<p><strong>编辑2013-12-11</strong>-这个答案很古老。它仍然有效且正确,但是人们看到它时应该更喜欢<a href="https://docs.python.org/2/library/string.html#format-string-syntax" rel="noreferrer">new format syntax</a>。</p>
<p>您可以像这样使用<a href="http://docs.python.org/library/stdtypes.html#string-formatting" rel="noreferrer">string formatting</a>:</p>
<pre><code>>>> print '%5s' % 'aa'
aa
>>> print '%5s' % 'aaa'
aaa
>>> print '%5s' % 'aaaa'
aaaa
>>> print '%5s' % 'aaaaa'
aaaaa
</code></pre>
<p>基本上:</p>
<ul>
<li><code>%</code>字符通知python它将不得不用某种东西替换令牌</li>
<li><code>s</code>字符通知python令牌将是一个字符串</li>
<li><code>5</code>(或您想要的任何数字)通知python用最多5个字符的空格填充字符串。</li>
</ul>
<p>在您的特定情况下,可能的实现可能如下所示:</p>
<pre><code>>>> dict_ = {'a': 1, 'ab': 1, 'abc': 1}
>>> for item in dict_.items():
... print 'value %3s - num of occurances = %d' % item # %d is the token of integers
...
value a - num of occurances = 1
value ab - num of occurances = 1
value abc - num of occurances = 1
</code></pre>
<p><strong>旁注:</strong>只是想知道您是否知道<a href="http://docs.python.org/library/itertools.html" rel="noreferrer">^{<cd4>} module</a>的存在。例如,您可以在一行中获得所有组合的列表:</p>
<pre><code>>>> [''.join(perm) for i in range(1, len(s)) for perm in it.permutations(s, i)]
['a', 'b', 'c', 'd', 'ab', 'ac', 'ad', 'ba', 'bc', 'bd', 'ca', 'cb', 'cd', 'da', 'db', 'dc', 'abc', 'abd', 'acb', 'acd', 'adb', 'adc', 'bac', 'bad', 'bca', 'bcd', 'bda', 'bdc', 'cab', 'cad', 'cba', 'cbd', 'cda', 'cdb', 'dab', 'dac', 'dba', 'dbc', 'dca', 'dcb']
</code></pre>
<p>您可以通过将<code>combinations</code>与<code>count()</code>结合使用来获得出现的次数。</p>