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BaseVariable
V2EX  ›  Python

遍历 list of lists of lists 的优化方法

  •  
  •   BaseVariable · Apr 29, 2020 · 2672 views
    This topic created in 2235 days ago, the information mentioned may be changed or developed.

    a = [1, 2, 3] b = [a for i in range(1000)] c = [b for i in range(1000)] 现在想从 c 取每个 int (每个 1 、2 、3 ),然后构成一个 3x1000x1000 的 numpy matrix 求问大佬们有什么除了遍历以外更快的方法。 感激不尽!

    3 replies    2020-05-08 14:16:11 +08:00
    chizuo
        1
    chizuo  
       Apr 29, 2020
    为什么要取呢?
    直接 x = np.asarray(c)
    不就行了?
    BaseVariable
        2
    BaseVariable  
    OP
       Apr 29, 2020
    @chizuo 对啊。。是我想错了
    necomancer
        3
    necomancer  
       May 8, 2020
    c = np.empty((1000,1000,3))
    c[...,:]=a
    c = np.transpose(c, (2,0,1))
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