擅长:python、mysql、java
<p>我想您应该使用<code>groupby</code>包中的<code>itertools</code></p>
<pre><code>from itertools import groupby
tideData = [
['Thursday 4 January',11.58,0.38],
['Thursday 4 January',16.95,0.73],
['Friday 5 January',6.48,0.83],
['Friday 5 January',12.42,0.33],
['Saturday 6 January',0.5,0.02],
['Saturday 6 January',7.18,0.85],
['Friday 2 February',23.52,0.04]
]
</code></pre>
<p>如果未对数据进行排序,则可以使用:</p>
<pre><code>tideData = sorted(tideData, key=lambda x: x[0])
</code></pre>
<p>在使用以下工具之前:</p>
<pre><code>[list(g) for _,g in groupby(tideData, key=lambda x: x[0])]
# returns:
[[['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73]],
[['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33]],
[['Saturday 6 January', 0.5, 0.02], ['Saturday 6 January', 7.18, 0.85]],
[['Friday 2 February', 23.52, 0.04]]]
</code></pre>