擅长:python、mysql、java
<p><strong>如果两个字典中都有键,则两个字典中的int值递增1</strong></p>
<ol>
<li>通过<a href="https://docs.python.org/2/library/sets.html" rel="nofollow">set intersection</a>方法从这两个字典中获取公共密钥。在</li>
<li>迭代公共键。在</li>
<li>每个常用字典的键值按1递增。在</li>
</ol>
<p><strong>演示</strong>:</p>
<pre><code>>>> a = {"a": 1, "b":2, "c":0}
>>> b = {"d": 1, "b":2, "c":0}
>>> ab = set(b.keys()).intersection(set(a.keys()))
>>> ab
set(['c', 'b'])
>>> for i in ab:
... a[i] = a[i] + 1
... b[i] = b[i] + 1
...
>>> a
{'a': 1, 'c': 1, 'b': 3}
>>> b
{'c': 1, 'b': 3, 'd': 1}
</code></pre>
<hr/>
<p><strong>代码问题:</strong></p>
<p>仅当<code>k==k2</code>时才需要递增。所以每当代码进入<code>else loop</code>时,当这个条件为false时,在其他情况下,我们将值递增1。在</p>
<p>只需在if循环中增加值。在</p>
<p><strong>尝试1</strong>:</p>
^{pr2}$
<p><strong>尝试2</strong>:</p>
<pre><code>for k, v in secondDict.iteritems():
if k in firstDict:
secondDict[k] = secondDict[k] + 1
</code></pre>