phrases = ['i am good', 'going to the market', 'eating cookies']dictionary = {'http://www.firsturl.com': 'i am going to the market and tomorrow will be eating cookies', 'http://www.secondurl.com': 'tomorrow is my birthday and i shall be', 'http://www.thirdurl.com': 'i am good and will go to sleep'}如果至少有一個匹配項：預(yù)期輸出：url phrasecount phrasehttp://www.firsturl.com 2 going to the market, eating cookieshttp://www.thirdurl.com 1 i am good如果所有 3 個 url 都沒有匹配項，則只返回第一次出現(xiàn)的 url，計數(shù)為零，預(yù)期輸出為空白短語：url phrasecount phrasehttp://www.firsturl.com 0
1 回答
拉風的咖菲貓
TA貢獻1995條經(jīng)驗 獲得超2個贊
df從相應(yīng)的設(shè)置初始數(shù)據(jù)幀dictionary：
df = pd.DataFrame({'urls': list(dictionary.keys()), 'strings': list(dictionary.values())})
pattern = '|'.join(phrases)
處理數(shù)據(jù)幀：
s = df.pop('strings').str.findall(pattern)
df = df.assign(phrasecount=s.str.len(), phrase=s.map(', '.join))
df = df.drop_duplicates(subset='phrasecount') if df['phrasecount'].eq(0).all() else df[df['phrasecount'].ne(0)]
結(jié)果：
# print(df)
urls phrasecount phrase
0 http://www.firsturl.com 2 going to the market, eating cookies
2 http://www.thirdurl.com 1 i am good
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