在词典列表中查找项目

Question

I have this data

data = [
    {
        'id': 'abcd738asdwe',
        'name': 'John',
        'mail': 'test@test.com',
    },
    {
        'id': 'ieow83janx',
        'name': 'Jane',
        'mail': 'test@foobar.com',
    }
]

The id's are unique, it's impossible that multiple dictonaries have the same id.

For example I want to get the item with the id "ieow83janx".

My current solution looks like this:

search_id = 'ieow83janx'
item = [x for x in data if x['id'] == search_id][0]

Do you think that's the be solution or does anyone know an alternative solution?

Answer 1

Since the id s are unique, you can store the items in a dictionary to achieve O(1) lookup.

lookup = {ele['id']: ele for ele in data}

then you can do

user_info = lookup[user_id]

to retrieve it

Answer 2

If you are going to get this kind of operations more than once on this particular object, I would recommend to translate it into a dictionary with id as a key.

data = [
    {
        'id': 'abcd738asdwe',
        'name': 'John',
        'mail': 'test@test.com',
    },
    {
        'id': 'ieow83janx',
        'name': 'Jane',
        'mail': 'test@foobar.com',
    }
]

data_dict = {item['id']: item for item in data}
#=> {'ieow83janx': {'mail': 'test@foobar.com', 'id': 'ieow83janx', 'name': 'Jane'}, 'abcd738asdwe': {'mail': 'test@test.com', 'id': 'abcd738asdwe', 'name': 'John'}}

data_dict['ieow83janx']
#=> {'mail': 'test@foobar.com', 'id': 'ieow83janx', 'name': 'Jane'}

In this case, this lookup operation will cost you some constant* O(1) time instead of O(N) .

Answer 3

How about the next built-in function ( docs ):

>>> data = [
...     {
...         'id': 'abcd738asdwe',
...         'name': 'John',
...         'mail': 'test@test.com',
...     },
...     {
...         'id': 'ieow83janx',
...         'name': 'Jane',
...         'mail': 'test@foobar.com',
...     }
... ]
>>> search_id = 'ieow83janx'
>>> next(x for x in data if x['id'] == search_id)
{'id': 'ieow83janx', 'name': 'Jane', 'mail': 'test@foobar.com'}

EDIT:

It raises StopIteration if no match is found, which is a beautiful way to handle absence:

>>> search_id = 'does_not_exist'
>>> try:
...     next(x for x in data if x['id'] == search_id)
... except StopIteration:
...     print('Handled absence!')
... 
Handled absence!

Answer 4

Without creating a new dictionary or without writing several lines of code, you can simply use the built-in filter function to get the item lazily, not checking after it finds the match.

next(filter(lambda d: d['id']==search_id, data))

should for just fine.

Answer 5

if any(item['id']=='ieow83janx' for item in data):
   #return item

As any function returns true if iterable (List of dictionaries in your case) has value present. While using Generator Expression there will not be need of creating internal List. As there will not be duplicate values for the id in List of dictionaries, any will stop the iteration until the condition returns true. ie the generator expression with any will stop iterating on shortcircuiting. Using List comprehension will create a entire List in the memory where as GE creates the element on the fly which will be better if you are having large items as it uses less memory.

Answer 6

Would this not achieve your goal?

for i in data:
    if i.get('id') == 'ieow83janx':
        print(i)

 (xenial)vash@localhost:~/python$ python3.7 split.py {'id': 'ieow83janx', 'name': 'Jane', 'mail': 'test@foobar.com'} 

Using comprehension:

[i for i in data if i.get('id') == 'ieow83janx']