保持出现N次或更多次的字符串

I have a list that is

我有一个列表

mylist = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd']

And I used Counter from collections on this list to get the result:

我在此列表中使用了来自集合的Counter来获得结果：

from collection import Counter
counts = Counter(mylist)

#Counter({'a': 3, 'c': 2, 'b': 2, 'd': 1})

Now I want to subset this so that I have all elements that occur some number of times, for example: 2 times or more - so that the output looks like this:

现在我想要对其进行子集化，以便我拥有多次出现的所有元素，例如：2次或更多次 - 这样输出如下所示：

['a', 'b', 'c']

This seems like it should be a simple task - but I have not found anything that has helped me so far.

这似乎应该是一个简单的任务 - 但到目前为止我还没有找到任何帮助我的东西。

Can anyone suggest somewhere to look? I am also not attached to using Counter if I have taken the wrong approach. I should note I am new to python so I apologise if this is trivial.

任何人都可以建议去看看吗？如果我采取了错误的方法，我也不会使用Counter。我应该注意我是python的新手，所以如果这是微不足道的话我会道歉。

5 个解决方案

#1

[s for s, c in counts.iteritems() if c >= 2]
# => ['a', 'c', 'b']

#2

Try this...

尝试这个...

def get_duplicatesarrval(arrval):
    dup_array = arrval[:]
    for i in set(arrval):
        dup_array.remove(i)       
    return list(set(dup_array))   



mylist = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd']
print get_duplicatesarrval(mylist)

Result:

结果：

[a, b, c]

#3

The usual way would be to use a list comprehension as @Adaman does.
In the special case of 2 or more, you can also subtract one Counter from another

通常的方法是使用@Adaman的列表理解。在2或更多的特殊情况下，您也可以从另一个计数器中减去一个计数器

>>> counts = Counter(mylist) - Counter(set(mylist))
>>> counts.keys()
['a', 'c', 'b']

#4

from itertools import groupby

mylist = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd']

res = [i for i,j in groupby(mylist) if len(list(j))>=2]

print res
['a', 'b', 'c']

#5

I think above mentioned answers are better, but I believe this is the simplest method to understand:

我认为上面提到的答案更好，但我相信这是最简单的理解方法：

mylist = ['a', 'a', 'a', 'b', 'b', 'c', 'c', 'd']
newlist=[]
newlist.append(mylist[0])
for i in mylist:
    if i in newlist:
        continue
    else:
        newlist.append(i)
print newlist

>>>['a', 'b', 'c', 'd']

#1

[s for s, c in counts.iteritems() if c >= 2]
# => ['a', 'c', 'b']

#2