Regex删除R中的前导零，除非最后（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中的字符串中删除前导零。由于此正则表达式，“ 000”和“ 0”均转换为“”。相反，我想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，但是我认为我在使用R中的解决方案时语法不是很正确。而且我不太了解下面的第一种和第二种解决方案之间的区别（如果他们确实有效）。

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中什么是正确的正则表达式才能得到我想要的？

gsub（“（？[1]” 5“” AB“”“”“ gsub（”（^ | [^ 0-9]）0+“，” \\ 1“，c（” 005“，” 0AB“，” 000“，” 0“），perl = TRUE）＃> [1]” 5“” AB“”“”“ The ...

Answer 1

您可以使用交替方式来匹配捕获组中字符串中的所有零，也可以匹配字符串开头的所有零。

Answer 2

省略前导零，除非它是单个零：