解析具有要列出的间隔的整数集的字符串

Question

我有"2,5,7-9,12"字符串.

我希望得到[2,5,7,8,9,12]列表.

在python中有没有内置函数？

谢谢.

UPD.我想,直接的答案是否定的.无论如何,谢谢你的"片段".使用一个,由Sven Marnach建议.

Answer 1

s = "2,5,7-9,12"
ranges = (x.split("-") for x in s.split(","))
print [i for r in ranges for i in range(int(r[0]), int(r[-1]) + 1)]

版画

[2, 5, 7, 8, 9, 12]

Answer 2

s = "2,5,7-9,12"
result = list()

for item in s.split(','):
    if '-' in item:
        x,y = item.split('-')
        result.extend(range(int(x), int(y)+1))
    else:
        result.append(int(item))

print result

Answer 3

此版本处理任意空白、重叠范围、无序范围和负整数：

from itertools import chain

def group_to_range(group):
  group = ''.join(group.split())
  sign, g = ('-', group[1:]) if group.startswith('-') else ('', group)
  r = g.split('-', 1)
  r[0] = sign + r[0]
  r = sorted(int(__) for __ in r)
  return range(r[0], 1 + r[-1])

def rangeexpand(txt):
  ranges = chain.from_iterable(group_to_range(__) for __ in txt.split(','))
  return sorted(set(ranges))


>>> rangeexpand('-6,-3--1,3-5,7-11,14,15,17-20')
[-6, -3, -2, -1, 3, 4, 5, 7, 8, 9, 10, 11, 14, 15, 17, 18, 19, 20]
>>> rangeexpand('1-4,6,3-2, 11, 8 - 12,5,14-14')
[1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14]