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Phishion
V2EX  ›  Python

请问有什么高效的方法,在保持顺序的前提下,做差集计算?

  •  
  •   Phishion · Apr 6, 2022 · 2007 views
    This topic created in 1482 days ago, the information mentioned may be changed or developed.

    大概代码如下,其中主列表 full_queue 有 5000 个元素,做差集的列表有 50 个左右,计算下来耗时大概 100 多 ms ,有什么方法或者第三方包可以优化这部分的执行时间?

    from datetime import datetime
    
    process_start = datetime.now().timestamp()
    
    full_queue = [4878214304830001, 4878214304830002, 4878214304830003, 4878214304830004, ...]
    diff_queue = [4878214548209665, 4878214308487169, 4878265720848385, ...]
    format_list = list(set(full_queue)-set(diff_queue))
    format_list.sort(key=full_queue.index)
    
    print(format_list[:5])
    process_elapsed = round(datetime.now().timestamp() - process_start, 4) * 1000
    print(process_elapsed)
    
    Phishion
        1
    Phishion  
    OP
       Apr 6, 2022
    行了,当我没问,我用列表解析遍历只要个位数 ms ,不知道为什么列表解析快那么多
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