Question

我有一堆数字，说如下：

[range(1, 5), range(6, 9), range(20, 33, 4)]

那里提供的信息可以用Python表示为范围：

1..4, 6..8, 20..32..4

在我的输出中，我写了user-select，但这只是一个介绍问题。

pen显示了如何为连续范围执行此操作。我不知道如何轻松地为上面的跨栏范围做这件事。对此有类似的伎俩吗？

Answer 1

这是解决问题的直接方法。

def get_ranges(ls):
    N = len(ls)
    while ls:
        # single element remains, yield the trivial range
        if N == 1:
            yield range(ls[0], ls[0] + 1)
            break

        diff = ls[1] - ls[0]
        # find the last index that satisfies the determined difference
        i = next(i for i in range(1, N) if i + 1 == N or ls[i+1] - ls[i] != diff)

        yield range(ls[0], ls[i] + 1, diff)

        # update variables
        ls = ls[i+1:]
        N -= i + 1

Answer 2

def ranges(data):
    result = []
    if not data:
        return result
    idata = iter(data)
    first = prev = next(idata)
    for following in idata:
        if following - prev == 1:
            prev = following
        else:
            result.append((first, prev + 1))
            first = prev = following
    # There was either exactly 1 element and the loop never ran,
    # or the loop just normally ended and we need to account
    # for the last remaining range.
    result.append((first, prev+1))
    return result

测试：

>>> data = range(1, 5) + range(6, 9) + range(20, 24)
>>> print ranges(data)
[(1, 5), (6, 9), (20, 24)]

Answer 3

您可以使用groupby模块中的count和itertools以及Counter模块中的collections，例如：

更新：请参阅评论，以了解此解决方案背后的逻辑及其局限性。

from itertools import groupby, count
from collections import Counter

def ranges_list(data=list, func=range, min_condition=1):
    # Sort in place the ranges list
    data.sort()

    # Find all the steps between the ranges's elements
    steps = [v-k for k,v in zip(data, data[1:])]

    # Find the repeated items's steps based on condition. 
    # Default: repeated more than once (min_condition = 1)
    repeated = [item for item, count in Counter(steps).items() if count > min_condition]

    # Group the items in to a dict based on the repeated steps
    groups = {k:[list(v) for _,v in groupby(data, lambda n, c = count(step = k): n-next(c))] for k in repeated}

    # Create a dict:
    # - keys are the steps
    # - values are the grouped elements
    sub = {k:[j for j in v if len(j) > 1] for k,v in groups.items()}

    # Those two lines are for pretty printing purpose:
    # They are meant to have a sorted output.
    # You can replace them by:
    # return [func(j[0], j[-1]+1,k) for k,v in sub.items() for j in v]
    # Otherwise:
    final = [(j[0], j[-1]+1,k) for k,v in sub.items() for j in v]
    return [func(*k) for k in sorted(final, key = lambda x: x[0])]

ranges1 = [1, 2, 3, 4, 6, 7, 8, 20, 24, 28, 32]
ranges2 = [1, 2, 3, 4, 6, 7, 10, 20, 24, 28, 50,51,59,60]

print(ranges_list(ranges1))
print(ranges_list(ranges2))

输出：

[range(1, 5), range(6, 9), range(20, 33, 4)]
[range(1, 5), range(6, 8), range(20, 29, 4), range(50, 52), range(59, 61)]

限制：

有了这种输入：

ranges3 = [1,3,6,10]
print(ranges_list(ranges3)
print(ranges_list(ranges3, min_condition=0))

将输出：

# Steps are repeated <= 1 with the condition: min_condition = 1
# Will output an empty list
[]
# With min_condition = 0
# Will output the ranges using: zip(data, data[1:])
[range(1, 4, 2), range(3, 7, 3), range(6, 11, 4)]

随意使用此解决方案并采用或修改它以满足您的需求。

Answer 4

它可能不是超短或优雅，但似乎有效：

def ranges(ls):
    li = iter(ls)
    first = next(li)
    while True:
        try:
            element = next(li)
        except StopIteration:
            yield range(first, first+1)
            return
        step = element - first
        last = element
        while True:
            try:
                element = next(li)
            except StopIteration:
                yield range(first, last+step, step)
                return
            if element - last != step:
                yield range(first, last+step, step)
                first = element
                break
            last = element

迭代遍历列表的迭代器，并产生范围对象：

>>> list(ranges([1, 2, 3, 4, 6, 7, 8, 20, 24, 28, 32]))
[range(1, 5), range(6, 9), range(20, 33, 4)]

它还处理负范围，以及只有一个元素的范围：

>>> list(ranges([9,8,7, 1,3,5, 99])
[range(9, 6, -1), range(1, 7, 2), range(99, 100)]

将数字列表转换为范围

4 个答案: