Question

我有一个数据框，其中的一列包含文本字符串：

1 Blue, Tall, leather, VA  
2 Green, Medium, VA*  
3 Pink, MD  
4 Yellow, MA

最后的2个，有时带有'*'的3个是状态名称。我希望能够提取每一行左侧或最后一个'，'的所有内容。在r中完成此操作的最佳方法是什么？

我是新来的，所以请帮助

我希望输出为：

1 Blue, Tall, leather  
2 Green, Medium  
3 Pink  
4 Yellow

Answer 1

使用正则表达式：

vector <- c("Blue, Tall, leather, VA", "Green, Medium, VA*", "Pink, MD", "Yellow, MA")

sub("^(.*),.*$", "\\1", vector)

Answer 2

split，然后用逗号分隔paste，vector <- c("Blue, Tall, leather, VA", "Green, Medium, VA*", "Pink, MD", "Yellow, MA") sapply(X = strsplit(x = vector, split = ","), FUN = function(x) paste(head(x, -1), collapse = ",")) #[1] "Blue, Tall, leather" "Green, Medium" "Pink" "Yellow"，除了最后一个用逗号分隔的内容

@Bean
 public WebClient webClient(ClientRegistrationRepository clientRegistrationRepository , OAuth2AuthorizedClientRepository authorizedClientRepository) {
        ServletOAuth2AuthorizedClientExchangeFilterFunction oauth =
             new ServletOAuth2AuthorizedClientExchangeFilterFunction (clientRegistrationRepository , authorizedClientRepository);
        return WebClient.builder().apply(oauth.oauth2Configuration()).build();
    }

Answer 3

具有sub且与,匹配的选项，后跟零个或多个不是,（[^,]*）的字符，直到结尾（$ ）并替换为空白（""）

sub(",[^,]*$", "", v1)
#[1] "Blue, Tall, leather" "Green, Medium"       "Pink"                "Yellow"

或者使用trimws（从R 3.6.0开始）

trimws(v1, whitespace = ",[^,]*")
#[1] "Blue, Tall, leather" "Green, Medium"       "Pink"                "Yellow"

或者使用str_remove中的stringr

library(stringr)
str_remove(v1, ",[^,]*$")

数据

v1 <- c("Blue, Tall, leather, VA", "Green, Medium, VA*", "Pink, MD", "Yellow, MA")

在r中的字符串中查找最后一个逗号的位置

3 个答案:

数据