擅长:python、mysql、java
<p>在Python3.2及更高版本中,<code>Counter</code>有{a1},它完全符合您的需要,保留0和负值,而实际的<code>-</code>运算符则没有。它在适当的位置运行,所以你应该:</p>
<pre><code>letter_counts = Counter(letters_to_work_with)
letter_counts.subtract(Counter(word))
# Alternatively, though it won't short-circuit, you could do:
# if min(letter_counts.values()) < 0:
if any(v < 0 for v in letter_counts.values()):
... not valid ...
</code></pre>