blocks|key|1168829|text|>>>+L+=+['apple','red','apple','red','red','pear']
>>>+from+collections+import+defaultdict
>>>+d+=+defaultdict(int)
>>>+for+i+in+L:
...+++d[i]+%2B=+1
>>>+d
defaultdict(<type+'int'>,+{'pear':+1,+'apple':+2,+'red':+3})|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1168830|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; L = ['apple','red','apple','red','red','pear']
&gt;&gt;&gt; from collections import defaultdict
&gt;&gt;&gt; d = defaultdict(int)
&gt;&gt;&gt; for i in L:
... d[i] += 1
&gt;&gt;&gt; d
defaultdict(&lt;type 'int'&gt;, {'pear': 1, 'apple': 2, 'red': 3})
</code></pre>

blocks|key|1168851|text|L+=+['apple','red','apple','red','red','pear']
d+=+{}
[d.__setitem__(item,1%2Bd.get(item,0))+for+item+in+L]
print+d+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1168852|提供{'pear':+1,+'apple':+2,+'red':+3}|unstyled|offset|length|style|CODE|1168853|entityMap^0|0|2|X|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>L = ['apple','red','apple','red','red','pear']
d = {}
[d.__setitem__(item,1+d.get(item,0)) for item in L]
print d 
</code></pre>

Gives <code>{'pear': 1, 'apple': 2, 'red': 3}</code>

blocks|key|954674|text|在2.7和3.1中，有专门的Counter字典用于此目的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|954675|>>>+from+collections+import+Counter
>>>+Counter(['apple','red','apple','red','red','pear'])
Counter({'red':+3,+'apple':+2,+'pear':+1})|code-block|syntax|javascript|954676|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/collections.html#collections.Counter^0|E|7|E|7|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

in 2.7 and 3.1 there is special <a href="https://docs.python.org/3/library/collections.html#collections.Counter" rel="noreferrer"><code>Counter</code></a> dict for this purpose.

<pre><code>&gt;&gt;&gt; from collections import Counter
&gt;&gt;&gt; Counter(['apple','red','apple','red','red','pear'])
Counter({'red': 3, 'apple': 2, 'pear': 1})
</code></pre>

blocks|key|954724|text|我一直认为，对于如此微不足道的任务，我不想导入任何东西。但我可能错了，这取决于collections.Counter是否更快。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|954725|items+=+"Whats+the+simpliest+way+to+add+the+list+items+to+a+dictionary+"

stats+=+{}
for+i+in+items:
++++if+i+in+stats:
++++++++stats[i]+%2B=+1
++++else:
++++++++stats[i]+=+1

#+bonus
for+i+in+sorted(stats,+key=stats.get):
++++print("%25d×'%25s'"+%25+(stats[i],+i))|code-block|syntax|javascript|954726|我认为这可能比使用count()更可取，因为它只会遍历可迭代一次，而count可能会在每次迭代中搜索整个对象。我使用这种方法解析了许多兆字节的统计数据，它总是相当快的。|954727|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I always thought that for a task that trivial, I wouldn't want to import anything. But i may be wrong, depending on collections.Counter being faster or not.

<pre><code>items = "Whats the simpliest way to add the list items to a dictionary "

stats = {}
for i in items:
 if i in stats:
 stats[i] += 1
 else:
 stats[i] = 1

# bonus
for i in sorted(stats, key=stats.get):
 print("%d×'%s'" % (stats[i], i))
</code></pre>

I think this may be preferable to using count(), because it will only go over the iterable once, whereas count may search the entire thing on every iteration. I used this method to parse many megabytes of statistical data and it always was reasonably fast.

blocks|key|1169000|text|我喜欢：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1169001|counts+=+dict()
for+i+in+items:
++counts[i]+=+counts.get(i,+0)+%2B+1|code-block|syntax|javascript|1169002|.get允许您在密钥不存在时指定默认值。|1169003|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I like:

<pre><code>counts = dict()
for i in items:
 counts[i] = counts.get(i, 0) + 1
</code></pre>

.get allows you to specify a default value if the key does not exist.

blocks|key|1169100|text|考虑一下collections.Counter+(从Python2.7开始可用)。https://docs.python.org/2/library/collections.html#collections.Counter|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1169101|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/collections.html#collections.Counter^0|15|1Y|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Consider collections.Counter (available from python 2.7 onwards). 
<a href="https://docs.python.org/2/library/collections.html#collections.Counter">https://docs.python.org/2/library/collections.html#collections.Counter</a>

blocks|key|1169151|text|只需使用list属性count\|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1169152|i+=+['apple','red','apple','red','red','pear']
d+=+{x:i.count(x)+for+x+in+i}
print+d|code-block|syntax|javascript|1169153|输出：|1169154|{'pear':+1,+'apple':+2,+'red':+3}|1169155|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|$]]

Simply use list property count\

<pre><code>i = ['apple','red','apple','red','red','pear']
d = {x:i.count(x) for x in i}
print d
</code></pre>

output : 

<pre><code>{'pear': 1, 'apple': 2, 'red': 3}
</code></pre>

I'm new to Python and I have a simple question, say I have a list of items:

<pre><code>['apple','red','apple','red','red','pear']
</code></pre>

Whats the simpliest way to add the list items to a dictionary and count how many times the item appears in the list.

So for the list above I would like the output to be:

<pre><code>{'apple': 2, 'red': 3, 'pear': 1}
</code></pre>

Using a dictionary to count the items in a list

Python

我是Python的新手，我有一个简单的问题，假设我有一个项目列表：['apple','red','apple','red','red','pear']将列表项添加到字典并计算该项在列表中出现的次数的最简单方法是什么。因此，对于上面的列表，我希望输出为：{'apple': 2, 'red': 3, 'pear': 1}

问使用字典计算列表中的条目
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