Question

可能重复：
How to convert sequence of numbers in an array to range of numbers

前几天我正在考虑这个问题。假设您有数字列表：

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 19, 21, 22, 23, 999, 1000, 1001

是否有任何剪辑的图书馆或代码会将以下内容转换为：

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

换句话说，将完整的数字列表减少到一堆范围左右。我找不到任何东西。如果不存在这样的事情，那么任何人都有一些有效实施的想法吗？

Answer 1

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]]

或

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']

Answer 2

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']

希望这有帮助

Answer 3

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)

给予：

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

从数字列表生成“高效”编码

3 个答案: