Python语句字符串滑动窗口

我正在寻找一个滑动窗口拆分器，由窗口大小为N的单词组成

输入：“我喜欢食物，我喜欢饮料”，窗口大小3

输出：[“我爱食物”，“爱食物和”，“食物和我”，“我喜欢”…]

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所有滑动窗口的建议都是围绕字符串的顺序，没有术语。有什么现成的吗？

您可以使用具有不同偏移量的迭代器并压缩所有偏移量

def token_sliding_window(str, size):
    tokens = str.split(' ')
    for i in range(len(tokens )- size + 1):
        yield tokens[i: i+size]

>>> arr = "I love food. blah blah".split()
>>> its = [iter(arr), iter(arr[1:]), iter(arr[2:])] #Construct the pattern for longer windowss
>>> zip(*its)
[('I', 'love', 'food.'), ('love', 'food.', 'blah'), ('food.', 'blah', 'blah')]

如果您有长句，或者可能是普通的旧循环（如另一个答案中所示），则可能需要使用。

一种基于订阅字符串序列的方法：

def split_on_window(sequence="I love food and I like drink", limit=4):
    results = []
    split_sequence = sequence.split()
    iteration_length = len(split_sequence) - (limit - 1)
    max_window_indicies = range(iteration_length)
    for index in max_window_indicies:
        results.append(split_sequence[index:index + limit])
    return results

样本输出：

>>> split_on_window("I love food and I like drink", 3)
['I', 'love', 'food']
['love', 'food', 'and']
['food', 'and', 'I']
['and', 'I', 'like']
['I', 'like', 'drink']

>>> list(split_on_window(s, 4))
[('I', 'love', 'food', 'and'), ('love', 'food', 'and', 'I'), 
('food', 'and', 'I', 'like'), ('and', 'I', 'like', 'drink')]

Sequence = I love food and I like drink, limit = 3
Repetitions = 1000000
Using subscripting -> 3.8326420784
Using izip -> 5.41380286217 # Modified to return a list for the benchmark.

以下是一个受@SuperSaiyan启发的备选答案：

from itertools import izip

def split_on_window(sequence, limit):
    split_sequence = sequence.split()
    iterators = [iter(split_sequence[index:]) for index in range(limit)]
    return izip(*iterators)

样本输出：

>>> split_on_window("I love food and I like drink", 3)
['I', 'love', 'food']
['love', 'food', 'and']
['food', 'and', 'I']
['and', 'I', 'like']
['I', 'like', 'drink']

>>> list(split_on_window(s, 4))
[('I', 'love', 'food', 'and'), ('love', 'food', 'and', 'I'), 
('food', 'and', 'I', 'like'), ('and', 'I', 'like', 'drink')]

Sequence = I love food and I like drink, limit = 3
Repetitions = 1000000
Using subscripting -> 3.8326420784
Using izip -> 5.41380286217 # Modified to return a list for the benchmark.

基准：

>>> split_on_window("I love food and I like drink", 3)
['I', 'love', 'food']
['love', 'food', 'and']
['food', 'and', 'I']
['and', 'I', 'like']
['I', 'like', 'drink']

>>> list(split_on_window(s, 4))
[('I', 'love', 'food', 'and'), ('love', 'food', 'and', 'I'), 
('food', 'and', 'I', 'like'), ('and', 'I', 'like', 'drink')]

Sequence = I love food and I like drink, limit = 3
Repetitions = 1000000
Using subscripting -> 3.8326420784
Using izip -> 5.41380286217 # Modified to return a list for the benchmark.

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下面是我最后做的：def find_ngrams（input_list，n）：返回zip（*[input_list[I:]表示范围（n）中的I）]