R 如何在矩阵中的行上复制相同的函数

我正在尝试编写一个循环，确定哪个单元格的值最大，并选择该单元格作为结果，使用高-中或低字符串。这是试用的数据

data <- matrix(c(0.3000003,0.3299896,0.3700101,
                 0.3299896,0.3700101,0.3000003,
                 0.3700101,0.3000003,0.3299896,
                 0.3000003,0.3299896,0.3700101,
                 0.3299896,0.3700101,0.3000003,
                 0.3700101,0.3000003,0.3299896),6,3)
colnames(data) <- c("Low","Medium","High")
rownames(data) <- paste("case",1:6)

> data
             Low    Medium      High
case 1 0.3000003 0.3700101 0.3299896
case 2 0.3299896 0.3000003 0.3700101
case 3 0.3700101 0.3299896 0.3000003
case 4 0.3299896 0.3000003 0.3700101
case 5 0.3700101 0.3299896 0.3000003
case 6 0.3000003 0.3700101 0.3299896

---

data以下是您可能更喜欢的矢量化解决方案：

colnames(data)[apply(data, 1, which.max)]
# [1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"

这是您尝试的简明版本：
将函数
which.max
应用于
数据的每一行（维度
1
），并获得相应的列名

就您的尝试而言，这里有一个更正版本：

assign.levels <- function(data) {
  scored.thetas.1 <- rep(NA, nrow(data))
  for (i in 1:nrow(data))
    scored.thetas.1[i] <- names(which.max(data[i, ]))
  scored.thetas.1
}
assign.levels(data)
# [1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"

assign.levels以下是您可能更喜欢的矢量化解决方案：

colnames(data)[apply(data, 1, which.max)]
# [1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"

这是您尝试的简明版本：
将函数
which.max
应用于
数据的每一行（维度
1
），并获得相应的列名

就您的尝试而言，这里有一个更正版本：

assign.levels <- function(data) {
  scored.thetas.1 <- rep(NA, nrow(data))
  for (i in 1:nrow(data))
    scored.thetas.1[i] <- names(which.max(data[i, ]))
  scored.thetas.1
}
assign.levels(data)
# [1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"

assign.levels这应该很快

colnames(data)[max.col(data)]
#[1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"


这里有一个小基准

n <- 1e6
set.seed(1)
data <- matrix(runif(n * 3), ncol = 3)
colnames(data) <- c("Low","Medium","High")

library(microbenchmark)

benchmark <- microbenchmark(
  OP = assign.levels(data), # as defined in Julius's answer
  Julius = colnames(data)[apply(data, 1, which.max)],
  markus = colnames(data)[max.col(data)], times = 20
)

autoplot(benchmark)

n这应该很快

colnames(data)[max.col(data)]
#[1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"


这里有一个小基准

n <- 1e6
set.seed(1)
data <- matrix(runif(n * 3), ncol = 3)
colnames(data) <- c("Low","Medium","High")

library(microbenchmark)

benchmark <- microbenchmark(
  OP = assign.levels(data), # as defined in Julius's answer
  Julius = colnames(data)[apply(data, 1, which.max)],
  markus = colnames(data)[max.col(data)], times = 20
)

autoplot(benchmark)

n我认为它实际上返回了最后一行的结果。我认为它实际上返回了最后一行的结果。感谢您对我的代码所做的努力和评论。非常有帮助！感谢您的努力和对我的代码的评论。非常有帮助！感谢您的快速解决方案！真的很感谢。这肯定会帮助我进行完整的模拟：）谢谢你的快速解决方案！非常感谢。这肯定会对我的完整模拟有所帮助：）