What is the best way to copy a list?
This question already has an answer here:
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)
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]]
你也可以这样做：
a = [1, 2, 3]
b = list(a)
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