只是一个递归的深度复制函数。

def deepcopy(A):
    rt = []
    for elem in A:
        if isinstance(elem,list):
            rt.append(deepcopy(elem))
        else:
            rt.append(elem)
    return rt

编辑:正如Cfreak提到的，这已经在复制模块中实现了。

下面是一个深度复制2D列表的例子:

  b = [x[:] for x in a]

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]]

如果你使用deepcopy给同一个列表赋值， 请使用临时变量。由于某些原因，copy.deepcopy()在尝试使用索引自我更新同一变量时，在对象列表上不起作用

S = S[idx] ->x
S = copy.deepcopy(S[idx])  -> x

VVVV，虽然这工作

Stemp = np.zeros(N,dtype=object)
for ii in range(N):
   Stemp[ii]=copy.deepcopy(S[idx[ii]])
S= copy.deepcopy(Stemp)

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]