从列表中删除重复和原始文件 - python [英] Remove duplicate and original from list - python
问题描述
,例如:
lst = ['a','b','c','c','c','d','e','e']
输出应该删除重复的
,所以这样的东西['a','b','d']
我不需要保留订单)
使用 collections.Counter()
对象,然后只保留计数为1的值:
从集合导入计数器
[k for k,v in Counter(lst).iteritems()if v = = 1]
这是一个O(N)算法;您只需要循环遍历N个项目列表一次,然后循环遍历较少的项目(< N),以提取出现只有一次的值。
如果顺序是重要的,分开步骤:
计数= Counter(lst)
[k for l in lst if [k] == 1]
演示:
>>>从集合导入计数器
>>>> lst = ['a','b','c','c','c','d','e','e']
>>> [k for k,v in Counter(lst).iteritems()if v == 1]
['a','b','d']
>>> count = Counter(lst)
>>>> [k for k in lst if count [k] == 1]
['a','b','d']
这两种方法的顺序相同是巧合;其他输入可能会导致不同的顺序。
given a list of string (i am not aware of list), i want to remove the duplicate and original word.
for example: lst = ['a', 'b', 'c', 'c', 'c', 'd', 'e', 'e']
the output should should remove the duplicates so something like this ['a', 'b', 'd']
I do not need to preserve the order)
Use a collections.Counter()
object, then keep only those values with a count of 1:
from collections import counter
[k for k, v in Counter(lst).iteritems() if v == 1]
This is a O(N) algorithm; you just need to loop through the list of N items once, then a second loop over fewer items (< N) to extract those values that appear just once.
If order is important, separate the steps:
counts = Counter(lst)
[k for k in lst if counts[k] == 1]
Demo:
>>> from collections import Counter
>>> lst = ['a', 'b', 'c', 'c', 'c', 'd', 'e', 'e']
>>> [k for k, v in Counter(lst).iteritems() if v == 1]
['a', 'b', 'd']
>>> counts = Counter(lst)
>>> [k for k in lst if counts[k] == 1]
['a', 'b', 'd']
That the order is the same for both approaches is a coincidence; other inputs may result in a different order.
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