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
如果列表的内容是基本数据类型，则可以使用理解式
new_list = [i for i in old_list]
你可以为多维列表嵌套它，比如:
new_grid = [[i for i in row] for row in grid]
其他回答
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。
如果列表的内容是基本数据类型，则可以使用理解式
new_list = [i for i in old_list]
你可以为多维列表嵌套它，比如:
new_grid = [[i for i in row] for row in grid]
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。
官方文档
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)
@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]]
