How to convert list of list into dictionary object?

Question

I have a list of lists that looks like this:

[[['1',
    '1@1`']],
  [['2', '[email protected]']],
  [['3', '[email protected]']],
  [['4', '[email protected]']],
  [['5', '[email protected]']],
  [['6', '6@6']],
  [['7', '7@7']],
  [['8', '8@8']],
  [['8.5', '[email protected]']],
  [['9', '9@9']],
  [['10', '10@10']],
  [['11', '11@11']],
  [['12', '12@12']],
  [['13', '[email protected]']],
  [['14', '[email protected]']],
  [['15', '[email protected]']],
  [['16', '[email protected]']],
  [['17', '[email protected]']],
  [['[email protected]', '18']],
  [['19', '[email protected]']]]

is there anyway I can clean up the list by making it into a dictionary object like so:

[{id:1,email:1@1},{id:2,email:[email protected]}]

Ideally if there are any emails in the id spot they flipped to the email spot?

yeah, sure there is a way. Iterate over your list, create a dict from the elements in your list (although, why it's a bunch of single-element lists is weird, but you can easily just handle that). — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jan 19, 2019 at 1:00
"Ideally if there are any emails in the id spot they flipped to the email spot?" This might be a little bit more tricky. But maybe just checking if @ is in it? — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Jan 19, 2019 at 1:01

iz_ · Accepted Answer · 2019-01-22 01:36:48Z

2

You can use a list comprehension:

In [1]: mylist = [[['1',
   ...:     '1@1`']],
   ...:   [['2', '[email protected]']],
   ...:   [['3', '[email protected]']],
   ...:   [['4', '[email protected]']],
   ...:   [['5', '[email protected]']],
   ...:   [['6', '6@6']],
   ...:   [['7', '7@7']],
   ...:   [['8', '8@8']],
   ...:   [['8.5', '[email protected]']],
   ...:   [['9', '9@9']],
   ...:   [['10', '10@10']],
   ...:   [['11', '11@11']],
   ...:   [['12', '12@12']],
   ...:   [['13', '[email protected]']],
   ...:   [['14', '[email protected]']],
   ...:   [['15', '[email protected]']],
   ...:   [['16', '[email protected]']],
   ...:   [['17', '[email protected]']],
   ...:   [['[email protected]', '18']],
   ...:   [['19', '[email protected]']]]

In [2]: [{'id': i, 'email': e} for i, e in (pair[0] if '@' not in pair[0][0] else reversed(pair[0]) for pair in mylist)]
Out[2]:
[{'id': '1', 'email': '1@1`'},
 {'id': '2', 'email': '[email protected]'},
 {'id': '3', 'email': '[email protected]'},
 {'id': '4', 'email': '[email protected]'},
 {'id': '5', 'email': '[email protected]'},
 {'id': '6', 'email': '6@6'},
 {'id': '7', 'email': '7@7'},
 {'id': '8', 'email': '8@8'},
 {'id': '8.5', 'email': '[email protected]'},
 {'id': '9', 'email': '9@9'},
 {'id': '10', 'email': '10@10'},
 {'id': '11', 'email': '11@11'},
 {'id': '12', 'email': '12@12'},
 {'id': '13', 'email': '[email protected]'},
 {'id': '14', 'email': '[email protected]'},
 {'id': '15', 'email': '[email protected]'},
 {'id': '16', 'email': '[email protected]'},
 {'id': '17', 'email': '[email protected]'},
 {'id': '18', 'email': '[email protected]'},
 {'id': '19', 'email': '[email protected]'}]

If you have arbitrary nesting, you can try this:

def flatten(lst):
    for sub in lst:
        if isinstance(sub, list):
            yield from flatten(sub)
        else:
            yield sub

[{'id': i, 'email': e} for i, e in (pair if '@' not in pair[0] else reversed(pair) for pair in zip(*[flatten(mylist)]*2))]

edited Jan 22, 2019 at 1:36

answered Jan 19, 2019 at 3:35

iz_

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9 Comments

RustyShackleford Over a year ago

thank you for the response. I am getting error : KeyError: 0. Not sure what this means.

iz_ Over a year ago

@RustyShackleford Can you confirm that your input is a list of lists? That KeyError likely occurred because you have a list of dictionaries.

RustyShackleford Over a year ago

confirmed its a list with 20 other lists inside of it.

iz_ Over a year ago

@RustyShackleford Can you share the full error via Pastebin or similar? Also, can you double check that each item is wrapped in two lists e.g. the whole list is a 3D structure.

RustyShackleford Over a year ago

I apologize used the wrong list but got this error with right list :ValueError: not enough values to unpack (expected 2, got 1). Confirmed each item is wrapped in two lists.

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