本文翻译自：How to return dictionary keys as a list in Python?
In Python 2.7 , I could get dictionary keys , values , or items as a list: 在Python 2.7中 ，我可以将字典键 ， 值或项作为列表获取：
>>> newdict = {1:0, 2:0, 3:0}
>>> newdict.keys()
[1, 2, 3]
Now, in Python >= 3.3 , I get something like this: 现在，在Python> = 3.3中 ，我得到如下信息：
>>> newdict.keys()
dict_keys([1, 2, 3])
So, I have to do this to get a list: 因此，我必须这样做以获得列表：
newlist = list()
for i in newdict.keys():
newlist.append(i)
I'm wondering, is there a better way to return a list in Python 3 ? 我想知道，是否有更好的方法在Python 3中返回列表？
#1楼
参考：https://stackoom.com/question/18ZRm/如何在Python中将字典键作为列表返回
#2楼
Try list(newdict.keys()) . 尝试list(newdict.keys()) 。
This will convert the dict_keys object to a list. 这会将dict_keys对象转换为列表。
On the other hand, you should ask yourself whether or not it matters. 另一方面，您应该问自己是否重要。 The Pythonic way to code is to assume duck typing ( if it looks like a duck and it quacks like a duck, it's a duck ). Python的编码方式是假设鸭子输入（ 如果看起来像鸭子，而像鸭子一样嘎嘎叫，那就是鸭子 ）。 The dict_keys object will act like a list for most purposes. 在dict_keys对象的作用类似于列表。 For instance: 例如：
for key in newdict.keys():
print(key)
Obviously, insertion operators may not work, but that doesn't make much sense for a list of dictionary keys anyway. 显然，插入运算符可能不起作用，但是对于字典关键字列表而言，这并没有多大意义。
#3楼
A bit off on the "duck typing" definition -- dict.keys() returns an iterable object, not a list-like object. 在“鸭子类型”定义上dict.keys()偏离dict.keys()返回一个可迭代的对象，而不是类似列表的对象。 It will work anywhere an iterable will work -- not any place a list will. 它可以在任何可迭代的地方都可以使用-列表不能在任何地方使用。 a list is also an iterable, but an iterable is NOT a list (or sequence...) 列表也是可迭代的，但可迭代的不是列表（或序列...）
In real use-cases, the most common thing to do with the keys in a dict is to iterate through them, so this makes sense. 在实际的用例中，与字典中的键有关的最常见的事情是遍历它们，因此这很有意义。 And if you do need them as a list you can call list() . 如果确实需要它们作为列表，则可以调用list() 。
Very similarly for zip() -- in the vast majority of cases, it is iterated through -- why create an entire new list of tuples just to iterate through it and then throw it away again? 与zip()非常相似-在大多数情况下，它都是经过迭代的-为什么要创建整个元组的新列表只是为了对其进行迭代，然后又将其丢弃？
This is part of a large trend in python to use more iterators (and generators), rather than copies of lists all over the place. 这是python中使用更多迭代器（和生成器）而不是在各处使用列表副本的一种大趋势的一部分。
dict.keys() should work with comprehensions, though -- check carefully for typos or something... it works fine for me: dict.keys()应该可以理解-仔细检查是否有错别字...对我来说很好用：
>>> d = dict(zip(['Sounder V Depth, F', 'Vessel Latitude, Degrees-Minutes'], [None, None]))
>>> [key.split(", ") for key in d.keys()]
[['Sounder V Depth', 'F'], ['Vessel Latitude', 'Degrees-Minutes']]
#4楼
list(newdict) works in both Python 2 and Python 3, providing a simple list of the keys in newdict . list(newdict)在Python 2和Python 3中均可使用，提供了newdict中的键的简单列表。 keys() isn't necessary. keys()不是必需的。 (: （：
#5楼
You can also use a list comprehension : 您还可以使用列表推导 ：
>>> newdict = {1:0, 2:0, 3:0}
>>> [k for k in newdict.keys()]
[1, 2, 3]
Or, shorter, 或更短一点
>>> [k for k in newdict]
[1, 2, 3]
Note: Order is not guaranteed on versions under 3.7 (ordering is still only an implementation detail with CPython 3.6). 注意：在3.7版以下的版本中，不能保证订购（订购仍然只是CPython 3.6的实现细节）。
#6楼
Converting to a list without using the keys method makes it more readable: 在不使用keys方法的情况下转换为列表使其更具可读性：
list(newdict)
and, when looping through dictionaries, there's no need for keys() : 并且，当遍历字典时，不需要keys() ：
for key in newdict:
print key
unless you are modifying it within the loop which would require a list of keys created beforehand: 除非您要在循环中进行修改，否则将需要预先创建的键列表：
for key in list(newdict):
del newdict[key]
On Python 2 there is a marginal performance gain using keys() . 在Python 2上，使用keys()获得少量性能提升 。
