blocks|key|2752059|text|这适用于任何长度的字典：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2752060|>>>+result+=+{}
>>>+for+d+in+L:
...++++result.update(d)
...+
>>>+result
{'a':1,'c':1,'b':2,'d':2}|code-block|syntax|javascript|2752061|作为comprehension|offset|length|2752062|#+Python+>=+2.7
{k:+v+for+d+in+L+for+k,+v+in+d.items()}

#+Python+<+2.7
dict(pair+for+d+in+L+for+pair+in+d.items())|2752063|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0274/^0|0|0|2|D|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

This works for dictionaries of any length:

<pre><code>&gt;&gt;&gt; result = {}
&gt;&gt;&gt; for d in L:
... result.update(d)
... 
&gt;&gt;&gt; result
{'a':1,'c':1,'b':2,'d':2}
</code></pre>

As a <a href="https://www.python.org/dev/peps/pep-0274/" rel="noreferrer">comprehension</a>:

<pre><code># Python &gt;= 2.7
{k: v for d in L for k, v in d.items()}

# Python &lt; 2.7
dict(pair for d in L for pair in d.items())
</code></pre>

blocks|key|892684|text|对于Python+3.3%2B，有一个ChainMap+collection|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|892685|>>>+from+collections+import+ChainMap
>>>+a+=+[{'a':1},{'b':2},{'c':1},{'d':2}]
>>>+dict(ChainMap(*a))
{'b':+2,+'c':+1,+'a':+1,+'d':+2}|code-block|syntax|javascript|892686|另请参阅：|892687|892688|What+is+the+purpose+of+collections.ChainMap?|unordered-list-item|892689|892690|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/collections.html#collections.ChainMap|1|https://stackoverflow.com/questions/23392976/what-is-the-purpose-of-collections-chainmap^0|H|8|H|J|0|0|0|0|0|0|18|1|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@$9|14|A|15|1|16]]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|Q|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]|Z|$5|U|V|W|E|$X|10]]]]

In case of Python 3.3+, there is a <a href="https://docs.python.org/3/library/collections.html#collections.ChainMap" rel="noreferrer"><code>ChainMap</code> collection</a>:

<pre><code>&gt;&gt;&gt; from collections import ChainMap
&gt;&gt;&gt; a = [{'a':1},{'b':2},{'c':1},{'d':2}]
&gt;&gt;&gt; dict(ChainMap(*a))
{'b': 2, 'c': 1, 'a': 1, 'd': 2}
</code></pre>

Also see:

<ul>
<li><a href="https://stackoverflow.com/questions/23392976/what-is-the-purpose-of-collections-chainmap">What is the purpose of collections.ChainMap?</a></li>
</ul>

blocks|key|1637646|text|这与@delnan类似，但提供了修改k/v+(键/值)项的选项，我认为更具可读性：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1637647|new_dict+=+{k:v+for+list_item+in+list_of_dicts+for+(k,v)+in+list_item.items()}|code-block|syntax|javascript|1637648|例如，按如下方式替换k/v元素：|1637649|new_dict+=+{str(k).replace("+","_"):v+for+list_item+in+list_of_dicts+for+(k,v)+in+list_item.items()}|1637650|从列表中拉出dict对象后，从字典.items()生成器中解包k，v元组|1637651|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This is similar to @delnan but offers the option to modify the k/v (key/value) items and I believe is more readable:

<pre><code>new_dict = {k:v for list_item in list_of_dicts for (k,v) in list_item.items()}
</code></pre>

for instance, replace k/v elems as follows:

<pre><code>new_dict = {str(k).replace(" ","_"):v for list_item in list_of_dicts for (k,v) in list_item.items()}
</code></pre>

unpacks the k,v tuple from the dictionary .items() generator after pulling the dict object out of the list

blocks|key|894869|text|用PEP+448的字典解包对@dietbuddha+answer做了一些小的改进，对我来说，这种方式更具可读性，而且，它是faster+as+well|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|894870|from+functools+import+reduce
result_dict+=+reduce(lambda+a,+b:+{**a,+**b},+list_of_dicts)|code-block|syntax|javascript|894871|但请记住，这只适用于Python+3.5%2B版本。|894872|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0448/|1|https://stackoverflow.com/users/247357/dietbuddha|2|https://gist.github.com/treyhunner/f35292e676efa0be1728^0|1|7|0|E|B|1|1P|E|2|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]|$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|18|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]|U|$5|N|O|P|C|$Q|V]]]]

Little improvement for <a href="https://stackoverflow.com/users/247357/dietbuddha">@dietbuddha</a> answer with dictionary unpacking from <a href="https://www.python.org/dev/peps/pep-0448/" rel="noreferrer">PEP 448</a>, for me, it`s more readable this way, also, it is <a href="https://gist.github.com/treyhunner/f35292e676efa0be1728" rel="noreferrer">faster as well</a>:

<pre><code>from functools import reduce
result_dict = reduce(lambda a, b: {**a, **b}, list_of_dicts)
</code></pre>

But keep in mind, this works only with Python 3.5+ versions.

blocks|key|2752100|text|对于平面字典，您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2752101|from+functools+import+reduce
reduce(lambda+a,+b:+dict(a,+**b),+list_of_dicts)|code-block|syntax|javascript|2752102|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|$]]

For flat dictionaries you can do this:

<pre><code>from functools import reduce
reduce(lambda a, b: dict(a, **b), list_of_dicts)
</code></pre>

blocks|key|2752033|text|>>>+L=[{'a':+1},+{'b':+2},+{'c':+1},+{'d':+2}]++++
>>>+dict(i.items()[0]+for+i+in+L)
{'a':+1,+'c':+1,+'b':+2,+'d':+2}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2752034|注意：“b”和“c”的顺序与您的输出不匹配，因为字典是无序的|unstyled|2752035|如果字典可以有多个键/值|2752036|>>>+dict(j+for+i+in+L+for+j+in+i.items())|2752037|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$B|C]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>&gt;&gt;&gt; L=[{'a': 1}, {'b': 2}, {'c': 1}, {'d': 2}] 
&gt;&gt;&gt; dict(i.items()[0] for i in L)
{'a': 1, 'c': 1, 'b': 2, 'd': 2}
</code></pre>

Note: the order of 'b' and 'c' doesn't match your output because dicts are unordered

if the dicts can have more than one key/value

<pre><code>&gt;&gt;&gt; dict(j for i in L for j in i.items())
</code></pre>

blocks|key|4160091|text|dict1.update(+dict2+)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4160092|这是不对称的，因为您需要选择如何处理重复的键；在这种情况下，dict2将覆盖dict1。换成另一种方式。|unstyled|offset|length|style|CODE|4160093|编辑:啊，对不起，我没看到。|4160094|可以在单个表达式中完成此操作：|4160095|>>>+from+itertools+import+chain
>>>+dict(+chain(+*map(+dict.items,+theDicts+)+)+)
{'a':+1,+'c':+1,+'b':+2,+'d':+2}|4160096|最后这件事不能归功于我！|4160097|然而，我认为用一个简单的for循环来做这件事可能更有Pythonic风格(显式>隐式，平面>嵌套)。YMMV.|4160098|entityMap^0|0|U|5|12|5|0|0|0|0|0|C|3|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@$G|Y|H|Z|I|J]|$G|10|H|11|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|F|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|16|8|@$G|17|H|18|I|J]]|9|@]|A|$]]|$1|U|3|-4|5|F|7|19|8|@]|9|@]|A|$]]]|V|$]]

<pre><code>dict1.update( dict2 )
</code></pre>

This is asymmetrical because you need to choose what to do with duplicate keys; in this case, <code>dict2</code> will overwrite <code>dict1</code>. Exchange them for the other way.

EDIT: Ah, sorry, didn't see that.

It is possible to do this in a single expression:

<pre><code>&gt;&gt;&gt; from itertools import chain
&gt;&gt;&gt; dict( chain( *map( dict.items, theDicts ) ) )
{'a': 1, 'c': 1, 'b': 2, 'd': 2}
</code></pre>

No credit to me for this last!

However, I'd argue that it might be more Pythonic (explicit > implicit, flat > nested ) to do this with a simple <code>for</code> loop. YMMV.

blocks|key|2752114|text|您可以使用funcy库中的join函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2752115|from+funcy+import+join
join(list_of_dicts)|code-block|syntax|javascript|2752116|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Suor/funcy|1|http://funcy.readthedocs.org/en/latest/colls.html#join^0|5|5|0|D|4|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

You can use <a href="http://funcy.readthedocs.org/en/latest/colls.html#join" rel="nofollow">join</a> function from <a href="https://github.com/Suor/funcy" rel="nofollow">funcy</a> library:

<pre><code>from funcy import join
join(list_of_dicts)
</code></pre>

blocks|key|4160394|text|如果您不再需要单例字典：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4160395|>>>+L+=+[{'a':1},+{'b':2},+{'c':1},+{'d':2}]
>>>+dict(map(dict.popitem,+L))
{'a':+1,+'b':+2,+'c':+1,+'d':+2}|code-block|syntax|javascript|4160396|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|$]]

If you don't need the singleton dicts anymore:
<pre><code>&gt;&gt;&gt; L = [{'a':1}, {'b':2}, {'c':1}, {'d':2}]
&gt;&gt;&gt; dict(map(dict.popitem, L))
{'a': 1, 'b': 2, 'c': 1, 'd': 2}
</code></pre>

blocks|key|2752089|text|>>>+dictlist+=+[{'a':1},{'b':2},{'c':1},{'d':2,+'e':3}]
>>>+dict(kv+for+d+in+dictlist+for+kv+in+d.iteritems())
{'a':+1,+'c':+1,+'b':+2,+'e':+3,+'d':+2}
>>>|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2752090|注意，我在最后一个字典中添加了第二个键/值对，以显示它可以处理多个条目。此外，列表中后面的字典中的键将覆盖较早的字典中的相同键。|unstyled|2752091|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>&gt;&gt;&gt; dictlist = [{'a':1},{'b':2},{'c':1},{'d':2, 'e':3}]
&gt;&gt;&gt; dict(kv for d in dictlist for kv in d.iteritems())
{'a': 1, 'c': 1, 'b': 2, 'e': 3, 'd': 2}
&gt;&gt;&gt;
</code></pre>

Note I added a second key/value pair to the last dictionary to show it works with multiple entries.
Also keys from dicts later in the list will overwrite the same key from an earlier dict.

How can I turn a list of dicts like this..
<pre><code>[{'a':1}, {'b':2}, {'c':1}, {'d':2}]
</code></pre>
...into a single dict like this:
<pre><code>{'a':1, 'b':2, 'c':1, 'd':2}
</code></pre>

How do I merge a list of dicts into a single dict?

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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我怎么能把像这样的字典列表..[{'a':1}, {'b':2}, {'c':1}, {'d':2}]...into单个字典，如下所示：{'a':1, 'b':2, 'c':1, 'd':2}

问如何将一个字典列表合并为一个字典？
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问如何将一个字典列表合并为一个字典？
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