英文:
Weird (to me) behavior of increment (+=) operator for lists
问题
以下是翻译好的部分：
下面的代码有效，并且将 `L` 设置为帕斯卡三角形中的前 `N` 行：
N = 5
L = [[]] * N
i = j = 0
while i < N:
if j == 0 or j == i:
a = 1
else:
a = L[i-1][j] + L[i-1][j-1]
L[i] = L[i] + [a] # ***
if j < i:
j += 1
else:
j = 0
i += 1
[print(x) for x in L]
这会产生如预期的结果：
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
但是，如果我将标有 # *** 的那一行替换为：
L[i] += [a]
它就不再起作用了... 它似乎将 [a] 添加到了 L 中的所有子列表，而不仅仅是 L[i]。我是否忽略了 += 运算符的某些行为？我不明白，因为以下代码片段按预期工作：
L = [[], []]
L[0] += [1]
L[1] += [2]
print(L)
结果是 [[1], [2]]
对于这个问题的任何见解将不胜感激。
英文:
The following code works and gives L as the first N lines in Pascal's triangle:
N = 5
L = [[]] * N
i = j = 0
while i < N:
if j == 0 or j == i:
a = 1
else:
a = L[i-1][j] + L[i-1][j-1]
L[i] = L[i] + [a] # ***
if j < i:
j += 1
else:
j = 0
i += 1
[print(x) for x in L]
which gives, as expected,
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
But, if I replace the line marked with # *** by
L[i] += [a]
it doesn't work anymore... It seems to append [a] to all sublists in L, and not just L[i]. Is there some behavior of the += operator I'm missing? I don't get it, because the following code fragment works as expected:
L = [[], []]
L[0] += [1]
L[1] += [2]
print(L)
resulting in [[1], [2]]
Any insights will be greatly appreciated.
答案1
得分: 1
以下是翻译好的部分：
"L[i] += [a]" 对应于 "L[i].extend([a])"，并且是对现有列表的变更操作，因此与以下表达式不等同：
"L[i] = L[i] + [a]"
在右侧表达式的评估中，将创建一个新的列表对象，并将该新对象绑定到 L[i]。
对 L 中的列表进行变更会影响它们所有，因为 L = [[]] * N 创建了一个包含 N 个对同一（空）列表对象的引用的列表。
一般来说，在Python中，等价关系
"a = a + b" <==> "a += b"
只能安全地假定适用于不可变类型。
英文:
L[i] += [a]
corresponds to
L[i].extend([a])
and is a mutation on the existing list and thus not equivalent to:
L[i] = L[i] + [a]
where the evaluation of the right hand side will create a new list object and bind that new object to L[i].
Mutating lists in in L would affect all of them, because L = [[]] * N creates a list of N references to one and the same (empty) list object.
In general in Python, the equivalence
a = a + b <==> a += b
can only be safely assumed for immutable types.
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