如何使用美丽的汤检索嵌入在标签中的 xml 文件中的特征

如何解决如何使用美丽的汤检索嵌入在标签中的 xml 文件中的特征

我正在尝试解析一系列 XML 文件，并使用美丽的汤使用这些函数来获取嵌入在标签中的某些值：

case_feature_keys = ['year','offenceCategory','offenceSubcategory']
person_feature_keys = ['gender','age','occupation','given']
outcome_key = 'verdictCategory'

case_feature_keys = ['year','given']
outcome_key = 'verdictCategory'


def get_person_features(trial_account,person_type: str):
    person_features = {}
    for key in person_feature_keys:
        matches = [x for x in trial_account.find_all(type=key) if person_type in 
x.parent.attrs.get("type","")]
        if matches:
            person_features[person_type + "_" + key] = matches[0]
    return person_features

def process_trial_account(trial_account) -> dict:
    """
    Takes in a single account and returns a dictionary representing a row of the table.
    """

    case_features = {key: trial_account.find(type=key) for key in case_feature_keys}
    defendant_features = get_person_features(trial_account,'defendant')
    victim_features = get_person_features(trial_account,'victim')
    outcome = trial_account.find(type=outcome_key)


    features = {**case_features,**defendant_features,**victim_features,"outcome": outcome or {}}
    return {key: value.get("value") for key,value in features.items()}

XML 文件如下所示：

</persName>
                
             .</p>
 <p>
 <persName id="t18100221-1-person52">
                   GEORGE 
                   ROSS
                <interp inst="t18100221-1-person52" type="surname" value="ROSS"/>
 <interp inst="t18100221-1-person52" type="given" value="GEORGE"/>
 <interp inst="t18100221-1-person52" type="gender" value="male"/>
 </persName>
             . Q. Were you in trade - A. Yes,as a <rs id="t18100221-1-viclabel3" type="occupation">merchant</rs>
 <join result="persNameOccupation" targOrder="Y" targets="t18100221-1-victim51 t18100221-1-viclabel3"/>; I lived in <placeName id="t18100221-1-crimeloc4">New Basinghall-street</placeName>
 <interp inst="t18100221-1-crimeloc4" type="placeName" value="New Basinghall-street"/>
 <interp inst="t18100221-1-crimeloc4" type="type" value="crimeLocation"/>
 <join result="offencePlace" targOrder="Y" targets="t18100221-1-off1 t18100221-1-crimeloc4"/>; the prisoner was my <rs id="t18100221-1-deflabel5" type="occupation">clerk</rs>

我遇到的问题是职业类别不像大多数其他特征那样包含在“值=”行中。如果你看下面，职业是嵌入在标签本身中的，就像这样：'id="t18100221-1-viclabel3" type="occupation">merchant' 而不是性别，例如，它包含在这样的行中：'type="gender" value="male"/>' 所以我可以使用上面的函数来获取这个属性，因为它包含在一个类型/值中。

有谁知道我如何为受害者和被告找回职业？

解决方法

要获取职业和人员类型，您可以使用以下示例：

from bs4 import BeautifulSoup

html_data = """
 <persName id="t18100221-1-person52">
                   GEORGE 
                   ROSS
                <interp inst="t18100221-1-person52" type="surname" value="ROSS"/>
 <interp inst="t18100221-1-person52" type="given" value="GEORGE"/>
 <interp inst="t18100221-1-person52" type="gender" value="male"/>
 </persName>
             . Q. Were you in trade - A. Yes,as a <rs id="t18100221-1-viclabel3" type="occupation">merchant</rs>
 <join result="persNameOccupation" targOrder="Y" targets="t18100221-1-victim51 t18100221-1-viclabel3"/>; I lived in <placeName id="t18100221-1-crimeloc4">New Basinghall-street</placeName>
 <interp inst="t18100221-1-crimeloc4" type="placeName" value="New Basinghall-street"/>
 <interp inst="t18100221-1-crimeloc4" type="type" value="crimeLocation"/>
 <join result="offencePlace" targOrder="Y" targets="t18100221-1-off1 t18100221-1-crimeloc4"/>; the prisoner was my <rs id="t18100221-1-deflabel5" type="occupation">clerk</rs>
 """

soup = BeautifulSoup(html_data,"html.parser")

for occupation in soup.select('[type="occupation"]'):
    id_ = occupation["id"]
    o = occupation.text
    person_type = "victim" if "vic" in id_ else "defendant"
    print("ID: {} Occupation: {} Person type: {}".format(id_,o,person_type))

打印：

ID: t18100221-1-viclabel3 Occupation: merchant Person type: victim
ID: t18100221-1-deflabel5 Occupation: clerk Person type: defendant