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

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)
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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))

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

A simple readable version is

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

It won't be efficient, as you need to walk the list checking every item in it (O(n)). If you want efficiency, you can use dict of dicts. On the question, here's one possible way to find it (though, if you want to stick to this data structure, it's actually more efficient to use a generator as Brent Newey has written in the comments; see also tokland's answer):

>>> L = [{'id':'1234','name':'Jason'},
...         {'id':'2345','name':'Tom'},
...         {'id':'3456','name':'Art'}]
>>> [i for i,_ in enumerate(L) if _['name'] == 'Tom'][0]
1

Seems most logical to use a filter/index combo:

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

And if you think there could be multiple matches:

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