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copied!<p>As @Cybis and others mentioned, you can't keep the O(N) complexity with Python lists; you'll have to create a linked list. At the risk of proving <a href="http://en.wikipedia.org/wiki/Greenspun%27s_Tenth_Rule" rel="nofollow noreferrer">Greenspun's 10th rule</a>, here is such a solution:</p> <pre><code>class cons(tuple): __slots__=() def __new__(cls, car, cdr): return tuple.__new__(cls, (car,cdr)) @classmethod def from_seq(class_, l): result = None for el in reversed(l): result = cons(el, result) return result @property def car(self): return self._getitem(0) @property def cdr(self): return self._getitem(1) def _getitem(self, i): return tuple.__getitem__(self, i) def __repr__(self): return '(%s %r)' % (self.car, self.cdr) def __iter__(self): v = self while v is not None: yield v.car v = v.cdr def __len__(self): return sum(1 for x in self) def __getitem__(self, i): v = self while i > 0: v = v.cdr i -= 1 return v.car def maplist(func, values): result = [ ] while values is not None: result.append(func(values)) values = values.cdr return result </code></pre> <p>Testing yields:</p> <pre><code>>>> l = cons.from_seq([1,2,3,4]) >>> print l (1 (2 (3 (4 None)))) >>> print list(l) [1, 2, 3, 4] >>> print maplistr(lambda l: list(reversed(l)), cons.from_seq([1,2,3])) [[3, 2, 1], [3, 2], [3]] </code></pre> <p>EDIT: Here is a generator-based solution that basically solves the same problem without the use of linked lists:</p> <pre><code>import itertools def mapiter(func, iter_): while True: iter_, iter2 = itertools.tee(iter_) iter_.next() yield func(iter2) </code></pre> <p>Testing yields:</p> <pre><code>>>> print list(mapiter(lambda l: list(reversed(list(l))), [1,2,3])) [[3, 2, 1], [3, 2], [3]] </code></pre>

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