Python：具有多个值的键的字典。 如何将字典打印为分隔值列表的字典集

Question

I have a dictionary having keys with multiple values like this:

my_dict = {'key0': (0, 1), 'key1': (a, b), 'key2': (x, y)}

I would like print my dictionary like

my_dict = {'key0': 0, 'key1': a, 'key2': x}, {'key0': 1, 'key1': b, 'key2': y}

How can I achieve this? Please help...

Answer 1

If all tuples are of the same length (ie 2) you could use a list comprehension:

[{key: value[i] for key, value in my_dict} for i in range(2)]

Here is an example:

my_dict = {'key0': (0, 1), 'key1': (1, 2), 'key2': ('a', 'b')}
[{key: value[i] for key, value in my_dict.iteritems()} for i in range(2)]

Output:

[{'key0': 0, 'key1': 1, 'key2': 'a'}, {'key0': 1, 'key1': 2, 'key2': 'b'}]

Answer 2

there is no data structure like a = {}, {} . But, you can split the initial dict into 2 dicts:

my_dict = {'key0': (0, 1), 'key1': ('a', 'b'), 'key2': ('x', 'y')}
my_dict_1 = dict((key, value[0]) for key, value in my_dict.iteritems())
my_dict_2 = dict((key, value[1]) for key, value in my_dict.iteritems())

Results:

print my_dict_1
{'key0': 0, 'key1': 'a', 'key2': 'x'}
print my_dict_1
{'key0': 1, 'key1': 'b', 'key2': 'y'}

Answer 3

Assuming all of your values are tuples/lists but could be of different sizes, you could do the following:

def expand_dict(input_dict):
    max_size = max(len(v) for v in input_dict.values())
    output_list = [dict() for _ in range(max_size)]
    for k, v in input_dict.items():
        for i, x in enumerate(v):
            output_list[i][k] = x
    return output_list

my_dict = {'key0': (0, 1), 'key1': ('a', 'b'), 'key2': ('x', 'y', 'z')}

print expand_dict(my_dict)

This prints:

[{'key2': 'x', 'key1': 'a', 'key0': 0}, {'key2': 'y', 'key1': 'b', 'key0': 1}, {'key2': 'z'}]

Answer 4

You could try like this:

>>> my_dict = {'key0': (0, 1), 'key1': (2, 3), 'key2': (4, 5)}
>>> dict_list = []
>>> for key in my_dict:
...    for index, item in enumerate(my_dict[key]):
...       if len(dict_list) >= index + 1:
...           dict_list[index][key] = item
...       else:
...           dict_list.append({key: item})
... 
>>> dict_list
[{'key2': 4, 'key1': 2, 'key0': 0}, {'key2': 5, 'key1': 3, 'key0': 1}]

This is quite efficient from the point of view of processing time and also covers all possible dict length cases (what if one of the tuples has 3 values not 2).

Answer 5

Extended solution with complex example:

Used functions: zip() and itertools.zip_longest()

import itertools

my_dict = {'key0': (0, 1), 'key1': ('a', 'b'), 'key2': ('x', 'y'), 'key3': ('z', 'v')}
result = [dict(t) for t in itertools.zip_longest(*[list(zip([k,k],v)) for k,v in my_dict.items()])]

print(result)

The output:

[{'key2': 'x', 'key0': 0, 'key1': 'a', 'key3': 'z'}, {'key2': 'y', 'key0': 1, 'key1': 'b', 'key3': 'v'}]