similar to: Faster way for weighted matching?

I received some great ideas (see below) from a number of people to whom I am grateful. Here is the code I put together from many of their suggestions: lx <- length(x) lw <- length(w) z <- matrix(abs( rep( x , lw ) - rep( w, each = lx ) ), nrow=lw, ncol=lx, byrow=TRUE) s <- pmax( abs( w - min(x) ), abs( w - max(x) ) ) z <- (1 - (z/rep(s,length=lx*lw))^3)^3 sums <-

Hi I am trying to get the tricube weights from the loess outputs as I need to calculate an error function which requires the weight. So I have used the following example from the R: cars.lo <- loess(dist ~ speed, cars, span=0.5, degree=1, family="symmetric") Then i try to get the weights: cars.lo$weights [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Hello, I need help with the details of loess prediction algorithm. I would like to get it implemented as a part of a measurement system programmed in LabView. My job is provide a detailed description of the algorithm. This is a simple one-dimensional problem - smoothing an (x, y) data set. I found quite a detailed description of the fitting procedure in the "white book". It is also

Hi again, I've got few very good options from the list and since they were not posted to the list, I will provide a summary. Thank you very much to all who answered and I hope this summary will benefit others interested in solving similar problems like that. Yasir Kaheil re-wrote my original code in a more streamlined way: pr<-array(0,c(dim(x)[2],dim(x)[2])); for (i in 1:dim(x)[2]) for

predict.lm(..., type="terms") gives wrong standard errors. Below, I have provided what I believe are the necessary fixes. However, there are subtleties, and the code needs careful checking. Some of the looping is surely not necessary, but it is surely best to begin with the minimum necessary changes. My tests, including checks against S-PLUS, have extended to fitting spline curves. I

Hi, First, when I try the example Prostate with bound 0.44 (as in the manual), I got a different result: > l1c.P <- l1ce(lpsa ~ ., Prostate, bound=0.44) > l1c.P .... Coefficients: (Intercept) lcavol lweight age lbph svi 1.0435803 0.4740831 0.1953156 0.0000000 0.0000000 0.3758199 lcp gleason pgg45 0.0000000 0.0000000

>From e980153 Tue Apr 25 14:42:27 2000 To: r-help@stat.math.ethz.ch Subject: Wrong SEs in predict.lm(..., type="terms") For what it is worth, I am using RW-1.0.0 under Windows 98. I submitted this earlier to r-help. There is one change below to my proposed corrected code: predict.lm(..., type="terms") gives wrong standard errors. Below, I have provided what I believe are

I've used barplot(), including the anmes.arg parameter, on data frames successfully, but I'm even newer to using zoo than I am to R. :-} I am working on a functon that accepts a data frame ("df") as its primary argument, extracts information from it to create a zoo, then generates a plot based on that. The data frame has a column called "time" which is a standard

I am trying to multiply a 3D array of 4x4x4 by the 4 3D arrays of a 4D array with dimensions 4x4x4x4 (the last dimension being the one that I want to split by). (4x4x4 array) > hiaAry , , a1 i1 i2 i3 i4 h1 9.5936098 6.001040 0.08772 0.3138600 h2 1.2003500 1.454570 2.79248 0.0000000 h3 0.1346500 0.201220 0.39256 0.5464000 h4 0.0109000 0.012270 0.16417 0.2766900 ,

I think one needs an LU decomposition rather than QR. However, I couldn't find anything off the shelf to do an LU, other than learning that determinant() now uses LU instead of QR or SVD, so the code to do it must be in there for those that want it. You'll probably need to divide rows of U by the first entry if you insist on the unique reduced REF. However, I can't see any reason

Does anyone have any "R" code for computing the entropy of a simple first or second order Markov chain, given a transition matrix something like the following (or the symbol vector from which it is computed)? AGRe ARIe CSRe DIRe DSCe eos HRMe SPTe TOBe AGRe 0.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000

Hi, I have run mars() with the default degree=1, and have constructed splines which are similar to what I would get from Salford System's MARS. I then use these splines as potential predictors, along with rpart( ) terminal nodes and other potential predictors, in logistic regression in SAS. The output of mars() and the splines I created are shown below (pardon the lengthy output): >

Hello all! I have a problem with ggplot2 library. I want to do an heat map and the y variables are the year months. If I use the following code, he y values are in alphabetical order, but I want it in month order. The code is: library(reshape) library(ggplot2) library(scales) p <- ggplot(data.m, aes(variable, Month)) + geom_tile(aes(fill = value),

Hi, I've this dataframes: > data1 1 2 3 4 5 6 5 0.4963017 0 0 0.2481509 1.9852069 0.4963017 10 0.0000000 0 0 0.0000000 0.6317266 0.0000000 15 0.0000000 0 0 0.0000000 0.0000000 0.0000000 20 0.0000000 0 0 3.3955301 0.0000000 0.0000000 25 0.0000000 0 0 0.0000000 0.0000000 0.0000000 30 0.0000000 0 0 0.0000000 0.0000000

Dear useRs, I'm having trouble with what is likely a very simple issue with a simple graph. I have data in the format pasted below - total from which this is a subset is > 3 million rows. I'm plotting "Window" on the x axis, and either of the other two columns on the y axis. In both the second and third columns, the values are in a range from 0 to 1. I want to be able to plot

I have the following matrix: > dat [,1] [,2] [,3] [,4] foo 0.7574657 0.2104075 0.02922241 0.002705617 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 foo 0.0000000 0.0000000 0.00000000 0.000000000 and given this:

eigen() seems to work for symmetric matrices only. This is out of sync with the help file. > trpr.37 0 1 2 3 4 0 1.00000000 0.0000000 0.0000000 0.0000000 0.0000000 1 0.44444444 0.5555556 0.0000000 0.0000000 0.0000000 2 0.02439024 0.2439024 0.7317073 0.0000000 0.0000000 3 0.00000000 0.0000000 0.2307692 0.7692308 0.0000000 4 0.00000000 0.0000000

Dear R helpers: I have tried many times to find some way to read a list into R. But I faid. Here is an example: I have a file 'List.txt' which includes data as follows: [[1]] [1] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.5 0.0 0.0 0.0 [19] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 [[2]] [1] 0.0000000 0.5000000 0.0000000 0.0000000

Hi, is there any direct R function to write an diagonal matrix in an opposite way? for example I want to get like:   > diag(rnorm(5))[,5:1]            [,1]       [,2]       [,3]      [,4]       [,5] [1,]  0.0000000  0.0000000  0.0000000  0.000000 -0.1504687 [2,]  0.0000000  0.0000000  0.0000000 -2.139669  0.0000000 [3,]  0.0000000  0.0000000 -0.2102133  0.000000  0.0000000 [4,]  0.0000000

I am doing analysis on hourly precipitation on a file that is disorganized. However, I managed to clean it up and store it in a dataframe (called CA1) which takes the form as followed: Station_ID Guage_Type Lat Long Date Time_Zone Time_Frame H0 H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 H11 H12 H13 H14 H15 H16 H17 H18 H19 H20 H21 H22 H23 1