Question

很难描述，但基本上，我正在尝试找到一种可以做到这一点的 general 方法：

    [1]" On The Grill(1)95 E Kennedy BlvdLakewood, NJ 08701(732) 942-6555Restaurants I had a business dinner at this restaurant with 5 other people. Everyone was pleased with their appetizers and main courses. We’ll be back for sure…" 
    [2]" Sushi Now231 3rd StLakewood, NJ 08701(732) 719-2275RestaurantsSushi BarsWebsiteMenuOrder Online"

对此：

    [1] "95 E Kennedy Blvd"
    [2] "231 3rd St"

使用R。我知道它涉及正则表达式，但是我并不像我想的那样流利。

谢谢！

Answer 1

您的预期输出没有很扎实的逻辑，但是查看预期数据，您可以使用此正则表达式实现您要尝试的工作，

^.*?(\d{2,}.*?[a-z])[A-Z].*

并用\1替换它，因为group1捕获了您想要的文本。

Regex Demo

R Code Demo

sub("^.*?(\\d{2,}.*?[a-z])[A-Z].*", "\\1", "On The Grill(1)95 E Kennedy BlvdLakewood, NJ 08701(732) 942-6555Restaurants I had a business dinner at this restaurant with 5 other people. Everyone was pleased with their appetizers and main courses. We’ll be back for sure…")
sub("^.*?(\\d{2,}.*?[a-z])[A-Z].*", "\\1", "Sushi Now231 3rd StLakewood, NJ 08701(732) 719-2275RestaurantsSushi BarsWebsiteMenuOrder Online")

按预期打印，

[1] "95 E Kennedy Blvd"
[1] "231 3rd St"

编辑： 好的，\d{2,}可能与数据有关，因此在这里我们可以使用另一种逻辑，在这里我将仅以一个或多个数字\d+开始捕获，然后以一个或多个空格开始捕获。由于比赛恰好在Lakewood之前停止，因此在正则表达式中也要使用积极的眼光(?=Lakewood)，并且可以使用的更新更好的正则表达式是这个

^.*?(\d+\s+.*?)(?=Lakewood).*

Regex Demo 2

现在，如果需要，您甚至可以使用str_match通过正则表达式\d+\s+.*?(?=Lakewood)使用以下代码行提取文本，

library(stringr)

str_match("On The Grill(1)95 E Kennedy BlvdLakewood, NJ 08701(732) 942-6555Restaurants I had a business dinner at this restaurant with 5 other people. Everyone was pleased with their appetizers and main courses. We’ll be back for sure…", "\\d+\\s+.*?(?=Lakewood)")
str_match("Sushi Now231 3rd StLakewood, NJ 08701(732) 719-2275RestaurantsSushi BarsWebsiteMenuOrder Online", "\\d+\\s+.*?(?=Lakewood)")

打印

     [,1]               
[1,] "95 E Kennedy Blvd"
     [,1]        
[1,] "231 3rd St"

Answer 2

Pushpesh Kumar Rajwanshi的answer很不错，也很笼统。但是，如果您觉得有帮助，请使用以下替代方法：

x <- c(" On The Grill(1)95 E Kennedy BlvdLakewood, NJ 08701(732) 942-6555Restaurants I had a business dinner at this restaurant with 5 other people. Everyone was pleased with their appetizers and main courses. We’ll be back for sure…",
       " Sushi Now231 3rd StLakewood, NJ 08701(732) 719-2275RestaurantsSushi BarsWebsiteMenuOrder Online")
street_types <- c("Blvd", "St")
address_pattern <- paste("\\d+ .+?", street_types, collapse = "|")
stringr::str_extract_all(string = x, pattern = address_pattern, simplify = TRUE)
#      [,1]               
# [1,] "95 E Kennedy Blvd"
# [2,] "231 3rd St"

这解决了1位地址号码的问题，并允许您指定街道类型，这可以帮助您防止其他类型的误报（尽管如果您不详尽地指定街道类型，则可能会产生一些误报）。 / p>

Answer 3

这种方法很好用

(\[\d])(?:.+[^\s\d])((?:\d+\s+)[^\R]+)

Regex Demo

Geshmak！

是否有一种通用的方法来删除以R开头的数字并以大写字母结尾的子字符串

3 个答案: