以列表理解方式更新对象属性 [英] Updating object properties in list comprehension way
问题描述
在Python中，是否有可能以列表理解或类似的方式更新对象列表? 例如，我想设置列表中所有对象的属性:
Is that possible, in Python, to update a list of objects in list comprehension or some similar way? For example, I'd like to set property of all objects in the list:
result = [ object.name = "blah" for object in objects]
或带有map功能
result = map(object.name = "blah", objects)
是否可以通过不带属性设置的循环来实现?
Could it be achieved without for-looping with property setting?
(注意:以上所有示例都是故意的错误，仅用于表达想法)
(Note: all above examples are intentionally wrong and provided only to express the idea)
推荐答案
最终，赋值是声明"，而不是表达式"，因此不能在lambda表达式或列表理解中使用.您需要一个常规函数来完成您要尝试的工作.
Ultimately, assignment is a "Statement", not an "Expression", so it can't be used in a lambda expression or list comprehension. You need a regular function to accomplish what you're trying.
有一个内置函数可以执行此操作(返回None的列表):
There is a builtin which will do it (returning a list of None):
[setattr(obj,'name','blah') for obj in objects]
但是请不要使用它.只需使用循环即可.我怀疑您会注意到效率方面的任何差异，而且循环是否清晰得多.
But please don't use it. Just use a loop. I doubt that you'll notice any difference in efficiency and a loop is so much more clear.
如果您确实需要1-liner(尽管我不知道为什么):
If you really need a 1-liner (although I don't see why):
for obj in objects: obj.name = "blah"
我发现大多数人都想使用列表理解，因为有人告诉他们快速".是正确的，但仅用于创建新列表.使用列表推导来产生副作用不太可能带来任何性能上的好处，并且代码的可读性会受到影响.确实，使用列表推导而不是.append的等效循环的最重要原因是因为它更易于阅读.
I find that most people want to use list-comprehensions because someone told them that they are "fast". That's correct, but only for creating a new list. Using a list comprehension for side-effects is unlikely to lead to any performance benefit and your code will suffer in terms of readability. Really, the most important reason to use a list comprehension instead of the equivalent loop with .append is because it is easier to read.
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