计算字符串中完全匹配的单词数答案

【问题标题】：Count number of exactly matching words in a string计算字符串中完全匹配的单词数
【发布时间】：2020-12-18 00:45:22
【问题描述】：

我有一个带有 id 列的 tibble 和一个捕获人们输入的一些 text_entry 的列。
目标：将每个人的 text_entry 与 key 和计算完美键入的单词的数量。
例如，如果我的输入是：

df <- tribble(~id, ~text_entry,
              1, "It was a Saturday night in December.",
              2, " It was a Saturday night",
              3, "It wuz a Sturday nite in",
              4, "IT WAS A SATURDAY",
              5, "was a Saturday"); df

key <- "It was a Saturday night in December."

那么我想要以下内容：

df2 <- tribble(~id, ~text_entry, ~words_correct, 
               1, "It was a Saturday night in December.", 7, # whole string perfect
               2, " It was a Saturday night", 5,             # first 5 words perfect
               3, "It wuz a Sturday nite in", 3,             # misspelled "was", "Saturday" and "night"
               4, "IT WAS A SATURDAY", 0,                    # case-sensitive
               5, "was a Saturday", 3); df2                  # ok to start several words into the key

我对@987654327@/stringi 解决方案非常满意。 tidyverse 总是首选，但我迫切需要任何解决方案。

非常感谢您提前提供的帮助和见解！

【问题讨论】：

标签： r string tidyverse stringr stringi

【解决方案1】：

您可以提取非空格部分并将它们传递给str_detect()。

library(tidyverse)

df %>%
  mutate(words_correct = map_dbl(str_extract_all(text_entry, "[^\\s]+"),
                                 ~ sum(str_detect(key, .))))

# # A tibble: 5 x 3
#      id text_entry                             words_correct
#   <dbl> <chr>                                          <dbl>
# 1     1 "It was a Saturday night in December."             7
# 2     2 " It was a Saturday night"                         5
# 3     3 "It wuz a Sturday nite in"                         3
# 4     4 "IT WAS A SATURDAY"                                0
# 5     5 "was a Saturday"                                   3

【讨论】：

这是一个完美的解决方案。谢谢@Darren Tsai！

【解决方案2】：

一种方法是将字符串拆分为空格，然后用key 计算常用词。

library(tidyverse)

keywords <- strsplit(key, '\\s+')[[1]]

df %>%
  mutate(text = str_split(text_entry, '\\s+'), 
         words_correct = map_dbl(text, ~sum(.x %in% keywords)))

# A tibble: 5 x 3
#     id text_entry                             words_correct
#  <dbl> <chr>                                          <dbl>
#1     1 "It was a Saturday night in December."             7
#2     2 " It was a Saturday night"                         5
#3     3 "It wuz a Sturday nite in"                         3
#4     4 "IT WAS A SATURDAY"                                0
#5     5 "was a Saturday"                                   3

我们也可以在基础 R 中做到这一点：

df$words_correct <- sapply(strsplit(df$text_entry, '\\s+'), 
                           function(x) sum(x %in% keywords))

【讨论】：