Question

有没有人可以使用gsub删除变量上的尾随空格？

以下是我的数据示例。如您所见，我在变量中嵌入了尾随空格和空格。

county <- c("mississippi ","mississippi canyon","missoula ",
            "mitchell ","mobile ", "mobile bay")

我可以使用以下逻辑删除所有空格，但我真正想要的只是在末尾移动空格。

county2 <- gsub(" ","",county)

非常感谢任何协助。

Answer 1

阅读?regex以了解正则表达式的工作原理。

gsub("[[:space:]]*$","",county)

[:space:]是一个预定义的字符类，它匹配您的语言环境中的空格字符。 *表示要重复匹配零次或多次，$表示匹配字符串的结尾。

Answer 2

您可以使用正则表达式：

 county <- c("mississippi ","mississippi canyon","missoula ",
        "mitchell ","mobile ", "mobile bay")  
 county2 <- gsub(" $","", county, perl=T)

$代表文本序列的结尾，因此只匹配尾随空格。 perl=T为匹配模式启用正则表达式。有关正则表达式的更多信息，请参阅?regex。

Answer 3

如果您不需要使用gsub命令 - str_trim函数对此很有用。

    library(stringr)
    county <- c("mississippi ","mississippi canyon","missoula ",
        "mitchell ","mobile ", "mobile bay")
    str_trim(county)

Answer 4

Above solution can not be generalized. Here is an example:


    a<-" keep business moving"
    str_trim(a) #Does remove trailing space in a single line string

However str_trim() from 'stringr' package works only for a vector of words   and a single line but does not work for multiple lines based on my testing as consistent with source code reference. 

    gsub("[[:space:]]*$","",a) #Does not remove trailing space in my example
    gsub(" $","", a, perl=T) #Does not remove trailing space in my example

Below code works for both term vectors and or multi-line character vectors   which was provided by the reference[1] below. 

    gsub("^ *|(?<= ) | *$", "", a, perl=T)


#Reference::

使用R中的gsub删除尾随空格

4 个答案: