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。
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。
@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]]
将列表视为树，python中的deep_copy可以最简洁地写成
def deep_copy(x):
if not isinstance(x, list):
return x
else:
return [deep_copy(elem) for elem in x]
它基本上是以深度优先的方式递归遍历列表。
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)
