blocks|key|1551812|text|您不能使用基础dict类做到这一点--它是按散列排序的。你可以构建你自己的字典，它实际上是一个键，值对或类似的列表，它将是有序的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1551813|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can't do this with the base dict class -- it's ordered by hash. You could build your own dictionary that is really a list of key,value pairs or somesuch, which would be ordered.

blocks|key|1551842|text|如果您使用的是CPython+dict+(对于Python的任何其他实现，则为Python+3.6%2B或Python+3.7%2B+)，则默认情况下，标准Python+3.6%2B会执行此操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1551843|在较旧版本的Python中，您可以使用collections.OrderedDict。|1551844|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/library/collections.html#ordereddict-objects^0|F|4|0|J|N|J|N|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|H|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

The standard Python <code>dict</code> does this by default if you're using CPython 3.6+ (or Python 3.7+ for any other implementation of Python).

On older versions of Python you can use <a href="https://docs.python.org/library/collections.html#ordereddict-objects" rel="noreferrer"><code>collections.OrderedDict</code></a>.

blocks|key|1766924|text|其他答案是正确的；这是不可能的，但你可以自己写。但是，如果您不确定如何实际实现这样的东西，这里有一个完整的、有效的实现，它是我刚刚编写和测试的子类dict。(请注意，传递给构造函数的值的顺序是未定义的，但将在稍后传递的值之前出现，并且您可以始终不允许使用值初始化有序的字典。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1766925|class+ordered_dict(dict):
++++def+__init__(self,+*args,+**kwargs):
++++++++dict.__init__(self,+*args,+**kwargs)
++++++++self._order+=+self.keys()

++++def+__setitem__(self,+key,+value):
++++++++dict.__setitem__(self,+key,+value)
++++++++if+key+in+self._order:
++++++++++++self._order.remove(key)
++++++++self._order.append(key)

++++def+__delitem__(self,+key):
++++++++dict.__delitem__(self,+key)
++++++++self._order.remove(key)

++++def+order(self):
++++++++return+self._order[:]

++++def+ordered_items(self):
++++++++return+[(key,self[key])+for+key+in+self._order]


od+=+ordered_dict()
od["hello"]+=+"world"
od["goodbye"]+=+"cruel+world"
print+od.order()++++++++++++#+prints+['hello',+'goodbye']

del+od["hello"]
od["monty"]+=+"python"
print+od.order()++++++++++++#+prints+['goodbye',+'monty']

od["hello"]+=+"kitty"
print+od.order()++++++++++++#+prints+['goodbye',+'monty',+'hello']

print+od.ordered_items()
#+prints+[('goodbye','cruel+world'),+('monty','python'),+('hello','kitty')]|code-block|syntax|javascript|1766926|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The other answers are correct; it's not possible, but you could write this yourself. However, in case you're unsure how to actually implement something like this, here's a complete and working implementation that subclasses dict which I've just written and tested. (Note that the order of values passed to the constructor is undefined but will come before values passed later, and you could always just not allow ordered dicts to be initialized with values.)

<pre><code>class ordered_dict(dict):
 def __init__(self, *args, **kwargs):
 dict.__init__(self, *args, **kwargs)
 self._order = self.keys()

 def __setitem__(self, key, value):
 dict.__setitem__(self, key, value)
 if key in self._order:
 self._order.remove(key)
 self._order.append(key)

 def __delitem__(self, key):
 dict.__delitem__(self, key)
 self._order.remove(key)

 def order(self):
 return self._order[:]

 def ordered_items(self):
 return [(key,self[key]) for key in self._order]


od = ordered_dict()
od["hello"] = "world"
od["goodbye"] = "cruel world"
print od.order() # prints ['hello', 'goodbye']

del od["hello"]
od["monty"] = "python"
print od.order() # prints ['goodbye', 'monty']

od["hello"] = "kitty"
print od.order() # prints ['goodbye', 'monty', 'hello']

print od.ordered_items()
# prints [('goodbye','cruel world'), ('monty','python'), ('hello','kitty')]
</code></pre>

blocks|key|1766972|text|如果您不需要dict功能，并且只需要按您插入元组的顺序返回元组，那么使用队列不是更好吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1766973|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

if you don't need the dict functionality, and only need to return tuples in the order you've inserted them, wouldn't a queue work better?

blocks|key|1767007|text|我以前使用过StableDict，并取得了很好的成功。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1767008|http://pypi.python.org/pypi/StableDict/0.2|offset|length|1767009|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|16|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

I've used StableDict before with good success.

<a href="http://pypi.python.org/pypi/StableDict/0.2" rel="nofollow noreferrer">http://pypi.python.org/pypi/StableDict/0.2</a>

blocks|key|1767053|text|或者，使用time.now()作为元组中的第一个字段，将键设置为元组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1767054|然后，您可以使用dictname.keys()、sort和voila检索键！|1767055|格里|1767056|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Or, just make the key a tuple with time.now() as the first field in the tuple.

Then you can retrieve the keys with dictname.keys(), sort, and voila!

Gerry

blocks|key|1767096|text|除非您将键存储在单独的列表中以供以后引用，否则这是不可能的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1767097|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It's not possible unless you store the keys in a separate list for referencing later.

blocks|key|1552159|text|或者使用PEP-372描述的here的任何实现，比如来自pythonutils的odict+module。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1552160|我成功地使用了pocoo.org实现，它就像替换您的|1552161|my_dict={}
my_dict["foo"]="bar"|code-block|syntax|javascript|1552162|使用|1552163|my_dict=odict.odict()
my_dict["foo"]="bar"|1552164|并且只需要this+file|1552165|entityMap|0|LINK|mutability|MUTABLE|url|http://www.python.org/dev/peps/pep-0372/|1|http://www.python.org/dev/peps/pep-0372/#reference-implementation|2|http://www.voidspace.org.uk/python/pythonutils.html|3|http://www.voidspace.org.uk/python/odict.html|4|http://dev.pocoo.org/hg/sandbox/raw-file/tip/odict.py^0|4|7|0|E|4|1|S|B|2|14|C|3|0|0|0|0|0|5|9|4|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@$A|17|B|18|1|19]|$A|1A|B|1B|1|1C]|$A|1D|B|1E|1|1F]|$A|1G|B|1H|1|1I]]|C|$]]|$1|D|3|E|5|6|7|1J|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|1K|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|1L|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|1M|8|@]|9|@]|C|$I|J]]|$1|O|3|P|5|6|7|1N|8|@]|9|@$A|1O|B|1P|1|1Q]]|C|$]]|$1|Q|3|-4|5|6|7|1R|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]|10|$5|T|U|V|C|$W|11]]|12|$5|T|U|V|C|$W|13]]|14|$5|T|U|V|C|$W|15]]]]

Or use any of the implementations for the <a href="http://www.python.org/dev/peps/pep-0372/" rel="nofollow noreferrer">PEP-372</a> described <a href="http://www.python.org/dev/peps/pep-0372/#reference-implementation" rel="nofollow noreferrer">here</a>, like the <a href="http://www.voidspace.org.uk/python/odict.html" rel="nofollow noreferrer">odict module</a> from the <a href="http://www.voidspace.org.uk/python/pythonutils.html" rel="nofollow noreferrer">pythonutils</a>.

I successfully used the pocoo.org implementation, it is as easy as replacing your

<pre><code>my_dict={}
my_dict["foo"]="bar"
</code></pre>

with

<pre><code>my_dict=odict.odict()
my_dict["foo"]="bar"
</code></pre>

and require just <a href="http://dev.pocoo.org/hg/sandbox/raw-file/tip/odict.py" rel="nofollow noreferrer">this file</a>

blocks|key|1767216|text|您可以做的是插入带有表示输入订单的键的值，然后对这些项调用sorted()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1767217|>>>+obj+=+{}
>>>+obj[1]+=+'Bob'
>>>+obj[2]+=+'Sally'
>>>+obj[3]+=+'Joe'
>>>+for+k,+v+in+sorted(obj.items()):
...+++++print+v
...+
Bob
Sally
Joe
>>>+|code-block|syntax|javascript|1767218|entityMap^0|T|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

What you can do is insert the values with a key representing the order inputted, and then call <code>sorted()</code> on the items.

<pre><code>&gt;&gt;&gt; obj = {}
&gt;&gt;&gt; obj[1] = 'Bob'
&gt;&gt;&gt; obj[2] = 'Sally'
&gt;&gt;&gt; obj[3] = 'Joe'
&gt;&gt;&gt; for k, v in sorted(obj.items()):
... print v
... 
Bob
Sally
Joe
&gt;&gt;&gt; 
</code></pre>

blocks|key|1552255|text|使用OrderedDict()，从2.7版开始提供|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1552256|只是出于好奇心：|1552257|from+collections+import+OrderedDict
a+=+{}
b+=+OrderedDict()
c+=+OrderedDict()

a['key1']+=+'value1'
a['key2']+=+'value2'

b['key1']+=+'value1'
b['key2']+=+'value2'

c['key2']+=+'value2'
c['key1']+=+'value1'

print+a+==+b++#+True
print+a+==+c++#+True
print+b+==+c++#+False|code-block|syntax|javascript|1552258|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Use OrderedDict(), available since version 2.7
Just a matter of curiosity: 
<pre><code>from collections import OrderedDict
a = {}
b = OrderedDict()
c = OrderedDict()

a['key1'] = 'value1'
a['key2'] = 'value2'

b['key1'] = 'value1'
b['key2'] = 'value2'

c['key2'] = 'value2'
c['key1'] = 'value1'

print a == b # True
print a == c # True
print b == c # False
</code></pre>

Is it possible to retrieve items from a Python dictionary in the order that they were inserted?

How do you retrieve items from a dictionary in the order that they're inserted?

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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是否可以按照插入的顺序从Python字典中检索条目？

问如何按照条目的插入顺序从字典中检索条目？
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