Question

我正在探索一个数据结构，它扩展到子元素并解析为最终元素。但我只想存储前两个级别。

示例：让我们说我从纽约开始闯入布朗克斯，国王，纽约，皇后区和里士满作为县，但最后他们终于以某种方式决定去美国。

我不确定这是否是一个很好的例子，但只是在这里说清楚是对这个问题的更清楚的解释。

A (expands to) B,C,D -> B (expands to) K,L,M -> K resolves to Z

我最初是用for循环编写的，然后使用递归但是在递归中我放弃了一些扩展的元素，因为我没有深入挖掘每个扩展元素。我把这两个递归版本和非递归版本。我正在寻找有关构建此数据结构的一些建议，以及最佳方法。

我为扩展版本中的每个元素调用数据库查询，该元素返回项目列表。直到它解析为单个元素。随着递归，我不会一直松动钻孔，直到其他人解决的最终元素。但随着递归它不一样。我也是python的新手，所以希望在这样的网站上提问这不是一个糟糕的问题。

returnCategoryQuery 是一种通过调用数据库查询来返回项目列表的方法。

没有递归

#Dictionary to save initial category with the rest of cl_to
baseCategoryTree = {};
#categoryResults = [];

# query get all the categories a category is linked to
categoryQuery = "select cl_to from categorylinks cl left join page p on cl.cl_from = p.page_id where p.page_namespace=14 and p.page_title ='";
cursor = db.cursor(cursors.SSDictCursor);

    for key, value in idTitleDictionary.iteritems():
        for startCategory in value[0]:
            #print startCategory + "End of Query";
            categoryResults = [];
            try:
                categoryRow = "";
                baseCategoryTree[startCategory] = [];
                print categoryQuery + startCategory + "'";
                cursor.execute(categoryQuery + startCategory + "'");
                done = False;
                while not done:
                    categoryRow = cursor.fetchone();
                    if not categoryRow:
                        done = True;
                        continue;
                    categoryResults.append(categoryRow['cl_to']);
                for subCategoryResult in categoryResults:
                    print startCategory.encode('ascii') + " - " +  subCategoryResult;
                    for item in returnCategoryQuery(categoryQuery + subCategoryResult + "'"):
                        print startCategory.encode('ascii') + " - " + subCategoryResult + " - "  + item;
                        for subItem in returnCategoryQuery(categoryQuery + item + "'"):
                            print startCategory.encode('ascii') + " - " + subCategoryResult + " - "  + item + " - " + subItem;
                            for subOfSubItem in returnCategoryQuery(categoryQuery + subItem + "'"):
                                 print startCategory.encode('ascii') + " - " + subCategoryResult + " - "  + item + " - " + subItem + " - " + subOfSubItem;
                                 for sub_1_subOfSubItem in returnCategoryQuery(categoryQuery + subOfSubItem + "'"):
                                      print startCategory.encode('ascii') + " - " + subCategoryResult + " - "  + item + " - " + subItem + " - " + subOfSubItem + " - " + sub_1_subOfSubItem;
                                      for sub_2_subOfSubItem in returnCategoryQuery(categoryQuery + sub_1_subOfSubItem + "'"):
                                          print startCategory.encode('ascii') + " - " + subCategoryResult + " - "  + item + " - " + subItem + " - " + subOfSubItem + " - " + sub_1_subOfSubItem + " - " + sub_2_subOfSubItem;
            except Exception, e:
                traceback.print_exc();

使用递归

def crawlSubCategory(subCategoryList):
    level = 1;
    expandedList = [];
    for eachCategory in subCategoryList:
        level = level + 1
        print "Level  " + str(level) + " " + eachCategory;
        #crawlSubCategory(returnCategoryQuery(categoryQuery + eachCategory + "'"));
        for subOfEachCategory in returnCategoryQuery(categoryQuery + eachCategory + "'"):
            level = level + 1
            print "Level  " + str(level) + " " + subOfEachCategory;
            expandedList.append(crawlSubCategory(returnCategoryQuery(categoryQuery + subOfEachCategory + "'")));
    return expandedList;


#Dictionary to save initial category with the rest of cl_to
baseCategoryTree = {};
#categoryResults = [];

# query get all the categories a category is linked to
categoryQuery = "select cl_to from categorylinks cl left join page p on cl.cl_from = p.page_id where p.page_namespace=14 and p.page_title ='";
cursor = db.cursor(cursors.SSDictCursor);

for key, value in idTitleDictionary.iteritems():
    for startCategory in value[0]:
        #print startCategory + "End of Query";
        categoryResults = [];
        try:
            categoryRow = "";
            baseCategoryTree[startCategory] = [];
            print categoryQuery + startCategory + "'";
            cursor.execute(categoryQuery + startCategory + "'");
            done = False;
            while not done:
                categoryRow = cursor.fetchone();
                if not categoryRow:
                    done = True;
                    continue;
                categoryResults.append(categoryRow['cl_to']);
            #crawlSubCategory(categoryResults);
        except Exception, e:
            traceback.print_exc();
        #baseCategoryTree[startCategory].append(categoryResults);
        baseCategoryTree[startCategory].append(crawlSubCategory(categoryResults));

Answer 1

您是否正在尝试查找“皇后区”并了解它位于美国？您是否尝试使用XML编码树，并使用lxml.etree查找元素，然后使用getpath以XPath格式返回路径？

这意味着在您的树中添加第四个顶级，即World，然后您将搜索Queens并了解Queens的路径为World/USA/NewYork/Queens。您问题的答案始终是XPath中的第二项。

当然，你总是可以从XML构建一个树并使用树搜索算法。

Python：扩展复杂的树数据结构

没有递归

使用递归

1 个答案: