从数字列表生成“efficent”编码

Question

Possible Duplicate:
How to convert sequence of numbers in an array to range of numbers

前几天我在想这个。假设你有一个数字列表：

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

或是将以下代码剪成以下代码：

^{pr2}$

换言之，将一个完整的数字列表减少到一组范围左右。我什么也没找到。如果不存在这样的东西，有人对有效的实现有一些想法吗？在

Answer 1

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：

^{pr2}$

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

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]]

或者

^{pr2}$