与动态 ndarray 作斗争

我是 MATLAB 用户。 将以下 MATLAB 脚本移植到 python 的最简单方法是什么

a = []
for i=1:10
    for j=1:10
        a(i,j) =  i*j
    end
end

问题是关于动态使用 np.ndarrays。在 MATLAB 中，我可以在不知道其大小的情况下将 a 初始化为通用 ndarray（通过 a=[]），然后使用显式索引 a(i,j)。 在Python中有.append方法，但是当使用ndarrays时它让我感到困惑，因为它需要首先打包行的元素，然后将行打包在一起，或者类似的东西。我更喜欢显式索引。这是否可以与动态数组一起完成，或者可以仅使用固定大小的数组来完成？

Thanks!

运行 MATLAB 时查看显示

a =  1
a =
   1   2
a =
   1   2   3
a =
   1   2   3   4
.... (so on for 100 iterations)

在八度中我可以这样做：

>> i=1:10
i =

    1    2    3    4    5    6    7    8    9   10

>> j=(1:10)'
j =

    1
    2
    3
    4
    5
    6
    7
    8
    9
   10

>> a=i+j
a =

    2    3    4    5    6    7    8    9   10   11
    3    4    5    6    7    8    9   10   11   12
    4    5    6    7    8    9   10   11   12   13
    5    6    7    8    9   10   11   12   13   14
    6    7    8    9   10   11   12   13   14   15
    7    8    9   10   11   12   13   14   15   16
    8    9   10   11   12   13   14   15   16   17
    9   10   11   12   13   14   15   16   17   18
   10   11   12   13   14   15   16   17   18   19
   11   12   13   14   15   16   17   18   19   20

这利用了广播，这是一个借用自 numpy 的概念

In [500]: i=np.arange(1,11)
In [501]: a = i[:,None] + i
In [502]: a
Out[502]: 
array([[ 2,  3,  4,  5,  6,  7,  8,  9, 10, 11],
       [ 3,  4,  5,  6,  7,  8,  9, 10, 11, 12],
       [ 4,  5,  6,  7,  8,  9, 10, 11, 12, 13],
       [ 5,  6,  7,  8,  9, 10, 11, 12, 13, 14],
       [ 6,  7,  8,  9, 10, 11, 12, 13, 14, 15],
       [ 7,  8,  9, 10, 11, 12, 13, 14, 15, 16],
       [ 8,  9, 10, 11, 12, 13, 14, 15, 16, 17],
       [ 9, 10, 11, 12, 13, 14, 15, 16, 17, 18],
       [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
       [11, 12, 13, 14, 15, 16, 17, 18, 19, 20]])

这是最佳实践 - 在 numpy 中，我敢说 MATLAB 和 Octave。

但如果你必须使用迭代做类似的事情

In [503]: a=np.zeros((10,10),int)
In [504]: for i in range(10):
     ...:     for j in range(10):
     ...:         a[i,j]=i+j

或者使用完整的 python 列表迭代：

In [512]: alist = []
In [513]: for i in range(10):
     ...:     sublist=[]
     ...:     for j in range(10):
     ...:         sublist.append(i+j)
     ...:     alist.append(sublist)
     ...:     
In [514]: alist
Out[514]: 
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
 [2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
 [3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
 [4, 5, 6, 7, 8, 9, 10, 11, 12, 13],
 [5, 6, 7, 8, 9, 10, 11, 12, 13, 14],
 [6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
 [7, 8, 9, 10, 11, 12, 13, 14, 15, 16],
 [8, 9, 10, 11, 12, 13, 14, 15, 16, 17],
 [9, 10, 11, 12, 13, 14, 15, 16, 17, 18]]
In [515]: np.array(alist)
Out[515]: 
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
       [ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10],
       [ 2,  3,  4,  5,  6,  7,  8,  9, 10, 11],
       [ 3,  4,  5,  6,  7,  8,  9, 10, 11, 12],
       [ 4,  5,  6,  7,  8,  9, 10, 11, 12, 13],
       [ 5,  6,  7,  8,  9, 10, 11, 12, 13, 14],
       [ 6,  7,  8,  9, 10, 11, 12, 13, 14, 15],
       [ 7,  8,  9, 10, 11, 12, 13, 14, 15, 16],
       [ 8,  9, 10, 11, 12, 13, 14, 15, 16, 17],
       [ 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]])

但我可以生成alist更紧凑地

alist=[[i+j for i in range(10)] for j in range(10)]

当您构建列表列表时，请确保子列表都具有相同的长度 - 否则您会带着问题回到SO。