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

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'}