擅长:python、mysql、java
<p>关于:</p>
<pre><code>vocab = ["foo", "bar", "baz"]
exception= ["no"]
s = "foo bar baz no bar quux foo bla bla"
wordcount = dict((x,0) for x in vocab)
words = s.split()
i = 0
while i < len(words):
cur_word = words[i]
if cur_word in exception:
i += 4
else:
if cur_word in vocab: wordcount[cur_word] += 1
i += 1
print wordcount # {'baz': 1, 'foo': 1, 'bar': 1}
</code></pre>
<p>这只是利用了一个事实,如果我们遇到“否”,我们可以跳过以下3个元素。你知道吗</p>