python帮助从列表中提取值

Question

lst = [{"id": 1, "name": "ernest"}, .... {"id": 16, name:"gellner"}]

for d in lst:
    if "id" in d and d["id"] == 16:
        return d

I want to extract the dictionary that the key "id" equals to "16" from the list.

Is there a more pythonic way of this?

Thanks in advance

Answer 1

Riffing on existing answers to get out of the comments:

Consider using a generator expression when you only need the first matching element out of a sequence:

d16 = (d for d in lst if d.get('id') == 16).next()

This is a pattern you'll see in my code often. This will raise a StopIteration if it turns out there weren't any items in lst that matched the condition. When that's expected to happen, you can either catch the exception:

try:
    d16 = (d for d in lst if d.get('id') == 16).next()
except StopIteration:
    d16 = None

Or better yet, just unroll the whole thing into a for-loop that stops

for d16 in lst:
    if d16.get('id') == 16:
        break
else:
    d16 = None

(the else: clause only gets run if the for loop exhausts its input, but gets skipped if the for loop ends because break was used)

Answer 2

This does it. Whether or not it's "cleaner" is a personal call.

d16 = [d for d in lst if d.get('id', 0) == 16][0]

This initial approach fails if there is no dictionary with id 0. You can overcome that somewhat like this:

d16 = [d for d in lst if d.get('id', 0) == 16]
if d16: d16 = d16[0]

This will prevent the index error, but d16 will be an empty list if there is no dictionary in the container that matches the criteria.

Answer 3

如果没有发现，则有一点改进：

d16 = ([d for d in lst if d.get('id', 0) == 16] + [None])[0]