blocks|key|1627486|text|使用sum()和每个字典值的长度：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1627487|count+=+sum(len(v)+for+v+in+d.itervalues())|code-block|syntax|javascript|1627488|如果你使用的是Python3，那么就使用d.values()。|1627489|使用您的输入示例和我的一个示例进行快速演示：|1627490|>>>+d+=+{'T1':+['eggs',+'bacon',+'sausage']}
>>>+sum(len(v)+for+v+in+d.itervalues())
3
>>>+d+=+{'T1':+['eggs',+'bacon',+'sausage'],+'T2':+['spam',+'ham',+'monty',+'python']}
>>>+sum(len(v)+for+v+in+d.itervalues())
7|1627491|Counter在这里帮不了你太多，你不是为每个条目创建一个计数，而是计算所有值的总长度。|1627492|entityMap^0|2|5|0|0|K|A|0|0|0|0|7|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

Use <code>sum()</code> and the lengths of each of the dictionary values:

<pre><code>count = sum(len(v) for v in d.itervalues())
</code></pre>

If you are using Python 3, then just use <code>d.values()</code>.

Quick demo with your input sample and one of mine:

<pre><code>&gt;&gt;&gt; d = {'T1': ['eggs', 'bacon', 'sausage']}
&gt;&gt;&gt; sum(len(v) for v in d.itervalues())
3
&gt;&gt;&gt; d = {'T1': ['eggs', 'bacon', 'sausage'], 'T2': ['spam', 'ham', 'monty', 'python']}
&gt;&gt;&gt; sum(len(v) for v in d.itervalues())
7
</code></pre>

A <code>Counter</code> won't help you much here, you are not creating a count per entry, you are calculating the total length of all your values.

blocks|key|3750913|text|>>>+d+=+{'T1':+['eggs',+'bacon',+'sausage'],+'T2':+['spam',+'ham',+'monty',+'python']}
>>>+sum(map(len,+d.values()))
7|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3750914|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>&gt;&gt;&gt; d = {'T1': ['eggs', 'bacon', 'sausage'], 'T2': ['spam', 'ham', 'monty', 'python']}
&gt;&gt;&gt; sum(map(len, d.values()))
7
</code></pre>

blocks|key|1637832|text|当我发现这个主题时，我一直在寻找这个问题的答案，直到我意识到我的代码中已经有了一些东西可以用来使用这个主题。这是我想出来的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1637833|count+=+0

for+key,+values+in+dictionary.items():
++++++++count+=+len(values)|code-block|syntax|javascript|1637834|如果要保存每个字典项的计数，可以创建一个新字典来保存每个键的计数。|1637835|count+=+{}

for+key,+values+in+dictionary.items():
++++++++count[key]+=+len(values)|1637836|我不能准确地找到这个方法在哪个版本中可用，但我认为.items方法只在Python3中可用。|1637837|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I was looking for an answer to this when I found this topic untill I realized I already had something in my code to use this for. This is what I came up with:

<pre><code>count = 0

for key, values in dictionary.items():
 count = len(values)
</code></pre>

If you want to save the count for every dictionary item you could create a new dictionary to save the count for each key.

<pre><code>count = {}

for key, values in dictionary.items():
 count[key] = len(values)
</code></pre>

I couldn't exactly find from which version this method is available but I think .items method is only available in Python 3.

blocks|key|496170|text|在做关于Treehouse的作业时，我想到了这个。它至少可以简化一步(据我所知)，但对于初学者(像我自己)来说，理解这个版本可能更容易。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|496171|dict+=+{'T1':+['eggs',+'bacon',+'sausage'],+'T2':+['bread',+'butter',+'tosti']}

total+=+0

for+value+in+dict:
++++value_list+=+dict[value]
++++count+=+len(value_list)
++++total+%2B=+count

print(total)|code-block|syntax|javascript|496172|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|$]]

Doing my homework on Treehouse I came up with this. It can be made simpler by one step at least (that I know of), but it might be easier for beginners (like myself) to onderstand this version.

<pre><code>dict = {'T1': ['eggs', 'bacon', 'sausage'], 'T2': ['bread', 'butter', 'tosti']}

total = 0

for value in dict:
 value_list = dict[value]
 count = len(value_list)
 total += count

print(total)
</code></pre>

blocks|key|1637614|text|对我来说，最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1637615|winners+=+{1931:+['Norman+Taurog'],+1932:+['Frank+Borzage'],+1933:+['Frank+Lloyd'],+1934:+['Frank+Capra']}

win_count_dict+=+{}

for+k,v+in+winners.items():
++++for+winner+in+v:
++++++++if+winner+not+in+win_count_dict:
++++++++++++win_count_dict[winner]=1
++++++++else:
++++++++++++win_count_dict[winner]%2B=1


print("win_count_dict+=+{}".format(win_count_dict))|code-block|syntax|javascript|1637616|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|$]]

For me the simplest way is:
<pre><code>winners = {1931: ['Norman Taurog'], 1932: ['Frank Borzage'], 1933: ['Frank Lloyd'], 1934: ['Frank Capra']}

win_count_dict = {}

for k,v in winners.items():
 for winner in v:
 if winner not in win_count_dict:
 win_count_dict[winner]=1
 else:
 win_count_dict[winner]+=1


print(&quot;win_count_dict = {}&quot;.format(win_count_dict))
</code></pre>

Python 3.3, a dictionary with key-value pairs in this form. 

<pre><code>d = {'T1': ['eggs', 'bacon', 'sausage']}
</code></pre>

The values are lists of variable length, and I need to iterate over the list items. This works: 

<pre><code>count = 0
for l in d.values():
 for i in l: count += 1
</code></pre>

But it's ugly. There must be a more Pythonic way, but I can't seem to find it.

<pre><code>len(d.values()) 
</code></pre>

produces 1. It's 1 list (DUH). Attempts with Counter from <a href="https://stackoverflow.com/questions/11751081/python-count-of-items-in-a-dictionary-of-lists">here</a> give 'unhashable type' errors.

Python count items in dict value that is a list

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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Python3.3，包含这种形式的键值对的字典。d = {'T1': ['eggs', 'bacon', 'sausage']}值是可变长度的列表，我需要迭代列表项。这是可行的：count = 0for l in d.values():   for i in l: count += 1但它很丑陋。一定有一种更有Pyth...

问Python计数为列表的dict值中的项目
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