在Python 2.7中，是否有一种快速比较ndarray中等效行的方法？ 我在一些坐标上应用对称操作，我将其存储在形状数组（N，4）的每一行中。我需要一种方法来判断我的转换是否将坐标映射回等效位置。需要注意的是，即使位置可能相同，它们也存储在数组的不同行中，因此这需要在比较之前对数组进行排序。如果我只需要调用一次就可以了，但是在我的代码中这个函数被称为~10,000次。
对此进行基准测试表明这需要约60μs：
%timeit structs_are_equiv_old(a,b)
The slowest run took 6.36 times longer than the fastest. This could mean that
an intermediate result is being cached.
10000 loops, best of 3: 59.6 µs per loop
有没有办法加快这种比较？
def structs_are_equiv(a, b):
"""
compares two numpy arrays row by row to determine if they contain the
coordinates after the application of a transformation operation.
"""
assert a.shape == b.shape
a_temp = a[ np.lexsort( (a[:,3], a[:,2], a[:,1], a[:,0]) ) ]
b_temp = b[ np.lexsort( (b[:,3], b[:,2], b[:,1], b[:,0]) ) ]
return np.allclose( a_temp, b_temp )
示例a和b（注意第一列不参与转换，只是表示存储在坐标处的对象类型的一种方式）：
a = array([[ 1. , 0. , 0.5 , 0.271149],
[ 1. , 0.5 , 0.5 , 0.271149],
[ 1. , 0. , 0. , 0.303063],
[ 1. , 0.5 , 0. , 0.303063],
[ 2. , 0.25 , 0. , 0.358071],
[ 2. , 0.75 , 0. , 0.358071],
[ 1. , 0.25 , 0.306215, 0.358071],
[ 1. , 0.75 , 0.306215, 0.358071],
[ 2. , 0. , 0.5 , 0.358071],
[ 2. , 0.5 , 0.5 , 0.358071],
[ 1. , 0.25 , 0.693785, 0.358071],
[ 1. , 0.75 , 0.693785, 0.358071],
[ 1. , 0. , 0. , 0.413078],
[ 1. , 0.5 , 0. , 0.413078],
[ 1. , 0. , 0.5 , 0.444992],
[ 1. , 0.5 , 0.5 , 0.444992],
[ 2. , 0. , 0. , 0.5 ],
[ 2. , 0.5 , 0. , 0.5 ],
[ 1. , 0.25 , 0.193785, 0.5 ],
[ 1. , 0.75 , 0.193785, 0.5 ],
[ 2. , 0.25 , 0.5 , 0.5 ],
[ 2. , 0.75 , 0.5 , 0.5 ],
[ 1. , 0.25 , 0.806215, 0.5 ],
[ 1. , 0.75 , 0.806215, 0.5 ],
[ 1. , 0. , 0.5 , 0.555008],
[ 1. , 0.5 , 0.5 , 0.555008],
[ 1. , 0. , 0. , 0.586922],
[ 1. , 0.5 , 0. , 0.586922],
[ 2. , 0.25 , 0. , 0.641929],
[ 2. , 0.75 , 0. , 0.641929],
[ 1. , 0.25 , 0.306215, 0.641929],
[ 1. , 0.75 , 0.306215, 0.641929],
[ 2. , 0. , 0.5 , 0.641929],
[ 2. , 0.5 , 0.5 , 0.641929],
[ 1. , 0.25 , 0.693785, 0.641929],
[ 1. , 0.75 , 0.693785, 0.641929],
[ 1. , 0. , 0. , 0.696937],
[ 1. , 0.5 , 0. , 0.696937],
[ 1. , 0. , 0.5 , 0.728851],
[ 1. , 0.5 , 0.5 , 0.728851],
[ 0. , 0.117635, 0.5 , 0.238728],
[ 0. , 0.617635, 0.5 , 0.238728],
[ 0. , 0. , 0.114216, 0.270642],
[ 0. , 0.5 , 0.114216, 0.270642],
[ 0. , 0. , 0. , 0.270642],
[ 0. , 0.5 , 0. , 0.270642],
[ 0. , 0.617635, 0.5 , 0.761272],
[ 0. , 0.117635, 0.5 , 0.761272],
[ 0. , 0.5 , 0.114216, 0.729358],
[ 0. , 0. , 0.114216, 0.729358],
[ 0. , 0.5 , 0. , 0.729358],
[ 0. , 0. , 0. , 0.729358],
[ 0. , 0.25 , 0.306215, 0.401299],
[ 0. , 0.75 , 0.306215, 0.401299],
[ 0. , 0.25 , 0.693785, 0.401299],
[ 0. , 0.75 , 0.693785, 0.401299],
[ 0. , 0.25 , 0.306215, 0.598701],
[ 0. , 0.75 , 0.306215, 0.598701],
[ 0. , 0.25 , 0.693785, 0.598701],
[ 0. , 0.75 , 0.693785, 0.598701],
[ 0. , 0.117635, 0.5 , 0.226923],
[ 0. , 0.117635, 0.5 , 0.773077],
[ 0. , 0. , 0.114216, 0.260279],
[ 0. , 0. , 0.114216, 0.739721],
[ 0. , 0. , 0.885784, 0.260279],
[ 0. , 0. , 0.885784, 0.739721],
[ 0. , 0.5 , 0.885784, 0.260279],
[ 0. , 0.5 , 0.885784, 0.739721],
[ 0. , 0.25 , 0.306215, 0.401299],
[ 0. , 0.25 , 0.306215, 0.598701],
[ 0. , 0.75 , 0.306215, 0.401299],
[ 0. , 0.75 , 0.306215, 0.598701],
[ 0. , 0.75 , 0.693785, 0.401299],
[ 0. , 0.75 , 0.693785, 0.598701]])
b = nparray([[ 1. , 0.5 , 0.5 , 0.271149],
[ 1. , 0. , 0.5 , 0.271149],
[ 1. , 0.5 , 0. , 0.303063],
[ 1. , 0. , 0. , 0.303063],
[ 2. , 0.75 , 0. , 0.358071],
[ 2. , 0.25 , 0. , 0.358071],
[ 1. , 0.75 , 0.306215, 0.358071],
[ 1. , 0.25 , 0.306215, 0.358071],
[ 2. , 0.5 , 0.5 , 0.358071],
[ 2. , 0. , 0.5 , 0.358071],
[ 1. , 0.75 , 0.693785, 0.358071],
[ 1. , 0.25 , 0.693785, 0.358071],
[ 1. , 0.5 , 0. , 0.413078],
[ 1. , 0. , 0. , 0.413078],
[ 1. , 0.5 , 0.5 , 0.444992],
[ 1. , 0. , 0.5 , 0.444992],
[ 2. , 0.5 , 0. , 0.5 ],
[ 2. , 0. , 0. , 0.5 ],
[ 1. , 0.75 , 0.193785, 0.5 ],
[ 1. , 0.25 , 0.193785, 0.5 ],
[ 2. , 0.75 , 0.5 , 0.5 ],
[ 2. , 0.25 , 0.5 , 0.5 ],
[ 1. , 0.75 , 0.806215, 0.5 ],
[ 1. , 0.25 , 0.806215, 0.5 ],
[ 1. , 0.5 , 0.5 , 0.555008],
[ 1. , 0. , 0.5 , 0.555008],
[ 1. , 0.5 , 0. , 0.586922],
[ 1. , 0. , 0. , 0.586922],
[ 2. , 0.75 , 0. , 0.641929],
[ 2. , 0.25 , 0. , 0.641929],
[ 1. , 0.75 , 0.306215, 0.641929],
[ 1. , 0.25 , 0.306215, 0.641929],
[ 2. , 0.5 , 0.5 , 0.641929],
[ 2. , 0. , 0.5 , 0.641929],
[ 1. , 0.75 , 0.693785, 0.641929],
[ 1. , 0.25 , 0.693785, 0.641929],
[ 1. , 0.5 , 0. , 0.696937],
[ 1. , 0. , 0. , 0.696937],
[ 1. , 0.5 , 0.5 , 0.728851],
[ 1. , 0. , 0.5 , 0.728851],
[ 0. , 0.617635, 0.5 , 0.238728],
[ 0. , 0.117635, 0.5 , 0.238728],
[ 0. , 0.5 , 0.114216, 0.270642],
[ 0. , 0. , 0.114216, 0.270642],
[ 0. , 0.5 , 0. , 0.270642],
[ 0. , 0. , 0. , 0.270642],
[ 0. , 0.117635, 0.5 , 0.761272],
[ 0. , 0.617635, 0.5 , 0.761272],
[ 0. , 0. , 0.114216, 0.729358],
[ 0. , 0.5 , 0.114216, 0.729358],
[ 0. , 0. , 0. , 0.729358],
[ 0. , 0.5 , 0. , 0.729358],
[ 0. , 0.75 , 0.306215, 0.401299],
[ 0. , 0.25 , 0.306215, 0.401299],
[ 0. , 0.75 , 0.693785, 0.401299],
[ 0. , 0.25 , 0.693785, 0.401299],
[ 0. , 0.75 , 0.306215, 0.598701],
[ 0. , 0.25 , 0.306215, 0.598701],
[ 0. , 0.75 , 0.693785, 0.598701],
[ 0. , 0.25 , 0.693785, 0.598701],
[ 0. , 0.117635, 0.5 , 0.226923],
[ 0. , 0.117635, 0.5 , 0.773077],
[ 0. , 0. , 0.114216, 0.260279],
[ 0. , 0. , 0.114216, 0.739721],
[ 0. , 0. , 0.885784, 0.260279],
[ 0. , 0. , 0.885784, 0.739721],
[ 0. , 0.75 , 0.306215, 0.401299],
[ 0. , 0.75 , 0.306215, 0.598701],
[ 0. , 0.25 , 0.306215, 0.401299],
[ 0. , 0.25 , 0.306215, 0.598701],
[ 0. , 0.75 , 0.693785, 0.401299],
[ 0. , 0.75 , 0.693785, 0.598701],
[ 0. , 0.25 , 0.693785, 0.401299],
[ 0. , 0.25 , 0.693785, 0.598701]])
答案 0 :(得分：2)
这是一种方法，考虑{{3}}将每一行减少为一个标量，然后简单地进行排序和比较，就像这样 -
def structs_are_equiv_dotreduc(a,b):
scale = 10000**np.arange(1,4)
a0 = np.sort(a[:,1:].dot(scale).astype(int))
b0 = np.sort(b[:,1:].dot(scale).astype(int))
return (a0 == b0).all()
运行时测试 -
In [538]: # Setup inputs with b array just a row-permuted version of a
...: a = np.random.rand(100,4)
...: b = a[np.random.permutation(a.shape[0])]
...:
In [539]: %timeit structs_are_equiv(a,b)
10000 loops, best of 3: 117 µs per loop
In [540]: %timeit structs_are_equiv_dotreduc(a,b)
10000 loops, best of 3: 42.7 µs per loop
答案 1 :(得分：0)
npi.sort应该比您当前的解决方案更快;虽然如果他的假设确实存在，那么divakar的解决方案应该更快。