r

问题 The following is my code. I am trying get the list of all the files (~20000) that end with .idat and read each file using the function illuminaio::readIDAT . library(illuminaio) library(parallel) library(data.table) # number of cores to use ncores = 8 # this gets all the files with .idat extension ~20000 files files <- list.files(path = './', pattern = "*.idat", full.names = TRUE) # function to read the idat file and create a data.table of filename, and two more columns # write out as csv

问题 The following is my code. I am trying get the list of all the files (~20000) that end with .idat and read each file using the function illuminaio::readIDAT . library(illuminaio) library(parallel) library(data.table) # number of cores to use ncores = 8 # this gets all the files with .idat extension ~20000 files files <- list.files(path = './', pattern = "*.idat", full.names = TRUE) # function to read the idat file and create a data.table of filename, and two more columns # write out as csv

问题 The following is my code. I am trying get the list of all the files (~20000) that end with .idat and read each file using the function illuminaio::readIDAT . library(illuminaio) library(parallel) library(data.table) # number of cores to use ncores = 8 # this gets all the files with .idat extension ~20000 files files <- list.files(path = './', pattern = "*.idat", full.names = TRUE) # function to read the idat file and create a data.table of filename, and two more columns # write out as csv

问题 So I'm using R to perform a DCA (detrended correspondence analysis) through Vegan package and everytime I plot my results I get a grid in the middle of the plot. I want to get rid of it. Here is my code: dca <- decorana(dados) plot(dca, type = "n", ann = FALSE, origin = TRUE) text(dca, display = "sites") points(dca, display = "species", pch = 21, col="red", bg = "red", cex=0.5) mtext(side = 3, text = "Análise de Correspondência Retificada (DCA)", font=2, line = 2.5, cex=1.8, font.lab=6, cex

问题 I want to write an R function arg2str that returns the names (that is a vector of strings) of the symbols that are fed as arguments. For the most simple case, I only have one input symbol: library ("rlang") arg2str.v0 <- function (arg) rlang::quo_name (enquo (arg)) arg2str.v0 (a) ## [1] "a" If I have multiple symbols, I can use the three-dots construct: arg2str.v1 <- function (...) sapply (enquos (...), rlang::quo_name) arg2str.v1 (a, b, c) ## ## "a" "b" "c" (Subsidiary question: why is the

问题 I am trying to turn the @ operator in R into a generic function for the S3 system. Based on the chapter in Writing R extensions: adding new generic I tried implementing the generic for @ like so: `@` <- function(object, name) UseMethod("@") `@.default` <- function(object, name) base::`@`(object, name) However this doesn't seem to work as it breaks the @ for the S4 methods. I am using Matrix package as an example of S4 instance: Matrix::Matrix(1:4, nrow=2, ncol=2)@Dim Error in @.default

问题 I'm playing around with R. I try to visualize the distribution of 1000 dice throws with the following R script: cases <- 1000 min <- 1 max <- 6 x <- as.integer(runif(cases,min,max+1)) mx <- mean(x) sd <- sd(x) hist( x, xlim=c(min - abs(mx/2),max + abs(mx/2)), main=paste(cases,"Samples"), freq = FALSE, breaks=seq(min,max,1) ) curve(dnorm(x, mx, sd), add = TRUE, col="blue", lwd = 2) abline(v = mx, col = "red", lwd = 2) legend("bottomleft", legend=c(paste('Mean (', mx, ')')), col=c('red'), lwd=2

问题 I'm working in a very unbalanced classification problem, and I'm using AUPRC as metric in caret. I'm getting very differents results for the test set in AUPRC from caret and in AUPRC from package PRROC. In order to make it easy, the reproducible example uses PimaIndiansDiabetes dataset from package mlbench: rm(list=ls()) library(caret) library(mlbench) library(PRROC) #load data, renaming it to 'datos' data(PimaIndiansDiabetes) datos=PimaIndiansDiabetes[,1:9] # training and test set.seed(998)

问题 I've been researching this for a while now but haven't come across any solution that fit my needs or that I can transform sufficiently to work in my case: I have a large car sharing data set for multiple cities in which I have the charging demand per location (e.g. row = carID, 55.63405, 12.58818, charging demand). I now would like to split the area over the city (example above is Copenhagen) up into a hexagonal grid and tag every parking location with an ID (e.g. row = carID, 55.63405, 12

问题 I'm working in a very unbalanced classification problem, and I'm using AUPRC as metric in caret. I'm getting very differents results for the test set in AUPRC from caret and in AUPRC from package PRROC. In order to make it easy, the reproducible example uses PimaIndiansDiabetes dataset from package mlbench: rm(list=ls()) library(caret) library(mlbench) library(PRROC) #load data, renaming it to 'datos' data(PimaIndiansDiabetes) datos=PimaIndiansDiabetes[,1:9] # training and test set.seed(998)