Python and Lisp : car and cdr
Franck Ditter
to mimic Lisp car and cdr in Python...
def length(L) :
if not L : return 0
return 1 + length(L[1:])
Should I think of the slice L[1:] as (cdr L) ? I mean, is the slice
a copy of a segment of L, or do I actually get a pointer to something
inside L ? Is the above function length O(n) or probably O(n^2) ?
Where are such implementation things (well) said ?
Thanks,
franck
Ian Kelly
> Hi, I'm just wondering about the complexity of some Python operations
> to mimic Lisp car and cdr in Python...
>
> def length(L) :
> if not L : return 0
> return 1 + length(L[1:])
>
> Should I think of the slice L[1:] as (cdr L) ? I mean, is the slice
> a copy of a segment of L, or do I actually get a pointer to something
> inside L ?
The slice is a copy of a segment of L.
> Is the above function length O(n) or probably O(n^2) ?
O(n^2). If you want to implement Lisp-style list processing in
Python, Python lists are not the most efficient data type to do it
with. I would suggest using 2-element tuples to represent cons cells
and building up from there.
Also note that Python does not do tail recursion optimization, so
recursion in general is inefficient and prone to stack overflow if the
data structure is large enough.
> Where are such implementation things (well) said ?
Nobody
> Hi, I'm just wondering about the complexity of some Python operations
> to mimic Lisp car and cdr in Python...
>
> def length(L) :
> if not L : return 0
> return 1 + length(L[1:])
Python's lists are arrays/vectors, not linked lists.
> Should I think of the slice L[1:] as (cdr L) ?
No.
> I mean, is the slice a copy of a segment of L,
Yes.
> or do I actually get a pointer to something inside L ?
No.
> Is the above function length O(n) or probably O(n^2) ?
O(n^2).
And Python doesn't do tail-call optimisation, so you're likely to run out
of stack for a large list.
Lie Ryan
Your function does not mimic Lisp's car/cdr. This one does:
def car(L):
return L[0]
def cdr(L):
return L[1]
def length(L):
if not L: return 0
return 1 + length(cdr(L))
L = (a, (b, (c, (d, None))))
length(L)
is O(n)
Ethan Furman
> On 06/18/11 00:45, Franck Ditter wrote:
>> Hi, I'm just wondering about the complexity of some Python operations
>> to mimic Lisp car and cdr in Python...
>>
>> def length(L) :
>> if not L : return 0
>> return 1 + length(L[1:])
>>
>> Should I think of the slice L[1:] as (cdr L) ? I mean, is the slice
>> a copy of a segment of L, or do I actually get a pointer to something
>> inside L ? Is the above function length O(n) or probably O(n^2) ?
>> Where are such implementation things (well) said ?
>>
>> Thanks,
>>
>> franck
>
> Your function does not mimic Lisp's car/cdr. This one does:
>
>
> def car(L):
> return L[0]
> def cdr(L):
> return L[1]
IANAL (I am not a Lisper), but shouldn't that be 'return L[1:]' ?
> def length(L):
> if not L: return 0
> return 1 + length(cdr(L))
How is this different from regular ol' 'len' ?
~Ethan~
Chris Angelico
> Lie Ryan wrote:
>> def cdr(L):
>> return L[1]
>
In LISP, a list is a series of two-item units (conses).
>> L = (a, (b, (c, (d, None))))
This represents the LISP equivalent of [a, b, c, d] in Python. A list
is a linked list, not an array (as in Python).
IANAL either though, someone else may wish to clarify the advantages
of this system.
ChrisA
Steven D'Aprano
> Lie Ryan wrote:
>> On 06/18/11 00:45, Franck Ditter wrote:
>>> Hi, I'm just wondering about the complexity of some Python operations
>>> to mimic Lisp car and cdr in Python...
>>>
>>> def length(L) :
>>> if not L : return 0
>>> return 1 + length(L[1:])
>>>
>>> Should I think of the slice L[1:] as (cdr L) ? I mean, is the slice a
>>> copy of a segment of L, or do I actually get a pointer to something
>>> inside L ? Is the above function length O(n) or probably O(n^2) ?
>>> Where are such implementation things (well) said ?
>>>
>>> Thanks,
>>>
>>> franck
>>
>> Your function does not mimic Lisp's car/cdr. This one does:
>>
>>
>> def car(L):
>> return L[0]
>> def cdr(L):
>> return L[1]
>
> IANAL (I am not a Lisper), but shouldn't that be 'return L[1:]' ?
No. Each cell in a Lisp-style linked list has exactly two elements, and
in Python are usually implemented as nested tuples:
(head, tail) # Annoyingly, this is also known as (car, cdr).
where head is the data value and tail is either another Lisp-style list
or a marker for empty (such as the empty tuple () or None).
So a one-element linked list might be given as:
(42, None)
A two element list: (42, (43, None))
Three element list: (42, (43, (44, None)))
and so forth. So while you could harmlessly use a slice L[1:], there is
no point, since L[1:] will have at most a single element.
>> def length(L):
>> if not L: return 0
>> return 1 + length(cdr(L))
>
> How is this different from regular ol' 'len' ?
The point is to duplicate Lisp's implementation, not to be useful :)
Regular len will return 2, no matter how many elements you have in the
linked list, because it doesn't look at the tail recursively.
--
Steven
Elias Fotinis
> Lie Ryan wrote:
>
>> def length(L):
>> if not L: return 0
>> return 1 + length(cdr(L))
>
> How is this different from regular ol' 'len' ?
It's better, because len() can't overflow the stack. ;)
Hrvoje Niksic
>> def car(L):
>> return L[0]
>> def cdr(L):
>> return L[1]
>
> IANAL (I am not a Lisper), but shouldn't that be 'return L[1:]' ?
Not for the linked list implementation he presented.
>> def length(L):
>> if not L: return 0
>> return 1 + length(cdr(L))
>
> How is this different from regular ol' 'len' ?
len would just return 2 for every linked list, and would raise an
exception for empty list (represented by None in Lie's implementation).
A more Pythonic implementation would represent the linked list as a
first-class objects with car and cdr being attributes, allowing for
fairly natural expression of __len__, __iter__, etc. For example:
class List(object):
__slots__ = 'car', 'cdr'
def __init__(self, it=()):
it = iter(it)
try:
self.car = it.next()
except StopIteration:
pass
else:
self.cdr = List(it)
def __len__(self):
if not hasattr(self, 'cdr'):
return 0
return 1 + len(self.cdr)
def __iter__(self):
head = self
while hasattr(head, 'cdr'):
yield head.car
head = head.cdr
def __repr__(self):
return "%s(%r)" % (type(self).__name__, list(self))
>>> l = List([1, 2, 3])
>>> l
List([1, 2, 3])
>>> l.car
1
>>> l.cdr
List([2, 3])
>>> l.cdr.cdr.car
3
>>> l.cdr.cdr.cdr
List([])
>>> tuple(l)
(1, 2, 3)
Ethan Furman
> IANAL (I am not a Lisper), but shouldn't that be 'return L[1:]' ?
Ah, thanks all for the clarification.
~Ethan~
Terry Reedy
> No. Each cell in a Lisp-style linked list has exactly two elements, and
> in Python are usually implemented as nested tuples:
>
> (head, tail) # Annoyingly, this is also known as (car, cdr).
>
> where head is the data value and tail is either another Lisp-style list
> or a marker for empty (such as the empty tuple () or None).
>
> So a one-element linked list might be given as:
>
> (42, None)
>
> A two element list: (42, (43, None))
> Three element list: (42, (43, (44, None)))
>
> and so forth. So while you could harmlessly use a slice L[1:], there is
> no point, since L[1:] will have at most a single element.
It should be noted that the head element of any 'list' can also be a
'list' (as with Python lists),
t = { { (1,None), (2,(3,None)) ), ( (4,(5,None)), (6,None) ) )
so that the structure is actually a tree, which is a much more general
data structure than a true sequence of atoms. But TREP (for
tree-processing) is not as catchy as LISP (for list processing).
--
Terry Jan Reedy
Teemu Likonen
> On 6/19/2011 9:24 AM, Steven D'Aprano wrote:
>> No. Each cell in a Lisp-style linked list has exactly two elements,
>> and in Python are usually implemented as nested tuples:
>>
>> (head, tail) # Annoyingly, this is also known as (car, cdr).
>>
>> where head is the data value and tail is either another Lisp-style
>> list or a marker for empty (such as the empty tuple () or None).
> It should be noted that the head element of any 'list' can also be a
> list' (as with Python lists),
Both the head and tail elements of a cons cell can refer to any Lisp
objects. Cons cell is a general-purpose primitive data type but it is
_often_ used to build lists and trees so the tail element often refers
to another cons cell (or nil that terminates the list).
Let's not forget that Lisp's program code itself is built on such trees
of cons cells. Lisp code itself is represented in this primitive Lisp
data type. That's why Lisp is so powerful in meta programming. It's
trivial to use Lisp functions to create Lisp code.