Question

我有一个词典列表，我需要使用其他词典列表中的信息进行更新。我当前的解决方案（如下）的工作方式是从第一个列表中取出每个词典，然后将其与第二个列表中的每个词典进行比较。它可以工作，但是有没有更快，更优雅的方法来达到相同的结果？

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height="50"/>
<br/>
<span class="quickview__icon" id="Haori jacket">
Quick View
</span>
<br/>
<span>
Haori jacket
</span>
<br/>
<span id="Haorijacketprice">
$210.00
</span>
<span id="Haorijacketnum" style="display: none;">1</span>
</div>
<br/>
<div class="clothing">
<img id="Linentopimg" src="https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcSHnKUQNqUcYm7fauoZSam3_c4gne4NemeUUyY2-RkvGWuOGR6O6g" width="50" height="50"/>
<br/>
<span class="quickview__icon" id="Linen top">Quick View</span>
<br/>
<span>Linen top</span>
<br/>
<span id="Linentopprice">
$170.00
</span>
<span id="Linentopnum" style="display: none;">2</span>
</div>
<div class="clothing">
<img id="T-shirtimg" src="data:image/jpeg;base64,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" width="50" height="50"/>
<br/>
<span class="quickview__icon" id="T-shirt">Quick View</span>
<br/>
<span>T-shirt</span>
<br/>
<span id="T-shirtprice">
$50.00
</span>
<span id="T-shirtnum" style="display: none;">3</span>
</div>
<div class="popup" style="display: none;" id="popup">
<span class="close" id="closepopup" title="Close">&times;</span>
<br/>
<img src="" id="popupimg" width="100" height= "100"/>
<br/>
<div style="width: 100%; position: relative">
<img id="previous" src="https://image.flaticon.com/icons/svg/60/60965.svg" height="50" width="50" style="position: absolute; left: 3px;"/>
<img id="next" src="https://image.flaticon.com/icons/svg/60/60758.svg" height="50" width="50" style="position: absolute; right: 3px;"/>
</div>
<div style="width: 100%; text-align: center;">
<span id="clothingname"></span>
<br/>
<span id="clothingprice"></span>
</div>
<br/>
</div>

Answer 1

使用第一个数组创建由id索引的字典。使用id遍历第二个数组。

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

这会将O(n^2)问题转换为O(n)问题。

Answer 2

将b转换为更易于使用的方法，而不是进行len（a）* len（b）循环：

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}]

请注意，{**somedict}语法要求使用现代版本的Python（> = 3.5。）

Answer 3

列表理解：

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

输出：

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

时间：

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

输出：

100000 loops, best of 3: 3.9 µs per loop

Answer 4

如果您愿意使用pandas，而a，b是熊猫数据帧，那么这里是一个单行纸

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

将a，b转换为数据帧很简单，只需使用pd.DataFrame.from_records

如果可以将“ newscore”更改为“ score”，则可以使用另一种方法

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

这是时间结果

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

首先创建分数更新字典：

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

然后遍历a并通过id更新：

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}]

哪个可能更简单：

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']]

用另一个词典列表更新词典列表。有没有更快的方法？

5 个答案: