在 python 中作为切片传递给 enumerate() 时是否形成了数组的新副本？

Question

作为切片传递给 enumerate() 时是否形成了数组的新副本？这是我在运行以下实验后的猜测。在我看来是这样，但我不确定如何用代码证明我的想法？

下面是一段代码和 2 个结果。 A选项和B选项对应的结果也如下

def selection_sort(arr):
    # for idx, item in enumerate(arr[:-1]):        # option A
    for idx, item in enumerate(arr):        # option B
        curr_min = item
        swap_ptr = idx
        
        print('========================================')
        print('curr_min: {}'.format(curr_min))
        print('arr before: {}'.format(arr))
        
        for j in range(idx+1, len(arr)):
            print('for item {}'.format(arr[j]))
            if arr[j] < curr_min:
                print('{} < {}'.format(arr[j], curr_min))
                curr_min = arr[j]
                swap_ptr = j
                print('swap_ptr now at {} pointing to {}'.format(swap_ptr, curr_min))
        
        (arr[idx], arr[swap_ptr]) = (arr[swap_ptr], arr[idx])
        print('arr after: {}'.format(arr))
        
    return arr
    
print(selection_sort([5,9,1,3,0,20,77,46]))

运行对选项 A 的结果（即注释掉选项 B）：

========================================
curr_min: 5
arr before: [5, 9, 1, 3, 0, 20, 77, 46]
for item 9
for item 1
1 < 5
swap_ptr now at 2 pointing to 1
for item 3
for item 0
0 < 1
swap_ptr now at 4 pointing to 0
for item 20
for item 77
for item 46
arr after: [0, 9, 1, 3, 5, 20, 77, 46]
========================================
curr_min: 9
arr before: [0, 9, 1, 3, 5, 20, 77, 46]
for item 1
1 < 9
swap_ptr now at 2 pointing to 1
for item 3
for item 5
for item 20
for item 77
for item 46
arr after: [0, 1, 9, 3, 5, 20, 77, 46]
========================================
curr_min: 1
arr before: [0, 1, 9, 3, 5, 20, 77, 46]
for item 3
for item 5
for item 20
for item 77
for item 46
arr after: [0, 1, 9, 3, 5, 20, 77, 46]
========================================
curr_min: 3
arr before: [0, 1, 9, 3, 5, 20, 77, 46]
for item 5
for item 20
for item 77
for item 46
arr after: [0, 1, 9, 3, 5, 20, 77, 46]
========================================
curr_min: 0
arr before: [0, 1, 9, 3, 5, 20, 77, 46]
for item 20
for item 77
for item 46
arr after: [0, 1, 9, 3, 5, 20, 77, 46]
========================================
curr_min: 20
arr before: [0, 1, 9, 3, 5, 20, 77, 46]
for item 77
for item 46
arr after: [0, 1, 9, 3, 5, 20, 77, 46]
========================================
curr_min: 77
arr before: [0, 1, 9, 3, 5, 20, 77, 46]
for item 46
46 < 77
swap_ptr now at 7 pointing to 46
arr after: [0, 1, 9, 3, 5, 20, 46, 77]
[0, 1, 9, 3, 5, 20, 46, 77]

选项 B 的运行结果（即注释掉选项 A）：

========================================
curr_min: 5
arr before: [5, 9, 1, 3, 0, 20, 77, 46]
for item 9
for item 1
1 < 5
swap_ptr now at 2 pointing to 1
for item 3
for item 0
0 < 1
swap_ptr now at 4 pointing to 0
for item 20
for item 77
for item 46
arr after: [0, 9, 1, 3, 5, 20, 77, 46]
========================================
curr_min: 9
arr before: [0, 9, 1, 3, 5, 20, 77, 46]
for item 1
1 < 9
swap_ptr now at 2 pointing to 1
for item 3
for item 5
for item 20
for item 77
for item 46
arr after: [0, 1, 9, 3, 5, 20, 77, 46]
========================================
curr_min: 9
arr before: [0, 1, 9, 3, 5, 20, 77, 46]
for item 3
3 < 9
swap_ptr now at 3 pointing to 3
for item 5
for item 20
for item 77
for item 46
arr after: [0, 1, 3, 9, 5, 20, 77, 46]
========================================
curr_min: 9
arr before: [0, 1, 3, 9, 5, 20, 77, 46]
for item 5
5 < 9
swap_ptr now at 4 pointing to 5
for item 20
for item 77
for item 46
arr after: [0, 1, 3, 5, 9, 20, 77, 46]
========================================
curr_min: 9
arr before: [0, 1, 3, 5, 9, 20, 77, 46]
for item 20
for item 77
for item 46
arr after: [0, 1, 3, 5, 9, 20, 77, 46]
========================================
curr_min: 20
arr before: [0, 1, 3, 5, 9, 20, 77, 46]
for item 77
for item 46
arr after: [0, 1, 3, 5, 9, 20, 77, 46]
========================================
curr_min: 77
arr before: [0, 1, 3, 5, 9, 20, 77, 46]
for item 46
46 < 77
swap_ptr now at 7 pointing to 46
arr after: [0, 1, 3, 5, 9, 20, 46, 77]
========================================
curr_min: 77
arr before: [0, 1, 3, 5, 9, 20, 46, 77]
arr after: [0, 1, 3, 5, 9, 20, 46, 77]
[0, 1, 3, 5, 9, 20, 46, 77]

请告诉我如何证明我对这个奇怪现象的看法。似乎当数组以切片方式传递给 enumerate() 时，它只会迭代函数参数列表中的 'old' 数组。但是当在没有任何切片的情况下传递时， enumerate() 迭代在循环内修改的新数组。我不确定我是否可以使用代码和文档而不是运行这个实验来证明这一点？为什么这个设计如此奇怪？ TIA.

Answer 1

我可以确认是这样的。来看看我的作品吧

我首先创建一个基本列表：

foo = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

然后，当我们迭代枚举时，我将编辑其中一个值：

>>> once = True
>>> for idx, val in enumerate(foo):
...  if once:
...   once = False
...   foo[2] = 666
...  print(idx, val)
... 
0 0
1 1
2 666
3 3
4 4
5 5
6 6
7 7
8 8
9 9

我们注意到枚举在到达其索引时显示编辑后的值。

现在我们在传递切片而不是原始切片时执行相同的操作：

>>> foo = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> once = True
>>> for idx, val in enumerate(foo[:]):
...  if once:
...   once = False
...   foo[2] = 42
...  print(idx, val)
... 
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
>>> foo
[0, 1, 42, 3, 4, 5, 6, 7, 8, 9]

我们看到虽然 foo 已经被编辑，枚举并不知道这个变化，因为它必须使用原始的未修改副本。

我调查了为什么会这样。当 using the bracket notation we are actually calling the slice class 在引擎盖下时。

这表明它 returns 一个 Slice 对象。这显然与您的原始列表不同。

有兴趣的还可以看the object creation in the source code

在 python 中作为切片传递给 enumerate() 时是否形成了数组的新副本？

Is a new copy of array formed when passed as a slice to enumerate() in python?

python

arrays

sorting

loops