blocks|key|5187651|text|sorted(dict.keys())返回一个仅包含键的排序列表。您的for循环错误地期望在同一列表中查找这些值。请尝试下面的代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5187652|for+k+in+ke:
++++print+k,+dict[k]|code-block|syntax|javascript|5187653|entityMap^0|0|J|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

<code>sorted(dict.keys())</code> returns a sorted list of just the keys. Your for loop is incorrectly expecting to find the values in the same list. Try the code below instead:

<pre><code>for k in ke:
 print k, dict[k]
</code></pre>

blocks|key|5278185|text|其他答案指出了问题所在。如果您使用的是python+2.7%2B，这里有一种更简单的方法，即使用collections模块中的Counter类。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5278186|from+collections+import+Counter

def+print_words(filename):
++++with+open(filename,+'rU')+as+f:
++++++++counts+=+Counter(f.read().split())
++++for+k,+v+in+sorted(counts.items()):
++++++++print+k,+v|code-block|syntax|javascript|5278187|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/collections.html#collections.Counter^0|1A|B|1P|7|1P|7|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Other answers point out the problem. If you're using python 2.7+, here's an easier way to do it, using the <a href="http://docs.python.org/library/collections.html#collections.Counter" rel="nofollow"><code>Counter</code></a> class from the <code>collections</code> module.

<pre><code>from collections import Counter

def print_words(filename):
 with open(filename, 'rU') as f:
 counts = Counter(f.read().split())
 for k, v in sorted(counts.items()):
 print k, v
</code></pre>

blocks|key|3065454|text|而不是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3065455|ke+=+sorted(dict.keys())
for+k,+v+in+ke:+print+k,+v|code-block|syntax|javascript|3065456|尝试：|3065457|for+k+in+sorted(dict.keys()):
++++print+k,+dict[k]|3065458|sorted(dict.keys())将返回一个仅包含键的排序列表(因为这就是您给它的全部内容)。|offset|length|style|CODE|3065459|entityMap^0|0|0|0|0|0|J|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Instead of 

<pre><code>ke = sorted(dict.keys())
for k, v in ke: print k, v
</code></pre>

Try:

<pre><code>for k in sorted(dict.keys()):
 print k, dict[k]
</code></pre>

<code>sorted(dict.keys())</code> will return a sorted list of only the keys (since that's all you're giving it).

blocks|key|5278294|text|对于2.5和2.6兼容的解决方案，请使用defaultdict|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5278295|from+collections+import+defaultdict
counter+=+defaultdict(int)

for+word+in+f.read().split():
++++counter[word]+%2B=+1|code-block|syntax|javascript|5278296|这将返回一个字典(实际上是一个子类，但使用方法相同)，您可以在其中只查找counter['someword']+(它返回一个整数)。|5278297|它的工作原理:如果请求的键不存在，则使用给定函数的返回值创建它。在本例中为int()，即0。另请参阅上面链接的文档中的示例部分。|5278298|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/release/2.6.8/library/collections.html#collections.defaultdict^0|K|B|K|B|0|0|0|10|J|0|11|5|18|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

For a 2.5 and 2.6 compatible solution, use <a href="http://docs.python.org/release/2.6.8/library/collections.html#collections.defaultdict" rel="nofollow"><code>defaultdict</code></a>:

<pre><code>from collections import defaultdict
counter = defaultdict(int)

for word in f.read().split():
 counter[word] += 1
</code></pre>

This returns a dictionary (a subclass actually, but to be used the same way), where you can just look up <code>counter['someword']</code> (that returns an integer).

How it works: If the requested key does not exist, it is created using the return value of the given function. In this case, <code>int()</code>, which is <code>0</code>. See also the examples section at the documentation linked above.

blocks|key|5187726|text|根据顺序和计数哪个更重要，您可以使用collections+module中的ordered+dictionary或counter+dictionary|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|5187727|OrderedDict会记住元素的添加顺序；对元素进行计数|5187728|对于前者，您可能会这样做。|5187729|>>>+words+=+open(filename).read().split()
>>>+counters+=+OrderedDict([(word,+words.count(word))+
++++++++++++++++++++++++++++for+word+in+sorted(list(set(words)))])|code-block|syntax|javascript|5187730|你将会得到带有计数器的排序字典--只有2行。|5187731|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/collections.html|1|http://docs.python.org/library/collections.html#collections.OrderedDict|2|http://docs.python.org/library/collections.html#collections.Counter^0|I|I|0|12|I|1|1L|I|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]|$A|14|B|15|1|16]|$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|6|7|1A|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|1B|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|1C|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|1D|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|1E|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|R|S|T|C|$U|Z]]]]

Depending on what is more important - order or counting - you may use either <a href="http://docs.python.org/library/collections.html#collections.OrderedDict" rel="nofollow">ordered dictionary</a> or <a href="http://docs.python.org/library/collections.html#collections.Counter" rel="nofollow">counter dictionary</a> from the <a href="http://docs.python.org/library/collections.html" rel="nofollow">collections module</a>

OrderedDict remembers the elements in the order they are added; Counter - counts elements

With the former, you may do something like that

<pre><code>&gt;&gt;&gt; words = open(filename).read().split()
&gt;&gt;&gt; counters = OrderedDict([(word, words.count(word)) 
 for word in sorted(list(set(words)))])
</code></pre>

You will get sorted dictionary with counters - just in 2 lines.

I am running the following code to count how many times a word occured in a text file.

<pre><code>def print_words(filename):
 f = open(filename, 'rU')
 dict={}
 for line in f:
 words=line.split()
 for word in words:
 if dict.get(word):
 dict[word]=dict[word]+1
 else: 
 dict[word]=1
 ke = sorted(dict.keys())
 for k, v in ke: print k, v
</code></pre>

The dictionary file should hold each word and its count. I was able to get that working fine. But I failed to sort the

Whats wrong with my dict?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在运行下面的代码来计算一个单词在文本文件中出现的次数。def print_words(filename):    f = open(filename, 'rU')    dict={}    for line in f:       words=line.split()       for word in words:          if dict.get(word):          

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