Python中是否有一个soundex函数？

Question

Python中是否有一个用于soundex的函数？如果没有，您将如何创建一个soundex代码？

Soundex
Code    Letters 
1   B, F, P, V  
2   C, G, J, K, Q, S, X, Z  
3   D, T    
4   L   
5   M, N    
6   R   
SKIP   A, E, H, I, O, U, W, Y, H, W, and Y

例如：

Jackson = J250

Washington = W252

Clement = C455

Ashcraft = A261

Wu = W000

- Josh C

---

这个问题不适合在StackOverflow上发布。（为了帮助你，我在谷歌搜索标题后找到了一个soundex包。） - Bhargav Rao

你尝试过谷歌搜索吗？https://pypi.python.org/pypi/Fuzzy - Goodies

@BhargavRao，为什么这个不是主题？ - Joey Baruch

Answer 1

是的，你可以使用Fuzzy库，它是一个实现一些语音算法的Python库。

sudo pip install fuzzy

>>> import fuzzy
>>> soundex = fuzzy.Soundex(4)
>>> soundex("Jackson")
'J250'
>>> soundex("Washington")
'W252'
>>> soundex("Clement")
'C453'
>>> soundex("Ashcraft")
'A261'
>>> soundex("Wu")
'W000'

- Abdullah Abdelmonem

Answer 2

你可以使用水母。

sudo pip install jellyfish

print "Soundex\t\t=", jellyfish.soundex("Ala ma kaca")
>Soundex                = A452
#...
>Metaphone              = AL M KK
>NYSIIS                 = AL
>Match rating codex     = ALMKC

- ravyr

Answer 3

直接使用下面的soundex()函数，无需安装任何软件包！

代码片段来自软件包Jellyfish > _jellyfish.py

示例

print(soundex('kent')) # K530
print(soundex('Paul')) # P400
print(soundex('amnesty')) # A523

代码

import unicodedata
def soundex(s):

    if not s:
        return ""

    s = unicodedata.normalize("NFKD", s)
    s = s.upper()

    replacements = (
        ("BFPV", "1"),
        ("CGJKQSXZ", "2"),
        ("DT", "3"),
        ("L", "4"),
        ("MN", "5"),
        ("R", "6"),
    )
    result = [s[0]]
    count = 1

    # find would-be replacment for first character
    for lset, sub in replacements:
        if s[0] in lset:
            last = sub
            break
    else:
        last = None

    for letter in s[1:]:
        for lset, sub in replacements:
            if letter in lset:
                if sub != last:
                    result.append(sub)
                    count += 1
                last = sub
                break
        else:
            if letter != "H" and letter != "W":
                # leave last alone if middle letter is H or W
                last = None
        if count == 4:
            break

    result += "0" * (4 - count)
    return "".join(result)

- Kent Pawar