python - 循环字典值

Question

我目前有一个程序可以读取一个文本文件并根据里面的输入回答一组查询。它有助于弄清楚被问及的 child 的母亲是谁。我现在更进一步，重新处理提供的这些输出以显示完整的家谱。

这是包含左侧 parent 和右侧 child 的文本文件。下面是被询问的查询，后面是输出。

Sue: Chad, Brenda, Harris
Charlotte: Tim
Brenda: Freddy, Alice
Alice: John, Dick, Harry

mother Sue
ancestors Harry
ancestors Bernard
ancestors Charlotte

>>> Mother not known
>>> Alice, Brenda, Sue
>>> Unknown Person
>>> No known ancestors

该程序能够计算出母亲是谁(感谢 Kampu 的帮助)，但我现在正试图了解如何获取此值并可能将其附加到一个新的无限循环查找任何可能的祖 parent 的列表或字典。

def lookup_ancestor(child):
    ancestor = REVERSE_MAP.get(child)
    if ancestor:
        ANCESTOR_LIST_ADD.append(ancestor)
    if ancestor not in LINES[0:]:
        print("Unknown person")
    else:
        print("No known ancestors")

这是我目前所拥有的，其中 REVERSE_MAP 将每个子项映射到字典中的父项。然后，我将 parent 放入一个新列表中，我计划再次运行该列表以确定他们的 parent 是谁。然而，我被困在这一点上，因为我无法找到一种优雅的方式来执行整个过程，而无需创建三个新列表来保持循环。目前的设置方式，我假设我需要通过 for 循环附加它们或只是 split() 之后的值以保留所有彼此的值(value)观。理想情况下，我想学习如何循环这个过程并找出每个问题的祖先是谁。

我觉得我似乎已经掌握了它应该看起来如何，但是我对 Python 的了解使我的试错法无法节省时间。

如有任何帮助，我们将不胜感激!

编辑: 链接 - http://pastebin.com/vMpT1GvX

编辑 2:

def process_commands(commands, relation_dict):
    '''Processes the output'''
    output = []
    for c in commands: 
        coms = c.split()
        if len(coms) < 2: 
            output.append("Invalid Command")
            continue
        action = coms[0]
        param = coms[1]

def mother_lookup(action, param, relation_dict):
    output_a = []
    if action == "mother": 
        name_found = search_name(relation_dict, param)
        if not name_found: 
            output_a.append("Unknown person")
        else: 
            person = find_parent(relation_dict, param)
            if person is None: 
                output_a.append("Mother not known")
            else: 
                output_a.append(person)
    return output_a

def ancestor_lookup(action, param, relation_dict):
    output_b = []
    if action == "ancestors": 
        name_found = search_name(relation_dict, param)
        if not name_found: 
            output_b.append("Unknown person")
        else: 
            ancestor_list = []
            person = param
            while True: 
                person = find_parent(relation_dict, person)
                if person == None: 
                    break
                else: 
                    ancestor_list.append(person)
                if ancestor_list: 
                    output_b.append(", ".join(ancestor_list))
                else: 
                    output_b.append("No known ancestors")
    return output_b

def main():
    '''Definining the file and outputting it'''
    file_name = 'relationships.txt'
    relations,commands = read_file(file_name)
    #Process Relqations to form a dictionary of the type 
    #{parent: [child1,child2,...]}
    relation_dict = form_relation_dict(relations)
    #Now process commands in the file
    action = process_commands(commands, relation_dict)
    param = process_commands(commands, relation_dict)
    output_b = ancestor_lookup(action, param, relation_dict)
    output_a = mother_lookup(action, param, relation_dict)
    print('\n'.join(output_a))
    print ('\n'.join(output_b))


if __name__ == '__main__': 
    main()

Answer 1

正如@NullUserException 所说，一棵树(或类似的东西)是一个不错的选择。我发布的答案与您为这个问题选择的答案完全不同。

您可以定义一个 Person 对象，该对象知道自己的名字并跟踪其父对象是谁。 parent 不是一个名字，而是另一个 Person 对象! (有点像链表)。然后，您可以将人员集合保存为一个列表。

在解析文件时，您不断将人员添加到列表中，同时使用正确对象更新他们的子级/父级属性。

以后给定任何人，只需要打印属性来查找关系

以下是一个可能的实现(在 Python-2.6 上)。本例中的文本文件仅包含关系。稍后使用交互式输入触发查询

class Person(object): 
    """Information about a single name"""
    def __init__(self,name): 
        self.name = name
        self.parent = None
        self.children = []

def search_people(people,name): 
    """Searches for a name in known people and returns the corresponding Person object or None if not found"""
    try: 
        return filter(lambda y: y.name == name,people)[0]
    except IndexError: 
        return None

def search_add_and_return(people,name): 
    """Search for a name in list of people. If not found add to people. Return the Person object in either case"""
    old_name = search_people(people,name)
    if old_name is None: 
        #First time entry for the name
        old_name = Person(name)
        people.append(old_name)
    return old_name

def read_file(file_name,people): 
    fp = open(file_name,'r')
    while True: 
        l = fp.readline()
        l.strip('\n').strip()
        if not l: 
            break
        names = l.split(':')
        mother = names[0].strip()
        children = [x.strip() for x in names[1].split(',')]
        old_mother = search_add_and_return(people,mother)
        #Now get the children objects
        child_objects = []
        for child in children: 
            old_child = search_add_and_return(people,child)
            child_objects.append(old_child)
        #All children have been gathered. Link them up
        #Set parent in child and add child to parent's 'children'
        old_mother.children.extend(child_objects)
        for c in child_objects: 
            c.parent = old_mother
    fp.close()


def main(): 
    file_name = 'try.txt'
    people = []
    read_file(file_name,people)

    #Now lets define the language and start a loop
    while True: 
        command = raw_input("Enter your command or 0 to quit\n")
        if command == '0': 
            break
        coms = command.split()
        if len(coms) < 2: 
            print "Wrong Command"
            continue
        action = coms[0]
        param = coms[1]
        if action == "mother": 
            person = search_people(people,param)
            if person == None: 
                print "person not found"
                continue
            else: 
                if person.parent is None: 
                    print "mother not known"
                else: 
                    print person.parent.name
        elif action == "ancestors": 
            person = search_people(people,param)
            if person == None: 
                print "person not found"
                continue
            else: 
                ancestor_list = []
                #Need to keep looking up parent till we don't reach a dead end
                #And collect names of each ancestor
                while True: 
                    person = person.parent
                    if person is None: 
                        break
                    ancestor_list.append(person.name)
                if ancestor_list: 
                    print ",".join(ancestor_list)    
                else: 
                    print "No known ancestors"

if __name__ == '__main__': 
    main()

编辑

既然你想让事情变得简单，这里有一种使用字典(单个字典)来做你想做的事情的方法

基本思路如下。您解析文件以形成字典，其中键是 Mother，值是 list of children。所以当你的示例文件被解析时，你会得到一个像

relation_dict = {'Charlotte': ['Tim'], 'Sue': ['Chad', 'Brenda', 'Harris'], 'Alice': ['John', 'Dick', 'Harry'], 'Brenda': ['Freddy', 'Alice']}

要查找父项，只需搜索该名称是否在字典值中，如果找到则返回键。如果没有找到妈妈，则返回None

mother = None
for k,v in relation_dict.items(): 
    if name in v: 
        mother = k
        break
return mother

如果你想找到所有的祖先，你只需要重复这个过程直到没有返回

ancestor_list = []
person = name
while True: 
    person = find_parent(relation_dict,person)
    if person == None: 
        #Top of the ancestor chain found
        break
    else: 
        ancestor_list.append(person)

这是 Python-2.6 中的一个实现。它假定您的文本文件的结构首先是所有关系，然后是一个空行，然后是所有命令。

def read_file(file_name): 
    fp = open(file_name,'r')
    relations = []
    commands = []
    reading_relations = True
    for l in fp: 
        l = l.strip('\n')
        if not l: 
            reading_relations = False
            continue
        if reading_relations:     
            relations.append(l.strip())
        else: 
            commands.append(l.strip())
    fp.close()
    return relations,commands

def form_relation_dict(relations): 
    relation_dict = {}
    for l in relations: 
        names = l.split(':')
        mother = names[0].strip()
        children = [x.strip() for x in names[1].split(',')]
        existing_children = relation_dict.get(mother,[])
        existing_children.extend(children)
        relation_dict[mother] = existing_children
    return relation_dict

def search_name(relation_dict,name): 
    #Returns True if name occurs anywhere in relation_dict
    #Else return False
    for k,v in relation_dict.items(): 
        if name ==k or name in v: 
            return True
    return False

def find_parent(relation_dict,param): 
    #Finds the parent of 'param' in relation_dict
    #Returns None if no mother found
    #Returns mother name otherwise
    mother = None
    for k,v in relation_dict.items(): 
        if param in v: 
            mother = k
            break
    return mother

def process_commands(commands,relation_dict): 
    output = []
    for c in commands: 
        coms = c.split()
        if len(coms) < 2: 
            output.append("Invalid Command")
            continue
        action = coms[0]
        param = coms[1]
        if action == "mother": 
            name_found = search_name(relation_dict,param)
            if not name_found: 
                output.append("person not found")
                continue
            else: 
                person = find_parent(relation_dict,param)
                if person is None: 
                    output.append("mother not known")
                else: 
                    output.append("mother - %s" %(person))
        elif action == "ancestors": 
            name_found = search_name(relation_dict,param)
            if not name_found: 
                output.append("person not found")
                continue
            else: 
                #Loop through to find the mother till dead - end (None) is not reached
                ancestor_list = []
                person = param
                while True: 
                    person = find_parent(relation_dict,person)
                    if person == None: 
                        #Top of the ancestor found
                        break
                    else: 
                        ancestor_list.append(person)
                if ancestor_list: 
                    output.append(",".join(ancestor_list))
                else: 
                    output.append("No known ancestors")
    return output

def main(): 
    file_name = 'try.txt'
    relations,commands = read_file(file_name)
    #Process Relqations to form a dictionary of the type {parent: [child1,child2,...]}
    relation_dict = form_relation_dict(relations)
    print relation_dict
    #Now process commands in the file
    output = process_commands(commands,relation_dict)
    print '\n'.join(output)


if __name__ == '__main__': 
    main()

样本输入的输出是

mother not known
Alice,Brenda,Sue
person not found
No known ancestors

EDIT2

如果你真的想把它进一步拆分成函数，下面是 process_commands 的样子

def process_mother(relation_dict,name): 
    #Processes the mother command
    #Returns the ouput string
    output_str = ''
    name_found = search_name(relation_dict,name)
    if not name_found: 
        output_str = "person not found"
    else: 
        person = find_parent(relation_dict,name)
        if person is None: 
            output_str = "mother not known"
        else: 
            output_str = "mother - %s" %(person)
    return output_str

def process_ancestors(relation_dict,name): 
    output_str = ''
    name_found = search_name(relation_dict,name)
    if not name_found: 
        output_str = "person not found"
    else: 
        #Loop through to find the mother till dead - end (None) is not reached
        ancestor_list = []
        person = name
        while True: 
            person = find_parent(relation_dict,person)
            if person == None: 
                #Top of the ancestor found
                break
            else: 
                ancestor_list.append(person)
        if ancestor_list: 
            output_str = ",".join(ancestor_list)
        else: 
            output_str = "No known ancestors"
    return output_str

def process_commands(commands,relation_dict): 
    output = []
    for c in commands: 
        coms = c.split()
        if len(coms) < 2: 
            output.append("Invalid Command")
            continue
        action = coms[0]
        param = coms[1]
        if action == "mother": 
            new_output = process_mother(relation_dict,param)
        elif action == "ancestors": 
            new_output = process_ancestors(relation_dict,param)
        if new_output: 
            output.append(new_output)    
    return output