优雅查找列表中的子列表

时间:2012-04-11 13:28:27

标签: python list design-patterns

给定一个包含噪声包围的已知模式的列表,是否有一种优雅的方法可以获得与模式相同的所有项目。请参阅下面的原始代码。

list_with_noise = [7,2,1,2,3,4,2,1,2,3,4,9,9,1,2,3,4,7,4,3,1,2,3,5]
known_pattern = [1,2,3,4]
res = []


for i in list_with_noise:
    for j in known_pattern:
        if i == j:
            res.append(i)
            continue

print res

我们会得到2, 1, 2, 3, 4, 2, 1, 2, 3, 4, 1, 2, 3, 4, 4, 3

奖励:如果不存在完整模式,请避免附加i(即,允许1,2,3,4但不允许1,2,3)

示例:

find_sublists_in_list([7,2,1,2,3,4,2,1,2,3,4,9,9,1,2,3,4,7,4,3,1,2,3,5],[1,2,3,4])

[1,2,3,4],[1,2,3,4],[1,2,3,4]


find_sublists_in_list([7,2,1,2,3,2,1,2,3,6,9,9,1,2,3,4,7,4,3,1,2,6],[1,2,3,4])

[1,2,3],[1,2,3],[1,2,3]

列表包含命名元组。

7 个答案:

答案 0 :(得分:35)

我知道这个问题是5个月大并且已经“被接受”了,但谷歌搜索一个非常相似的问题带我到这个问题,所有的答案似乎有几个相当重要的问题,加上我很无聊,想要试试我的答案,所以我只是滔滔不绝地说出我发现的东西。

问题的第一部分,据我所知,非常简单:只需返回原始列表,其中所有元素都不会被“模式”过滤掉。按照这个想法,我想到的第一个代码使用了filter()函数:

def subfinder(mylist, pattern):
    return list(filter(lambda x: x in pattern, mylist))

我会说这个解决方案肯定比原始解决方案更简洁,但它不是更快,或者至少不是很明显,如果没有很好的理由使用它们,我会尽量避免使用lambda表达式。事实上,我能提出的最佳解决方案是简单的列表理解:

def subfinder(mylist, pattern):
    pattern = set(pattern)
    return [x for x in mylist if x in pattern]

这个解决方案比原版解决方案更优雅,速度更快:理解速度比原版速度快120%,同时将模式投射到第一个碰撞中,在我的测试中速度提高了320%。

现在获得奖金:我会直接进入它,我的解决方案如下:

def subfinder(mylist, pattern):
    matches = []
    for i in range(len(mylist)):
        if mylist[i] == pattern[0] and mylist[i:i+len(pattern)] == pattern:
            matches.append(pattern)
    return matches

这是Steven Rumbalski的“低效单线程”的变体,通过添加“mylist [i] == pattern [0]”检查,并且由于python的短路评估,显着快于两者原始语句和itertools版本(以及我所能提供的所有其他解决方案)它甚至支持重叠模式。你去吧。

答案 1 :(得分:4)

这将获得您问题的“奖励”部分:

pattern = [1, 2, 3, 4]
search_list = [7,2,1,2,3,4,2,1,2,3,4,9,9,1,2,3,4,7,4,3,1,2,3,5]
cursor = 0
found = []
for i in search_list:
    if i == pattern[cursor]:
        cursor += 1
        if cursor == len(pattern):
            found.append(pattern)
            cursor = 0
    else:
        cursor = 0

对于非奖金:

pattern = [1, 2, 3, 4]
search_list = [7,2,1,2,3,4,2,1,2,3,4,9,9,1,2,3,4,7,4,3,1,2,3,5]
cursor = 0
found = []
for i in search_list:
    if i != pattern[cursor]:
        if cursor > 0:
            found.append(pattern[:cursor])
        cursor = 0
    else:
        cursor += 1

最后,这个处理重叠:

def find_matches(pattern_list, search_list):
    cursor_list = []
    found = []
    for element in search_list:
        cursors_to_kill = []
        for cursor_index in range(len(cursor_list)):
            if element == pattern_list[cursor_list[cursor_index]]:
                cursor_list[cursor_index] += 1
                if cursor_list[cursor_index] == len(pattern_list):
                    found.append(pattern_list)
                    cursors_to_kill.append(cursor_index)
            else:
                cursors_to_kill.append(cursor_index)
        cursors_to_kill.reverse()
        for cursor_index in cursors_to_kill:
            cursor_list.pop(cursor_index)
        if element == pattern_list[0]:
            cursor_list.append(1)
    return found

答案 2 :(得分:1)

list_with_noise = [7,2,1,2,3,4,2,1,2,3,4,9,9,1,2,3,4,7,4,3,1,2,3,5]
string_withNoise = "".join(str(i) for i in list_with_noise)
known_pattern = [1,2,3,4]
string_pattern = "".join(str(i) for i in known_pattern)
string_withNoise.count(string_pattern)

答案 3 :(得分:1)

基于迭代器的方法仍然基于朴素算法,但是尝试使用.index()来进行尽可能多的隐式循环:

def find_pivot(seq, subseq):
    n = len(seq)
    m = len(subseq)
    stop = n - m + 1
    if n > 0:
        item = subseq[0]
        i = 0
        try:
            while i < stop:
                i = seq.index(item, i)
                if seq[i:i + m] == subseq:
                    yield i
                i += 1
        except ValueError:
            return

与其他几种具有不同程度的显式循环的方法相比:

def find_loop(seq, subseq):
    n = len(seq)
    m = len(subseq)
    for i in range(n - m + 1):
        if all(seq[i + j] == subseq[j] for j in (range(m))):
            yield i
def find_slice(seq, subseq):
    n = len(seq)
    m = len(subseq)
    for i in range(n - m + 1):
        if seq[i:i + m] == subseq:
            yield i
def find_mix(seq, subseq):
    n = len(seq)
    m = len(subseq)
    for i in range(n - m + 1):
        if seq[i] == subseq[0] and seq[i:i + m] == subseq:
            yield i

一个会得到:

a = list(range(10))
print(a)
# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

b = list(range(5, 10))
print(b)
# [5, 6, 7, 8, 9]

funcs = find_pivot, find_loop, find_slice, find_mix, 
for func in funcs:
    print()
    print(func.__name__)
    print(list(func(a * 10, b)))
    aa = a * 100
    %timeit list(func(aa, b))
    random.shuffle(aa)
    %timeit list(func(aa, b))

# find_pivot
# [5, 15, 25, 35, 45, 55, 65, 75, 85, 95]
# 10000 loops, best of 3: 49.6 µs per loop
# 10000 loops, best of 3: 50.1 µs per loop

# find_loop
# [5, 15, 25, 35, 45, 55, 65, 75, 85, 95]
# 1000 loops, best of 3: 712 µs per loop
# 1000 loops, best of 3: 680 µs per loop

# find_slice
# [5, 15, 25, 35, 45, 55, 65, 75, 85, 95]
# 10000 loops, best of 3: 162 µs per loop
# 10000 loops, best of 3: 162 µs per loop

# find_mix
# [5, 15, 25, 35, 45, 55, 65, 75, 85, 95]
# 10000 loops, best of 3: 82.2 µs per loop
# 10000 loops, best of 3: 83.9 µs per loop


请注意,这比当前接受测试的答案快30%。

答案 4 :(得分:0)

假设:

a_list = [7,2,1,2,3,4,2,1,2,3,4,9,9,1,2,3,4,7,4,3,1,2,3,5]
pat = [1,2,3,4]

这是一个效率低下的一个班轮:

res = [pat for i in range(len(a_list)) if a_list[i:i+len(pat)] == pat]

这是一个更高效的itertools版本:

from itertools import izip_longest, islice

res = []
i = 0  

while True:
    try:
        i = a_list.index(pat[0], i)
    except ValueError:
        break
    if all(a==b for (a,b) in izip_longest(pat, islice(a_list, i, i+len(pat)))):
        res.append(pat)
        i += len(pat)
    i += 1

答案 5 :(得分:0)

解决问题的惯用,可组合的解决方案。

首先,我们需要借用an itertools recipeconsume(它从迭代器中消耗并丢弃给定数量的元素。然后我们采用itertools pairwise的方法},并使用nwise将其扩展为consume函数:

import itertools

def nwise(iterable, size=2):
    its = itertools.tee(iterable, size)
    for i, it in enumerate(its):
        consume(it, i)  # Discards i elements from it
    return zip(*its)

现在我们已经有了,解决红利问题非常简单:

def find_sublists_in_list(biglist, searchlist):
    searchtup = tuple(searchlist)
    return [list(subtup) for subtup in nwise(biglist, len(searchlist)) if subtup == searchtup]

    # Or for more obscure but faster one-liner:
    return map(list, filter(tuple(searchlist).__eq__, nwise(biglist, len(searchlist))))

同样,对主要问题的解决方案更简洁,更快速(如果不那么漂亮)取代:

def subfinder(mylist, pattern):
    pattern = set(pattern)
    return [x for x in mylist if x in pattern]

使用:

def subfinder(mylist, pattern):
    # Wrap filter call in list() if on Python 3 and you need a list, not a generator
    return filter(set(pattern).__contains__, mylist)

此行为方式相同,但避免需要将临时set存储到名称,并将所有过滤工作推送到C.

答案 6 :(得分:0)

def sublist_in_list(sub,lis):     在str(lis).strip('[]')中返回str(sub).strip('[]')