Question

我想将元素列表分成嵌套列表，每个子列表都有连续的元素。如果一个元素没有连续的元素，则应该在单个列表中。

输入：

l1 = [1, 2, 3, 11, 12, 13, 23, 33, 34, 35, 45]
l2 = [11, 12, 13, 22, 23, 24, 33, 34]
l3 = [1, 2, 3, 11, 12, 13, 32, 33, 34, 45]

预期输出：

l1 = [[1, 2, 3], [11, 12, 13], [23], [33, 34, 35], [45]]
l2 = [[11, 12, 13], [22, 23, 24], [33, 34]]
l3 = [[1, 2, 3], [11, 12, 13], [32, 33, 34], [45]]

我尝试了下面的代码，但未给出预期的结果，打印了一个空列表：

def split_into_list(l):

    t = []
    for i in range(len(l) - 1):

        if abs(l[i] - l[i + 1]) == 0:
            t.append(l[i])

        elif abs(l[i] - l[i + 1]) != 0 and abs(l[i - 1] - l[i]) == 0:
            t.append(l[i])
            yield t
            split_into_list(l[i:])
        if i + 1 == len(l):
            t.append(l[i])
            yield t

l = [1, 2, 3, 11, 12, 13, 32, 33, 34, 45]
li = []
li.append(split_into_list(l))

for i in li:
    print(i, list(i))

Answer 1

使用自定义split_adjacent函数的更短方法：

def split_adjacent(lst):
    res = [[lst[0]]]    # start/init with the 1st item/number
    for i in range(1, len(lst)):
        if lst[i] - res[-1][-1] > 1:  # compare current and previous item
            res.append([])
        res[-1].append(lst[i])
    return res


l1 = [1, 2, 3, 11, 12, 13, 23, 33, 34, 35, 45]
l2 = [11, 12, 13, 22, 23, 24, 33, 34]
l3 = [1, 2, 3, 11, 12, 13, 32, 33, 34, 45]

print(split_adjacent(l1))
print(split_adjacent(l2))
print(split_adjacent(l3))

最终输出：

[[1, 2, 3], [11, 12, 13], [23], [33, 34, 35], [45]]
[[11, 12, 13], [22, 23, 24], [33, 34]]
[[1, 2, 3], [11, 12, 13], [32, 33, 34], [45]]

Answer 2

def split_into_list(l):
    result = [[]]
    for i, elt in enumerate(l[1:]):
        diff = abs(elt - l[i])
        if diff == 1:
            # still the same group
            result[-1].append(elt)
        else:
            # new group
            result.append([elt])
    return result



l = [1,2,3,11,12,13,32,33,34,45]
print(split_into_list(l))

收益

[[2, 3], [11, 12, 13], [32, 33, 34], [45]]

Answer 3

def split_into_list(l):
    t = []
    temp = [l[0]]
    prev = l[0]
    for i in l[1:]:
        if i == prev+1:
            temp.append(i)
        else:
            t.append(temp)
            temp = [i]
        prev = i
    return t

请记住，到目前为止，所有解决方案都依赖于排序列表，您在问题中未明确指定。

Answer 4

这很适合使用diff和split的numpy解决方案：

def consecutives(x):
    np.split(x, np.flatnonzero(np.diff(x) != 1) + 1)

例如，consecutives(l1)将导致

[array([1, 2, 3]),
 array([11, 12, 13]),
 array([23]),
 array([33, 34, 35]),
 array([45])]

如果您需要嵌套列表，则可以应用list或ndarray.tolist：

def consecutives(x):
    return [a.tolist() for a in np.split(x, np.flatnonzero(np.diff(x) != 1) + 1)]

现在consecutives(l1)的结果是

[[1, 2, 3], [11, 12, 13], [23], [33, 34, 35], [45]]

将列表制作为嵌套列表，并在单独的列表中包含连续的元素

4 个答案: