Assuming your data is stored in d, there are several ways to do this:

List Comprehension

For every iterable you can use a list comprehension, so iterate over the items with [it for it in iterable] (you can use list comprehensions in dictionaries, too).

In [6]: [{dic['answer_type']: dic['answer__count']} for dic in d]
Out[6]: [{'Yes': 5}, {'No': 3}, {'Maybe': 2}]

Functional Approach

If you are more the functional programming guy, consider using a map, which applies a function to each element of an iterable (this is my preferred approach, if I have to do this again and again with a list, I only have to define the function once then):

In [2]: map(lambda dic: {dic['answer_type']: dic['answer__count']}, d)
Out[2]: [{'Yes': 5}, {'No': 3}, {'Maybe': 2}]

For some people this is not as good to read as list comprehensions, for some it's better, depends on personal feelings. This will work with python2, in py3 it gives you an iterator, so you have to use list around the map to get the list.

This is as simple as continuously making a dictionary with those values. We can use dict.items() to iterate through the items (key value pair).

new_dict_list = [
    {'name': name, 'time': time}
    for name, time in dict.items()
]

sample_dict = [{"id":1, "count":10},
               {"id":2, "count":20},
               {"id":3, "count":30}]

output = []
for elem in sample_dict:
    new_dict = {elem["id"]: elem["count"]}
    output.append(new_dict)

Printing output will return

print(output)

[{1: 10}, {2: 20}, {3: 30}]

my_dict = { my_dict[k]:k for k in my_dict}

new_dict = dict(zip(my_dict.values(), my_dict.keys()))

I am assuming that your order is always the same, i.e., in groups of 4. The idea is to split the strings using : and then create key/value pairs and use nested for loops. The .strip() is to get rid of whitespace

lst = ['name: test1', 'email: [email protected]', 'role: test', 'description: test', 
       'name: test2', 'email: [email protected]', 'role: test2', 'description: test2', 
       'name: test3', 'email: [email protected]', 'role: test3', 'description: test3']

answer = []

for i in range(0, len(lst), 4):
    dic = {}
    for j in lst[i:i+4]:
        dic[j.split(':')[0]] = j.split(':')[1].strip() 
    answer.append(dic)

# [{'name': 'test1',  'email': '[email protected]',  'role': 'test',  'description': 'test'},
    #  {'name': 'test2',  'email': '[email protected]',  'role': 'test2',  'description': 'test2'},
    #  {'name': 'test3',  'email': '[email protected]',  'role': 'test3',  'description': 'test3'}]

A list comprehension would look like

answer = [{j.split(':')[0]:j.split(':')[1].strip() for j in lst[i:i+4]} for i in range(0, len(lst), 4)]