合併2個字典在Python這樣

Question

我想合併兩個字典是這樣的：

來自：

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']} 

到：

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

我需要一個函數合併兩個字典

Answer 1

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']} 

Answer 2

使用set,sorted將兩個列表合併爲一個並排序：

>>> 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, []) + d[key])) 
...   # ^^ 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']} 

---

UPDATE

使用字典解析

替代版本（假設合併只有兩個詞典）：

>>> {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']} 

Answer 3

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 

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

Answer 4

它看起來可怕，但它的工作原理：

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

我舒爾有更聰明和Python的方式，雖然:)

Answer 5

我得出這個選項（在這裏張貼的答案）：

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

它建立的集而不用列表級聯。

它產生所需的結果：

>>> 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']} 

Answer 6

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