Question

我想找到更好的方法来找到我正在比较的两个字符串的更大的nchar。

假设我在 sentenceMatch data.frame中有字符串，我需要创建一个max（nchar（string1），nchar（string2））的矩阵，但是没有for循环这是非常慢的方法。

sentenceMatch <- data.frame(Sentence=c("hello how are you",
                                   "hello how are you friend",
                                   "im fine and how about you",
                                   "good thanks",
                                   "great to hear that"))

sentenceMatch$Sentence <- as.character(sentenceMatch$Sentence)

overallMatrix_nchar <- matrix(, nrow = dim(sentenceMatch)[1], ncol = dim(sentenceMatch)[1])

for (k in 1:dim(sentenceMatch)[1]) {
  for (l in 1:dim(sentenceMatch)[1]) {
    overallMatrix_nchar[k, l] <- max(nchar(sentenceMatch[k, ]), nchar(sentenceMatch[l, ]))
  }
}

有没有更好的解决方案如何才能加快计算速度？非常感谢您对前进的任何帮助。

Answer 1

使用outer：

nc <- nchar(sentenceMatch[[1]])
outer(nc, nc, pmax)

，并提供：

     [,1] [,2] [,3] [,4] [,5]
[1,]   17   24   25   17   18
[2,]   24   24   25   24   24
[3,]   25   25   25   25   25
[4,]   17   24   25   11   18
[5,]   18   24   25   18   18

Answer 2

sentences <- c("hello how are you",
               "hello how are you friend",
               "im fine and how about you",
               "good thanks",
               "great to hear that")
sn <- nchar(sentences)
n <- length(sn)
M1 <- matrix(sn, n, n)
M2 <- t(M1)
(M1 + M2 + abs(M1 - M2)) / 2
#      [,1] [,2] [,3] [,4] [,5]
# [1,]   17   24   25   17   18
# [2,]   24   24   25   24   24
# [3,]   25   25   25   25   25
# [4,]   17   24   25   11   18
# [5,]   18   24   25   18   18

我使用的事实是max（x，y）=（x + y + abs（x-y））/ 2.性能非常相似：

set.seed(1)
sentences <- replicate(paste0(rep("a", rpois(1, 3000)), collapse = ""), n = 1000)

f1 <- function(sentences) {
  sn <- nchar(sentences)
  n <- length(sn)
  M1 <- matrix(sn, n, n)
  M2 <- t(M1)
  (M1 + M2 + abs(M1 - M2)) / 2
}

f2 <- function(sentences) {
  nc <- nchar(sentences)
  outer(nc, nc, pmax)
}

library(microbenchmark)
microbenchmark(f1(sentences), f2(sentences))
# Unit: milliseconds
#           expr      min       lq    mean   median       uq      max neval cld
#  f1(sentences) 33.39924 37.66673 57.9912 42.45684 82.01905 122.5075   100   b
#  f2(sentences) 31.59887 34.97866 50.5065 37.82217 77.82042 103.6342   100  a

矩阵中两个字符串的最大nchar

2 个答案: