正则表达式删除R中的前导零，除非最终(或唯一)字符为零

Question

gsub("(?<![0-9])0+", "", c("005", "0AB", "000", "0"), perl = TRUE)
#> [1] "5"  "AB" ""   ""
gsub("(^|[^0-9])0+", "\\1", c("005", "0AB", "000", "0"), perl = TRUE)
#> [1] "5"  "AB" ""   ""

is from this SO thread上方的正则表达式解释了如何从R中的字符串中删除所有前导零。由于此正则表达式，“000”和“0”都转换为“”。相反，我希望从字符串中删除所有前导零，但最终字符恰好为零或唯一字符为零的情况除外。

"005" would become "5"
"0AB" would become "AB"
"000" would become "0"
"0"   would become "0"

This other SO thread解释了如何做我想做的事，但我认为我在使用R中的解决方案时语法不是很正确。而且我不太理解下面的第一和第二解决方案之间的区别(如果它们确实工作)。

gsub("s/^0*(\d+)$/$1/;", "", c("005", "0AB", "000", "0"), perl = TRUE)  # 1st solution
# Error: '\d' is an unrecognized escape in character string starting ""s/^0*(\d"
gsub("s/0*(\d+)/$1/;", "", c("005", "0AB", "000", "0"), perl = TRUE)    # 2nd solution
# Error: '\d' is an unrecognized escape in character string starting ""s/0*(\d"

R中正确的正则表达式是什么才能得到我想要的？

Answer 1

您可以从字符串的开头删除所有零，但不能删除最后一个:

sub("^0+(?!$)", "", x, perl=TRUE)

参见 regex demo。

详细信息

^-字符串开始

0+-一个或多个零

(?!$)-如果当前位置右侧的字符串位置的末尾紧邻字符串，则匹配失败的负前行

参见 R demo:

x <- c("005", "0AB", "000", "0")
sub("^0+(?!$)", "", x, perl=TRUE)
## => [1] "5"  "AB" "0"  "0"