我有以下代码（顺便说一句，这会产生语法错误）。有人可以帮我修复它，以便我可以获得产生预期结果的版本吗？
al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
a, b, c = [i.a, i.b, i.c for i in al]
预期结果：
a = [1, 4, 7, 10]
b = [2, 5, 8, 11]
c = [3, 6, 9, 12]
如果您事先知道密钥：
>>> al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
>>> a, b, c = [[d[key] for d in al] for key in 'abc'] # ('a', 'b', 'c')
>>> a
[1, 4, 7, 10]
>>> b
[2, 5, 8, 11]
>>> c
[3, 6, 9, 12]
如果您事先不知道密钥：
>>> d = {key: [d[key] for d in al] for key in al[0]}
>>> d
{'a': [1, 4, 7, 10], 'c': [3, 6, 9, 12], 'b': [2, 5, 8, 11]}
>>> a, b, c = map(d.get, 'abc') # OR map(d.get, ('a', 'b', 'c'))
>>> a
[1, 4, 7, 10]
>>> b
[2, 5, 8, 11]
>>> c
[3, 6, 9, 12]
如果您的密钥未知，您可以简单地转置数据并创建另一个转置字典，您可以简单地访问它keys而不是创建独立变量
>>> al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
>>> keys = al[0].keys()
>>> #Given your list of dictionary
>>> al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
>>> #determine the keys
>>> keys = al[0].keys()
>>> #and using itemgetter
>>> from operator import itemgetter
>>> #create a transpose dictionary
>>> al_transpose = dict(zip(keys,zip(*map(itemgetter(*keys),al))))
>>> al_transpose['a']
(1, 4, 7, 10)
>>> al_transpose['b']
(2, 5, 8, 11)
>>> al_transpose['c']
(3, 6, 9, 12)
注意不推荐
如果您确实想创建独立变量，可以通过将字典添加到locals
locals().update(al_transpose)
>>> a
(1, 4, 7, 10)
>>> b
(2, 5, 8, 11)
>>> c
(3, 6, 9, 12)
如果您对单线感兴趣，并且如果您保证 dicts 中键的顺序和存在al，那么：
>>> al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
>>> dict(zip(al[0].keys(),zip(*[i.values() for i in al])))
{'a': (1, 4, 7, 10), 'c': (3, 6, 9, 12), 'b': (2, 5, 8, 11)}
你可以这样做：
al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
result={}
for dic in al:
for key,val in dic.items():
result.setdefault(key,[]).append(val)
print result
# {'a': [1, 4, 7, 10], 'c': [3, 6, 9, 12], 'b': [2, 5, 8, 11]}
a,b,c=result['a'],result['b'],result['c']
print a,b,c
# [1, 4, 7, 10] [2, 5, 8, 11] [3, 6, 9, 12]
如果要将字典中的名称绑定到当前名称空间中的名称，可以这样做：
al = [{'a': 1, 'b': 2, 'c': 3, 'd':4},
{'a': 4, 'b': 5, 'c': 6},
{'a': 7, 'b': 8, 'c': 9}]
result={}
for dic in al:
for k in dic:
result.setdefault(k,[]).append(dic[k])
print result
# {'a': [1, 4, 7, 10], 'c': [3, 6, 9, 12], 'b': [2, 5, 8, 11], 'd': [4]}
for k in result:
code=compile('{}=result["{}"]'.format(k,k), '<string>', 'exec')
exec code
print a,b,c,d
# [1, 4, 7, 10] [2, 5, 8, 11] [3, 6, 9, 12] [4]
尝试这个
a,b,c = [map(lambda x:x[ele], al) for ele in 'abc']
>>> al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
>>> from operator import itemgetter
>>> a, b, c = zip(*map(itemgetter(*'abc'),(al)))
>>> a
(1, 4, 7, 10)
>>> b
(2, 5, 8, 11)
>>> c
(3, 6, 9, 12)
>>> al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
>>> a, b, c = zip(*(sorted(x.values(), key=x.get) for x in al))
>>> a
(1, 4, 7, 10)
>>> b
(3, 6, 9, 12)
>>> c
(2, 5, 8, 11)
这就是您需要修改方法的方式
>>> al = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}, {'a': 7, 'b': 8, 'c': 9}, {'a': 10, 'b': 11, 'c': 12}]
>>> a, b, c = zip(*((i['a'], i['b'], i['c']) for i in al))
首先，您需要使用i['a']而不是i.a等，因为a它是一个键，而不是一个属性
所以那会给你
>>> [(i['a'], i['b'], i['c']) for i in al]
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12)]
啊，行和列是错误的方式。他们交换行和列的标准技巧是zip(*A)
>>> zip(*[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12)])
[(1, 4, 7, 10), (2, 5, 8, 11), (3, 6, 9, 12)]
结合这两个想法给出了这个答案顶部的表达
更直接的方法是嵌套列表推导
>>> a, b, c = [[x[k] for x in al] for k in 'abc']
如果键不是单个字符，则需要将它们写成元组
>>> a, b, c = [[x[k] for x in al] for k in ('a', 'b', 'c')]