如何在python中从一个集转换为一个数组

Question

The question:

I need to convert a set to an array. If I try to just do np.array(X), I get an array that contains a set, which isn't very helpful. If I convert my set to a list and then my list to an array, everything works fine, but that seems needlessly complex (the line of code in question is a convoluted mess of needless complexity already). Is there a way to go straight from set to array?

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Code context, if helpful:

rad = 0.944
n = 5
pairs = np.array(list(set(chain.from_iterable(((x,y),(x,-y),(-x,y),(-x,-y)) for x in np.linspace(0, rad, n) for y in np.linspace(0, rad, n) if math.hypot(x,y) <= rad))))

The point of the set is to remove duplicates from the chain. I know there are ways of doing this with an array, but it doesn't seem possible to convert an itertools.chain object directly into an array either.

Holistically, this code just models a circle using a uniform distribution of x,y points, and speeds up the process by evoking symmetry between the four quadrants.

Answer 1

Here is a pure Numpythonic approach:

In [244]: lins = np.unique(np.linspace(0, rad, n))
# Create the prucuct of lins
In [245]: arr = np.array((np.repeat(lins, 5), np.tile(lins, 5))).T
# remove extra items based on your condition
In [246]: new = np.compress(np.sqrt(np.power(arr, 2).sum(1)) <= 0.944, arr, 0)

In [247]: a = np.array([ 1, -1])
# Create a product of (-1, 1) for creating the total expected result by multiplying the product on each row of the compressed array
In [248]: perm = np.array((np.repeat(a, 2), np.tile(a, 2))).T

In [249]: total = new[:,None] * perm

Note that if you make sure that there is no duplicate item in your linspace array there won't be any one in the total combinations.

And here is the total result:

array([[[ 0.   ,  0.   ],
        [ 0.   , -0.   ],
        [-0.   ,  0.   ],
        [-0.   , -0.   ]],

       [[ 0.   ,  0.236],
        [ 0.   , -0.236],
        [-0.   ,  0.236],
        [-0.   , -0.236]],

       [[ 0.   ,  0.472],
        [ 0.   , -0.472],
        [-0.   ,  0.472],
        [-0.   , -0.472]],

       [[ 0.   ,  0.708],
        [ 0.   , -0.708],
        [-0.   ,  0.708],
        [-0.   , -0.708]],

       [[ 0.   ,  0.944],
        [ 0.   , -0.944],
        [-0.   ,  0.944],
        [-0.   , -0.944]],

       [[ 0.236,  0.   ],
        [ 0.236, -0.   ],
        [-0.236,  0.   ],
        [-0.236, -0.   ]],

       [[ 0.236,  0.236],
        [ 0.236, -0.236],
        [-0.236,  0.236],
        [-0.236, -0.236]],

       [[ 0.236,  0.472],
        [ 0.236, -0.472],
        [-0.236,  0.472],
        [-0.236, -0.472]],

       [[ 0.236,  0.708],
        [ 0.236, -0.708],
        [-0.236,  0.708],
        [-0.236, -0.708]],

       [[ 0.472,  0.   ],
        [ 0.472, -0.   ],
        [-0.472,  0.   ],
        [-0.472, -0.   ]],

       [[ 0.472,  0.236],
        [ 0.472, -0.236],
        [-0.472,  0.236],
        [-0.472, -0.236]],

       [[ 0.472,  0.472],
        [ 0.472, -0.472],
        [-0.472,  0.472],
        [-0.472, -0.472]],

       [[ 0.472,  0.708],
        [ 0.472, -0.708],
        [-0.472,  0.708],
        [-0.472, -0.708]],

       [[ 0.708,  0.   ],
        [ 0.708, -0.   ],
        [-0.708,  0.   ],
        [-0.708, -0.   ]],

       [[ 0.708,  0.236],
        [ 0.708, -0.236],
        [-0.708,  0.236],
        [-0.708, -0.236]],

       [[ 0.708,  0.472],
        [ 0.708, -0.472],
        [-0.708,  0.472],
        [-0.708, -0.472]],

       [[ 0.944,  0.   ],
        [ 0.944, -0.   ],
        [-0.944,  0.   ],
        [-0.944, -0.   ]]])

Answer 2

My conclusion is that there is no easier way to convert a set to an array other than going through a list first.

In terms of my particular situation, I ended up using the following:

rad = 0.944
n = 5
pairs = np.array([item for item in itertools.product(np.linspace(-rad, rad, 2*n-1), repeat=2) if math.hypot(item[0], item[1]) <= rad])

Answer 3

You can use np.fromiter to construct an array directly from the generator output of chain.from_iterable , and np.unique to remove duplicates:

pairs = np.unique(np.fromiter(itertools.chain.from_iterable(data), dtype=float))

Note that the dtype argument of the np.fromiter function is required.