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例如，句子是"The corporate balance sheets data are available on an annual basis"，我需要标记"corporate balance sheets"从给定句子中找到的子字符串。

所以，我需要找到的模式是：

"corporate balance sheets"

给定字符串：

"The corporate balance sheets data are available on an annual basis".

我想要的输出标签序列将是：

[0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]

有一堆句子（超过 2GB），还有一堆我需要找到的模式。我不知道如何在 python 中有效地做到这一点。有人可以给我一个好的算法吗？

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2 回答 2

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列表理解和使用拆分：

import re
lst=[]
search_word = 'corporate balance sheets'
p = re.compile(search_word)
sentence="The corporate balance sheets data are available on an annual basis"

lst=[1 for i in range(len(search_word.split()))]
vect=[ lst if items == '__match_word' else 0 for items in re.sub(p,'__match_word',sentence).split()]
vectlstoflst=[[vec] if isinstance(vec,int) else vec for vec in vect]
flattened = [val for sublist in vectlstoflst for val in sublist]

输出：

 [0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]

Sentence ="公司资产负债表数据可在年表上获得"

输出

[0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0]
于 2018-11-26T04:33:30.420 回答
1

由于子字符串中的所有单词都必须匹配，因此您可以all在遍历句子时检查并更新适当的索引：

def encode(sub, sent):
    subwords, sentwords = sub.split(), sent.split()
    res = [0 for _ in sentwords]    
    for i, word in enumerate(sentwords[:-len(subwords) + 1]):
        if all(x == y for x, y in zip(subwords, sentwords[i:i + len(subwords)])):
            for j in range(len(subwords)):
                res[i + j] = 1
    return res


sub = "corporate balance sheets"
sent = "The corporate balance sheets data are available on an annual basis"
print(encode(sub, sent))
# [0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]

sent = "The corporate balance data are available on an annual basis sheets"
print(encode(sub, sent))
# [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
于 2018-11-26T04:53:59.870 回答