Find the index of a dict within a list, by matching the dict's value

Question

I have a list of dicts:

list = [{\'id\':\'1234\',\'name\':\'Jason\'},
        {\'id\':\'2345\',\'name\':\'Tom\'},
        {\'id\':\'3456\',\'name\':\'Art\'}]         


        

Answer 1

A simple readable version is

def find(lst, key, value):
    for i, dic in enumerate(lst):
        if dic[key] == value:
            return i
    return -1

Answer 2

One liner!?

elm = ([i for i in mylist if i['name'] == 'Tom'] or [None])[0]

Answer 3

Seems most logical to use a filter/index combo:

names=[{}, {'name': 'Tom'},{'name': 'Tony'}]
names.index(filter(lambda n: n.get('name') == 'Tom', names)[0])
1

And if you think there could be multiple matches:

[names.index(n) for item in filter(lambda n: n.get('name') == 'Tom', names)]
[1]

Answer 4

lst = [{'id':'1234','name':'Jason'}, {'id':'2345','name':'Tom'}, {'id':'3456','name':'Art'}]

tom_index = next((index for (index, d) in enumerate(lst) if d["name"] == "Tom"), None)
# 1

If you need to fetch repeatedly from name, you should index them by name (using a dictionary), this way get operations would be O(1) time. An idea:

def build_dict(seq, key):
    return dict((d[key], dict(d, index=index)) for (index, d) in enumerate(seq))

info_by_name = build_dict(lst, key="name")
tom_info = info_by_name.get("Tom")
# {'index': 1, 'id': '2345', 'name': 'Tom'}

Answer 5

def search(itemID,list):
     return[i for i in list if i.itemID==itemID]

Answer 6

I needed a more general solution to account for the possibility of multiple dictionaries in the list having the key value, and a straightforward implementation using list comprehension:

dict_indices = [i for i, d in enumerate(dict_list) if d[dict_key] == key_value]