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Accepted
mr.T
Asked: 2023-12-02 21:02:27 +0800 CST

有效地计算表达式序列

  • 772

我有一些数据

set.seed(1)
n <- 100
df <- data.frame(
  x = sample(1:30, n, replace = T),
  y = sample(1:30, n, replace = T),
  z = sample(1:30, n, replace = T)
)

和带有表达式的向量，它们可能不同。

rules <- c("df$x[i] < df$y[i-2] - df$x[i]", 
           "df$y[i] >= mean(df$x)",
           "df$y[i] == 20",
           "df$z[i-30] >= df$x[5]",
           "df$y[i-5] == 16",
           "df$x[10] > sd(as.matrix(df[(i-5):i,]))")

接下来，我有一个函数可以顺序搜索第一个表达式的触发，然后是第二个表达式，依此类推

seq_rules <- function(df, rules, show=T){
  
ln <- length(rules)
res <- matrix(0,nrow = ln, ncol = 2, dimnames = list(NULL, c("row","res")))
n <- 1

  for(i in 30:nrow(df)){
    if(eval(str2expression(rules[n]))){
      res[n,"row"] <- i
      res[n,"res"] <- 1
      if(show) print( cbind.data.frame(df[i,], rule=rules[n], row=i))
      n <- n+1
    }
    if(n>ln) break
  }
res
}

我想加快我的代码速度。您将如何编写这段代码以使其尽可能快？我也希望您的解决方案在不同的情况下与我的解决方案相同seeds

=========================================

如果规则表示为已评估的函数

Frules <- lapply(rules,\(x) eval(str2expression(paste("function(i) {", x ,"}"))))

然后由于没有eval(str2expression..))循环，我可以获得一点速度

新功能

Fseq_rules <- function(df, rules){
  ln <- length(rules)
  res <- matrix(0,nrow = ln, ncol = 2, dimnames = list(NULL, c("row","res")))
  n <- 1
for(i in 30:nrow(df)){
    if(rules[[n]](i)){
        res[n,"row"] <- i
        res[n,"res"] <- 1
        n <- n+1
      }
    if(n>ln) break
  }
  res
}

microbenchmark::microbenchmark(Fseq_rules(df, Frules),
                                seq_rules(df, rules,show = F),times = 100)
Unit: milliseconds
                           expr      min       lq     mean   median       uq      max neval
          Fseq_rules(df, Frules) 1.083315 1.118951 1.283135 1.156011 1.247808 5.601309   100
 seq_rules(df, rules, show = F) 2.495045 2.545790 2.779712 2.607938 2.861662 6.243315   100

1 个回答

  1. Best Answer

    比原来的快不了多少：

    rules2 <- c(
  "x[i] < y[i-2] - x[i]", 
  "y[i] >= mean(x)",
  "y[i] == 20",
  "z[i-30] >= x[5]",
  "y[i-5] == 16",
  "x[10] > sd(as.matrix(df[(i-5):i,]))"
)

seq_rules2 <- function(df, rules) {
  rules <- sapply(rules, str2expression)
  M <- length(rules)
  res <- matrix(0L, nrow = M, ncol = 2L, dimnames = list(NULL, c("row", "res")))
  j <- 1L
  
  for (i in 30:nrow(df)) {
    if (eval(rules[[j]], envir = df)) {
      res[j, ] <- c(i, 1L)
      j <- j + 1L
    }
    if(j > M) break
  }
  res
}

bench::mark(seq_rules(df, rules), seq_rules2(df, rules2))

    在此输入图像描述

    df如果用矩阵替换，你会获得很多速度。并相应地更改规则：

    M <- as.matrix(df)

rules_matrix <- c(
  "df[i, 'x'] < y[i-2] - x[i]", 
  "df[i, 'y'] >= mean(df[, 'x'])",
  "df[i, 'y'] == 20",
  "df[i-30, 'z'] >= df[5, 'x']",
  "df[i-5, 'y'] == 16",
  "df[10, 'x'] > sd(df[(i-5):i, ])"
)

seq_rules_matrix <- function(df, rules) {
  rules <- sapply(rules, str2expression)
  M <- length(rules)
  res <- matrix(0L, nrow = M, ncol = 2L, dimnames = list(NULL, c("row", "res")))
  j <- 1L
  
  for (i in 30:nrow(df)) {
    if (eval(rules[[j]])) {
      res[j, ] <- c(i, 1L)
      j <- j + 1L
    }
    if(j > M) break
  }
  res
}

bench::mark(
  mat = seq_rules_matrix(M, rules_matrix),
  df = seq_rules2(df, rules2)
)

    在此输入图像描述

    • 2

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