python - 当外部和内部循环包含计算和删除时，对其进行矢量化

Question

我一直在研究如何矢量化外部和内部 for 循环。这些有一些计算，并且内部还有一个删除 - 这似乎使它变得不那么直接。

如何最好地对其进行矢量化？

import numpy as np

flattenedArray = np.ndarray.tolist(someNumpyArray)

#flattenedArray is a python list of lists.
c = flattenedArray[:]
for a in range (len(flattenedArray)):
    for b in range(a+1, len(flattenedArray)):
        if a == b:
            continue
        i0 = flattenedArray[a][0]
        j0 = flattenedArray[a][1]
        z0 = flattenedArray[a][2]
        i1 = flattenedArray[b][0]
        i2 = flattenedArray[b][1]
        z1 = flattenedArray[b][2]

        if ((np.square(z0-z1)) <= (np.square(i0-i1) + (np.square(j0-j2)))):
            if (np.square(i0-i1) + (np.square(j0-j1))) <= (np.square(z0+z1)):
                        c.remove(flattenedArray[b])

Answer 1

@MSeifert 当然，通常都是正确的。因此，以下完整矢量化只是为了展示“它是如何完成的”

import numpy as np

N = 4
data = np.random.random((N, 3))

# vectorised code
j, i = np.tril_indices(N, -1) # chose tril over triu to have contiguous columns
                              # useful later
sqsum = np.square(data[i,0]-data[j,0]) + np.square(data[i,1]-data[j,1])
cond = np.square(data[i, 2] + data[j, 2]) >= sqsum
cond &= np.square(data[i, 2] - data[j, 2]) <= sqsum
# because equal 'b's are grouped together we can use reduceat:
cond = np.r_[False, np.logical_or.reduceat(
    cond, np.add.accumulate(np.arange(N-1)))]
left = data[~cond, :]


# original code (modified to make it run)
flattenedArray = np.ndarray.tolist(data)

#flattenedArray is a python list of lists.
c = flattenedArray[:]
for a in range (len(flattenedArray)):
    for b in range(a+1, len(flattenedArray)):
        if a == b:
            continue
        i0 = flattenedArray[a][0]
        j0 = flattenedArray[a][1]
        z0 = flattenedArray[a][2]
        i1 = flattenedArray[b][0]
        j1 = flattenedArray[b][1]
        z1 = flattenedArray[b][2]

        if ((np.square(z0-z1)) <= (np.square(i0-i1) + (np.square(j0-j1)))):
            if (np.square(i0-i1) + (np.square(j0-j1))) <= (np.square(z0+z1)):
                try:
                        c.remove(flattenedArray[b])
                except:
                        pass

# check they are the same
print(np.alltrue(c == left))