blocks|key|1442749|text|如果每个值都是每个字母的字数，那么求和的代码应该是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1442750|sum+=+0
for+(key,+value)+in+dictionary:
++++sum+=+sum+%2B+value|code-block|syntax|javascript|1442751|那么这就是一个例子：|1442752|def+get_letter_percentage(dictionary,+letter):
++++return+dictionary[letter]+/+sum|1442753|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If each value is the number of words for each letter, then your code to find the sum would be:

<pre><code>sum = 0
for (key, value) in dictionary:
 sum = sum + value
</code></pre>

Then it's a case of:

<pre><code>def get_letter_percentage(dictionary, letter):
 return dictionary[letter] / sum
</code></pre>

blocks|key|5483972|text|也许你可以在你的函数中尝试这样的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5483973|dictionary+=+{'a':+5,+'b':+8,+'c':15}
sum+=+0
for+(key,+value)+in+dictionary.items():+sum+%2B=+value
percentage+=+dictionary['a']+/+(sum+%2B+0.0)
print+"percentage+of+'%25s'+is+%25.2f+%25%25"+%25+('a'+,+percentage*100)|code-block|syntax|javascript|5483974|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Maybe you could try something like this inside your function:

<pre><code>dictionary = {'a': 5, 'b': 8, 'c':15}
sum = 0
for (key, value) in dictionary.items(): sum += value
percentage = dictionary['a'] / (sum + 0.0)
print "percentage of '%s' is %.2f %%" % ('a' , percentage*100)
</code></pre>

blocks|key|1442812|text|def+get_letter_percentage(dictionary,+s):
+++'''(dict+of+{str+:+int},+str)+->+float'''

+++return+dictionary[s]+*+1.0+/+sum(dictionary.values())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1442813|百分比是出现的次数除以总出现的次数。请注意乘以1.0以避免整数除法。|unstyled|1442814|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>def get_letter_percentage(dictionary, s):
 '''(dict of {str : int}, str) -&gt; float'''

 return dictionary[s] * 1.0 / sum(dictionary.values())
</code></pre>

The percentage is the amount of occurrences divided by the total occurrences.
Note the multiplication by 1.0 to avoid int division.

blocks|key|1442819|text|我想出了这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1442820|mydict+=+{"a":5,+"b":1,+"c":4,+"d":3,+"e":6}


def+get_letter_percentage(dictionary,+s):
++++sum+=+0
++++for+key+in+dictionary:
++++++++sum+%2B=+dictionary[key]

++++return+float(dictionary[s])+/+float(sum)

print+get_letter_percentage(mydict,+"b")|code-block|syntax|javascript|1442821|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I came up with this:

<pre><code>mydict = {"a":5, "b":1, "c":4, "d":3, "e":6}


def get_letter_percentage(dictionary, s):
 sum = 0
 for key in dictionary:
 sum += dictionary[key]

 return float(dictionary[s]) / float(sum)

print get_letter_percentage(mydict, "b")
</code></pre>

blocks|key|1442823|text|def+get_letter_percentage(stats,+letter):
++return+stats[letter]+/+(sum(stats.values())+%2B+0.0)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1442824|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def get_letter_percentage(stats, letter):
 return stats[letter] / (sum(stats.values()) + 0.0)
</code></pre>

In the given dictionary, each key is a single lowercase letter and each value is the number of lowercase words that start with that letter. The str parameter is a single lowercase letter. Based on the values in the dictionary, return the percentage of words that start with that letter. 

Note: use floating-point division.

<pre><code>def get_letter_percentage(dictionary, s):
 '''(dict of {str : int}, str) -&gt; float'''

# Start with a counter for the sum of the values.
count = 0
# And then look at the key and value of the dictionary
for (key, value) in dictionary.items(): 
</code></pre>

This is where I got stuck. I know I need to create a sum of the values in order to do a floating point division.

<pre><code># guessing it is something along these lines
count = len(values) #??
</code></pre>

Get percentage of a letter in python dictionary

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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在给定的字典中，每个键都是一个小写字母，每个值都是以该字母开头的小写单词的数量。str参数是一个小写字母。根据字典中的值，返回以该字母开头的单词的百分比。注意:使用浮点除法。def get_letter_percentage(dictionary, s):   '''(dict of {str : int}, str)...

问获取python字典中字母的百分比
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