从数字列表中生成“有效”编码[重复]

def get_groups(lst):
    slices = [i+1 for i, v in enumerate(zip(lst, l[1:])) if v[0] != v[1]-1] 
    slices = [0] + slices + [len(lst)]
    for start, end in zip(slices, slices[1:]):
        yield lst[start:end]

>>> list(get_groups([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 19, 21, 22, 23, 999, 1000, 1001]))
[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [19], [21, 22, 23], [999, 1000, 1001]]

def get_ranges(lst):
    slices = [i+1 for i, v in enumerate(zip(lst, l[1:])) if v[0] != v[1]-1]
    slices = [0] + slices + [len(lst)]
    for start, end in zip(slices, slices[1:]):
        yield \"%d-%d\" % (lst[start], lst[end-1])

>>> list(get_ranges([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 19, 21, 22, 23, 999, 1000, 1001]))
[\'1-13\', \'19-19\', \'21-23\', \'999-1001\']

def compress(nums):
    nums = [int(i) for i in nums.strip().split(\',\')]
    answer = []
    start = nums[0]
    prev = nums[0]
    for num in nums:
        if num-prev != 1:
            answer.append(\"%d-%d\" %(start, prev))
            start = num
        prev = num
    answer.append(\"%d-%d\" %(start, prev))
    return answer[1:]

>>> compress(\"1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 19, 21, 22, 23, 999, 1000, 1001\")
[\'1-13\', \'19-19\', \'21-23\', \'999-1001\']

l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 19, 21, 22, 23, 999, 1000, 1001]
range_pos = []
start=0
end=0
for i in range(1,len(l)):
    if l[i] - l[i-1] == 1:
        end = i
    else:
        range_pos.append((start, end))
        start = end = i
range_pos.append((start, end))
ranges = [\"%s-%s\" % (l[s], l[e]) if s < e else str(l[s]) for (s, e) in range_pos]
print \', \'.join(ranges)

1-13, 19, 21-23, 999-1001