Transform a list of dict to an simpler dict - python-3.x

Question

I have list of dict like this:
[{
'attr': 'bla',
'status': '1',
'id': 'id1'
}, {
'attr': 'bla',
'status': '1',
'id': 'id2'
}, {
'attr': 'bli',
'status': '0',
'id': 'id1'
}, {
'attr': 'bli',
'status': '1',
'id': 'id2'
}]
I wan't to get a simpler results dict like this:
result = {
'bla' : True,
'bli' : False
}
If the two id have a 1 for an attr, the value will be True. else, it will False.
I've tried with
for elem in dict:
for key, value in enumerate(elem):
# ???
But i don't see how to do. I've alos tried something like
if all( val == '1' for val in list ):
# ..

Answer 1

Here you go:
dicts = [{
'attr': 'bla',
'status': '1',
'id': 'id1'
}, {
'attr': 'bla',
'status': '1',
'id': 'id2'
}, {
'attr': 'bli',
'status': '0',
'id': 'id1'
}, {
'attr': 'bli',
'status': '1',
'id': 'id2'
}]
# First run is to create all nessecary items in the
# new Dictionary so i can use the and operator on them later.
newDict = {}
for dictio in dicts:
for key, value in dictio.items():
if key == 'attr':
newDict[value] = True
# The second run uses the and operator
for dictio in dicts:
for key, value in dictio.items():
if key == 'attr':
tmpAttr = value
if key == 'status':
newDict[tmpAttr] = newDict[tmpAttr] and (value == '1')
print(newDict)
Have a nice day!