Python List& for-each访问（在内置列表中查找/替换） [英] Python List & for-each access (Find/Replace in built-in list)
问题描述
来自C / C ++的我不禁想到内存管理，而且它也是一个纯粹的传递引用的语言。很难把它从我的头上。所以我试图从Java的角度来思考它，并把除了原始数据之外的所有东西都视为通过引用来传递。
问题：我有一个列表，包含一堆
如果我使用for-each语法，例如： $ p> 用于my_list中的成员：
print（member.str）;
是成员相当于一个实际的参考对象是什么？
这是相当于做：
i = 0
，而我< len（my_list）：
print（my_list [i]）
i + = 1
<我认为这不是，因为当我想要做一个替换，它不起作用，也就是说，这是行不通的：
for my_list中的成员：
if member == some_other_obj：
member = some_other_obj
一个简单的查找和替换列表。那可以在for-each循环中完成，如果是的话，怎么样？否则，我只需要使用随机访问语法（方括号），或将无法工作，我需要删除条目，并插入一个新的？即：
i = 0
for my_list中的成员：
if member == some_other_obj：
my_list.remove（i）
my_list.insert（i，member）
i + = 1
<解决方案
解决方案
解决方案
成员成员成员变量绑定到每个连续的列表元素。但是，在循环内重新分配该变量不会直接影响列表本身。例如，这段代码不会改变列表：
my_list = [1,2,3]
成员在my_list中：
member = 42
print my_list
输出：
$ block $
$ 1,2,3 $]如果您想更改包含不可变类型的列表，您需要执行以下操作：
my_list = [1,2,3 ]
for ndx，枚举成员（my_list）：
my_list [ndx] + = 42
print my_list
$ b $输出：
lockquote
<43,44,45>
如果您的列表包含可变对象，则可以直接修改当前的成员对象：
class C：
def __init __（self，n）：
self.num = n
def __repr __（self ）：
return str（self.num）
my_list = [C（i）for my xrange（3）]
my_list中的成员：
成员.num + = 42
print my_list
<42，43， 44
请注意，您仍然不会更改列表，只需修改列表中的对象即可。
您可以通过阅读命名和绑定。
I originally thought Python was a pure pass-by-reference language.
Coming from C/C++ I can't help but think about memory management, and it's hard to put it out of my head. So I'm trying to think of it from a Java perspective and think of everything but primitives as a pass by reference.
Problem: I have a list, containing a bunch of instances of a user defined class.
If I use the for-each syntax, ie.:
for member in my_list:
print(member.str);
Is member the equivalent of an actual reference to the object?
Is it the equivalent of doing:
i = 0
while i < len(my_list):
print(my_list[i])
i += 1
I think it's NOT, because when I'm looking to do a replace, it doesn't work, that is, this doesn't work:
for member in my_list:
if member == some_other_obj:
member = some_other_obj
A simple find and replace in a list. Can that be done in a for-each loop, if so, how? Else, do I simply have to use the random access syntax (square brackets), or will NEITHER work and I need to remove the entry, and insert a new one? I.e.:
i = 0
for member in my_list:
if member == some_other_obj:
my_list.remove(i)
my_list.insert(i, member)
i += 1
Answering this has been good, as the comments have led to an improvement in my own understanding of Python variables.
As noted in the comments, when you loop over a list with something like for member in my_list the member variable is bound to each successive list element. However, re-assigning that variable within the loop doesn't directly affect the list itself. For example, this code won't change the list:
my_list = [1,2,3]
for member in my_list:
member = 42
print my_list
Output:
[1, 2, 3]
If you want to change a list containing immutable types, you need to do something like:
my_list = [1,2,3]
for ndx, member in enumerate(my_list):
my_list[ndx] += 42
print my_list
Output:
[43, 44, 45]
If your list contains mutable objects, you can modify the current member object directly:
class C:
def __init__(self, n):
self.num = n
def __repr__(self):
return str(self.num)
my_list = [C(i) for i in xrange(3)]
for member in my_list:
member.num += 42
print my_list
[42, 43, 44]
Note that you are still not changing the list, simply modifying the objects in the list.
You might benefit from reading Naming and Binding.
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