Question

给出<section class="soon"> <h2>LLORUM ISPUS DOR!</h2> <p>Lorem ipsum dolor sit amet, consectetur adipisicing elit. Voluptate dolorem reprehenderit illo officia<br> ex quidem aut veniam modi numquam iusto, ipsam placeat dolorum quae eum vero! Eveniet esse<br>Lorem ipsum dolor sit amet, consectetur adipisicing elit. Voluptate dolorem reprehenderit illo officia<br> enim molestiae delectus modi officiis ab porro, maiores, dolores consequuntur ipsum expedita!Lorem <br> ipsum dolor sit amet, consectetur adipisicing elit. Voluptate dolorem reprehenderit illo officia <br> ex quidem aut veniam modi numquam iusto, ipsam placeat dolorum quae eum vero! Eveniet esse<br>Lorem ipsum dolor sit amet, consectetur adipisicing elit. Voluptate dolorem reprehenderit illo officia<br> enim molestiae delectus modi officiis ab porro, maiores, dolores consequuntur ipsum expedita!</p> </section> <div class="leafPic"> <img src="https://via.placeholder.com/350" alt="Blooming flower"> </div>个以n个字典为元素的列表，我想生成一个新列表，其中包含一组联接（左联接）的字典。每个字典都保证有一个名为m的键，但除此之外可以有任意键集。例如，想象以下两个列表：

index

我想生成一个联合列表：

l1 = [{"index":1, "b":2}, 
      {"index":2, "b":3},
      {"index":3, "b":"10"},
      {"index":4, "c":"7"}]

l2 = [{"index":1, "c":4},
      {"index":2, "c":5},
      {"index":6, "c":8},
      {"index":7, "c":9}]

在Python中最有效的方法是什么？

当前我有这段代码，但是它只做一个内部联接，如何修改它以给我左联接？

l3 = [{"index":1, "b":2, "c":4}, 
      {"index":2, "b":3, "c":5}, 
      {"index":3, "b":10},
      {"index":4, "c":7}]

Answer 1

为了提高效率，您可以先构建一个以索引为键的字典，并以l2的对应字典为值，这样您就不必每次都经过l2在其中寻找匹配的字典。

然后，您可以构建新的词典列表：对于l1中的每个词典，我们都会对其进行复制以保留原始副本，并使用l2中匹配的词典对其进行更新

l1 = [{"index":1, "b":2}, {"index":2, "b":3}, {"index":3, "b":"10"}, {"index":4, "c":"7"}]

l2 = [{"index":1, "c":4}, {"index":2, "c":5}, {"index":6, "c":8}, {"index":7, "c":9}]

dict2 = {dct['index']:dct for dct in l2}

out = []
for d1 in l1:
    d = dict(**d1)
    d.update(dict2.get(d1['index'], {}))
    out.append(d)

print(out)
# [{'index': 1, 'b': 2, 'c': 4}, {'index': 2, 'b': 3, 'c': 5}, {'index': 3, 'b': '10'}, {'index': 4, 'c': '7'}]

Answer 2

下面的代码片段只是将它们转换为字典以实现更快的合并，然后将合并的字典重新转换为列表以匹配您的预期输出。

l1_dict = {item['index']: item for item in l1}
l2_dict = {item['index']: item for item in l2}

for item in l1_dict:
    l1_dict[item].update(l2_dict.get(item, {}))

l3 = list(l1_dict.values())
print(l3)

左键将两个字典列表连接到一个键上

2 个答案: