从向量中删除作为另一个字符串的子字符串的元素。

Question

有更好的方法吗？我想从这个向量中删除所有作为其他元素的子字符串存在的字符串。

words = c("please can you", 
  "please can", 
  "can you", 
  "how did you", 
  "did you",
  "have you")
> words
[1] "please can you" "please can"     "can you"        "how did you"    "did you"        "have you"

library(data.table)
library(stringr)
dt = setDT(expand.grid(word1 = words, word2 = words, stringsAsFactors = FALSE))
dt[, found := str_detect(word1, word2)]
setdiff(words, dt[found == TRUE & word1 != word2, word2])
[1] "please can you" "how did you"    "have you" 

这个方法可行，但看起来有些冗长，我想知道更加优雅的解决方案。

- Akhil Nair

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CJ 是比 data.table 更快的 expand.grid 函数。 - Rorschach

只是想为那些跟进的人提供更多信息。CJ速度快得多。我拿了12431行，平均每行15.69个单词，总共195065个单词，并通过system.time(dt <- setDT(expand.grid(word1 = words, word2 = words, stringsAsFactors = FALSE)))运行它，在user system elapsed 8.414 3.387 13.854内完成，而在user system elapsed 0.932 0.365 1.320内完成system.time(dt1 <- CJ(words,words,unique = TRUE))。数量级的差异。 - Shawn Mehan

太棒了，感谢您进行基准测试。 - Akhil Nair

Answer 1

在 words 中搜索每个组件，保留只出现一次的组件：

words[colSums(sapply(words, grepl, words, fixed = TRUE)) == 1]

提供：

[1] "please can you" "how did you"    "have you"   

- G. Grothendieck

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太棒了 - 非常感谢！ - Akhil Nair