我想用Python2.7从一个大的整数列表中提取列表/子列表，使用开始和结束模式。我希望用一个函数来实现，但我找不到解决这个问题的库、算法或正则表达式。
def myFunctionForSublists(data, startSequence, endSequence):
# ... todo
data = [99, 99, 1, 2, 3, 99, 99, 99, 4, 5, 6, 99, 99, 1, 2, 3, 99, 4, 5, 6, 99]
startSequence = [1,2,3]
endSequence = [4,5,6]
sublists = myFunctionForSublists(data, startSequence, endSequence)
print sublists[0] # [1, 2, 3, 99, 99, 99, 4, 5, 6]
print sublists[1] # [1, 2, 3, 99, 4, 5, 6]
from collections import deque
def gen(l, start, stop):
start_deque = deque(start)
end_deque = deque(stop)
curr_deque = deque(maxlen=len(start))
it = iter(l)
for c in it:
curr_deque.append(c)
if curr_deque == start_deque:
potential = list(curr_deque)
curr_deque = deque(maxlen=len(stop))
for c in it:
potential.append(c)
curr_deque.append(c)
if curr_deque == end_deque:
yield potential
curr_deque = deque(maxlen=len(start))
break
print(list(gen([99, 99, 1, 2, 3, 99, 99, 99, 4, 5, 6, 99, 99, 1, 2, 3, 99, 4, 5, 6, 99], [1,2,3], [4,5,6])))
# [[1, 2, 3, 99, 99, 99, 4, 5, 6], [1, 2, 3, 99, 4, 5, 6]]
collections.deque的itertools方法，它使用长度有限的双向队列来保持适当大小的最后元素缓冲区。它假设你的子列表不重叠，并且你的起始和结束序列也不重叠。
它适用于任何数据、起始、结束序列（甚至是生成器）。
from collections import deque
from itertools import islice
def sublists(data, start, end):
it = iter(data)
start, end = deque(start), deque(end)
while True:
x = deque(islice(it, len(start)), len(start))
# move forward until start is found
while x != start:
x.append(next(it))
out = list(x)
x = deque(islice(it, len(end)), len(end))
# move forward until end is found, storing the sublist
while x != end:
out.append(x[0])
x.append(next(it))
out.extend(end)
yield out
data = [99, 99, 1, 2, 3, 99, 99, 99, 4, 5, 6, 99, 99, 1, 2, 3, 99, 4, 5, 6, 99]
startSequence = [1,2,3]
endSequence = [4,5,6]
print(list(sublists(data, startSequence, endSequence)))
# [[1, 2, 3, 99, 99, 99, 4, 5, 6], [1, 2, 3, 99, 4, 5, 6]]
import re
def find_span(numbers, start, end):
# Create strings from the start and end lists.
start_pattern = ''.join(map(chr, start))
end_pattern = ''.join(map(chr, end))
# convert the list to search into one string.
s = ''.join(map(chr, numbers))
# Create a pattern that starts and ends with the correct sublists,
# and match all sublists. Then convert each match back to a list of
# integers
# The '?' is to make the regex non-greedy
return [
[ord(c) for c in match]
for match in re.findall(rf'{start_pattern}.*?{end_pattern}', s, re.DOTALL)
]
>>> find_span(search, start, end) # Using OP's sample values
[[1, 2, 3, 99, 99, 99, 4, 5, 6], [1, 2, 3, 99, 4, 5, 6]]
re.DOTALL，否则它不会匹配任何包含10（即换行符的ASCII编码）的内容。但是，如果你真的想使用正则表达式，这个方法可以工作。?，这使得正则表达式变得非贪婪。加上这个符号后，它会按照 OP 想要的输出结果进行匹配。 - Edward Minnixint替换为ord（在更新非贪婪符号时忘记复制该更改）。 - Edward Minnix只需在列表中迭代所有索引，并分别将切片与startSequence或endSequence进行比较。假设子列表不应重叠，您可以为两个循环使用相同的迭代器。
def myFunctionForSublists(data, startSequence, endSequence):
positions = iter(range(len(data)))
for start in positions:
if data[start:start+len(startSequence)] == startSequence:
for end in positions:
if data[end:end+len(endSequence)] == endSequence:
yield data[start:end+len(endSequence)]
break
这样，start 循环将从 end 循环结束的地方继续执行。如果它们可以重叠，则为循环使用两个单独的迭代器，即 for start in range(len(data)): 和 for end in range(start+1, len(data)):
def find_sub_list(sl,l):
sll=len(sl)
for ind in (i for i,e in enumerate(l) if e==sl[0]):
if l[ind:ind+sll]==sl:
return ind,ind+sll-1
find_sub_list([1,2,3], data)
>>>(2, 4)
find_sub_list([4,5,6], data)
>>>(8, 10)
data[2:10+1]
>>>[1, 2, 3, 99, 99, 99, 4, 5, 6]
您可以采用类似的方法处理 sublists [1]
def myFunctionForSublists(data, startSequence, endSequence):
start,end = tuple(startSequence), tuple(endSequence)
l1, l2 = len(start), len(end)
s = -1
result = []
for i,v in enumerate(zip(*[data[i:] for i in range(0,l1)])):
if v == start:
s = i
if v == end and s != -1:
result.append(data[s:i+l2])
s = -1
return result
print (myFunctionForSublists(data, startSequence, endSequence))
# [[1, 2, 3, 99, 99, 99, 4, 5, 6], [1, 2, 3, 99, 4, 5, 6]]