将数据解析为列表列表答案

【问题标题】：Parsing data to lists of lists将数据解析为列表列表
【发布时间】：2017-09-02 13:51:23
【问题描述】：

我在尝试将数据解析为列表时遇到了问题。我正在尝试收集有关部门及其主题的信息。但是，由于每个部门的科目数量不同，我需要创建一个列表列表，以便以后可以将数据链接在一起。我设法解决了索引错误，问题似乎来自编译主题列表。

from lxml import html
import requests

page = requests.get('URL')

page_source_code = html.fromstring(page.text)


departments_list = []
subject_list = []

for dep in range(1,3):
    departments = page_source_code.xpath('tag'
                                         +str(dep)+']tag/text()')

    ### print(dep, departments)
    if departments == []:
        pass
    else:
        departments_list.append(departments[0])


    for sub in range(1,20):
        subjects = page_source_code.xpath('tag'
                                      +str(dep)+']tag'
                                      +str(sub)+']tag/text()')
        ### print(sub, subjects)
        if subjects == []:
            pass
        else:
            subject_list.append(subjects[0])

print('Department list ------ ', len(departments_list), departments_list, '\n')
print('Subject list ------ ', len(subject_list), subject_list)

我的输出如下所示：

Department list ------  2 ['Department_1', 'Department_2'] 

Subject list ------  7 ['Subject_1'(dep_1), 'Subject_2 '(dep_1), 'Subject_3 '(dep_1), 'Subject_4'(dep_1), 'Subject_5'(dep_2), 'Subject_6 '(dep_2), 'Subject_7 '(dep_2)']

此代码似乎将所有主题放在一个列表中。我希望如下：

Subject list ------  7 [['Subject_1'(dep_1), 'Subject_2 '(dep_1), 'Subject_3 '(dep_1), 'Subject_4'(dep_1)], ['Subject_5'(dep_2), 'Subject_6 '(dep_2), 'Subject_7 '(dep_2)']]

【问题讨论】：

标签： python list multidimensional-array xml-parsing

【解决方案1】：

您需要两个将两个列表全局添加为主题列表并找出 subject[0] 字符串中的 'dep_1' 或 'dep_2' 词。

     #declare the list for subject 
     sub_list1 = [] sub_list2 = []

     #this code is under the second for loop
     if subjects.find('dep_1') == -1 :
         sub_list2.append(subjects[0])
     else:
         sub_list1.append(subjects[0])

     #Please remove the subjectList.append statement 
     #from second for loop
     #and put it end of both loop like that .
     subjectList = [sub_list1,sub_list2]

【讨论】：