使用另一个词典列表更新词典列表。有更快的方法吗？

Question

I have a list of dictionaries that I need to update with information from another list of dictionaries. My current solution (below) works by taking every dictionary from the first list and comparing it to every dictionary in the second list. It works, but is there a faster, more elegant way of achieving the same result?

a = [ { "id": 1, "score":200 }, { "id": 2, "score":300 }, { "id":3, "score":400 } ]
b = [ { "id": 1, "newscore":500 }, { "id": 2, "newscore":600 } ]
# update a with data from b
for item in a:
    for replacement in b:
        if item["id"]==replacement["id"]:
            item.update({"score": replacement["newscore"]})

Answer 1

Create a dictionary indexed by id using the first array. Loop through the second array using the id .

for replacement in b:
   v = lookup.get(replacement['id'], None)
   if v is not None:
      v['score'] = replacement['newscore']

This converts an O(n^2) problem to an O(n) problem.

Answer 2

Instead of doing a len(a) * len(b) loop, process b into something easier to work with:

In [48]: replace = {d["id"]: {"score": d["newscore"]} for d in b}

In [49]: new_a = [{**d, **replace.get(d['id'], {})} for d in a]

In [50]: new_a
Out[50]: [{'id': 1, 'score': 500}, {'id': 2, 'score': 600}, {'id': 3, 'score': 400}]

Note that the {**somedict} syntax requires a modern version of Python (>= 3.5.)

Answer 3

List Comprehension:

[i.update({"score": x["newscore"]}) for x in b for i in a if i['id']==x['id']]
print(a)

Output:

[{'id': 1, 'score': 500}, {'id': 2, 'score': 600}, {'id': 3, 'score': 400}]

Timing:

%timeit [i.update({"score": x["newscore"]}) for x in b for i in a if i['id']==x['id']]

Output:

100000 loops, best of 3: 3.9 µs per loop

Answer 4

If you are open to using pandas and a, b are pandas dataframes then here is a oneliner

a.loc[a.id.isin(b.id), 'score'] = b.loc[b.id.isin(a.id), 'newscore']

Converting a, b to dataframes is simple, just use pd.DataFrame.from_records

---

Another way of doing this if you can change "newscore" to "score"

a = pd.DataFrame.from_records(a, index="id")
b = pd.DataFrame.from_records(b, index="id")
a.update(b)

Here are the timeit results

In [10]: %timeit c = a.copy(); c.update(b)
702 µs ± 37.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answer 5

First create a dict of scores to update:

>>> new_d={d['id']:d for d in b}
>>> new_d
{1: {'id': 1, 'newscore': 500}, 2: {'id': 2, 'newscore': 600}}

Then iterate over a and update by id:

for d in a:
    if d['id'] in new_d:
        d['score']=new_d[d['id']]['newscore']

>>> a
[{'id': 1, 'score': 500}, {'id': 2, 'score': 600}, {'id': 3, 'score': 400}]

Which could be even simpler as:

new_d={d['id']:d['newscore'] for d in b}
for d in a:
    if d['id'] in new_d:
        d['score']=new_d[d['id']]