在 Python 中，什么决定了迭代 kwargs 时的顺序？

发布于 2024-12-28 13:22:43 字数 420 浏览 1 评论 0原文

在 python 中，我编写了这个函数来教自己 **kwargs 在 Python 中如何工作：

def fxn(a1, **kwargs):
    print a1
    for k in kwargs:
        print k, " : ", kwargs[k]

然后我使用以下命令调用该函数

fxn(3, a2=2, a3=3, a4=4)

这是我的 Python 解释器打印的输出：

3
a3  :  3
a2  :  2
a4  :  4

Why did the for loop print the value a3 在 a2 之前，即使我先将 a2 输入到我的函数中？

原文

In python, I wrote this function to teach myself how **kwargs works in Python:

def fxn(a1, **kwargs):
    print a1
    for k in kwargs:
        print k, " : ", kwargs[k]

I then called this function with

fxn(3, a2=2, a3=3, a4=4)

Here was the output that my Python interpreter printed:

3
a3  :  3
a2  :  2
a4  :  4

Why did the for loop print the value of a3 before that of a2 even though I fed a2 into my function first?

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荆棘i 2025-01-04 13:22:43

kwargs 是一本字典。字典是无序的——简单地说，顺序是未指定的，并且是实现细节。仔细观察会发现，顺序会根据项目的哈希值、插入顺序等而发生很大变化，因此您最好不要依赖与之相关的任何内容。

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丢了幸福的猪 2025-01-04 13:22:43

这是一本字典。并且，正如文档中提到的，字典没有顺序（来自 http ://docs.python.org/tutorial/datastructs.html#dictionaries）：

最好将字典视为一组无序的键：值对，并要求键是唯一的（在一个字典内）。

但是您可以按某种顺序处理它，例如：

使用sorted()：

def fxn(a1, **kwargs):
    打印a1
    for k in Sorted(kwargs): # 注意“kwargs”被“sorted(kwargs)”替换
        打印 k, " : ", kwargs[k]

或使用OrderedDict类型（您可以将OrderedDict对象作为参数传递包含所有键值对）：

从集合导入 OrderedDict

def fxn(a1, ordkwargs):
    打印a1
    对于 ordkwargs 中的 k：
        打印 k, " : ", ordkwargs[k]

fxn(3, OrderedDict((('a2',2), ('a3',3), ('a4',4))))

This is a dictionary. And, as mentioned in documentation, dictionary has no order (from http://docs.python.org/tutorial/datastructures.html#dictionaries):

It is best to think of a dictionary as an unordered set of key: value pairs, with the requirement that the keys are unique (within one dictionary).

But you can make it processed in some order, like that:

using sorted():

def fxn(a1, **kwargs):
    print a1
    for k in sorted(kwargs): # notice "kwargs" replaced by "sorted(kwargs)"
        print k, " : ", kwargs[k]

or by using OrderedDict type (you can pass OrderedDict object as parameter containing all the key-value pairs):

from collections import OrderedDict

def fxn(a1, ordkwargs):
    print a1
    for k in ordkwargs:
        print k, " : ", ordkwargs[k]

fxn(3, OrderedDict((('a2',2), ('a3',3), ('a4',4))))

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晨曦÷微暖 2025-01-04 13:22:43

这最终在 3.6 版本中引入：dict 现在已排序，因此保留了关键字参数顺序。

Python 3.6.0 (default, Jan 13 2017, 13:27:48) 
>>> def print_args(**kwargs):
...     print(kwargs.keys())
... 
>>> print_args(first=1, second=2, third=3)
dict_keys(['first', 'second', 'third'])

This has finally been introduced in the 3.6 release: dicts are now ordered, therefore the keyword argument order is preserved.

Python 3.6.0 (default, Jan 13 2017, 13:27:48) 
>>> def print_args(**kwargs):
...     print(kwargs.keys())
... 
>>> print_args(first=1, second=2, third=3)
dict_keys(['first', 'second', 'third'])

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拿命拼未来 2025-01-04 13:22:43

不幸的讽刺是 **kwargs 的字典化意味着以下内容将不起作用（至少不是人们期望的方式）：

od = OrderedDict(a=1, b=2, c=3)

由于关键字参数首先构建到无序字典中，因此您不能依赖它们将按照列出的顺序插入到 OrderedDict 中。 :(

The unfortunate irony is that the dict-ification of **kwargs means that the following will not work (at least not the way one would expect):

od = OrderedDict(a=1, b=2, c=3)

Since the keyworded args are first built into an unordered dict, you cannot depend that they will be inserted into the OrderedDict in the order they are listed. :(

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