大量选择以产生具有所有唯一值的2D数组

问题描述：

所以我想知道在使用np.random.choice生成二维数组时，是否存在一种更有效的解决方案，其中每一行都有唯一的值.

So I am wondering if there's a more efficient solution in generating a 2-D array using np.random.choice where each row has unique values.

例如，对于形状为(3,4)的数组，我们期望输出为:

For example, for an array with shape (3,4), we expect an output of:

# Expected output given a shape (3,4)
array([[0, 1, 3, 2],
       [2, 3, 1, 0],
       [1, 3, 2, 0]])

这意味着每行的值在列数方面必须是唯一的.因此，对于out中的每一行，整数应仅介于0到3之间.

This means that the values for each row must be unique with respect to the number of columns. So for each row in out, the integers should only fall between 0 to 3.

我知道我可以通过将False传递给replace参数来实现.但是我只能为每一行而不是整个矩阵执行此操作.例如，我可以这样做:

I know that I can achieve it by passing False to the replace argument. But I can only do it for each row and not for the whole matrix. For instance, I can do this:

>>> np.random.choice(4, size=(1,4), replace=False)
array([[0,2,3,1]])

但是当我尝试这样做时:

But when I try to do this:

>>> np.random.choice(4, size=(3,4), replace=False)

我收到这样的错误:

 File "<stdin>", line 1, in <module>
 File "mtrand.pyx", line 1150, in mtrand.RandomState.choice 
 (numpy\random\mtrand\mtrand.c:18113)
 ValueError: Cannot take a larger sample than population when 
 'replace=False'

我认为这是因为由于矩阵的大小，它试图绘制3 x 4 = 12样本而不进行替换，但我只给出了4的限制.

I assume it's because it's trying to draw 3 x 4 = 12 samples due to the size of the matrix without replacement but I'm only giving it a limit of 4.

我知道我可以使用for-loop来解决它:

I know that I can solve it by using a for-loop:

 >>> a = (np.random.choice(4,size=4,replace=False) for _ in range(3))
 >>> np.vstack(a)
 array([[3, 1, 2, 0],
        [1, 2, 0, 3],
        [2, 0, 3, 1]])

但是我想知道是否有不使用任何for循环的解决方法? (我有点假设，如果行数大于1000，添加for循环可能会使速度变慢.但是正如您所看到的，我实际上是在a中创建一个生成器，因此我也不确定是否毕竟有效果.)

But I wanted to know if there's a workaround without using any for-loops? (I'm kinda assuming that adding for-loops might make it slower if I have a number of rows greater than 1000. But as you can see I am actually creating a generator in a so I'm also not sure if it has an effect after all.)

答

我经常使用的一个技巧是生成一个随机数组，并使用argsort获取唯一索引作为所需的唯一数字.因此，我们可以做-

One trick I have used often is generating a random array and using argsort to get unique indices as the required unique numbers. Thus, we could do -

def random_choice_noreplace(m,n, axis=-1):
    # m, n are the number of rows, cols of output
    return np.random.rand(m,n).argsort(axis=axis)

样品运行-

In [98]: random_choice_noreplace(3,7)
Out[98]: 
array([[0, 4, 3, 2, 6, 5, 1],
       [5, 1, 4, 6, 0, 2, 3],
       [6, 1, 0, 4, 5, 3, 2]])

In [99]: random_choice_noreplace(5,7, axis=0) # unique nums along cols
Out[99]: 
array([[0, 2, 4, 4, 1, 0, 2],
       [1, 4, 3, 2, 4, 1, 3],
       [3, 1, 1, 3, 2, 3, 0],
       [2, 3, 0, 0, 0, 2, 4],
       [4, 0, 2, 1, 3, 4, 1]])

运行时测试-

# Original approach
def loopy_app(m,n):
    a = (np.random.choice(n,size=n,replace=False) for _ in range(m))
    return np.vstack(a)

时间-

In [108]: %timeit loopy_app(1000,100)
10 loops, best of 3: 20.6 ms per loop

In [109]: %timeit random_choice_noreplace(1000,100)
100 loops, best of 3: 3.66 ms per loop

大量选择以产生具有所有唯一值的2D数组

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