blocks|key|1800952|text|如果要删除出现频率至少相同两次的所有名称：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1800953|import+collections

lst+=+['Mike',+'Mike',+'Mike',+'Jhon',+'Jhon',+'Rob',+'Rob',+'Carl',+'Carl',
+++++++"foo","foo","foo","foo","bar",+"bar",+"bar",+"bar"]
list_counter+=+collections.Counter(lst)

#+count+the+frequency+
times+=+collections.Counter(list_counter.values())

#+keep+names+that+do+not+appear+at+the+same+frequency+as+any+other+names
lst[:]+=+[name+for+name,v++in+list_counter.items()+if+times[v]+<+2]
print(lst)|code-block|syntax|javascript|1800954|这将会给你留下：|1800955|['Mike']|1800956|如果您还想确保计数不是1：|1800957|[name+for+name,+v++in+list_counter.items()+if+v+==+1+or+times[v]+<+2]|1800958|如果你想保留所有的Mike，那么遍历列表：|1800959|import+collections

lst+=+['Mike',+'Mike',+'Mike',+'Jhon',+'Jhon',+'Rob',+'Rob',+'Carl',+'Carl',
+++++++"foo","foo","foo","foo","bar",+"bar",+"bar",+"bar"]
list_counter+=+collections.Counter(lst)
times+=+collections.Counter(list_counter.values())

lst[:]+=+[name+for+name+in+lst+if+times[list_counter[name]]+<+2]
print(lst)|1800960|这将为您提供：|1800961|+['Mike',+'Mike',+'Mike']|1800962|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

If you want to remove all names that appear at the same frequency at least two times:

<pre><code>import collections

lst = ['Mike', 'Mike', 'Mike', 'Jhon', 'Jhon', 'Rob', 'Rob', 'Carl', 'Carl',
 "foo","foo","foo","foo","bar", "bar", "bar", "bar"]
list_counter = collections.Counter(lst)

# count the frequency 
times = collections.Counter(list_counter.values())

# keep names that do not appear at the same frequency as any other names
lst[:] = [name for name,v in list_counter.items() if times[v] &lt; 2]
print(lst)
</code></pre>

which would leave you with:

<pre><code>['Mike']
</code></pre>

If you want to also makes sure the count is not 1:

<pre><code>[name for name, v in list_counter.items() if v == 1 or times[v] &lt; 2]
</code></pre>

If you want to keep all the Mikes then iterate over the list:

<pre><code>import collections

lst = ['Mike', 'Mike', 'Mike', 'Jhon', 'Jhon', 'Rob', 'Rob', 'Carl', 'Carl',
 "foo","foo","foo","foo","bar", "bar", "bar", "bar"]
list_counter = collections.Counter(lst)
times = collections.Counter(list_counter.values())

lst[:] = [name for name in lst if times[list_counter[name]] &lt; 2]
print(lst)
</code></pre>

Which would give you:

<pre><code> ['Mike', 'Mike', 'Mike']
</code></pre>

blocks|key|1502221|text|如果我没弄错，你想保留那些出现了两次以上的名字。在这种情况下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1502222|[x+for+x,y+in+list_counter.iteritems()+if+y>2]|code-block|syntax|javascript|1502223|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|$]]

if I get it right you want to keep those names what occure more than 2 times.
In this case:

<pre><code>[x for x,y in list_counter.iteritems() if y&gt;2]
</code></pre>

blocks|key|1801025|text|import+collections
your_list+=+['Sam',+'Sam',+'Sam',+'Sam',+'Mike',+'Mike',+'Mike',+'Jhon',+'Jhon',+'Rob',+'Rob']
list_counter+=+collections.Counter(your_list)
filtered_list+=+[item+for+item+in+your_list+if+list_counter.get(item)+!=+2]
result+=+list(set(filtered_list))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1801026|过滤后的列表是一种python列表理解方式，用于比较列表项是否有计数器!=+2。|unstyled|1801027|在示例二的情况下。|1801028|'Mike'，'Mike'，'Mike‘|1801029|现在正在做|1801030|list(set(filtered_list))|1801031|“迈克”|1801032|你不应该像'list‘那样命名你的变量。|1801033|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|F|7|12|8|@]|9|@]|A|$]]]|T|$]]

<pre><code>import collections
your_list = ['Sam', 'Sam', 'Sam', 'Sam', 'Mike', 'Mike', 'Mike', 'Jhon', 'Jhon', 'Rob', 'Rob']
list_counter = collections.Counter(your_list)
filtered_list = [item for item in your_list if list_counter.get(item) != 2]
result = list(set(filtered_list))
</code></pre>

The filtered list is an python list comprehension way to compare if your list item has counter != 2.

In case of example two.

['Mike', 'Mike', 'Mike']

now doing 

list(set(filtered_list))

['Mike']

You should not name your variables like 'list'.

blocks|key|4118040|text|我的方法是首先将具有相同count的所有元素分组到defaultdict对象中，然后基于过滤条件构建结果字典：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4118041|>>>+from+collections+import+Counter,+defaultdict++
>>>
>>>+lst+=+['Mike',+'Mike',+'Mike',+'Jhon',+'Jhon',+'Rob',+'Rob',+'Carl',+'Carl']
>>>+c+=+Counter(lst)
>>>+c
Counter({'Mike':+3,+'Carl':+2,+'Rob':+2,+'Jhon':+2})
>>>
>>>+d+=+defaultdict(list)
>>>+
>>>+for+k,v+in+c.items():
++++++++d[v].append(k)
>>>+
>>>+d
defaultdict(<class+'list'>,+{2:+['Jhon',+'Rob',+'Carl'],+3:+['Mike']})
>>>
>>>+result+=+dict((*v,k)+for+k,v+in+d.items()+if+len(v)+<+2)
>>>+result
{'Mike':+3}
>>>+|code-block|syntax|javascript|4118042|entityMap^0|C|5|P|B|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

My approach was to first group all elements with same <code>count</code> into <code>defaultdict</code> object, then build your resulting dictionary based on a filtering condition:

<pre><code>&gt;&gt;&gt; from collections import Counter, defaultdict 
&gt;&gt;&gt;
&gt;&gt;&gt; lst = ['Mike', 'Mike', 'Mike', 'Jhon', 'Jhon', 'Rob', 'Rob', 'Carl', 'Carl']
&gt;&gt;&gt; c = Counter(lst)
&gt;&gt;&gt; c
Counter({'Mike': 3, 'Carl': 2, 'Rob': 2, 'Jhon': 2})
&gt;&gt;&gt;
&gt;&gt;&gt; d = defaultdict(list)
&gt;&gt;&gt; 
&gt;&gt;&gt; for k,v in c.items():
 d[v].append(k)
&gt;&gt;&gt; 
&gt;&gt;&gt; d
defaultdict(&lt;class 'list'&gt;, {2: ['Jhon', 'Rob', 'Carl'], 3: ['Mike']})
&gt;&gt;&gt;
&gt;&gt;&gt; result = dict((*v,k) for k,v in d.items() if len(v) &lt; 2)
&gt;&gt;&gt; result
{'Mike': 3}
&gt;&gt;&gt; 
</code></pre>

blocks|key|1502289|text|通过使用.values()，然后对出现两次>=的值进行列表理解，您可以找到所有不同的金额：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1502290|values+=+collections.Counter(list_counter.values())
result+=+[item+for+item+in+list+if+values[item]+>=+2]|code-block|syntax|javascript|1502291|但是，我不会使用list作为变量名，因为它隐藏了内置类型。|1502292|entityMap^0|4|9|L|2|0|0|8|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

You can find all of the different amounts by using <code>.values()</code> and then making a list comprehension of the values that appear <code>&gt;=</code> 2 times:

<pre><code>values = collections.Counter(list_counter.values())
result = [item for item in list if values[item] &gt;= 2]
</code></pre>

I wouldn't use <code>list</code> as a variable name, however, because it shadows the built-in type.

Example 1:

<pre><code>import collections

list = ['Mike', 'Mike', 'Mike', 'Jhon', 'Jhon', 'Rob', 'Rob', 'Carl', 'Carl']
list_counter = collections.Counter(list)
&gt;&gt; Counter({'Mike': 3, 'Rob': 2, 'Jhon': 2, 'Carl': 2})
</code></pre>

Example 2:

<pre><code>import collections
list = ['Sam', 'Sam', 'Sam', 'Sam', 'Mike', 'Mike', 'Mike', 'Jhon', 'Jhon', 'Rob', 'Rob']
list_counter = collections.Counter(list)
&gt;&gt; Counter({'Sam': 4, 'Mike': 3, 'Rob': 2, 'Jhon': 2})
</code></pre>

All I want to do is to verify if in the list occours 2 or more elements the same number of times, and delete them. 

In example 1 Rob, Jhon and Carl would be removed, they are 3 elements that appears 2 times in the list. In example 2 Rob and Jhon would be removed, while Sam and Mike would be ok and left in the list. If in the list appears only 1 element (for example Jhon n times), nothing would be removed.

Python / Collections - Remove elements that appear same number of times

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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示例1：import collectionslist = ['Mike', 'Mike', 'Mike', 'Jhon', 'Jhon', 'Rob', 'Rob', 'Carl', 'Carl']list_counter = collections.Counter(list)>> Counter({'Mike': 3...

问Python / Collections -删除出现相同次数的元素
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