Question

在python列表中，我想删除所有重复的少于'k'的元素。例如，如果k == 3，则我们的列表为：

l = [a,b,c,c,c,a,d,e,e,d,d]

然后输出必须是：

[c,c,c,d,d,d]

什么是快速的方法（我的数据很大），有什么很好的pythonic建议吗？

这是我编写的代码，但我认为这不是最快，最Python化的方式：

from collections import Counter

l = ['a', 'b', 'c', 'c', 'c', 'a', 'd', 'e', 'e', 'd', 'd']

counted = Counter(l)

temp = []
for i in counted:
    if counted[i] < 3:
        temp.append(i)

new_l = []
for i in l:
    if i not in temp:
        new_l.append(i)

print(new_l)

Answer 1

You can use collections.Counter to construct a dictionary mapping values to counts. Then use a list comprehension to filter for counts larger than a specified value.

from collections import Counter

L = list('abcccadeedd')
c = Counter(L)
res = [x for x in L if c[x] >=3]

# ['c', 'c', 'c', 'd', 'd', 'd']

Answer 2

I would use a Counter from collections:

from collections import Counter
count_dict = Counter(l)
[el for el in l if count_dict[el]>2]

Answer 3

A brute-force option would be to get the number of occurrences per item, then filter that output. The collections.Counter object works nicely here:

l = [a,b,c,c,c,a,d,e,e,d,d]
c = Counter(l)

# Counter looks like {'a': 2, 'b': 1, 'c': 3...}

l = [item for item in l if c[item]>=3]

Under the hood, Counter acts as a dictionary, which you can build yourself like so:

c = {}
for item in l:
    # This will check if item is in the dictionary
    # if it is, add to current count, if it is not, start at 0
    # and add 1
    c[item] = c.get(item, 0) + 1

# And the rest of the syntax follows from here
l = [item for item in l if c[item]>=3]

Answer 4

此选项有任何缺点吗？

l = ['a','b','c','c','c','a','d','e','e','d','d']

res = [ e for e in l if l.count(e) >= 3]

#=> ['c', 'c', 'c', 'd', 'd', 'd']

如何删除列表中少于k个重复项

4 个答案: