擅长:python、mysql、java
<p>根据您上面提到的要求,您可以简单地编写列表,为单个列表整平一行:</p>
<pre><code>first_dict = {'address1': [1], 'address2': [2]}
second_dict = {'address1': [3], 'address2': [4]}
outlet_list1 = [item for sublist in first_dict.values() for item in sublist]
outlet_list2 = [item for sublist in second_dict.values() for item in sublist]
result=[outlet_list1, outlet_list2]
print(result)
</code></pre>
<p>这给了你:</p>
<pre><code>[[1, 2], [3, 4]]
</code></pre>
<p>一个班轮基本上意味着:</p>
<pre><code>for sublist in first_dict.values():
for item in sublist:
outlet_list1.append(item)
</code></pre>
<p>希望这能正确回答你的问题</p>
<p><strong>更有组织的解决方案</strong></p>
<p>如果您有一个DICT列表,您可以按如下方式进行操作:</p>
<pre><code>first_dict = {'address1': [1], 'address2': [2]}
second_dict = {'address1': [3], 'address2': [4]}
list_of_dict = [first_dict, second_dict]
result = []
for dict_item in list_of_dict:
result.append([item for sublist in dict_item.values() for item in sublist])
print(result)
</code></pre>
<p>你也会得到同样的结果</p>