擅长:python、mysql、java
<p>您可以为<code>origin</code>键存储<code>uuid</code>列表。你知道吗</p>
<p>我建议有以下两种方法:</p>
<ol>
<li>创建空的<code>list</code>以便首先访问<code>origin</code>,然后附加到它:</li>
</ol>
<pre class="lang-py prettyprint-override"><code>backup = {}
for u in backup_list['storages']['storage']:
srvuuidorg = u['origin']
backup_uuid = u['uuid']
if not backup.get(srvuuidorg):
backup[srvuuidorg] = []
backup[srvuuidorg].append(backup_uuid)
</code></pre>
<ol start=“2”>
<li>使用<code>defaultdict</code>集合,基本上在引擎盖下对您也有相同的作用:</li>
</ol>
<pre class="lang-py prettyprint-override"><code>from collections import defaultdict
backup = defaultdict(list)
for u in backup_list['storages']['storage']:
srvuuidorg = u['origin']
backup_uuid = u['uuid']
backup[srvuuidorg].append(backup_uuid)
</code></pre>
<p>在我看来,最后一条路更优雅。
如果您需要存储<code>uuid</code>唯一列表,您应该使用saem方法来处理<code>set</code>,而不是<code>list</code>。你知道吗</p>