我试图在Python中实现具有无限深度子类别的类别树,我有多个列表元素,我必须从中进行此操作。
让我详细解释一下,这是我的清单清单。
>mylists = [
>['home', 'desktop', 'mouse', 'wireless'],
>['home', 'desktop', 'mouse', 'wired'],
>['home', 'laptop', 'mouse'],
>['home', 'laptop', '13-inch'],
>]
我希望输出为:
>home
> desktop
> mouse
> wireless
> wired
> laptop
> mouse
> 13-inch
我明白我应该使用递归函数迭代列表并制作一些神奇的东西。
为实现这一目标,我将分两步完成此任务: 1.将此嵌套列表转换为嵌套字典(仅用于保持层次结构) 2.将嵌套的dict转换为上面解释的所需格式。
Step1:这是我将嵌套列表转换为嵌套字典的代码:
>def make_rec_dict(dict):
> d = {}
> for path in dict:
> current_level = d
> for part in path:
> if part not in current_level:
> current_level[part] = {}
> current_level = current_level[part]
> #print part
> return d
>
>make_rec_dict(mylists)
>{'home': {'laptop': {'mouse': {}, '13-inch': {}}, 'desktop': {'mouse': {'wireless': {}, 'wired': {}}}}}
步骤2: 要以所需的格式显示,
spaces = { 1 : '', 2 : '>>>>', 3 : '>>>>>>>>', 4 : '>>>>>>>>>>>>', 5 : '>>>>>>>>>>>>>>>>>>>>'}
def display_recusively(dictionary, level=0):
if type(dictionary) is dict:
values = [] # get all the values and parse each again
for key, value in dictionary.iteritems():
if value != '':
print spaces[level], key
values.append(value)
level = level + 1
return display_recusively(values, level)
elif value == '': # this is the last child
print spaces[level], key
elif type(dictionary) is list:
for d in dictionary:
return display_recusively(d, level)
else:
print dictionary
但是代码的缺点是,我无法获得与父母相关的子元素的链接。我的意思是鼠标和鼠标应该是不同的,上面代码的缺点是它出现了循环..
所以请建议我或纠正我更好的方法:
答案 0 :(得分:0)
对于任何有同样问题的人..我想出了实现这个的方法:),这里是display_recursively()的改变代码:
def display_recusively(dictionary, level=0):
if type(dictionary) is dict:
values = [] # get all the values and parse each again
for key, value in dictionary.iteritems():
parent = key
if value != '': # recurse only when value is dict
print spaces[level], key
values.append(value)
level = level + 1
display_recusively(values, level)
level = level -1
values = [] #sanitise the list
elif value == '': # this is the last child
print spaces[level], key , "<>"
elif type(dictionary) is list:
for d in dictionary:
display_recusively(d, level)
level = level +1
else:
print dictionary
答案 1 :(得分:0)
通过这样做可以获得相同的输出 -
my_lists = [['home', 'desktop', 'mouse', 'wireless'], ['home', 'desktop', 'mouse', 'wired'],
['home', 'laptop', 'mouse'], ['home', 'laptop', '13-inch']]
path_list = []
for lists in my_lists:
path = ''
for i in range(len(lists)):
path = path + lists[i]
if path not in path_list:
print ' '*i + lists[i]
path_list.append(path)