R正则表达式向后看 [英] R Regular Expression Lookbehind

问题描述

我有一个填充有以下格式字符串的向量:<year1><year2><id1><id2>

I have a vector filled with strings of the following format: <year1><year2><id1><id2>

向量的第一项如下:

199719982001
199719982002
199719982003
199719982003

对于第一个条目，我们有:year1 = 1997，year2 = 1998，id1 = 2，id2 =001.

For the first entry we have: year1 = 1997, year2 = 1998, id1 = 2, id2 = 001.

我想写一个正则表达式，提取出year1，id1和id2的数字不为零.因此，对于第一个条目，正则表达式应输出:199721.

I want to write a regular expression that pulls out year1, id1, and the digits of id2 that are not zero. So for the first entry the regex should output: 199721.

我尝试使用stringr软件包执行此操作，并创建了以下正则表达式:

I have tried doing this with the stringr package, and created the following regex:

"^\\d{4}|\\d{1}(?<=\\d{3}$)"

拉出year1和id1，但是在使用lookbehind时出现无效的正则表达式"错误.这让我有些困惑，R不能处理前瞻和后顾之忧吗?

to pull out year1 and id1, however when using the lookbehind i get a "invalid regular expression" error. This is a bit puzzling to me, can R not handle lookaheads and lookbehinds?

推荐答案

由于这是固定格式，为什么不使用substr?使用substr(s,1,4)提取year1，使用substr(s,9,9)提取id1，将id2用作as.numeric(substr(s,10,13)).在最后一种情况下，我使用as.numeric消除了零.

Since this is fixed format, why not use substr? year1 is extracted using substr(s,1,4), id1 is extracted using substr(s,9,9) and the id2 as as.numeric(substr(s,10,13)). In the last case I used as.numeric to get rid of the zeroes.

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