blocks|key|309785|text|要获取这些值，请使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|309786|sorted(data.values())|code-block|syntax|javascript|309787|要获取匹配的密钥，请使用key函数|offset|length|style|CODE|309788|sorted(data,+key=data.get)|309789|获取按值排序的元组列表|309790|sorted(data.items(),+key=lambda+x:x[1])|309791|相关:请参阅此处的讨论：Dictionaries+are+ordered+in+Python+3.6%2B|309792|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/39980323/174728^0|0|0|C|3|0|0|0|0|C|13|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1A|8|@]|9|@$I|1B|J|1C|1|1D]]|A|$]]|$1|U|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

To get the values use

<pre><code>sorted(data.values())
</code></pre>

To get the matching keys, use a <code>key</code> function

<pre><code>sorted(data, key=data.get)
</code></pre>

To get a list of tuples ordered by value

<pre><code>sorted(data.items(), key=lambda x:x[1])
</code></pre>

Related: see the discussion here: <a href="https://stackoverflow.com/q/39980323/174728">Dictionaries are ordered in Python 3.6+</a>

blocks|key|92811|text|如果您实际上想对字典进行排序，而不仅仅是获得一个排序列表，那么可以使用collections.OrderedDict|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|92812|>>>+from+collections+import+OrderedDict
>>>+from+operator+import+itemgetter
>>>+data+=+{1:+'b',+2:+'a'}
>>>+d+=+OrderedDict(sorted(data.items(),+key=itemgetter(1)))
>>>+d
OrderedDict([(2,+'a'),+(1,+'b')])
>>>+d.values()
['a',+'b']|code-block|syntax|javascript|92813|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/collections.html#collections.OrderedDict^0|Z|N|Z|N|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

If you actually want to sort the dictionary instead of just obtaining a sorted list use <a href="http://docs.python.org/2/library/collections.html#collections.OrderedDict"><code>collections.OrderedDict</code></a>

<pre><code>&gt;&gt;&gt; from collections import OrderedDict
&gt;&gt;&gt; from operator import itemgetter
&gt;&gt;&gt; data = {1: 'b', 2: 'a'}
&gt;&gt;&gt; d = OrderedDict(sorted(data.items(), key=itemgetter(1)))
&gt;&gt;&gt; d
OrderedDict([(2, 'a'), (1, 'b')])
&gt;&gt;&gt; d.values()
['a', 'b']
</code></pre>

blocks|key|309904|text|从你的评论到gnibbler的回答，我想说你想要一个按值排序的键值对列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|309905|sorted(data.items(),+key=lambda+x:x[1])|code-block|syntax|javascript|309906|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|$]]

From your comment to gnibbler answer, i'd say you want a list of pairs of key-value sorted by value:

<pre><code>sorted(data.items(), key=lambda x:x[1])
</code></pre>

blocks|key|309810|text|对值进行排序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|309811|sorted(data.values())|code-block|syntax|javascript|309812|返回|309813|['a','b']|309814|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Sort the values:

<pre><code>sorted(data.values())
</code></pre>

returns

<pre><code>['a','b']
</code></pre>

blocks|key|92864|text|谢谢你所有的回答。你们都是我的英雄;-)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|92865|最后做了一些类似这样的事情：|92866|d+=+sorted(data,+key+=+data.get)

for+key+in+d:
++++text+=+data[key]|code-block|syntax|javascript|92867|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|$]]

Thanks for all answers.
You are all my heros ;-)
Did in the end something like this:
<pre><code>d = sorted(data, key = data.get)

for key in d:
 text = data[key]
</code></pre>

blocks|key|92791|text|我还认为重要的是要注意到，Python+dict对象类型是一个哈希表(more+on+this+here)，因此如果不将其键/值转换为列表，则无法对其进行排序。这允许在固定时间O(1)中检索dict项，而不管字典中元素的大小/数量。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|92792|话虽如此，一旦您对其键-+sorted(data.keys())或值-+sorted(data.values())进行了排序，您就可以使用该列表来访问设计模式中的键/值，如下所示：|92793|for+sortedKey+in+sorted(dictionary):
++++print+dictionary[sortedKeY]+#+gives+the+values+sorted+by+key

for+sortedValue+in+sorted(dictionary.values()):
++++print+sortedValue+#+gives+the+values+sorted+by+value|code-block|syntax|javascript|92794|希望这能有所帮助。|92795|entityMap|0|LINK|mutability|MUTABLE|url|http://mail.python.org/pipermail/python-list/2000-March/048085.html^0|K|4|2G|4|2N|4|Z|H|0|0|D|J|10|L|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1C|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

I also think it is important to note that Python <code>dict</code> object type is a hash table (<a href="http://mail.python.org/pipermail/python-list/2000-March/048085.html" rel="noreferrer">more on this here</a>), and thus is not capable of being sorted without converting its keys/values to lists. What this allows is <code>dict</code> item retrieval in constant time <code>O(1)</code>, no matter the size/number of elements in a dictionary.

Having said that, once you sort its keys - <code>sorted(data.keys())</code>, or values - <code>sorted(data.values())</code>, you can then use that list to access keys/values in design patterns such as these:

<pre><code>for sortedKey in sorted(dictionary):
 print dictionary[sortedKeY] # gives the values sorted by key

for sortedValue in sorted(dictionary.values()):
 print sortedValue # gives the values sorted by value
</code></pre>

Hope this helps.

blocks|key|92925|text|在回复John的评论中，您建议您需要字典的键和值，而不仅仅是值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|92926|PEP+256建议按值对字典进行排序。|offset|length|92927|import+operator
sorted(d.iteritems(),+key=operator.itemgetter(1))|code-block|syntax|javascript|92928|如果您想要降序，请执行以下操作|92929|sorted(d.iteritems(),+key=itemgetter(1),+reverse=True)|92930|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0265/^0|0|0|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@$D|Y|E|Z|1|10]]|A|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

In your comment in response to John, you suggest that you want the keys and values of the dictionary, not just the values.

<a href="https://www.python.org/dev/peps/pep-0265/" rel="nofollow noreferrer">PEP 256</a> suggests this for sorting a dictionary by values.

<pre><code>import operator
sorted(d.iteritems(), key=operator.itemgetter(1))
</code></pre>

If you want descending order, do this

<pre><code>sorted(d.iteritems(), key=itemgetter(1), reverse=True)
</code></pre>

blocks|key|310026|text|您可以根据值创建排序列表并重新构建字典：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|310027|myDictionary={"two":"2",+"one":"1",+"five":"5",+"1four":"4"}

newDictionary={}

sortedList=sorted(myDictionary.values())

for+sortedKey+in+sortedList:
++++for+key,+value+in+myDictionary.items():
++++++++if+value==sortedKey:
++++++++++++newDictionary[key]=value|code-block|syntax|javascript|310028|输出:+newDictionary={'one'：'1'，'two'：'2'，'1four'：'4'，'five'：'5'}|310029|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|$]]

You could created sorted list from Values and rebuild the dictionary:

<pre><code>myDictionary={"two":"2", "one":"1", "five":"5", "1four":"4"}

newDictionary={}

sortedList=sorted(myDictionary.values())

for sortedKey in sortedList:
 for key, value in myDictionary.items():
 if value==sortedKey:
 newDictionary[key]=value
</code></pre>

Output: newDictionary={'one': '1', 'two': '2', '1four': '4', 'five': '5'}

blocks|key|92969|text|没有lambda方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|92970|#+sort+dictionary+by+value
d+=+{'a1':+'fsdfds',+'g5':+'aa3432ff',+'ca':'zz23432'}
def+getkeybyvalue(d,i):
++++for+k,+v+in+d.items():
++++++++if+v+==+i:
++++++++++++return+(k)

sortvaluelist+=+sorted(d.values())
sortresult+={}
for+i1+in+sortvaluelist:+++
++++key+=+getkeybyvalue(d,i1)
++++sortresult[key]+=+i1
print+('=====sort+by+value=====')
print+(sortresult)
print+('=======================')|code-block|syntax|javascript|92971|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|$]]

no lambda method

<pre><code># sort dictionary by value
d = {'a1': 'fsdfds', 'g5': 'aa3432ff', 'ca':'zz23432'}
def getkeybyvalue(d,i):
 for k, v in d.items():
 if v == i:
 return (k)

sortvaluelist = sorted(d.values())
sortresult ={}
for i1 in sortvaluelist: 
 key = getkeybyvalue(d,i1)
 sortresult[key] = i1
print ('=====sort by value=====')
print (sortresult)
print ('=======================')
</code></pre>

Assume that I have a dict.

<pre><code>data = {1:'b', 2:'a'}
</code></pre>

And I want to sort data by 'b' and 'a' so I get the result

<pre><code>'a','b'
</code></pre>

How do I do that? 
Any ideas?

sort dict by value python

Python

假设我有一个字典。data = {1:'b', 2:'a'}我想按“b”和“a”对数据进行排序，这样就可以得到结果。'a','b'我该怎么做？有什么想法吗？

问按值对dict排序python
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