如何从单词列表中查找 DF 中的匹配单词并在新列中返回匹配的单词 [重复]

Posted 2023-02-19

【中文标题】如何从单词列表中查找 DF 中的匹配单词并在新列中返回匹配的单词 [重复]

【英文标题】：How to find matching words in a DF from list of words and returning the matched words in new column [duplicate]

【发布时间】：2018-10-13 12:25:44

【问题描述】：

我有一个 2 列的 DF，我有一个单词列表。

list_of_words <- c("tiger","elephant","rabbit", "hen", "dog", "Lion", "camel", "horse")

df <- tibble::tibble(page=c(12,6,9,18,2,15,81,65),
               text=c("I have two pets: a dog and a hen",
                      "lion and Tiger are dangerous animals",
                      "I have tried to ride a horse",
                      "Why elephants are so big in size",
                      "dogs are very loyal pets",
                      "I saw a tiger in the zoo",
                      "the lion was eating a buffalo",
                      "parrot and crow are very clever birds"))

animals <- c("dog,hen", "lion,tiger", "horse", FALSE, "dog", "tiger", "lion", FALSE)

cbind(df, animals)
#>   page                                  text    animals
#> 1   12      I have two pets: a dog and a hen    dog,hen
#> 2    6  lion and Tiger are dangerous animals lion,tiger
#> 3    9          I have tried to ride a horse      horse
#> 4   18      Why elephants are so big in size      FALSE
#> 5    2              dogs are very loyal pets        dog
#> 6   15              I saw a tiger in the zoo      tiger
#> 7   81         the lion was eating a buffalo       lion
#> 8   65 parrot and crow are very clever birds      FALSE

我需要找出列表中的任何单词是否存在于 DF 的一列中。如果是，则将单词/单词返回到 DF 中的新列。这是单词列表 ->(tiger,elephant,rabbit, hen, dog, Lion, camel, horse)。 This is how my DF Looks like I want something like this

【问题讨论】：

请将您的示例数据添加为代码，而不是图像。

是的，部分正确。但我想从列表中找出哪些匹配的单词出现在 DF 中，并将这些单词返回到同一 DF 的新列中。

这 4 个步骤将起作用：首先在您的列上使用 strsplit df$text 和 " " 作为拆分参数，就像这样 test <- strsplit(df$text, " ")。然后使用grepl 和tolower 得到与你的向量匹配的词：test2 <- lapply(test, function(x) x[grepl(tolower(paste(words, collapse = "|")), tolower(x))])。现在将它们放在每一行中，并使用df$animals <- unlist(lapply(test2, paste, collapse = ", ")) 取消列出它们，然后使用df$animals[nchar(df$animals) == 0] <- FALSE 将所有空字符设置为FALSE。

@LAP 不起作用

【参考方案1】：

library(dplyr)

df %>% 
  rowwise() %>%
  mutate(animals = paste(list_of_words[unlist(
    lapply(list_of_words, function(x) grepl(x, text, ignore.case = T)))], collapse=",")) %>%
  data.frame()

输出为：

  page                                  text    animals
1   12                       pets: dog & hen    hen,dog
2    6 Lions and tigers are dangerous animal tiger,Lion
3    9          I have tried to ride a horse      horse
4   65   parrot & crow are very clever birds           

样本数据：

df <- structure(list(page = c(12, 6, 9, 65), text = structure(c(4L, 
2L, 1L, 3L), .Label = c("I have tried to ride a horse", "Lions and tigers are dangerous animal", 
"parrot & crow are very clever birds", "pets: dog & hen"), class = "factor")), .Names = c("page", 
"text"), row.names = c(NA, -4L), class = "data.frame")

list_of_words <- c("tiger", "elephant", "rabbit", "hen", "dog", "Lion", "camel", "horse")

**另一种方法：**

library(data.table)
setDT(df)[, animals := paste(list_of_words[unlist(lapply(list_of_words, function(x) grepl(x, text, ignore.case = T)))], collapse = ","), by = 1:nrow(df)]

#> df
#   page                                  text    animals
#1:   12                       pets: dog & hen    hen,dog
#2:    6 Lions and tigers are dangerous animal tiger,Lion
#3:    9          I have tried to ride a horse      horse
#4:   65   parrot & crow are very clever birds