正则表达式匹配字符串除以 'and'

Question

我需要解析一个字符串以从字符串中获取所需的数字和位置，例如：

2 Better Developers and 3 Testers
5 Mechanics and chef
medic and 3 nurses

目前我正在使用这样的代码，其中 returns 元组列表，例如 [('2', 'Better Developers'), ('3', 'Testers')]:

def parse_workers_list_from_str(string_value: str) -> [(str, str)]:
    result: [(str, str)] = []
    if string_value:
        for part in string_value.split('and'):
            result.append(re.findall(r'(?: *)(\d+|)(?: |)([\w ]+)', part.strip())[0])
    return result

我可以不用 .split() 只使用正则表达式吗？

Answer 1

与 re.MULTILINE 一起，您可以在一个正则表达式中完成所有操作，这也将正确拆分所有内容：

>>> s = """2 Better Developers and 3 Testers
5 Mechanics and chef
medic and 3 nurses"""
>>> re.findall(r"\s*(\d*)\s*(.+?)(?:\s+and\s+|$)", s, re.MULTILINE)
[('2', 'Better Developers'), ('3', 'Testers'), ('5', 'Mechanics'), ('', 'chef'), ('', 'medic'), ('3', 'nurses')]

有空 '' 到 1 的解释和转换：

import re

s = """2 Better Developers and 3 Testers
5 Mechanics and chef
medic and 3 nurses"""

results = re.findall(r"""
    # Capture the number if one exists
    (\d*)
    # Remove spacing between number and text
    \s*
    # Caputre the text
    (.+?)
    # Attempt to match the word 'and' or the end of the line
    (?:\s+and\s+|$\n?)
    """, s, re.MULTILINE|re.VERBOSE)

results = [(int(n or 1), t.title()) for n, t in results]
results == [(2, 'Better Developers'), (3, 'Testers'), (5, 'Mechanics'), (1, 'Chef'), (1, 'Medic'), (3, 'Nurses')]

Answer 2

如果你想处理多个 and 分离器，那么你应该考虑使用 PyPi regex 模块，它允许我们使用 分支重置组 即 (?!...) 提供 Subpatterns 在此构造的每个备选方案中声明将从相同的索引重新开始。

(?|(\d*) *(\b[a-z]+(?: [a-z]+)*?)(?= and )|(?<= and )(\d*) *(\b[a-z]+(?: [a-z]+)*))

RegEx Demo

import regex
rx = regex.compile(r'(?|(\d*) *(\b[a-z]+(?: [a-z]+)*?)(?= and )|(?<= and )(\d*) *(\b[a-z]+(?: [a-z]+)*))', regex.I)

arr = ['2 Better Developers and 3 Testers', '5 Mechanics and chef', 'medic and 3 nurses', '5 foo', '5 Mechanics and 2 chefs and tester']
for s in arr: print (rx.findall(s), ':', s)

输出：

[('2', 'Better Developers'), ('3', 'Testers')] : 2 Better Developers and 3 Testers
[('5', 'Mechanics'), ('', 'chef')] : 5 Mechanics and chef
[('', 'medic'), ('3', 'nurses')] : medic and 3 nurses
[] : 5 foo
[('5', 'Mechanics'), ('2', 'chefs'), ('', 'tester')] : 5 Mechanics and 2 chefs and tester

根据原始问题发布的早期答案，存在单个 and.

您可以使用这个正则表达式：

(\d*) *(\S+(?: \S+)*?) and (\d*) *(\S+(?: \S+)*)

这里我们匹配 and 并在两侧用单个 space 包围。在 and 之前和之后，我们使用这个子模式进行匹配：

(\d*) *(\S+(?: \S+)*?)

匹配可选的 0+ 数字开头，后跟 0 个或多个 space 后跟 1 个或多个由 space 分隔的非白色 space 字符串。

RegEx Demo

代码：

import re
arr = ['2 Better Developers and 3 Testers', '5 Mechanics and chef', 'medic and 3 nurses', '5 foo']

rx = re.compile(r'(\d*) *(\S+(?: \S+)*?) and (\d*) *(\S+(?: \S+)*)')

for s in arr: print (rx.findall(s))

输出：

[('2', 'Better Developers', '3', 'Testers')]
[('5', 'Mechanics', '', 'chef')]
[('', 'medic', '3', 'nurses')]
[]

正则表达式匹配字符串除以 'and'

Regex match strings divided by 'and'

python

regex

regex-group