Question

最近I asked a question here我希望在更大的列表中找到子列表。我有一个类似但略有不同的问题。假设我有这个列表：

 [['she', 'is', 'a', 'student'],
 ['she', 'is', 'a', 'lawer'],
 ['she', 'is', 'a', 'great', 'student'],
 ['i', 'am', 'a', 'teacher'],
 ['she', 'is', 'a', 'very', 'very', 'exceptionally', 'good', 'student']]

我希望使用matches = ['she', 'is', 'student']进行查询，并打算从查询列表中查看包含matches元素的所有子列表。链接中问题的唯一区别是我想向range函数添加find_gappy参数，因此它将避免检索元素之间的间隙超出指定范围的子列表。例如，在上面的例子中，我想要一个像这样的函数：

matches = ['she', 'is', 'student']
x = [i for i in x if find_gappy(i, matches, range=2)]

会返回：

[['she', 'is', 'a', 'student'], ['she', 'is', 'a', 'great', 'student']]

最后一个元素没有显示，因为在she is a very very exceptionally good student句中，a和good之间的距离超出了范围限制。

我怎样才能写出这样的功能？

之间的差距

Answer 1

以下是将match列表中的项目顺序考虑在内的一种方法：

In [102]: def find_gappy(all_sets, matches, gap_range=2):
     ...:     zip_m = list(zip(matches, matches[1:]))
     ...:     for lst in all_sets:
     ...:         indices = {j: i for i, j in enumerate(lst)}
     ...:         try:
     ...:             if all(0 <= indices[j]-indices[i] - 1 <= gap_range for i, j in zip_m):
     ...:                 yield lst
     ...:         except KeyError:
     ...:             pass
     ...:         
     ...:

演示：

In [110]: lst = [['she', 'is', 'a', 'student'],
     ...:  ['student', 'she', 'is', 'a', 'lawer'],  # for order check
     ...:  ['she', 'is', 'a', 'great', 'student'],
     ...:  ['i', 'am', 'a', 'teacher'],
     ...:  ['she', 'is', 'a', 'very', 'very', 'exceptionally', 'good', 'student']] 
     ...:  

In [111]: 

In [111]: list(find_gappy(lst, ['she', 'is', 'student'], gap_range=2))
Out[111]: [['she', 'is', 'a', 'student'], ['she', 'is', 'a', 'great', 'student']]

如果您的子列表中有重复的字词，您可以使用defaultdict()来跟踪所有索引，并使用itertools.prodcut来比较所有已订购字对的产品的差距。

In [9]: from collections import defaultdict
In [10]: from itertools import product

In [10]: def find_gappy(all_sets, matches, gap_range=2):
    ...:     zip_m = list(zip(matches, matches[1:]))
    ...:     for lst in all_sets:
    ...:         indices = defaultdict(list)
    ...:         for i, j in enumerate(lst):
    ...:             indices[j].append(i)
    ...:         try:
    ...:             if all(any(0 <= v - k - 1 <= gap_range for k, v in product(indices[j], indices[i])) for i, j in zip_m):
    ...:                 yield lst
    ...:         except KeyError:
    ...:             pass

Answer 2

链接问题中的技术足够好，你只需要在途中添加间隙，并且由于你不想要全局计数，所以每当遇到匹配时重置计数器。类似的东西：

import collections

def find_gappy(source, matches, max_gap=-1):
    matches = collections.deque(matches)
    counter = max_gap  # initialize as -1 if you want to begin counting AFTER the first match
    for word in source:
        if word == matches[0]:
            counter = max_gap  # or remove this for global gap counting
            matches.popleft()
            if not matches:
                return True
        else:
            counter -= 1
            if counter == -1:
                return False
    return False

data = [['she', 'is', 'a', 'student'],
        ['she', 'is', 'a', 'lawer'],
        ['she', 'is', 'a', 'great', 'student'],
        ['i', 'am', 'a', 'teacher'],
        ['she', 'is', 'a', 'very', 'very', 'exceptionally', 'good', 'student']]

matches = ['she', 'is', 'student']
x = [i for i in data if find_gappy(i, matches, 2)]
# [['she', 'is', 'a', 'student'], ['she', 'is', 'a', 'great', 'student']]

作为奖励，您可以将其用作原始函数，仅当您将正数传递为max_gap时才应用间隙计数。

在一定范围内找到gappy子列表

2 个答案: