在Python中的数组过滤器？ [英] array filter in python?

问题描述

例如，我有两个列表：

  A = [6，7，8，9，10，11，12 ] 
 subset_of_A = [6，9，12]; ＃A 
 
 
的子集，结果应该是[7,8,10,11];其余的元素
  

Python中是否有内置函数？ b $ b

解决方案

如果顺序不重要，应该使用 set.difference 。然而，如果你想保留订单，只需要简单的列表理解。

  result = [a for a in A如果一个不是在subset_of_A] 
  

编辑：如果 subset_of_A 是一个实际的 set ，因为检查集合为O（1），而O（n）为列表。

  A = [6 ，7，8，9，10，11，12] 
 subset_of_A = set（[6,9,12]）＃A 
 
的子集结果= [a for a in A如果一个不在[subset_of_A] 
  

For example, I have two lists

 A           = [6, 7, 8, 9, 10, 11, 12]
subset_of_A  = [6, 9, 12]; # the subset of A


the result should be [7, 8, 10, 11]; the remaining elements 

Is there a built-in function in python to do this?

解决方案

If the order is not important, you should use set.difference. However, if you want to retain order, a simple list comprehension is all it takes.

result = [a for a in A if a not in subset_of_A]

EDIT: As delnan says, performance will be substantially improved if subset_of_A is an actual set, since checking for membership in a set is O(1) as compared to O(n) for a list.

A = [6, 7, 8, 9, 10, 11, 12]
subset_of_A = set([6, 9, 12]) # the subset of A

result = [a for a in A if a not in subset_of_A]

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