从R中的文本中提取年龄[关闭]

Question

我有一个 .csv 文件，其中有一列包含从网络上抓取的书籍描述，我将其导入 R 以进行进一步分析。我的目标是从R中的这个专栏中提取主角的年龄，所以我想象的是：

1. 使用正则表达式匹配“age”和“-year-old”等字符串
2. 将包含这些字符串的句子复制到一个新列中（这样我就可以确保该句子不是，例如“In the middle age 50 people living in xy”
3. 从该列中提取数字（如果可能，还包括一些数字单词）到一个新列中。

结果表（或者可能是 data.frame）希望看起来像这样

|Description             |Sentence           |Age
|YY is a novel by Mr. X  |The 12-year-old boy| 12
|about a boy. The 12-year|is named Dave.     |
|-old boy is named Dave..|                   |

如果你能帮上忙，那就太好了，因为我的 R 技能仍然非常有限，而且我还没有找到解决这个问题的方法！

Answer 1

另一个选项，如果字符串包含除年龄之外的其他数字/描述，但您只需要年龄。

library(stringr)
description <- "YY is a novel by Mr. X about a boy. The boy is 5 feet tall. The 12-year-old boy is named Dave. Dave is happy. Dave lives at 42 Washington street."
sentence <- str_split(description, "\\.")[[1]][which(grepl("-year-old", unlist(str_split(description, "\\."))))]
> sentence 
[1] " The 12-year-old boy is named Dave"

age <- as.numeric(str_extract(description, "\\d+(?=-year-old)"))
> age
[1] 12

这里我们使用字符串“-year-old”告诉我们要提取哪个句子，然后我们提取该字符串后面的年龄。

Answer 2

你可以试试下面的

library(stringr)

description <- "YY is a novel by Mr. X about a boy. The 12-year-old boy is named Dave. Dave is happy."

sentence <- str_extract(description, pattern = "\\.[^\\.]*[0-9]+[^\\.]*.") %>% 
  str_replace("^\\. ", "")
> sentence
[1] "The 12-year-old boy is named Dave."

age <- str_extract(sentence, pattern = "[0-9]+")
> age
[1] "12"