUE)) pero da fallo. ¿Cómo se podría hacer para que devolviera además la media obviando los NA y que contara el numero de NA por categoria? > Date: Wed, 28 Oct 2015 00:13:45 +0100 > From: cof

Hello, Another way would be n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D5 <- data.frame(distance=integer(n),difference=integer(n)) D5[] <- do.call(rbind, lapply(seq_len(nrow(D1)), function(i) t(sapply(seq_len(nrow(D2)), function(j){ c(distance=sqrt(sum((D1[i,1:2]-D2[j,1:2])^2)),difference=(D1[i,3]-D2[j,3])^2) } )))) identical(D3, D5) In my first answer I forgot to say that

Hi Berend, I see you are one of the contributors to the rbecnhmark package. I am sorry that I am bothering you again. I have tried to run your code (slightly tweaked) involving the benchmark function, and I am getting the following error message. What am I doing wrong? Error in benchmark(d1 <- s1(df), d2 <- s2(df), d3 <- s3(df), d4 <- s4(df), : could not find function

Hello, The obvious way would be to preallocate the resulting data.frame, to expand an empty one on each iteration being a time expensive operation. n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D4 <- data.frame(distance=integer(n),difference=integer(n)) k <- 0 for (i in 1:nrow(D1)){ for (j in 1:nrow(D2)) { k <- k + 1 D4[k, ] <-

Dear All, on rbind:ing together a number of data.frames, I found that character variables are converted into factors. Since this occurred for a data identifier, it was a little inconvenient and, to me, unexpected. (The help page explains the general procedure used. I also found that on forming a data frame, character variables are converted to factors. The help page on read.table has the

Dear Sir, Please excuse my akwardness as I a new to R and computers, but would kindly appreciate help. { a <- sample (1:10,100,replace=T ) b <-sample(10:20,100,replace=T) c <- sample(20:30,100,replace=T) d <- sample(30:40,100,replace=T) e <- sample(40:50,100,replace=T) } d1 <- a d2 <- b d3 <-c d4 <- d d5 <- e data.frame(d1,d2,d3,d4,d5) dd <-

Sounds like some other program is holding the port. Have you stopped other NUT drivers for the device (e.g. via auto-resuscitating services) before starting this one? Does udev, ugen or similar facility have the configuration to hand off this device to NUT run-time user? (BTW, if you are now testing a custom build - was it configured to use same accounts as pre-packaged variant)? On Fri, Mar 24,

Hi List l am interested in developing price model. I have found a research paper related to price model of corn in US market where it has taken demand & supply forces into consideration. Following are the equation: Supply equation: St= a0+a1Pt-1+a2Rt-1+a3St-1+a5D1+a6D2+a7D3+U1 -(1) Where D1,D2,D3=Quarterly Dummy Variables(Since quarterly data are considered) Here, Supply

Hello I am working on pass which has dependency on multiple passes. Say D1,D2,D3 I used INITIALIZE_PASS_BEGIN INITIALIZE_PASS_DEPENDENCY(D1) INITIALIZE_PASS_DEPENDENCY(D2) INITIALIZE_PASS_DEPENDENCY(D3) INITIALIZE_PASS_END. While running it through opt tool it, I had to specify this D1,D2,D3 pass names to get this pass executed before my pass. Is there way, to let llvm pass manager to know

Hi R help, Hi R help, Which is the easiest (most elegant) way to force "aov" to treat numerical variables as categorical ? Sincerely, Andrea Bernasconi DG PROBLEM EXAMPLE I consider the latin squares example described at page 157 of the book: Statistics for Experimenters: Design, Innovation, and Discovery by George E. P. Box, J. Stuart Hunter, William G. Hunter. This example use

For certain reason, the content was not visible in the last mail, so posting it again. Dear Members, I have two different dataframes with a different number of rows. I need to apply a set of functions to each possible combination of rows with one row coming from 1st dataframe and other from 2nd dataframe. Though I am able to perform this task using for loops, I feel that there must be a more

Hi All I am trying to use the round function on some columns of a dataframe while leaving others unchanged. I wish to specify those columns to leave unchanged. My attempt is below - here, I would like the column d3 to be left but columns d1, d2 and d4 to be rounded to 0 decimal places. I would welcome any suggestions for a nicer way of doing this. d1= rnorm(10,10) d2= rnorm(10,6) d3=

I am new to R and have a problem that I haven't been able to find the answer to in the guides or online. I have multiple datasets, D1, D2, D3, D4, D5, D6, D7 and D8, and I would like to produce two plots side by side using mfrow.  The first plot should contain data from D1-D4, the second should contain D5-D8. I can plot these separately using the code, par(mfrow=c(1,1))

Hello, I have a longitudinal dataset where each individual has a different number of entries. Thus, it is of the following structure: x <- runif(12) id.var <- factor(c(rep("D1",4),rep("D2",2),rep("D3",3),rep("D4",3))) dat <- as.data.frame(x) dat$id.var <- id.var dat > dat x id.var 1 0.9611269 D1 2 0.6738606 D1 3

Hi Hongbin I am not quite familiar with AnalysisUsage, let me correct question a bit. I have read Writing Pass <http://llvm.org/docs/WritingAnLLVMPass.html#specifying-interactions-between-passes>, All examples that i see here are based on collecting information .i.e Analysis Passes. I wonder if this applies to Transformation passes also. e.g. void MyInliner::getAnalysisUsage(AnalysisUsage