Question

如果我有一个字符串：

moon <- "The cow jumped over the moon with a silver plate in its mouth"

有没有办法可以提取"moon"附近的单词。邻居可能是“月亮”周围的2或3个单词。

所以，如果我的

"The cow jumped over the moon with a silver plate in its mouth"

我希望我的输出只是：

"jumped over the moon with a silver"

我知道如果我想通过字符提取我可以使用str_locate，但不知道如何使用“单词”来完成它。这可以在R？

中完成

谢谢＆amp;问候，西马克

Answer 1

使用strsplit：

x <- strsplit(str, " ")[[1]]
i <- which(x == "moon")
paste(x[seq(max(1, (i-2)), min((i+2), length(x)))], collapse= " ")

Answer 2

我是这样做的：

keyword <- "moon"
lookaround <- 2
pattern <- paste0("([[:alpha:]]+ ){0,", lookaround, "}", keyword, 
                "( [[:alpha:]]+){0,", lookaround, "}")

regmatches(str, regexpr(pattern, str))[[1]]
# [1] "The cow jumped over"

想法：搜索任意字符后跟空格至少出现0次，最多出现“环视”（此处为2次），然后是“关键字”（此处为“月亮”） “），然后是空间和一堆字符模式在0和”环视“时间之间重复。 regexpr函数给出了这种模式的开始和结束。 regmatches包装此函数然后从此开始/停止位置获取子字符串。

注意：如果您要搜索同一模式的多次出现，regexpr可以替换为gregexpr。

以下是比较Hong和这个答案的大数据基准：

str <- "The cow jumped over the moon with a silver plate in its mouth" 
ll <- rep(str, 1e5)
hong <- function(str) {
    str <- strsplit(str, " ")
    sapply(str, function(y) {
        i <- which(y=="moon")
        paste(y[seq(max(1, (i-2)), min((i+2), length(y)))], collapse= " ")
    })
}

arun <- function(str) {
    keyword <- "moon"
    lookaround <- 2
    pattern <- paste0("([[:alpha:]]+ ){0,", lookaround, "}", keyword, 
                    "( [[:alpha:]]+){0,", lookaround, "}")

    regmatches(str, regexpr(pattern, str))
}

require(microbenchmark)
microbenchmark(t1 <- hong(ll), t2 <- arun(ll), times=10)
# Unit: seconds
#            expr      min       lq   median       uq      max neval
#  t1 <- hong(ll) 6.172986 6.384981 6.478317 6.654690 7.193329    10
#  t2 <- arun(ll) 1.175950 1.192455 1.200674 1.227279 1.326755    10

identical(t1, t2) # [1] TRUE

Answer 3

这是使用tm套餐的方法（当你所有人都拿到锤子时......）

moon <- "The cow jumped over the moon with a silver plate in its mouth"

require(tm)
my.corpus <- Corpus(VectorSource(moon))
# Tokenizer for n-grams and passed on to the term-document matrix constructor
library(RWeka)
neighborhood  <- 3 # how many words either side of word of interest
neighborhood1 <- 2 + neighborhood  * 2 
ngramTokenizer <- function(x) NGramTokenizer(x, Weka_control(min = neighborhood1, max = neighborhood1))
dtm <- TermDocumentMatrix(my.corpus, control = list(tokenize = ngramTokenizer))
inspect(dtm)

#  find ngrams that have the word of interest in them
word <- 'moon'
subset_ngrams <- dtm$dimnames$Terms[grep(word, dtm$dimnames$Terms)]

# keep only ngrams with the word of interest in the middle. This
# removes duplicates and lets us see what's on either side
# of the word of interest

subset_ngrams <- subset_ngrams[sapply(subset_ngrams, function(i) {
  tmp <- unlist(strsplit(i, split=" "))
  tmp <- tmp[length(tmp) - span]
  tmp} == word)]

# inspect output
subset_ngrams
[1] "jumped over the moon with a silver plate"

substring +获取关键字周围的单词

3 个答案:

以下是比较Hong和这个答案的大数据基准：