擅长:python、mysql、java
<p>你的理解列表中的逻辑很好。您可以通过更改</p>
<pre><code>print {x:y for x,z in scrabble_scores for y in z}
</code></pre>
<p>至</p>
<pre><code>print [(x,y) for x,z in scrabble_scores for y in z]
</code></pre>
<p>打印出来的</p>
<pre><code>[(1, 'E'), (1, 'A'), (1, 'O'), (1, 'I'), (1, 'N'), (1, 'R'), (1, 'T'), (1, 'L'), (1, 'S'), (1, 'U'), (2, 'D'), (2, 'G'), (3, 'B'), (3, 'C'), (3, 'M'), (3, 'P'), (4, 'F'), (4, 'H'), (4, 'V'), (4, 'W'), (4, 'Y'), (5, 'K'), (8, 'J'), (8, 'X'), (10, 'Q'), (10, 'Z')]
</code></pre>
<p>实现不起作用的原因是字典的每个键都必须是唯一的。因此,在设置<code>1:A</code>时,会覆盖<code>1:E</code>的前一对<code>key:value</code>。你知道吗</p>
<p>也许你在找那封信作为钥匙?如果是这样,那么就交换x和y:</p>
<pre><code>print {y:x for x,z in scrabble_scores for y in z}
</code></pre>