python – 比较名称之间的相似性

我必须根据名称对某些数据进行交叉验证.

我面临的问题是,根据来源,名称有轻微的变化,例如：

L & L AIR CONDITIONING   vs L & L AIR CONDITIONING Service

BEST ROOFING vs ROOFING INC

我有几千条记录,所以手动操作会非常耗时,我希望尽可能自动化这个过程.

由于还有其他单词,因此小写名称是不够的.

哪个是处理这个问题的好算法？

也许计算相关性给予“INC”或“服务”等词语较低的权重

编辑：

我尝试了difflib库

difflib.SequenceMatcher(None,name_1.lower(),name_2.lower()).ratio()

我用它获得了不错的结果.

解决方法:

我会用余弦相似来达到同样的效果.它会给你一个匹配得分,表示字符串的接近程度.

下面的代码可以帮助您(我记得几个月前从Stackoverflow本身获取此代码 – 现在找不到链接)

import re, math
from collections import Counter

WORD = re.compile(r'\w+')

def get_cosine(vec1, vec2):
    # print vec1, vec2
    intersection = set(vec1.keys()) & set(vec2.keys())
    numerator = sum([vec1[x] * vec2[x] for x in intersection])

    sum1 = sum([vec1[x]**2 for x in vec1.keys()])
    sum2 = sum([vec2[x]**2 for x in vec2.keys()])
    denominator = math.sqrt(sum1) * math.sqrt(sum2)

    if not denominator:
        return 0.0
    else:
        return float(numerator) / denominator

def text_to_vector(text):
    return Counter(WORD.findall(text))

def get_similarity(a, b):
    a = text_to_vector(a.strip().lower())
    b = text_to_vector(b.strip().lower())

    return get_cosine(a, b)

get_similarity('L & L AIR CONDITIONING', 'L & L AIR CONDITIONING Service') # returns 0.9258200997725514

我发现有用的另一个版本是基于NLP的,我创作了它.

import re, math
from collections import Counter
from nltk.corpus import stopwords
from nltk.stem.porter import *
from nltk.corpus import wordnet as wn

stop = stopwords.words('english')

WORD = re.compile(r'\w+')
stemmer = Porterstemmer()

def get_cosine(vec1, vec2):
    # print vec1, vec2
    intersection = set(vec1.keys()) & set(vec2.keys())
    numerator = sum([vec1[x] * vec2[x] for x in intersection])

    sum1 = sum([vec1[x]**2 for x in vec1.keys()])
    sum2 = sum([vec2[x]**2 for x in vec2.keys()])
    denominator = math.sqrt(sum1) * math.sqrt(sum2)

    if not denominator:
        return 0.0
    else:
        return float(numerator) / denominator

def text_to_vector(text):
    words = WORD.findall(text)
    a = []
    for i in words:
        for ss in wn.synsets(i):
            a.extend(ss.lemma_names())
    for i in words:
        if i not in a:
            a.append(i)
    a = set(a)
    w = [stemmer.stem(i) for i in a if i not in stop]
    return Counter(w)

def get_similarity(a, b):
    a = text_to_vector(a.strip().lower())
    b = text_to_vector(b.strip().lower())

    return get_cosine(a, b)

def get_char_wise_similarity(a, b):
    a = text_to_vector(a.strip().lower())
    b = text_to_vector(b.strip().lower())
    s = []

    for i in a:
        for j in b:
            s.append(get_similarity(str(i), str(j)))
    try:
        return sum(s)/float(len(s))
    except: # len(s) == 0
        return 0

get_similarity('I am a good boy', 'I am a very disciplined guy')
# Returns 0.5491201525567068

您可以调用get_similarity或get_char_wise_similarity来查看更适合您的用例的内容.我使用了两者 – 正常的相似性与非常接近的杂草,然后是明显的相似性,以排除足够接近的相似性.然后其余的必须手动处理.