Python：在包含区间的列表中查找字符串，并用其中的每个数字替换该区间

Question

我想知道是否有任何漂亮/干净的方式来做我想做的事情:)。 （我确定有） 所以我的函数接收一个字符串列表，可以包含两种格式的字符串： “12,13,14,15”或“12 到 15” 目标是解析第二种类型并将“to”替换为区间中的数字。

数字之间的分隔符无关紧要，正则表达式将在之后完成这项工作。 这是伪代码和丑陋的实现

这个想法是用区间中的数字替换列表中的“to”，以便之后我可以使用正则表达式轻松解析数字

# The list is really inconsistent, separators may change and it's hand filled so some comments like in the last example might be present
l = ["12,13,14,15",
     "12 to 18",
     "10,21,22 to 42",
     "14,48,52",
     "12,14,22;45 and also 24 to 32"
]

def process_list(l):
  for x in l:
     if "to" in x:
         # Find the 2 numbers around the to and replace the "to" by ",".join(list([interval of number]))
  final_list = numero_regex.findall(num)
  return final_list

Answer 1

这是一种解决方案：

from itertools import chain
def split(s):
    return list(chain(*(list(range(*list(map(int, x.split(' to ')))))+[int(x.split(' to ')[1])]
                        if ' to ' in x else
                        [int(x)]
                        for x in s.split(',')
                       )))
    
[split(e) for e in l]

输出：

[[12, 13, 14, 15],
 [12, 13, 14, 15, 16, 17, 18],
 [10, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42],
 [14, 48, 52]]

编辑：我将上述解决方案改编为与正则表达式一起使用：

from itertools import chain
def split(s):
    regex = re.compile('(\d+\s*to\s*\d+|\d+)')
    return list(chain(*([int(x)] if x.isnumeric() else 
                        list(range(*map(int, re.split('\s+to\s+', x))))
                        +[int(re.split('\s+to\s+', x)[-1])]
                        for x in regex.findall(s)
                       )))

Answer 2

我认为你不需要正则表达式：

def process_list(l):
    final_list = []
    for s in l:
        l2 = []
        for n in s.split(','):
            params = n.split(' to ')
            nums = list(range(int(params[0]), int(params[-1])+1))
            l2.extend(nums)
        final_list.append(l2)
    return final_list

输出：

>>> process_list(l)

[[12, 13, 14, 15],
 [12, 13, 14, 15, 16, 17, 18],
 [10, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42],
 [14, 48, 52]]

更新：

我想要这样的输出 ["12,21,32;14, and the 12,13,14,15,[...],40"]。我可以很容易地用正则表达式解析

如果你只是想替换'number1 to number2'，你可以这样做：

def process_list(l):
    def to_range(m):
        return ','.join([str(i) for i in range(int(m.group('start')),
                                               int(m.group('stop'))+1)])
    return [re.sub(pat, to_range, s) for s in l]

输出：

# l = ["12,21,18 to 20;32;14, and the 12 to 16"]
>>> process_list(l)
['12,21,18,19,20;32;14, and the 12,13,14,15,16']