在Python3中按索引访问dict_keys元素-Python 实用宝典

在Python3中按索引访问dict_keys元素

我正在尝试通过其索引访问dict_key的元素: test = {'foo': 'bar', 'hello': 'world'} keys = test.keys() # dict_keys object keys.index(0) AttributeError: 'dict_keys' object has no attribute 'index' 我想得到foo。 与: keys[0] TypeError: 'dict_keys' object does not support indexing 我怎样才能做到这一点?

问题:在Python3中按索引访问dict_keys元素

我正在尝试通过其索引访问dict_key的元素:

test = {'foo': 'bar', 'hello': 'world'}
keys = test.keys()  # dict_keys object

keys.index(0)
AttributeError: 'dict_keys' object has no attribute 'index'

我想得到foo

与:

keys[0]
TypeError: 'dict_keys' object does not support indexing

我怎样才能做到这一点?

I'm trying to access a dict_key's element by its index:

test = {'foo': 'bar', 'hello': 'world'}
keys = test.keys()  # dict_keys object

keys.index(0)
AttributeError: 'dict_keys' object has no attribute 'index'

I want to get foo.

same with:

keys[0]
TypeError: 'dict_keys' object does not support indexing

How can I do this?


回答 0

list()而是调用字典:

keys = list(test)

在Python 3中,该dict.keys()方法返回一个字典视图对象,它作为一个集合。直接迭代字典也会产生键,因此将字典转换为列表会得到所有键的列表:

>>> test = {'foo': 'bar', 'hello': 'world'}
>>> list(test)
['foo', 'hello']
>>> list(test)[0]
'foo'

Call list() on the dictionary instead:

keys = list(test)

In Python 3, the dict.keys() method returns a dictionary view object, which acts as a set. Iterating over the dictionary directly also yields keys, so turning a dictionary into a list results in a list of all the keys:

>>> test = {'foo': 'bar', 'hello': 'world'}
>>> list(test)
['foo', 'hello']
>>> list(test)[0]
'foo'

回答 1

不是完整的答案,但可能是有用的提示。如果它确实是您想要的第一项*,那么

next(iter(q))

比快得多

list(q)[0]

对于大词典,因为不必将整个内容存储在内存中。

对于10.000.000物品,我发现它快了将近40.000倍。

*如果dict只是Python 3.6之前的伪随机项目,则第一项(此后它在标准实现中已订购,尽管不建议依赖它)。

Not a full answer but perhaps a useful hint. If it is really the first item you want*, then

next(iter(q))

is much faster than

list(q)[0]

for large dicts, since the whole thing doesn't have to be stored in memory.

For 10.000.000 items I found it to be almost 40.000 times faster.

*The first item in case of a dict being just a pseudo-random item before Python 3.6 (after that it's ordered in the standard implementation, although it's not advised to rely on it).


回答 2

我想要第一个字典项的“键”和“值”对。我用下面的代码。

 key, val = next(iter(my_dict.items()))

I wanted "key" & "value" pair of a first dictionary item. I used the following code.

 key, val = next(iter(my_dict.items()))

回答 3

test = {'foo': 'bar', 'hello': 'world'}
ls = []
for key in test.keys():
    ls.append(key)
print(ls[0])

将键附加到静态定义的列表然后对其进行索引的常规方式

test = {'foo': 'bar', 'hello': 'world'}
ls = []
for key in test.keys():
    ls.append(key)
print(ls[0])

Conventional way of appending the keys to a statically defined list and then indexing it for same


回答 4

在许多情况下,这可能是XY问题。为什么要按位置索引字典键?您真的需要吗?直到最近,字典甚至还没有在Python中排序,因此访问第一个元素是任意的。

我刚刚将一些Python 2代码翻译为Python 3:

keys = d.keys()
for (i, res) in enumerate(some_list):
    k = keys[i]
    # ...

这不是很漂亮,但也不是很糟糕。起初,我正要用可怕的东西代替它

    k = next(itertools.islice(iter(keys), i, None))

在我意识到这一切写成更好之前

for (k, res) in zip(d.keys(), some_list):

效果很好。

我相信在许多其他情况下,可以避免按位置索引字典关键字。尽管字典在Python 3.7中是有序的,但是依靠它并不是很漂亮。上面的代码仅起作用,因为的内容some_list是最近从的内容中产生的d

如果您确实需要disk_keys按索引访问元素,请仔细看一下代码。也许您不需要。

In many cases, this may be an XY Problem. Why are you indexing your dictionary keys by position? Do you really need to? Until recently, dictionaries were not even ordered in Python, so accessing the first element was arbitrary.

I just translated some Python 2 code to Python 3:

keys = d.keys()
for (i, res) in enumerate(some_list):
    k = keys[i]
    # ...

which is not pretty, but not very bad either. At first, I was about to replace it by the monstrous

    k = next(itertools.islice(iter(keys), i, None))

before I realised this is all much better written as

for (k, res) in zip(d.keys(), some_list):

which works just fine.

I believe that in many other cases, indexing dictionary keys by position can be avoided. Although dictionaries are ordered in Python 3.7, relying on that is not pretty. The code above only works because the contents of some_list had been recently produced from the contents of d.

Have a hard look at your code if you really need to access a disk_keys element by index. Perhaps you don't need to.


回答 5

试试这个

keys = [next(iter(x.keys())) for x in test]
print(list(keys))

结果看起来像这样。['foo','hello']

您可以在此处找到更多可能的解决方案。

Try this

keys = [next(iter(x.keys())) for x in test]
print(list(keys))

The result looks like this. ['foo', 'hello']

You can find more possible solutions here.


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