Question

我正在尝试创建一个评分函数（称为evalFunc）。为了获得分数，我试图计算生成模型的R平方值。我如何知道如何将值传递给＆＃39; evalFunc＆＃39;来自rbga.bin函数？

library(genalg)
library(ggplot2)
set.seed(1)
df_factored<-data.frame(colA=runif(30),colB=runif(30),colC=runif(30),colD=runif(30))
dataset <- colnames(df_factored)[2:length(df_factored)]
chromosome = sample(c(0,1),length(dataset),replace=T)
#dataset[chromosome == 1]

evalFunc <- function(x) {
  #My end goal is to have the values passed into evalFunc be evaluated as the IV's in a linear model
  res<-summary(lm(as.numeric(colA)~x,
                  data=df_factored))$r.squared 
  return(res)
}

iter = 10
GAmodel <- rbga.bin(size = 2, popSize = 200, iters = iter, mutationChance = 0.01, elitism = T, evalFunc = evalFunc)
cat(summary(GAmodel))

Answer 1

您可以通过键入$sql = "SELECT * FROM uk_postcodes"; $result = mysqli_query($connection, $sql) or die("Error " . mysqli_error($connection)); $dname_list = array(); while($row = mysqli_fetch_array($result)) { // [ { label: "Choice1", value: "value1" }, { label: "Choice2", value: "value2" } ] $dname_list[] = "{label:".$row['postcode'].","."value:".$row['town']."}"; } header('Content-Type: application/json'); echo json_encode($dname_list);来查看来源，但最好是运行rbga.bin，然后在下次调用该功能时，它允许您单步执行该功能。在这种情况下，第一次进入函数就在这一行（大约是函数的第82行）：

debug(rbga.bin)

此时，evalVals[object] = evalFunc(population[object, ])是由0和1组成的200x2矩阵：

population

而head(population) # [1,] 0 1 # [2,] 0 1 # [3,] 0 1 # [4,] 1 0 # [5,] 0 1 # [6,] 1 0是数字1，因此object是向量population[object,]。

完成c(0,1)后，您可debug，并且每次拨打undebug(rbga.bin)时都不会进入调试模式。

查找提供给未知矢量化函数的参数

1 个答案: