在一定范围内找到有间隙的子列表

Question

最近我想在更大的列表中查找子列表。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']]

并且我想使用 matches = ['she', 'is', 'student'] 查询它，目的是从查询列表中以相同的顺序获取包含 matches 元素的所有子列表。与 link 中问题的唯一区别是我想向 find_gappy 函数添加一个 range 参数，这样它就不会检索元素之间存在间隙的子列表超出指定范围。例如，在上面的例子中，我想要一个这样的函数：

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

这会 return:

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

最后一个元素没有出现，因为在句子 she is a very very exceptionally good student 中，a 和 good 之间的距离超出了范围限制。

这样的函数怎么写呢？差距

Answer 1

链接问题中的技术足够好，您只需要沿途添加间隙计数，并且由于您不想要全局计数，因此在遇到匹配项时重置计数器。类似于：

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，仅当您将正数作为 max_gap 传递时才应用间隙计数。

Answer 2

这是一种将 match 列表中的项目顺序考虑在内的方法：

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']]

如果你的子列表中有重复的单词，你可以使用defaultdict()来跟踪所有索引并使用itertools.prodcut来比较所有单词对有序产品的差距。

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

在一定范围内找到有间隙的子列表

finding gappy sublists within a certain range

python

list

pattern-matching

sublist

python-3.x