blocks|key|383889|text|Python3改变了dict.keys的行为，现在它返回一个可迭代但不可索引的dict_keys对象(就像旧的dict.iterkeys，现在已经不存在了)。您可以通过显式调用list来获取Python2结果|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|383890|>>>+b+=+{+'video':0,+'music':23+}
>>>+k+=+list(b.keys())
>>>+k
['music',+'video']|code-block|syntax|javascript|383891|或者只是|383892|>>>+list(b)
['music',+'video']|383893|entityMap^0|A|9|13|9|1J|D|2G|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

Python 3 changed the behavior of <code>dict.keys</code> such that it now returns a <code>dict_keys</code> object, which is iterable but not indexable (it's like the old <code>dict.iterkeys</code>, which is gone now). You can get the Python 2 result back with an explicit call to <code>list</code>:

<pre><code>&gt;&gt;&gt; b = { 'video':0, 'music':23 }
&gt;&gt;&gt; k = list(b.keys())
&gt;&gt;&gt; k
['music', 'video']
</code></pre>

or just

<pre><code>&gt;&gt;&gt; list(b)
['music', 'video']
</code></pre>

blocks|key|383895|text|如果您像这样分配k：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|383896|k+=+list(b.keys())|code-block|syntax|javascript|383897|您的代码将会正常工作。|383898|如错误所述，dict_keys类型不支持索引。|383899|entityMap^0|8|1|0|0|0|6|9|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

If you assigned <code>k</code> like so:

<pre><code>k = list(b.keys())
</code></pre>

your code will work.

As the error says, the <code>dict_keys</code> type does not support indexing.

blocks|key|4342058|text|这是Python+2和3之间的突破性变化之一。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4342059|在Python+2中：|4342060|>>>+help(dict.keys)
keys(...)
++++D.keys()+->+list+of+D's+keys|code-block|syntax|javascript|4342061|在Python+3中：|4342062|>>>+help(dict.keys)
keys(...)
++++D.keys()+->+a+set-like+object+providing+a+view+on+D's+keys|4342063|这种行为的改变很有意义，因为字典在语义上是无序的，它的键是唯一的-就像一个集合。|4342064|这一改变意味着你不必每次都创建一个新的关键点列表，每次你想要与dict的关键点进行某种集合比较时。|4342065|在2和3中获得相同的行为|4342066|为了帮助过渡到Python3，Python2.7有另一个dict方法，viewkeys。viewkeys方法与Python3的dict.keys方法非常相似：|offset|length|style|CODE|4342067|>>>+d
{'a':+None,+'c':+None,+'b':+None,+'d':+None}
>>>+for+k+in+d.viewkeys():+print+k
...+
a
c
b
d
>>>+d.viewkeys()+&+set('abc')
set(['a',+'c',+'b'])|4342068|在Python3中，与旧行为最相似的是将dict.keys()传递给list|4342069|>>>+d
{'d':+None,+'a':+None,+'c':+None,+'b':+None}
>>>+list(d.keys())
['d',+'a',+'c',+'b']|4342070|或者直接将dict传递给list，因为dict无论如何都会遍历它的键：|4342071|>>>+list(d)
['d',+'a',+'c',+'b']|4342072|您可以创建一个实用函数来对2和3的行为进行抽象：|4342073|if+hasattr(dict,+'viewkeys'):+#+Python+2.7
++++def+keys(d):
++++++++return+d.viewkeys()
else:++#+Python+3
++++def+keys(d):
++++++++return+d.keys()|4342074|并将dict传递给list以获得列表形式，在2和3中，您将获得相同的输出：|4342075|>>>+d
{'b':+None,+'a':+None,+'c':+None,+'d':+None}
>>>+keys(d)
dict_keys(['b',+'a',+'c',+'d'])
>>>+list(d)
['b',+'a',+'c',+'d']|4342076|entityMap^0|0|0|0|0|0|0|0|0|Z|8|18|8|1R|9|0|0|K|B|Y|4|0|0|C|4|0|0|0|0|9|4|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1J|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1L|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1M|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1N|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1O|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1P|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1Q|8|@$U|1R|V|1S|W|X]|$U|1T|V|1U|W|X]|$U|1V|V|1W|W|X]]|9|@]|A|$]]|$1|Y|3|Z|5|F|7|1X|8|@]|9|@]|A|$G|H]]|$1|10|3|11|5|6|7|1Y|8|@$U|1Z|V|20|W|X]|$U|21|V|22|W|X]]|9|@]|A|$]]|$1|12|3|13|5|F|7|23|8|@]|9|@]|A|$G|H]]|$1|14|3|15|5|6|7|24|8|@$U|25|V|26|W|X]]|9|@]|A|$]]|$1|16|3|17|5|F|7|27|8|@]|9|@]|A|$G|H]]|$1|18|3|19|5|6|7|28|8|@]|9|@]|A|$]]|$1|1A|3|1B|5|F|7|29|8|@]|9|@]|A|$G|H]]|$1|1C|3|1D|5|6|7|2A|8|@$U|2B|V|2C|W|X]]|9|@]|A|$]]|$1|1E|3|1F|5|F|7|2D|8|@]|9|@]|A|$G|H]]|$1|1G|3|-4|5|6|7|2E|8|@]|9|@]|A|$]]]|1H|$]]

This is one of the breaking changes between Python 2 and 3.

In Python 2:

<pre><code>&gt;&gt;&gt; help(dict.keys)
keys(...)
 D.keys() -&gt; list of D's keys
</code></pre>

In Python 3:

<pre><code>&gt;&gt;&gt; help(dict.keys)
keys(...)
 D.keys() -&gt; a set-like object providing a view on D's keys
</code></pre>

This change in behavior makes a lot of sense since a dict is semantically unordered and its keys are unique - just like a set. 

This change means that you don't have to create a new list of keys every time you want to do some kind of set comparison with a dict's keys.

<h1>Getting the same behavior in 2 and 3</h1>

To help transition to Python 3, Python 2.7 has another dict method, <code>viewkeys</code>. The <code>viewkeys</code> method is most similar to Python 3's <code>dict.keys</code> method:

<pre><code>&gt;&gt;&gt; d
{'a': None, 'c': None, 'b': None, 'd': None}
&gt;&gt;&gt; for k in d.viewkeys(): print k
... 
a
c
b
d
&gt;&gt;&gt; d.viewkeys() &amp; set('abc')
set(['a', 'c', 'b'])
</code></pre>

In Python 3, the closest analog to the old behavior is to pass <code>dict.keys()</code> to <code>list</code>:

<pre><code>&gt;&gt;&gt; d
{'d': None, 'a': None, 'c': None, 'b': None}
&gt;&gt;&gt; list(d.keys())
['d', 'a', 'c', 'b']
</code></pre>

Or just pass the dict to <code>list</code>, since a dict will iterate over its keys anyways:

<pre><code>&gt;&gt;&gt; list(d)
['d', 'a', 'c', 'b']
</code></pre>

You could create a utility functions to abstract the behavior over 2 and 3:

<pre><code>if hasattr(dict, 'viewkeys'): # Python 2.7
 def keys(d):
 return d.viewkeys()
else: # Python 3
 def keys(d):
 return d.keys()
</code></pre>

And pass a dict to <code>list</code> to get the list form, and in both 2 and 3, you'll get the same output:

<pre><code>&gt;&gt;&gt; d
{'b': None, 'a': None, 'c': None, 'd': None}
&gt;&gt;&gt; keys(d)
dict_keys(['b', 'a', 'c', 'd'])
&gt;&gt;&gt; list(d)
['b', 'a', 'c', 'd']
</code></pre>

blocks|key|301822|text|如果你只是想要一个字典中的键列表，你可以直接这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|301823|b+=+{"name":+"xyz",+"class":"abc",+"college":+"qwert"}++
key_list+=+list(b)|code-block|syntax|javascript|301824|key_list将以列表的形式包含所有键名称，但是，如果发现多次，则不会重复一个键。重复的密钥将计入一个密钥。|301825|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you simply want a list of keys from a dictionary you can directly do like this:

<pre><code>b = {"name": "xyz", "class":"abc", "college": "qwert"} 
key_list = list(b)
</code></pre>

key_list will contain all the key names as a list, though, this will not repeats a key, if found more than once. Duplicate keys will be counted as one.

blocks|key|382773|text|import+random
b+=+{+'video':0,+'music':23,"picture":12+}
random.choice(tuple(b.items()))

#+Returns+a+random+dictionary+entry+as+a+tuple:
#+('music',+23)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|382774|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import random
b = { 'video':0, 'music':23,"picture":12 }
random.choice(tuple(b.items()))

# Returns a random dictionary entry as a tuple:
# ('music', 23)
</code></pre>

I am trying to use the <code>.keys()</code> and instead of getting a list of the keys like 
always have in the past. However I get this.

<pre><code>b = { 'video':0, 'music':23 }
k = b.keys()
print( k[0] )

&gt;&gt;&gt;TypeError: 'dict_keys' object does not support indexing

print( k )
dict_keys(['music', 'video'])
</code></pre>

it should just print ['music', 'video'] unless I'm going crazy. 

What's going on?

Python dictionary.keys() error

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试使用.keys()，而不是像过去一样获取密钥列表。然而，我得到了这个。b = { 'video':0, 'music':23 }k = b.keys()print( k[0] )>>>TypeError: 'dict_keys' object does not support indexingprint( k...

问Python dictionary.keys()错误
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问Python dictionary.keys()错误EN