如何在R中使用gsub()函数替换“+”

Question

我正在尝试删除数据帧的字符串元素中的“+”字符。但我找不到出路。

下面是数据帧。

txtdf <- structure(list(ID = 1:9, Var1 = structure(c(1L, 1L, 1L, 1L, 4L, 
            5L, 5L, 2L, 3L), .Label = c("government", "parliament", "parliment", 
            "poli+tician", "politician"), class = "factor")), .Names = c("ID", 
            "Var1"), class = "data.frame", row.names = c(NA, -9L))
#  ID   Var1
#  1    government
#  2    government
#  3    government
#  4    government
#  5    poli+tician
#  6    politician
#  7    politician
#  8    parliament
#  9    parliment

我尝试了两种方法，它们都没有给出预期的结果：

Way1

txtdf <- gsub("[:punct:]","", txtdf)
# [1] "goverme" "goverme" "goverme" "goverme" "oli+iia" "oliiia"  "oliiia" 
# [8] "arliame" "arlime" 

我不明白这是怎么回事。我希望将'+‘字符替换为仅用于第5个元素的值，但所有元素都按上述方式编辑。

Way2

txtdf<-gsub("*//+","",txtdf)
# [1] "government"  "government"  "government"  "government"  "poli+tician"
# [6] "politician"  "politician"  "parliament"  "parliment" 

这里一点变化也没有。我想我试过的是，我试图用双斜杠来转义+字符。

Answer 1

只需用fixed = TRUE替换它(不需要使用正则表达式)，但是您必须通过指定列名来替换data.frame的每个“列”：

txtdf <- data.frame(job = c("government", "poli+tician", "parliament"))
txtdf

给出

          job
1  government
2 poli+tician
3  parliament

现在替换"+"：

txtdf$job <- gsub("+", "", txtdf$job, fixed = TRUE)
txtdf

结果是：

         job
1 government
2 politician
3 parliament

Answer 2

您需要转义您的加号，"+“有一个特殊的含义(它是一个量词)，当涉及到正则表达式时，因此不能被视为标点符号，从文档：?regex

"+“前面的项目将匹配一次或多次。

为了匹配这些特殊的字符，你需要转义这些字符，这样它们的意思就可以被字面理解，因此它们的特殊意义就不会被翻译出来。在R中，您需要两个反斜杠()来转义。所以在你的例子中，这应该是这样的：

gsub("\\+","",df$job)

以上运行将从数据中删除所有加号，从而给出所需的结果。

因此，假设您的df是：

df <- data.frame(job = c("government", "poli+tician","politician", "parliament"))

那么，您的输出将是：

> gsub("\\+","",df$job)
[1] "government" "politician" "politician"
[4] "parliament"