python列表迭代期间的元素修改
任何人都可以向“元素”解释以下代码在迭代过程中究竟发生了什么？
array = [2,3,4]
for element in array:
element = 3
print(array)
>>>[2, 3, 4]
输出是 [2, 3, 4] 而不是 [3, 3, 3]
我是否正确理解在使用“for element in l”语法时，我们只能引用但不能修改此处发生的数组的每个元素？
PS我看到了一个名为“为什么你不应该像'for element in array'那样迭代”的问题，但是我找不到那个问题，所以我是这样问的。似乎我发现了以这种方式迭代的缺点之一。如果可能，请将我重定向到上述问题。
慕森卡浏览 10385回答 9
9回答
-
慕侠9296154
解释在上面的示例中，不可能对循环中的变量元素进行任何更改。代码要获得预期的输出，请尝试以下操作：array = [2,3,4]for i in range(len(array)): array[i] = 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翻翻过去那场雪
解释在上面的示例中，不可能对循环中的变量元素进行任何更改。代码要获得预期的输出，请尝试以下操作：array = [2,3,4]for i in range(len(array)): array[i] = 3print(array) -
慕标琳琳
当迭代开始元素变量已经在数组中具有当前值时。当您为其分配 3 时，它将包含它，直到下一次迭代时它将再次采用数组的当前值，依此类推。要获得 [3, 3, 3] 您需要按照以下方式进行操作：array = [2,3,4]for i in range(0,len(array)): array[i]=3print(array) -
慕村9548890
那是因为elementis 是for循环范围内的局部变量。运行这个片段。我希望它可以不言自明（我用e代替element），我也用来enumerate获取索引：array = [2,3,4]for i, e in enumerate(array): print('i =', i, 'e =', e) e = 100 print('e = 100-->','e =', e, 'but array[i]=',array[i]) array[i] = e print('array[i] = e --> array[i]=',array[i]) print('-'*10)print(array) #=> [100,100,100]快速解释e和i是局部变量，它们array在每次迭代时接收元素的值和 的索引。在循环内部，您可以更改 的值，e但不会影响array. 要更改其中的值array，需要通过索引 ( array[i])访问它。
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