blocks|key|1397599|text|使用set、sorted将两个列表组合为一个列表并进行排序：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1397600|>>>+a1+=+{u'2016-03-11':+[u'20:00',+u'22:10']}
>>>+a2+=+{u'2016-03-11':+[u'20:00',+u'23:10'],u'2016-03-12':+[u'20:00',+u'22:10']}


>>>+an+=+{}
>>>+for+d+in+a1,+a2:
...+++++for+key+in+d:
...+++++++++an[key]+=+sorted(set(an.get(key,+[])+%2B+d[key]))
...+++++++++#+%5E%5E++Merge+two+lists+(dictionary+values)+into+one,+and+sort
...+
>>>+print+an
{u'2016-03-11':+[u'20:00',+u'22:10',+u'23:10'],
+u'2016-03-12':+[u'20:00',+u'22:10']}|code-block|syntax|javascript|1397601|更新|BOLD|1397602|使用字典理解的替代版本(假设只合并两个字典)：|1397603|>>>+{key:+sorted(set(a1.get(key,[])+%2B+a2.get(key,[])))+for+key+in+set(a1)%7Cset(a2)}
{u'2016-03-11':+[u'20:00',+u'22:10',+u'23:10'],
+u'2016-03-12':+[u'20:00',+u'22:10']}|1397604|entityMap^0|2|3|6|6|0|0|0|2|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Z|8|@$9|10|A|11|B|M]]|D|@]|E|$]]|$1|N|3|O|5|6|7|12|8|@]|D|@]|E|$]]|$1|P|3|Q|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|R|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|S|$]]

Using <code>set</code>, <code>sorted</code> to combine to two list into one and to sort:

<pre><code>&gt;&gt;&gt; a1 = {u'2016-03-11': [u'20:00', u'22:10']}
&gt;&gt;&gt; a2 = {u'2016-03-11': [u'20:00', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']}


&gt;&gt;&gt; an = {}
&gt;&gt;&gt; for d in a1, a2:
... for key in d:
... an[key] = sorted(set(an.get(key, []) + d[key]))
... # ^^ Merge two lists (dictionary values) into one, and sort
... 
&gt;&gt;&gt; print an
{u'2016-03-11': [u'20:00', u'22:10', u'23:10'],
 u'2016-03-12': [u'20:00', u'22:10']}
</code></pre>

<hr>

UPDATE

alternative version using dictionary comprehension (assuming merging only two dictionaries):

<pre><code>&gt;&gt;&gt; {key: sorted(set(a1.get(key,[]) + a2.get(key,[]))) for key in set(a1)|set(a2)}
{u'2016-03-11': [u'20:00', u'22:10', u'23:10'],
 u'2016-03-12': [u'20:00', u'22:10']}
</code></pre>

blocks|key|5027213|text|a1+=+{u'2016-03-11':+[u'20:00',+u'22:10']}
a2+=+{u'2016-03-11':+[u'20:00',+u'23:10'],u'2016-03-12':+[u'20:00',+u'22:10']}


import+copy
an+=+copy.deepcopy(a1)

for+key,+value+in+a2.iteritems():
++++if+key+in+an:
++++++++an[key]+=+list(set(an[key]+%2B+a2[key]))
++++else:
++++++++an[key]+=+value

print+an

>>+{u'2016-03-11':+[u'22:10',+u'20:00',+u'23:10'],+u'2016-03-12':+[u'20:00',+u'22:10']}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5027214|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>a1 = {u'2016-03-11': [u'20:00', u'22:10']}
a2 = {u'2016-03-11': [u'20:00', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']}


import copy
an = copy.deepcopy(a1)

for key, value in a2.iteritems():
 if key in an:
 an[key] = list(set(an[key] + a2[key]))
 else:
 an[key] = value

print an

&gt;&gt; {u'2016-03-11': [u'22:10', u'20:00', u'23:10'], u'2016-03-12': [u'20:00', u'22:10']}
</code></pre>

blocks|key|5027227|text|它看上去很糟糕，但效果很好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5027228|an+=+{}
for+key+in+set(list(a1.keys())+%2B+list(a2.keys())):
++++an[key]+=+list(set((a1.get(key)+or+[])+%2B+(a2.get(key)+or+[])))|code-block|syntax|javascript|5027229|不过，我认为还有更聪明、更有节奏的方法:)|5027230|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|$]]

It looks awful but it works:

<pre><code>an = {}
for key in set(list(a1.keys()) + list(a2.keys())):
 an[key] = list(set((a1.get(key) or []) + (a2.get(key) or [])))
</code></pre>

I am shure there is more clever and pythonic way, though :)

blocks|key|1397627|text|from+collections+import+defaultdict


a1+=+{u'2016-03-11':+[u'20:00',+u'22:10']}
a2+=+{u'2016-03-11':+[u'20:00',+u'23:10'],u'2016-03-12':+[u'20:00',+u'22:10']}++++

dd+=+defaultdict(set)
for+d+in+a1,+a2:
++++for+k,+v+in+d.items():
++++++++dd[k]+%7C=+set(v)
res+=+{k:+sorted(v)+for+k,+v+in+dd.items()}

print(res)
#+{'2016-03-12':+['20:00',+'22:10'],+'2016-03-11':+['20:00',+'22:10',+'23:10']}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1397628|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>from collections import defaultdict


a1 = {u'2016-03-11': [u'20:00', u'22:10']}
a2 = {u'2016-03-11': [u'20:00', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']} 

dd = defaultdict(set)
for d in a1, a2:
 for k, v in d.items():
 dd[k] |= set(v)
res = {k: sorted(v) for k, v in dd.items()}

print(res)
# {'2016-03-12': ['20:00', '22:10'], '2016-03-11': ['20:00', '22:10', '23:10']}
</code></pre>

blocks|key|5027272|text|我导出了这个选项(在这里发布作为一个答案)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5027273|an+=+{}
for+key+in+set().union(a1,+a2):+
++++an[key]+=+sorted(set().union(a1.get(key,+[]),+a2.get(key,+[])))|code-block|syntax|javascript|5027274|它构建集合时不需要列表连接。|5027275|它产生了所需的结果：|5027276|>>>+a1+=+{u'2016-03-11':+[u'20:00',+u'22:10']}
>>>+a2+=+{u'2016-03-11':+[u'20:00',+u'23:10'],u'2016-03-12':+[u'20:00',+u'22:10']}
>>>+an+=+{}
>>>+for+key+in+set().union(a1,+a2):+
...+++++an[key]+=+sorted(set().union(a1.get(key,+[]),+a2.get(key,+[])))
...+
>>>+an
{u'2016-03-11':+[u'20:00',+u'22:10',+u'23:10'],+u'2016-03-12':+[u'20:00',+u'22:10']}|5027277|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I derived this option (posted here as an answer):

<pre><code>an = {}
for key in set().union(a1, a2): 
 an[key] = sorted(set().union(a1.get(key, []), a2.get(key, [])))
</code></pre>

It builds the sets without list concatenations.

It produces the required result:

<pre><code>&gt;&gt;&gt; a1 = {u'2016-03-11': [u'20:00', u'22:10']}
&gt;&gt;&gt; a2 = {u'2016-03-11': [u'20:00', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']}
&gt;&gt;&gt; an = {}
&gt;&gt;&gt; for key in set().union(a1, a2): 
... an[key] = sorted(set().union(a1.get(key, []), a2.get(key, [])))
... 
&gt;&gt;&gt; an
{u'2016-03-11': [u'20:00', u'22:10', u'23:10'], u'2016-03-12': [u'20:00', u'22:10']}
</code></pre>

blocks|key|328998|text|a1+=+{u'2016-03-11':+[u'20:00',+u'22:10']}
a2+=+{u'2016-03-11':+[u'20:00',+u'23:10'],u'2016-03-12':+[u'20:00',+u'22:10']}


an+=+a1

for+key+in+a2+:
++++if+key+in+an+:
++++++++an[key]+=+list(set(a2[key]%2B+a2[key]))
++++++++an[key].sort()
++++else+:
++++++++x+=+{key+:+a2[key]}
++++++++an.update(x)

keyList+=+an.keys()
keyList.sort()
temp+=+{}
for+key+in+keyList:
++++temp.update({key:an[key]})

an+=+temp+
print+an|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|328999|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>a1 = {u'2016-03-11': [u'20:00', u'22:10']}
a2 = {u'2016-03-11': [u'20:00', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']}


an = a1

for key in a2 :
 if key in an :
 an[key] = list(set(a2[key]+ a2[key]))
 an[key].sort()
 else :
 x = {key : a2[key]}
 an.update(x)

keyList = an.keys()
keyList.sort()
temp = {}
for key in keyList:
 temp.update({key:an[key]})

an = temp 
print an
</code></pre>

I want merge two dict like this:

from:

<pre><code> a1 = {u'2016-03-11': [u'20:00', u'22:10']}
 a2 = {u'2016-03-11': [u'20:00', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']}
</code></pre>

to:

<pre><code> an = {u'2016-03-11': [u'20:00',u'22:10', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']}
</code></pre>

i need a function merge two dict

merger two dict in python like this

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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我想把两个混蛋合并成这样：发自：  a1 = {u'2016-03-11': [u'20:00', u'22:10']}  a2 = {u'2016-03-11': [u'20:00', u'23:10'],u'2016-03-12': [u'20:00', u'22:10']}至：  an = {u'2016-03-...

问像这样合并python中的两个小块
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问像这样合并python中的两个小块EN