python如何在字符串中动态查找人名

Question

我正在做一个项目，我必须使用语音到文本作为输入来确定呼叫谁，但是使用语音到文本会产生一些意想不到的结果，所以我想要对字符串进行一点动态匹配，我我从小开始尝试匹配 1 个单一的名字，我的名字是 Nick Vaes，我尝试将我的名字与口语文本匹配，但我也希望它匹配例如某些文本是 Nik 或其他东西，理想情况下我如果只有 1 个字母是错误的，那么希望有一些可以匹配所有内容的东西，所以

尼克 ick 尼克 网卡 恩克

都符合我的名字，我目前的简单代码是：

  def user_to_call(s):
  if "NICK" or "NIK" in s.upper(): redirect = "Nick"
  if redirect: return redirect

对于 4 个字母的名称，可以将所有可能性都放在过滤器中，但对于 12 个字母的名称，这有点矫枉过正，因为我很确定它可以更有效地完成。

Answer 1

你需要使用Levenshtein_distance

一个python实现是nltk

import nltk
nltk.edit_distance("humpty", "dumpty")

Answer 2

你基本上需要的是模糊字符串匹配，见：

https://en.wikipedia.org/wiki/Approximate_string_matching

https://www.datacamp.com/community/tutorials/fuzzy-string-python

基于此，您可以检查输入与您的字典相比有多相似：

 from fuzzywuzzy import fuzz

 name = "nick"
 tomatch = ["Nick", "ick", "nik", "nic", "nck", "nickey", "njick", "nickk", "nickn"]
 for str in tomatch:
    ratio = fuzz.ratio(str.lower(), name.lower())
    print(ratio)

此代码将产生以下输出：

100
86
86
86
86
80
89
89
89

您必须尝试不同的比率并检查哪个符合您的要求，以免错过一个字母

Answer 3

据我了解，您没有看到任何模糊匹配。 （因为您没有支持其他回复）。 如果您只是想评估您在请求中指定的内容，这里是代码。我在打印相应消息的地方添加了一些附加条件。随意删除它们。

def wordmatch(baseword, wordtoMatch, lengthOfMatch):
    lis_of_baseword = list(baseword.lower())
    lis_of_wordtoMatch = list(wordtoMatch.lower()) 
    sum = 0
    for index_i, i in enumerate(lis_of_wordtoMatch):
        for index_j, j in enumerate(lis_of_baseword):
            if i in lis_of_baseword:
                if i == j and index_i <= index_j:
                    sum = sum + 1
                    break
                else:
                    pass
            else:
                print("word to match has characters which are not in baseword")
                return 0
    if sum >= lengthOfMatch and len(wordtoMatch) <= len(baseword):
        return 1
    elif sum >= lengthOfMatch and len(wordtoMatch) > len(baseword):
        print("word to match has no of characters more than that of baseword")
        return 0
    else:
        return 0

base = "Nick"
tomatch = ["Nick", "ick", "nik", "nic", "nck", "nickey","njick","nickk","nickn"]
wordlength_match = 3 # this says how many words to match in the base word. In your case, its 3

for t_word in tomatch:
    print(wordmatch(base,t_word,wordlength_match))

输出如下所示

1
1
1
1
1
word to match has characters which are not in baseword
0
word to match has characters which are not in baseword
0
word to match has no of characters more than that of baseword
0
word to match has no of characters more than that of baseword
0

让我知道这是否符合您的目的。