通过单词在URL熊猫中的位置查找

Question

基本上，我有一个数据框，其中一列是名称列表，另一列是通过某种方式与名称相关的关联网址（示例df）：

   Name                    Domain
'Apple Inc'             'https://mapquest.com/askjdnas387y1/apple-inc', 'https://linkedin.com/apple-inc/askjdnas387y1/', 'https://www.apple-inc.com/asdkjsad542/'     
'Aperture Industries'   'https://www.cakewasdelicious.com/aperture/run-away/', 'https://aperture-incorporated.com/aperture/', 'https://www.buzzfeed.com/aperture/the-top-ten-most-evil-companies=will-shock-you/'
'Umbrella Corp'         'https://www.umbrella-corp.org/were-not-evil/', 'https://umbrella.org/experiment-death/', 'https://www.most-evil.org/umbrella-corps/'

我试图在以下任一情况下直接找到具有关键字或至少与关键字部分匹配的网址：

'https://NAME.whateverthispartdoesntmatter'

或

'https://www.NAME.whateverthispartdoesntmatter' <- not a real link

现在我正在使用Fuzzywuzzy获取部分匹配项：

fuzz.token_set_ratio(name, value)

它对于部分匹配非常有效，但是匹配不依赖于位置，因此我会得到一个完美的关键字匹配，但它位于URL中间的某个地方，不是我想要的：

https://www.bloomberg.com/profiles/companies/aperture-inc/0117091D 

Answer 1

使用explode/unnest string，str.extract和fuzzywuzzy

首先，我们将使用this函数将您的字符串嵌套到行中：

df = explode_str(df, 'Domain', ',').reset_index(drop=True)

然后，我们使用正则表达式查找带有或不带有www的两个模式，并从中提取名称：

m = df['Domain'].str.extract('https://www.(.*)\.|https://(.*)\.')
df['M'] = m[0].fillna(m[1])
print(df)


                  Name                                             Domain                      M
0            Apple Inc       https://mapquest.com/askjdnas387y1/apple-inc               mapquest
1            Apple Inc      https://linkedin.com/apple-inc/askjdnas387y1/               linkedin
2            Apple Inc             https://www.apple-inc.com/asdkjsad542/              apple-inc
3  Aperture Industries  https://www.cakewasdelicious.com/aperture/run-...       cakewasdelicious
4  Aperture Industries        https://aperture-incorporated.com/aperture/  aperture-incorporated
5  Aperture Industries   https://www.buzzfeed.com/aperture/the-top-ten...               buzzfeed
6        Umbrella Corp       https://www.umbrella-corp.org/were-not-evil/          umbrella-corp
7        Umbrella Corp             https://umbrella.org/experiment-death/               umbrella
8        Umbrella Corp          https://www.most-evil.org/umbrella-corps/              most-evil

然后，我们使用fuzzywuzzy来过滤匹配度高于80的行：

from fuzzywuzzy import fuzz

m2 = df.apply(lambda x: fuzz.token_sort_ratio(x['Name'], x['M']), axis=1)

df[m2>80]

---

            Name                                        Domain              M
2      Apple Inc        https://www.apple-inc.com/asdkjsad542/      apple-inc
6  Umbrella Corp  https://www.umbrella-corp.org/were-not-evil/  umbrella-corp

---

注意，我使用token_sort_ratio而不是token_set_ratio来捕捉umbrella和umbrella-corp的区别

---

链接答案中使用的功能：

def explode_str(df, col, sep):
    s = df[col]
    i = np.arange(len(s)).repeat(s.str.count(sep) + 1)
    return df.iloc[i].assign(**{col: sep.join(s).split(sep)})