E0_copy = list(E0)后，我猜E0_copy是E0的深度拷贝，因为id(E0)不等于id(E0_copy)。然后我在循环中修改了E0_copy，但是为什么E0之后不一样了?
E0 = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
for k in range(3):
E0_copy = list(E0)
E0_copy[k][k] = 0
#print(E0_copy)
print E0 # -> [[0, 2, 3], [4, 0, 6], [7, 8, 0]]
当前回答
如果不允许直接导入模块，可以将自己的deepcopy函数定义为-
def copyList(L):
if type(L[0]) != list:
return [i for i in L]
else:
return [copyList(L[i]) for i in range(len(L))]
它的工作很容易被看作是
>>> x = [[1,2,3],[3,4]]
>>> z = copyList(x)
>>> x
[[1, 2, 3], [3, 4]]
>>> z
[[1, 2, 3], [3, 4]]
>>> id(x)
2095053718720
>>> id(z)
2095053718528
>>> id(x[0])
2095058990144
>>> id(z[0])
2095058992192
>>>
其他回答
E0_copy不是深度拷贝。你不能使用list()进行深度复制。(list(…)和testList[:]都是浅拷贝。)
你可以使用copy.deepcopy(…)来深度复制列表。
deepcopy(x, memo=None, _nil=[])
Deep copy operation on arbitrary Python objects.
请参阅下面的代码片段-
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = list(a)
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0][1] = 10
>>> a
[[1, 10, 3], [4, 5, 6]]
>>> b # b changes too -> Not a deepcopy.
[[1, 10, 3], [4, 5, 6]]
现在看深度复制操作
>>> import copy
>>> b = copy.deepcopy(a)
>>> a
[[1, 10, 3], [4, 5, 6]]
>>> b
[[1, 10, 3], [4, 5, 6]]
>>> a[0][1] = 9
>>> a
[[1, 9, 3], [4, 5, 6]]
>>> b # b doesn't change -> Deep Copy
[[1, 10, 3], [4, 5, 6]]
为了解释，list(…)不会递归地复制内部对象。它只生成最外层列表的副本，同时仍然引用相同的内部列表，因此，当您改变内部列表时，更改会反映在原始列表和浅副本中。您可以看到，通过检查id(a[0]) == id(b[0]) where b = list(a)，浅复制引用了内部列表。
@Sukrit卡尔拉
No.1: list()， [:]， copy.copy()都是浅拷贝。如果一个对象是复合的，它们都不合适。你需要使用copy.deepcopy()。
2 . b直接= a, a和b有相同的参照物，改变a等于改变b。
将a设为b
如果直接将a赋给b，则a和b共享一个引用。
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = a
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0] = 1
>>> a
[1, [4, 5, 6]]
>>> b
[1, [4, 5, 6]]
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = a
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0][1] = 10
>>> a
[[1, 10, 3], [4, 5, 6]]
>>> b
[[1, 10, 3], [4, 5, 6]]
影子复制
通过列表()
List()和[:]是相同的。除了第一层更改外，其他所有层的更改都将被传输。
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = list(a)
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0] = 1
>>> a
[1, [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = list(a)
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0][1] = 10
>>> a
[[1, 10, 3], [4, 5, 6]]
>>> b
[[1, 10, 3], [4, 5, 6]]
由[:]
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = a[:]
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0] = 1
>>> a
[1, [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = a[:]
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0][1] = 10
>>> a
[[1, 10, 3], [4, 5, 6]]
>>> b
[[1, 10, 3], [4, 5, 6]]
List()和[:]改变除了第一层之外的其他层
# =========== [:] ===========
>>> a = [[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b = a[:]
>>> a
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> a[0][2] = 4
>>> a
[[1, 2, 4], [4, 5, 6]]
>>> b
[[1, 2, 4], [4, 5, 6]]
>>> a = [[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b = a[:]
>>> a
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> a[0][2][0] = 999
>>> a
[[1, 2, [999, 6]], [4, 5, 6]]
>>> b
[[1, 2, [999, 6]], [4, 5, 6]]
# =========== list() ===========
>>> a = [[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b = list(a)
>>> a
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> a[0][2] = 4
>>> a
[[1, 2, 4], [4, 5, 6]]
>>> b
[[1, 2, 4], [4, 5, 6]]
>>> a = [[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b = list(a)
>>> a
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> b
[[1, 2, [3.5, 6]], [4, 5, 6]]
>>> a[0][2][0] = 999
>>> a
[[1, 2, [999, 6]], [4, 5, 6]]
>>> b
[[1, 2, [999, 6]], [4, 5, 6]]
通过复制()
你会发现copy()函数与list()和[:]相同。他们都是肤浅的拷贝。
有关浅复制和深度复制的更多信息，可以参考这里。
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = copy.copy(a)
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0][1] = 10
>>> a
[[1, 10, 3], [4, 5, 6]]
>>> b
[[1, 10, 3], [4, 5, 6]]
通过深度复制（）
>>> import copy
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = copy.deepcopy(a)
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0] = 1
>>> a
[1, [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a = [[1, 2, 3], [4, 5, 6]]
>>> b = copy.deepcopy(a)
>>> a
[[1, 2, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
>>> a[0][1] = 10
>>> a
[[1, 10, 3], [4, 5, 6]]
>>> b
[[1, 2, 3], [4, 5, 6]]
如果你的列表元素是不可变对象，那么你可以使用这个，否则你必须从copy模块中使用deepcopy。
你也可以使用最短的方法来深度复制列表，就像这样。
a = [0,1,2,3,4,5,6,7,8,9,10]
b = a[:] #deep copying the list a and assigning it to b
print id(a)
20983280
print id(b)
12967208
a[2] = 20
print a
[0, 1, 20, 3, 4, 5, 6, 7, 8, 9,10]
print b
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10]
如果不允许直接导入模块，可以将自己的deepcopy函数定义为-
def copyList(L):
if type(L[0]) != list:
return [i for i in L]
else:
return [copyList(L[i]) for i in range(len(L))]
它的工作很容易被看作是
>>> x = [[1,2,3],[3,4]]
>>> z = copyList(x)
>>> x
[[1, 2, 3], [3, 4]]
>>> z
[[1, 2, 3], [3, 4]]
>>> id(x)
2095053718720
>>> id(z)
2095053718528
>>> id(x[0])
2095058990144
>>> id(z[0])
2095058992192
>>>
在Python中，有一个名为copy的模块，它有两个有用的函数:
import copy
copy.copy()
copy.deepcopy()
Copy()是一个浅拷贝函数。如果给定的参数是一个复合数据结构，例如一个列表，那么Python将创建另一个相同类型的对象(在这种情况下，是一个新列表)，但对于旧列表中的所有内容，只有它们的引用被复制。可以这样想:
newList = [elem for elem in oldlist]
直观地，我们可以假设deepcopy()将遵循相同的范例，唯一的区别是，对于每个elem，我们将递归地调用deepcopy，(就像mbguy的答案一样)
但这是错误的!
Deepcopy()实际上保留了原始复合数据的图形结构:
a = [1,2]
b = [a,a] # there's only 1 object a
c = deepcopy(b)
# check the result
c[0] is a # False, a new object a_1 is created
c[0] is c[1] # True, c is [a_1, a_1] not [a_1, a_2]
这是棘手的部分:在deepcopy()过程中，使用哈希表(Python中的字典)将每个旧对象ref映射到每个新对象ref，这防止了不必要的重复，从而保留了复制的复合数据的结构。
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