在 R 中，如何找到一个单词在字符串中的位置？

Question

如何找到数据框单元格中特定单词的第一个位置，并将输出保存在同一数据框的新列中？

理想情况下，我想要字典中每个单词的第一个匹配项。

df <- data.frame(text = c("omg coke is so awsme","i always preferred pepsi", "mozart is so overrated by yeah fanta makes my day, always"))

dict <- c("coke", "pepsi", "fanta")

位置可以是字典单词之前的 N 个字符或单词。

我一直在研究找到的代码 here，但我无法让它工作。

例如，这段代码完成了这项工作，但只针对一个词和一个字符串（而不是 df 和字典）

my_string = "omg coke is so awsme"
unlist(gregexpr("coke", my_string))[1]

期望的输出：

                                                       text  location
1                                      omg coke is so awsme         2
2                                  i always preferred pepsi         4
3 mozart is so overrated by yeah fanta makes my day, always         7

就像我说的，位置也可以是字符串而不是单词，如果这样更容易的话。

Answer 1

就运行

c(regexpr(paste0(dict,collapse = '|'), df$text))

[1]  5 20 32

编辑：

如果你想要单词的位置：

library(tidyverse)
pat <-  sprintf(".*(%s)", paste0(dict,collapse = '|'))
df %>%
  mutate(loc = str_count(str_extract(text,pat), "\w+"))

                                                       text loc
1                                      omg coke is so awsme   2
2                                  i always preferred pepsi   4
3 mozart is so overrated by yeah fanta makes my day, always   7

Answer 2

这是一个简单的 for 循环：

for(i in dict) {
  df[[i]] = stringi::stri_locate_first_fixed(df$text, i)[, 1]
}
df
#                                                        text coke pepsi fanta
# 1                                      omg coke is so awsme    5    NA    NA
# 2                                  i always preferred pepsi   NA    20    NA
# 3 mozart is so overrated by yeah fanta makes my day, always   NA    NA    32

或使用 regexpr（基础的一部分，因此没有依赖性）：

for(i in dict) {
  df[[i]] = regexpr(i, df$text, fixed = TRUE)
}
df
#                                                        text coke pepsi fanta
# 1                                      omg coke is so awsme    5    -1    -1
# 2                                  i always preferred pepsi   -1    20    -1
# 3 mozart is so overrated by yeah fanta makes my day, always   -1    -1    32

下面是单词编号的解决方案，但我建议在使用之前删除所有标点符号：

df$words = strsplit(df$text, split = " ")
for(i in dict) {
  df[[i]] = sapply(df$words, \(x) match(i, unlist(x)))
}
df
#                                                        text coke pepsi fanta
# 1                                      omg coke is so awsme    2    NA    NA
# 2                                  i always preferred pepsi   NA     4    NA
# 3 mozart is so overrated by yeah fanta makes my day, always   NA    NA     7
#                                                                 words
# 1                                            omg, coke, is, so, awsme
# 2                                         i, always, preferred, pepsi
# 3 mozart, is, so, overrated, by, yeah, fanta, makes, my, day,, always

Answer 3

require(quanteda)
#> Loading required package: quanteda
#> Package version: 3.2.1
#> Unicode version: 13.0
#> ICU version: 69.1
#> Parallel computing: 8 of 8 threads used.
#> See https://quanteda.io for tutorials and examples.

df <- data.frame(text = c("omg coke is so awsme",
                          "i always preferred pepsi", 
                          "mozart is so overrated by yeah fanta makes my day, always"))
dict <- c("coke", "pepsi", "fanta")

corp <- corpus(df)
toks <- tokens(corp)
index(toks, dict)
#>   docname from to pattern
#> 1   text1    2  2    coke
#> 2   text2    4  4   pepsi
#> 3   text3    7  7   fanta

^{由 reprex package (v2.0.1)}

创建于 2022-05-27

在 R 中，如何找到一个单词在字符串中的位置？

In R, how to find the location of a word in a string?

r

text

nlp

text-mining

quanteda