How do I prepend to a short python list?

This creates a new list with x prepended to it, rather than modifying an existing list:

new_list = [x] + old_list

If someone finds this question like me, here are my performance tests of proposed methods:

Python 2.7.8

In [1]: %timeit ([1]*1000000).insert(0, 0)
100 loops, best of 3: 4.62 ms per loop

In [2]: %timeit ([1]*1000000)[0:0] = [0]
100 loops, best of 3: 4.55 ms per loop

In [3]: %timeit [0] + [1]*1000000
100 loops, best of 3: 8.04 ms per loop

As you can see, insert and slice assignment are as almost twice as fast than explicit adding and are very close in results. As Raymond Hettinger noted insert is more common option and I, personally prefer this way to prepend to list.

What’s the idiomatic syntax for prepending to a short python list?

You don’t usually want to repetitively prepend to a list in Python.

If the list is short, and you’re not doing it a lot… then ok.

list.insert

The list.insert can be used this way.

list.insert(0, x)

But this is inefficient, because in Python, a list is an array of pointers, and Python must now take every pointer in the list and move it down by one to insert the pointer to your object in the first slot, so this is really only efficient for rather short lists, as you ask.

Here’s a snippet from the CPython source where this is implemented – and as you can see, we start at the end of the array and move everything down by one for every insertion:

for (i = n; --i >= where; )
    items[i+1] = items[i];

If you want a container/list that’s efficient at prepending elements, you want a linked list. Python has a doubly linked list, which can insert at the beginning and end quickly – it’s called a deque.

deque.appendleft

A collections.deque has many of the methods of a list. list.sort is an exception, making deque definitively not entirely Liskov substitutable for list.

>>> set(dir(list)) - set(dir(deque))
{'sort'}

The deque also has an appendleft method (as well as popleft). The deque is a double-ended queue and a doubly-linked list – no matter the length, it always takes the same amount of time to preprend something. In big O notation, O(1) versus the O(n) time for lists. Here’s the usage:

>>> import collections
>>> d = collections.deque('1234')
>>> d
deque(['1', '2', '3', '4'])
>>> d.appendleft('0')
>>> d
deque(['0', '1', '2', '3', '4'])

deque.extendleft

Also relevant is the deque’s extendleft method, which iteratively prepends:

>>> from collections import deque
>>> d2 = deque('def')
>>> d2.extendleft('cba')
>>> d2
deque(['a', 'b', 'c', 'd', 'e', 'f'])

Note that each element will be prepended one at a time, thus effectively reversing their order.

Performance of list versus deque

First we setup with some iterative prepending:

import timeit
from collections import deque


def list_insert_0(prepends: int):
    l = []
    for i in range(prepends):
        l.insert(0, i)

def list_slice_insert(prepends):
    l = []
    for i in range(prepends):
        l[:0] = [i]      # semantically same as list.insert(0, i)

def list_add(prepends):
    l = []
    for i in range(prepends):
        l = [i] + l      # caveat: new list each time

def deque_appendleft(prepends):
    d = deque()
    for i in range(prepends):
        d.appendleft(i)  # semantically same as list.insert(0, i)

def deque_extendleft(prepends):
    d = deque()
    d.extendleft(range(prepends)) # semantically same as deque_appendleft above

And a function for analysis, so that we can fairly compare all operations across a range of usages:

def compare_prepends(n, runs_per_trial):
    results = {}
    for function in (
        list_insert_0, list_slice_insert,
        list_add, deque_appendleft, deque_extendleft,
        ):
        shortest_time = min(timeit.repeat(
            lambda: function(n), number=runs_per_trial))
        results[function.__name__] = shortest_time
    ranked_methods = sorted(results.items(), key=lambda kv: kv[1])
    for name, duration in ranked_methods:
        print(f'{name} took {duration} seconds')

and performance (adjusting the number of runs per trial down to compensate for longer running times of more prepends – repeat does three trials by default):

compare_prepends(20, 1_000_000)
compare_prepends(100, 100_000)
compare_prepends(500, 100_000)
compare_prepends(2500, 10_000)

>>> compare_prepends(20, 1_000_000)
deque_extendleft took 0.6490256823599339 seconds
deque_appendleft took 1.4702797569334507 seconds
list_insert_0 took 1.9417422469705343 seconds
list_add took 2.7092894352972507 seconds
list_slice_insert took 3.1809083241969347 seconds
>>> compare_prepends(100, 100_000)
deque_extendleft took 0.1177942156791687 seconds
deque_appendleft took 0.5385235995054245 seconds
list_insert_0 took 0.9471780974417925 seconds
list_slice_insert took 1.4850486349314451 seconds
list_add took 2.1660344172269106 seconds
>>> compare_prepends(500, 100_000)
deque_extendleft took 0.7309095915406942 seconds
deque_appendleft took 2.895373275503516 seconds
list_slice_insert took 8.782583676278591 seconds
list_insert_0 took 8.931685039773583 seconds
list_add took 30.113558700308204 seconds
>>> compare_prepends(2500, 10_000)
deque_extendleft took 0.4839253816753626 seconds
deque_appendleft took 1.5615574326366186 seconds
list_slice_insert took 6.712615916505456 seconds
list_insert_0 took 13.894083382561803 seconds
list_add took 72.1727528590709 seconds

The deque is much faster. As the lists get longer, deques perform even better. If you can use deque’s extendleft you’ll probably get the best performance that way.

If you must use lists, keep in mind that for small lists, list.insert works faster, but for larger lists, inserting using slice notation becomes faster.

Don’t prepend to lists

Lists were meant to be appended to, not prepended to. If you have a situation where this kind of prepending is a hurting the performace of your code, either switch to a deque or, if you can reverse your semantics and accomplish the same goal, reverse your list and append instead.

In general, avoid prepending to the built-in Python list object.

In my opinion, the most elegant and idiomatic way of prepending an element or list to another list, in Python, is using the expansion operator * (also called unpacking operator),

# Initial list
l = [4, 5, 6]

# Modification
l = [1, 2, 3, *l]

Where the resulting list after the modification is [1, 2, 3, 4, 5, 6]

I also like simply combining two lists with the operator +, as shown,

# Prepends [1, 2, 3] to l
l = [1, 2, 3] + l

# Prepends element 42 to l
l = [42] + l

I don’t like the other common approach, l.insert(0, value), as it requires a magic number. Moreover, insert() only allows prepending a single element, however the approach above has the same syntax for prepending a single element or multiple elements.

Lets go over 4 methods

1. Using insert()

>>> 
>>> l = list(range(5))
>>> l
[0, 1, 2, 3, 4]
>>> l.insert(0, 5)
>>> l
[5, 0, 1, 2, 3, 4]
>>> 

2. Using [] and +

>>> 
>>> l = list(range(5))
>>> l
[0, 1, 2, 3, 4]
>>> l = [5] + l
>>> l
[5, 0, 1, 2, 3, 4]
>>> 

3. Using Slicing

>>> 
>>> l = list(range(5))
>>> l
[0, 1, 2, 3, 4]
>>> l[:0] = [5]
>>> l
[5, 0, 1, 2, 3, 4]
>>> 

4. Using collections.deque.appendleft()

>>> 
>>> from collections import deque
>>> 
>>> l = list(range(5))
>>> l
[0, 1, 2, 3, 4]
>>> l = deque(l)
>>> l.appendleft(5)
>>> l = list(l)
>>> l
[5, 0, 1, 2, 3, 4]
>>> 

I would have done something quite fast forward in python >= 3.0

list=[0,*list]

It maybe be not the most efficient way, but it’s the most Pythonic in my opinion.

from collections import deque
l = list(range(5))
l
[0, 1, 2, 3, 4]
l = deque(l)
l.appendleft(5)
l = list(l)
l
[5, 0, 1, 2, 3, 4]