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我得到了一个非常奇怪的数据。我有带有键和值的字典，我想在其中使用该字典来搜索这些关键字是否仅是文本的开头和/或结尾而不是句子的中间。我尝试在下面创建简单的数据框来显示我迄今为止尝试过的问题案例和 python 代码。我如何让它只搜索句子的开头或结尾？这个搜索整个文本子字符串。

代码：

d = {'apple corp':'Company','app':'Application'} #dictionary
l1 = [1, 2, 3,4]
l2 = [
    "The word Apple is commonly confused with Apple Corp which is a business",
    "Apple Corp is a business they make computers",
    "Apple Corp also writes App",
    "The Apple Corp also writes App"
]
df = pd.DataFrame({'id':l1,'text':l2})
df['text'] = df['text'].str.lower()
df

原始数据框：

id   text 
1    The word Apple is commonly confused with Apple Corp which is a business         
2    Apple Corp is a business they make computers                                    
3    Apple Corp also writes App                                                      
4    The Apple Corp also writes App

代码试用：

def matcher(k):
    x = (i for i in d if i in k)
    # i.startswith(k) getting error
    return ';'.join(map(d.get, x))
df['text_value'] = df['text'].map(matcher)
df

错误： TypeError: 'in <string>' requires string as left operand, not bool 当我使用它时x = (i for i in d if i.startswith(k) in k)

如果我尝试这个，则为空值x = (i for i in d if i.startswith(k) == True in k)

TypeError: sequence item 0: expected str instance, NoneType found 当我使用这个x = (i.startswith(k) for i in d if i in k)

上述代码的结果...创建新字段'text_value'：

id   text                                                                            text_value
1    The word Apple is commonly confused with Apple Corp which is a business         Company;Application
2    Apple Corp is a business they make computers                                    Company;Application
3    Apple Corp also writes App                                                      Company;Application
4    The Apple Corp also writes App                                                  Company;Application

试图获得这样的最终输出：

id   text                                                                            text_value
1    The word Apple is commonly confused with Apple Corp which is a business         NaN
2    Apple Corp is a business they make computers                                    Company
3    Apple Corp also writes App                                                      Company;Application
4    The Apple Corp also writes App                                                  Application
4

2 回答 2

1

您需要一个matcher可以接受flag然后调用两次以获取 和 的结果startswith的函数endswith。

def matcher(s, flag="start"):
    if flag=="start":
        for i in d:
            if s.startswith(i):
                return d[i]
    else:
        for i in d:
            if s.endswith(i):
                return d[i]
    return None

df['st'] = df['text'].apply(matcher)
df['ed'] = df['text'].apply(matcher, flag="end")
df['text_value'] = df[['st', 'ed']].apply(lambda x: ';'.join(x.dropna()),1)
df = df[['id','text', 'text_value']]

该text_value列如下所示：

0                       
1                Company
2    Company;Application
3            Application
Name: text_value, dtype: object
于 2019-08-27T19:51:25.660 回答
0 
joined = "|".join(d.keys())

pat = '(?i)^(?:the\\s*)?(' + joined + ')\\b.*?|.*\\b(' + joined + ')$'+'|.*'

get = lambda x: d.get(x.group(1),"") + (';' +d.get(x.group(2),"") if x.group(2) else '')

df.text.str.replace(pat,get)


0                       
1                Company
2    Company;Application
3    Company;Application
Name: text, dtype: object
于 2019-08-27T20:39:26.850 回答