提取段落中与列表中的单词相似的单词

Question

我有以下字符串：

"The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"

要提取的单词列表：

["town","teddy","chicken","boy went"]

注：town和teddy在给定的句子中拼写错误

我尝试了以下方法，但我得到的其他词语不是答案的一部分：

import difflib

sent = "The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"

list1 = ["town","teddy","chicken","boy went"]

[difflib.get_close_matches(x.lower().strip(), sent.split()) for x in list1 ]

我得到以下结果：

[['twn', 'to'], ['tddy'], ['chicken.', 'picked'], ['went']]

而不是：

'twn', 'tddy', 'chicken','boy went'

Answer 1

在^{}的文档中注意：

difflib.get_close_matches(word, possibilities, n=3, cutoff=0.6)

Return a list of the best "good enough" matches. word is a sequence for which close matches are desired (typically a string), and possibilities is a list of sequences against which to match word (typically a list of strings).

Optional argument n (default 3) is the maximum number of close matches to return; n must be greater than 0.

Optional argument cutoff (default 0.6) is a float in the range [0, 1]. Possibilities that don’t score at least that similar to word are ignored.

---

目前，您正在使用默认的n和cutoff参数

您可以指定其中一个（或两个），以缩小返回的匹配项的范围

例如，您可以使用0.75的cutoff分数：

result = [difflib.get_close_matches(x.lower().strip(), sent.split(), cutoff=0.75) for x in list1]

或者，您可以指定最多只返回1个匹配项：

result = [difflib.get_close_matches(x.lower().strip(), sent.split(), n=1) for x in list1]

在任何一种情况下，您都可以使用列表理解来展平列表列表（因为difflib.get_close_matches()总是返回一个列表）：

matches = [r[0] for r in result]

---

由于您还希望检查bigram的紧密匹配，因此可以通过提取相邻“单词”的配对，并将它们作为possibilities参数的一部分传递给difflib.get_close_matches()

下面是一个完整的工作示例：

import difflib
import re

sent = "The boy went to twn and bought sausage and chicken. He then picked a tddy for his sister"

list1 = ["town", "teddy", "chicken", "boy went"]

# this extracts overlapping pairings of "words"
# i.e. ['The boy', 'boy went', 'went to', 'to twn', ...
pairs = re.findall(r'(?=(\b[^ ]+ [^ ]+\b))', sent)

# we pass the sent.split() list as before
# and concatenate the new pairs list to the end of it also
result = [difflib.get_close_matches(x.lower().strip(), sent.split() + pairs, n=1) for x in list1]

matches = [r[0] for r in result]

print(matches)
# ['twn', 'tddy', 'chicken.', 'boy went']

Answer 2

如果您阅读了关于ifflib.get_close_matches（）的Python文档 https://docs.python.org/3/library/difflib.html 它返回所有可能的最佳匹配。 方法签名： difflib.get_close_匹配（单词，可能性，n=3，截止值=0.6）

这里n是要返回的最大接近匹配数。所以我想你可以把这个作为1通过

>>> [difflib.get_close_matches(x.lower().strip(), sent.split(),1)[0] for x in list1]
['twn', 'tddy', 'chicken.', 'went']