如何对嵌套列表中的每个元素都符合的嵌套列表进行排序

Question

I have a model Page which stores an optional parent page (instance of itself). In my views I made a function which will return a nested list of all pages and their parents.

For example, my page architecture is

CCC
  AAA
    DDD
    KKK
      EEE
ZZZ
  BBB

So DDD has the parent page AAA, which has it's own parent page CCC. The CCC is top page and has no parent page.

The function will first get a queryset of all instances of Pages and sort them alphabetically. Then it will proceed to recursively generate a "full parent architecture" list, where each element on that list is another list of all parent pages, including the page itself. From the example above if we take a slice of list for page DDD, it would return [CCC, AAA, DDD].

My function currently returns a list like this for the above stated example:

[ 
  [CCC, AAA],
  [ZZZ, BBB],
  [CCC],
  [CCC, AAA, DDD],
  [CCC, AAA, KKK, EEE],
  [CCC, AAA, KKK],
  [ZZZ],
]

As you can see from the that list, all elements are sorted alphabetically according to the last element on that list. Now I want to display all those parent pages on my front end to basically look like a sitemap and show the proper parent architecture of all Pages on my site that is sorted alphabetically according to each element in each nested list. The end result would be:

[ 
  [CCC],
  [CCC, AAA],
  [CCC, AAA, DDD],
  [CCC, AAA, KKK],
  [CCC, AAA, KKK, EEE],
  [ZZZ],
  [ZZZ, BBB],
]

To put it simple, I want to go through each first element on each list and sort it alphabetically, then each second element and sort them as well, then third, then forth, and so on. Is there a way to do this?

EDIT: To avoid confusion here's my view:

def page_list(request):


    # Fetch all pages and sort them alphabetically
    queryset = Page.objects.all().order_by("title")
    output = []

    # Generate list of pages and their parents
    for page in queryset:
        output.append(get_list_of_parents(page))

    context = {
        "title": "Page List",
        "page_list": output,
    }

    return render(request, template + '/page_list.html', context)

# Get an array of all parent instances of a Page model
def get_list_of_parents(page, list=None):

    current_page = page 
    parent_list = []    

    if list is not None:
        parent_list = list

    parent_list.append(current_page)

    if current_page.parent is not None:
        parent_list = get_list_of_parents(current_page.parent, parent_list)
    else:
        # if this is the last parent page, reverse the order of list to display list in form of parent path to child
        parent_list.reverse()

    return parent_list

Answer 1

Using a tree of Pages:

I define this simple tree structure:

class Page(object):
    def __init__(self, name, children=None):
        self.name = name
        self.children = children or []

    def display(self, indent=""):
        child_repr = [child.display(indent=indent + "  ") for child in self.children]
        return indent + self.name + "\n" + "".join(child_repr)

    def __str__(self):
        return self.display()

The display and __str__ methods are used for printing.

I can build your tree as a list of Pages:

tree = [
    Page("CCC", [
        Page("AAA", [
            Page("DDD"),
            Page("KKK", [
                Page("EEE")])])]),
    Page("ZZZ", [
        Page("BBB")])]

I can display the structure like this:

for item in tree:
    print(item)

I get:

CCC
  AAA
    DDD
    KKK
      EEE

ZZZ
  BBB

To traverse the tree, I define the following method:

    def traverse(self, result, stack=None):
        stack = stack or []
        stack.append(self.name)
        result.append(list(stack))
        for child in self.children:
            child.traverse(result, stack)
        stack.pop()

Where result is the list to fill.

Usage:

result = []
for item in tree:
    item.traverse(result)

import pprint

pprint.pprint(result)

I'll get:

[['CCC'],
 ['CCC', 'AAA'],
 ['CCC', 'AAA', 'DDD'],
 ['CCC', 'AAA', 'KKK'],
 ['CCC', 'AAA', 'KKK', 'EEE'],
 ['ZZZ'],
 ['ZZZ', 'BBB']]

Great!

Answer 2

Take advantage of helper functions in itertools and do your sorting with a simple and highly efficient one-line comparison function.

>>> import pprint
>>> from itertools import dropwhile, izip_longest
>>> pages = [['CCC', 'AAA'],
...          ['ZZZ', 'BBB'],
...          ['CCC'],
...          ['CCC', 'AAA', 'DDD'],
...          ['CCC', 'AAA', 'DDD', 'EEE'],
...          ['CCC', 'AAA', 'KKK'],
...          ['ZZZ']]
>>> pprint.pprint(sorted(pages, cmp=lambda a, b: cmp(*next(dropwhile(lambda x: not cmp(x[0], x[1]), izip_longest(a, b))))))
[['CCC'],
 ['CCC', 'AAA'],
 ['CCC', 'AAA', 'DDD'],
 ['CCC', 'AAA', 'DDD', 'EEE'],
 ['CCC', 'AAA', 'KKK'],
 ['ZZZ'],
 ['ZZZ', 'BBB']]
>>>

This works not only for strings, but also Page instances, as long as they are comparable, ie that you have defined your __cmp__ method in your Page class.