I have following code (which produces a syntax error, btw). Can someone help me fixing it so I can get a version that produces the expected result?
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]
Expected result:
a = [1, 4, 7, 10]
b = [2, 5, 8, 11]
c = [3, 6, 9, 12]
If you know keys in advance:
>>> 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]
If you don't know keys in advance:
>>> 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]
[dict_.values() for dict_ in al]
If your keys are unknown, you can simply transpose the data and create another transpose dictionary, which you can simply access by the keys instead of creating standalone variables
>>> 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)
Note Not Recommended
If you actually want to create standalone variables, you can do so by adding the dictionary to the locals
locals().update(al_transpose)
>>> a
(1, 4, 7, 10)
>>> b
(2, 5, 8, 11)
>>> c
(3, 6, 9, 12)
If you're interested in a one-liner, and if you guarantee the order and presence of the keys in the dicts of al, then:
>>> 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)}
Try this
a,b,c = [map(lambda x:x[ele], al) for ele in 'abc']
You could do something like this:
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]
If you want to bind the names in the dictionary to names in the current name space, you can do this:
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]
>>> 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)
This is how you need to modify your method
>>> 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))
Firstly, you need to use i['a'] instead of i.a etc. since a is a key, not an attribute
So that would give you
>>> [(i['a'], i['b'], i['c']) for i in al]
[(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12)]
Ah, the rows and columns are the wrong way around. They standard trick to swap the rows and columns is 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)]
Combining those two ideas gives the expression at the top of this answer
A more direct approach is a nested list comprehension
>>> a, b, c = [[x[k] for x in al] for k in 'abc']
If the keys are not single characters, you need to write them out as a tuple longhand
>>> a, b, c = [[x[k] for x in al] for k in ('a', 'b', 'c')]