将字典转换为字典python列表

Question

I would like to convert the following dictionary:

{'Stockholm': ['235', '35'], 'Helsinki': ['285', '15'], 
'Copenhagen': ['180', '60'], 'Berlin': ['185', '85'], 
'Prague': ['190', '115'], 'Warsaw': ['260', '105'], 
'Moscow': ['370', '80'], 'Sofia': ['275', '170'], 
'Ankara': ['340', '190'], 'Athens': ['285', '205']}

into:

[{'Stockholm':['235', '35']},{'Helsinki': ['285', '15']},
{'Copenhagen':   ['180', '60']},...,{'Athens': ['285', '205']}] 

I'm out of ideas how to convert it.

Answer 1

You can use a list comprehension over the dict.items() pairs:

[{k: v} for k, v in d.items()]  # where d is your dictionary

This creates a separate dictionary for each key-value pair.

You can't rely on any order in the input dictionary, so the output order will depend on the insertion and deletion history of the dictionary (unless you are using Python 3.6 or up).

Demo:

>>> d = {'Stockholm': ['235', '35'], 'Helsinki': ['285', '15'],
... 'Copenhagen': ['180', '60'], 'Berlin': ['185', '85'],
... 'Prague': ['190', '115'], 'Warsaw': ['260', '105'],
... 'Moscow': ['370', '80'], 'Sofia': ['275', '170'],
... 'Ankara': ['340', '190'], 'Athens': ['285', '205']}
>>> [{k: v} for k, v in d.items()]
[{'Stockholm': ['235', '35']}, {'Helsinki': ['285', '15']}, {'Copenhagen': ['180', '60']}, {'Berlin': ['185', '85']}, {'Prague': ['190', '115']}, {'Warsaw': ['260', '105']}, {'Moscow': ['370', '80']}, {'Sofia': ['275', '170']}, {'Ankara': ['340', '190']}, {'Athens': ['285', '205']}]

Answer 2

The simple way:

d = {'Stockholm': ['235', '35'], 'Helsinki': ['285', '15'], 
'Copenhagen': ['180', '60'], 'Berlin': ['185', '85'], 
'Prague': ['190', '115'], 'Warsaw': ['260', '105'], 
'Moscow': ['370', '80'], 'Sofia': ['275', '170'], 
'Ankara': ['340', '190'], 'Athens': ['285', '205']}
d = [d]