python - What is the best way to copy a list?

Question

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List changes unexpectedly after assignment. Why is this and how can I prevent it? (22 answers)

Closed 7 years ago.

What is the best way to copy a list? I know the following ways, which one is better? Or is there another way?

lst = ['one', 2, 3]

lst1 = list(lst)

lst2 = lst[:]

import copy
lst3 = copy.copy(lst)

Answer 1

If you want a shallow copy (elements aren't copied) use:

lst2=lst1[:]

If you want to make a deep copy then use the copy module:

import copy
lst2=copy.deepcopy(lst1)

Answer 2

I often use:

lst2 = lst1 * 1

If lst1 it contains other containers (like other lists) you should use deepcopy from the copy lib as shown by Mark.

---

UPDATE: Explaining deepcopy

>>> a = range(5)
>>> b = a*1
>>> a,b
([0, 1, 2, 3, 4], [0, 1, 2, 3, 4])
>>> a[2] = 55 
>>> a,b
([0, 1, 55, 3, 4], [0, 1, 2, 3, 4])

As you may see only a changed... I'll try now with a list of lists

>>> 
>>> a = [range(i,i+3) for i in range(3)]
>>> a
[[0, 1, 2], [1, 2, 3], [2, 3, 4]]
>>> b = a*1
>>> a,b
([[0, 1, 2], [1, 2, 3], [2, 3, 4]], [[0, 1, 2], [1, 2, 3], [2, 3, 4]])

Not so readable, let me print it with a for:

>>> for i in (a,b): print i   
[[0, 1, 2], [1, 2, 3], [2, 3, 4]]
[[0, 1, 2], [1, 2, 3], [2, 3, 4]]
>>> a[1].append('appended')
>>> for i in (a,b): print i

[[0, 1, 2], [1, 2, 3, 'appended'], [2, 3, 4]]
[[0, 1, 2], [1, 2, 3, 'appended'], [2, 3, 4]]

You see that? It appended to the b[1] too, so b[1] and a[1] are the very same object. Now try it with deepcopy

>>> from copy import deepcopy
>>> b = deepcopy(a)
>>> a[0].append('again...')
>>> for i in (a,b): print i

[[0, 1, 2, 'again...'], [1, 2, 3, 'appended'], [2, 3, 4]]
[[0, 1, 2], [1, 2, 3, 'appended'], [2, 3, 4]]

Answer 3

You can also do:

a = [1, 2, 3]
b = list(a)

Answer 4

I like to do:

lst2 = list(lst1)

The advantage over lst1[:] is that the same idiom works for dicts:

dct2 = dict(dct1)

Answer 5

Short lists, [:] is the best:

In [1]: l = range(10)

In [2]: %timeit list(l)
1000000 loops, best of 3: 477 ns per loop

In [3]: %timeit l[:]
1000000 loops, best of 3: 236 ns per loop

In [6]: %timeit copy(l)
1000000 loops, best of 3: 1.43 us per loop

For larger lists, they're all about the same:

In [7]: l = range(50000)

In [8]: %timeit list(l)
1000 loops, best of 3: 261 us per loop

In [9]: %timeit l[:]
1000 loops, best of 3: 261 us per loop

In [10]: %timeit copy(l)
1000 loops, best of 3: 248 us per loop

For very large lists (I tried 50MM), they're still about the same.

Answer 6

You can also do this:

import copy
list2 = copy.copy(list1)

This should do the same thing as Mark Roddy's shallow copy.

Answer 7

In terms of performance, there is some overhead to calling list() versus slicing. So for short lists, lst2 = lst1[:] is about twice as fast as lst2 = list(lst1).

In most cases, this is probably outweighed by the fact that list() is more readable, but in tight loops this can be a valuable optimization.