Python list of dictionaries search

Question

Assume I have this:

[
{"name": "Tom", "age": 10},
{"name": "Mark", "age": 5},
{"name": "Pam", "age": 7}
]

and by searching "Pam" as name, I want to retrieve the related dictionary: {name: "Pam", age: 7}

How to achieve this ?

Answer 1

This is a general way of searching a value in a list of dictionaries:

def search_dictionaries(key, value, list_of_dictionaries):
    return [element for element in list_of_dictionaries if element[key] == value]

Answer 2

I tested various methods to go through a list of dictionaries and return the dictionaries where key x has a certain value.

Results:

• Speed: list comprehension > generator expression >> normal list iteration >>> filter.
• All scale linear with the number of dicts in the list (10x list size -> 10x time).
• The keys per dictionary does not affect speed significantly for large amounts (thousands) of keys. Please see this graph I calculated: https://imgur.com/a/quQzv (method names see below).

All tests done with Python 3.6.4, W7x64.

from random import randint
from timeit import timeit


list_dicts = []
for _ in range(1000):     # number of dicts in the list
    dict_tmp = {}
    for i in range(10):   # number of keys for each dict
        dict_tmp[f"key{i}"] = randint(0,50)
    list_dicts.append( dict_tmp )



def a():
    # normal iteration over all elements
    for dict_ in list_dicts:
        if dict_["key3"] == 20:
            pass

def b():
    # use 'generator'
    for dict_ in (x for x in list_dicts if x["key3"] == 20):
        pass

def c():
    # use 'list'
    for dict_ in [x for x in list_dicts if x["key3"] == 20]:
        pass

def d():
    # use 'filter'
    for dict_ in filter(lambda x: x['key3'] == 20, list_dicts):
        pass

Results:

1.7303 # normal list iteration 
1.3849 # generator expression 
1.3158 # list comprehension 
7.7848 # filter

Answer 3

people = [
{'name': "Tom", 'age': 10},
{'name': "Mark", 'age': 5},
{'name': "Pam", 'age': 7}
]

def search(name):
    for p in people:
        if p['name'] == name:
            return p

search("Pam")

Answer 4

To add just a tiny bit to @FrédéricHamidi.

In case you are not sure a key is in the the list of dicts, something like this would help:

next((item for item in dicts if item.get("name") and item["name"] == "Pam"), None)

Answer 5

I found this thread when I was searching for an answer to the same question. While I realize that it's a late answer, I thought I'd contribute it in case it's useful to anyone else:

def find_dict_in_list(dicts, default=None, **kwargs):
    """Find first matching :obj:`dict` in :obj:`list`.

    :param list dicts: List of dictionaries.
    :param dict default: Optional. Default dictionary to return.
        Defaults to `None`.
    :param **kwargs: `key=value` pairs to match in :obj:`dict`.

    :returns: First matching :obj:`dict` from `dicts`.
    :rtype: dict

    """

    rval = default
    for d in dicts:
        is_found = False

        # Search for keys in dict.
        for k, v in kwargs.items():
            if d.get(k, None) == v:
                is_found = True

            else:
                is_found = False
                break

        if is_found:
            rval = d
            break

    return rval


if __name__ == '__main__':
    # Tests
    dicts = []
    keys = 'spam eggs shrubbery knight'.split()

    start = 0
    for _ in range(4):
        dct = {k: v for k, v in zip(keys, range(start, start+4))}
        dicts.append(dct)
        start += 4

    # Find each dict based on 'spam' key only.  
    for x in range(len(dicts)):
        spam = x*4
        assert find_dict_in_list(dicts, spam=spam) == dicts[x]

    # Find each dict based on 'spam' and 'shrubbery' keys.
    for x in range(len(dicts)):
        spam = x*4
        assert find_dict_in_list(dicts, spam=spam, shrubbery=spam+2) == dicts[x]

    # Search for one correct key, one incorrect key:
    for x in range(len(dicts)):
        spam = x*4
        assert find_dict_in_list(dicts, spam=spam, shrubbery=spam+1) is None

    # Search for non-existent dict.
    for x in range(len(dicts)):
        spam = x+100
        assert find_dict_in_list(dicts, spam=spam) is None

Answer 6

You have to go through all elements of the list. There is not a shortcut!

Unless somewhere else you keep a dictionary of the names pointing to the items of the list, but then you have to take care of the consequences of popping an element from your list.