我意识到我有点过于复杂化,手动计数比使用稍微复杂的生成器要容易得多:
def ranges(seq):
start, end = seq[0], seq[0]
count = start
for item in seq:
if not count == item:
yield start, end
start, end = item, item
count = item
end = item
count += 1
yield start, end
print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))
制作:
[(1, 5), (8, 11), (200, 202)]
这个方法非常快:
这个方法(以及旧方法,它们的表现几乎完全相同):
python -m timeit -s "from test import ranges" "ranges([1,2,3,4,5,8,9,10,11,200,201,202])"
1000000 loops, best of 3: 0.47 usec per loop
Jeff Mercado的方法:
python -m timeit -s "from test import as_range; from itertools import groupby, count" "[as_range(g) for _, g in groupby([1,2,3,4,5,8,9,10,11,200,201,202], key=lambda n, c=count(): n-next(c))]"
100000 loops, best of 3: 11.1 usec per loop
这个方法比原来的快了20倍以上 - 当然,除非速度很重要,否则这不是一个真正的问题。
我的旧解决方案使用生成器:
import itertools
def resetable_counter(start):
while True:
for i in itertools.count(start):
reset = yield i
if reset:
start = reset
break
def ranges(seq):
start, end = seq[0], seq[0]
counter = resetable_counter(start)
for count, item in zip(counter, seq):
if not count == item:
yield start, end
start, end = item, item
counter.send(item)
end = item
yield start, end
print(list(ranges([1,2,3,4,5,8,9,10,11,200,201,202])))
生成:
[(1, 5), (8, 11), (200, 202)]