Python list of dictionaries search
You can use a generator expression:
>>> dicts = [... { "name": "Tom", "age": 10 },... { "name": "Mark", "age": 5 },... { "name": "Pam", "age": 7 },... { "name": "Dick", "age": 12 }... ]>>> next(item for item in dicts if item["name"] == "Pam"){'age': 7, 'name': 'Pam'}
If you need to handle the item not being there, then you can do what user Matt suggested in his comment and provide a default using a slightly different API:
next((item for item in dicts if item["name"] == "Pam"), None)
And to find the index of the item, rather than the item itself, you can enumerate() the list:
next((i for i, item in enumerate(dicts) if item["name"] == "Pam"), None)
This looks to me the most pythonic way:
people = [{'name': "Tom", 'age': 10},{'name': "Mark", 'age': 5},{'name': "Pam", 'age': 7}]filter(lambda person: person['name'] == 'Pam', people)
result (returned as a list in Python 2):
[{'age': 7, 'name': 'Pam'}]
Note: In Python 3, a filter object is returned. So the python3 solution would be:
list(filter(lambda person: person['name'] == 'Pam', people))
@Frédéric Hamidi's answer is great. In Python 3.x the syntax for .next()
changed slightly. Thus a slight modification:
>>> dicts = [ { "name": "Tom", "age": 10 }, { "name": "Mark", "age": 5 }, { "name": "Pam", "age": 7 }, { "name": "Dick", "age": 12 } ]>>> next(item for item in dicts if item["name"] == "Pam"){'age': 7, 'name': 'Pam'}
As mentioned in the comments by @Matt, you can add a default value as such:
>>> next((item for item in dicts if item["name"] == "Pam"), False){'name': 'Pam', 'age': 7}>>> next((item for item in dicts if item["name"] == "Sam"), False)False>>>