如何将具有特定条件的列表合并并创建到python中的嵌套列表?

你写的

b[3]和b[4]之间的间隙大于3

但不是更大。它是更大的或相等的。

a = [0, 0, 2, 2, 2, 2]
b = [1, 3, 2, 3, 6, 7]
c = []

for s in set(a):
    i = a.index(s)
    count = a.count(s)
    pom = []

    for j in range(i, i + count):
        if not pom:
            pom.append(b[j])
        elif abs(pom[-1] - b[j]) < 3:
                pom.append(b[j])
                if j + 1 == count:
                    pom.insert(0, s)
                    c.append(pom)
                    pom = []
                else:
                    pom.insert(0, s)
                    c.append(pom)
                    pom = [b[j]]

print(c)

首先根据 a

c = zip(a, b)

c = {k: [bi for ai, bi in g] for k, g in groupby(c, lambda i: i[0])}

现在连接到列表列表(按顺序):

c = [v for k, v in sorted(c.items())]

现在您需要一个函数来按值差距进行拆分:

def split_max_gap(l, max_gap=2):
    acc = [l[0]]
    for x, y in zip(l, l[1:]):
        if abs(x - y) > max_gap:
            yield acc
            acc = [y]
            continue
        acc.append(y)
    if acc:
        yield acc

c = map(split_max_gap, c)

扁平化:

c = list(chain.from_iterable(c))

c 现在应该保持:

[[1, 3], [2, 3], [6, 7]]