linked list in Python

linked_list.py

import functools

class LList:
    @classmethod
    def fromArray(cls, array):
        return functools.reduce(lambda ll, val: ll.append(val), array, cls())

    def __init__(self):                   self.isEmpty = True
    def append(self, val):                return LNode(val, self)
    def __iter__(self):                   return self
    def next(self):                       raise StopIteration
    def __len__(self):                    return 0
    def __repr__(self):                   return 'LList()'
    def __str__(self, pre='(', delim=''): return pre + ')'

class LNode:
    def __init__(self, val, rest):
        self.val     = val
        self.rest    = rest

    def append(self, val):
        return LNode(self.val, self.rest.append(val))

    def __iter__(ll):
        yield ll.val
        for val in ll.rest: yield val
        # is there not a way I can just pass execution to `ll.next`?
        # I think this "for val in" thing means we've got a stack of iterators
        # each yielding their value up through the stack to the top one which
        # finally yields it to the the thing receiving it.
        # In Ruby, you would do it like this:
        #   def each(&block)
        #     block.call val
        #     rest.each &block
        #   end
        # where `block.call` is equivalent to `yield`, so you can pass the thing
        # that is being yielded to.

    def __len__(self):
        return 1 + len(self.rest)

    def __str__(self, pre='(', delim=''):
        return pre + delim + str(self.val) + self.rest.__str__('', '->')

    def __repr__(self):
        return 'LNode({0!r}, {1!r})'.format(self.val, self.rest)

def show(ll):
    print('ll = {0!r}'.format(ll))
    print('len(ll)   # => {0}'.format(len(ll)))
    print('print(ll) # => {0}'.format(ll))
    print('')

show(LList.fromArray([]))
show(LList.fromArray([11,22,33]))