使用itertools和numpy的几种不同方式：
from itertools import groupby, tee, cycle
x = [17, 17, 19, 20, 21, 22, 0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 14, 14, 28, 29, 30, 31, 32, 33, 34, 35,
36, 1, 2, 3, 4,34,54]
def sequences(l):
x2 = cycle(l)
next(x2)
grps = groupby(l, key=lambda j: j + 1 == next(x2))
for k, v in grps:
if k:
yield tuple(v) + (next((next(grps)[1])),)
print(list(sequences(x)))
[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), (28, 29, 30, 31, 32, 33, 34, 35, 36), (1, 2, 3, 4)]
或者使用python3并从中产生：
def sequences(l):
x2 = cycle(l)
next(x2)
grps = groupby(l, key=lambda j: j + 1 == next(x2))
yield from (tuple(v) + (next((next(grps)[1])),) for k,v in grps if k)
print(list(sequences(x)))
在这里使用我的答案的变化与numpy.split：
out = [tuple(arr) for arr in np.split(x, np.where(np.diff(x) != 1)[0] + 1) if arr.size > 1]
print(out)
[(19, 20, 21, 22), (0, 1, 2), (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), (28, 29, 30, 31, 32, 33, 34, 35, 36), (1, 2, 3, 4)]
类似于ekhumoro的回答：
def sequences(x):
it = iter(x)
prev, temp = next(it), []
while prev is not None:
start = next(it, None)
if prev + 1 == start:
temp.append(prev)
elif temp:
yield tuple(temp + [prev])
temp = []
prev = start
def sequences(l):
x2 = cycle(l)
next(x2)
grps = groupby(l, key=lambda j: j + 1 == next(x2))
for k, v in grps:
if k:
t = tuple(v) + (next(next(grps)[1]),)
yield t, len(t)
def sequences(l):
x2 = cycle(l)
next(x2)
grps = groupby(l, lambda j: j + 1 == next(x2))
yield from ((t, len(t)) for t in (tuple(v) + (next(next(grps)[1]),)
for k, v in grps if k))
def sequences(x):
it = iter(x)
prev, temp = next(it), []
while prev is not None:
start = next(it, None)
if prev + 1 == start:
temp.append(prev)
elif temp:
yield tuple(temp + [prev]), len(temp) + 1
temp = []
prev = start
这三个输出都将相同：
[((19, 20, 21, 22), 4), ((0, 1, 2), 3), ((4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14), 11)
, ((28, 29, 30, 31, 32, 33, 34, 35, 36), 9), ((1, 2, 3, 4), 4)]
602392714
清零编程群