r-如何为数据帧中的每一行提取句子中的最后3个元素

Question

我有以下数据帧：

df <- structure(list(matrix.unlist.all_dates...nrow...230..byrow...T. = c(
"Willem F. Duisenberg, President of the European Central Bank, Christian Noyer,  Vice-President of the European Central Bank,  Frankfurt am Main,  14 December 2000", 
"Willem F. Duisenberg,  President of the European Central  Bank,  Christian Noyer,  Vice-President of the European Central Bank,  Frankfurt am Main,  2 November 2000", 
"Willem F. Duisenberg,  President of the European Central  Bank,  Christian Noyer,  Vice-President of the European Central Bank,  Paris,  19 October 2000", 
"Willem F. Duisenberg,  President of the European Central  Bank,  Christian Noyer,  Vice-President of the European Central Bank,  Frankfurt am Main,  5 October 2000", 
"Willem F. Duisenberg,  President of the European Central Bank,  Christian Noyer,  Vice-President of the European Central Bank,  Frankfurt am Main,  14 September 2000", 
"Willem F. Duisenberg,  President of the European Central Bank,  Lucas Papademos,  Vice-President of the European Central Bank,  Frankfurt,  10 July 2003.", 
"Willem F. Duisenberg,  President of the European Central Bank,  Lucas Papademos,  Vice-President of the European Central Bank,    Frankfurt,  5 June 2003."
)), class = "data.frame", row.names = c(NA, -7L))

正如你所看到的，每一行的文本都遵循一个清晰的模式，最后三个单词是日期。我只想提取这三个"；单词"；(基本上是日期(。

你会怎么做？我尝试了substr，但由于每行的长度不同，我没有成功

Answer 1

您可以使用正则表达式提取日期。

gsub(".* (\d+ \w+ \d+)\.?$", "\1", df[, 1])

图案(\d+ \w+ \d+)与匹配

1. 一个或多个数字(\d+(，后跟
2. 一个空间，后面
3. 一个或多个字母(\w+(，后跟
4. 一个空间，后面
5. 一个或多个数字(\d+(

因此，在括号内可以捕捉日期。然后用日期替换整个字符串：\1表示括号内匹配的内容。

Answer 2

一个选项是使用word函数从包stringr(属于tidyverse世界(中直接选择最后三个单词

library(stringr)
str_replace_all(word(df[,1], -3, -1), fixed("."), "")
# [1] "14 December 2000"  "2 November 2000"   "19 October 2000"   "5 October 2000"    "14 September 2000" "10 July 2003"      "5 June 2003"

str_replace_all函数用于替换字符串末尾可能出现的点。fixed辅助函数表示.是一个实际的点，而不是正则表达式标记。