如何优化以下列表操作？

Question

代码注释中描述的操作花费了我实施它的时间。我可以通过哪些方法优化流程？列表由6个数字组成的行组成，即[[float] * 6，...]。数字以[x1，x2，y1，y2，z1，z2]的格式位于每个多维数据集的对角。

# Merging by creating a list with the first cube and another with the rest.
# If the cube in the first list can be merged with a cube in the second list,
# the cube in the second list is removed and the cube in the first list is
# replaced with the merged shape.
# If there is no suitable merge to be done, the lowest positioned cube in
# the second list is moved to the first list and then we check if
# anything in the second list can be merged with it and so on.

# loops until qbs has been emptied into css
# the condition for merging is matching faces
def xmerge(qbs, css):
    tot = len(qbs)
    j = 0
    while len(qbs) > 0:
        i = 0
        k = 0
        printProgressBar(tot - len(qbs)+1, tot, prefix = ' Merging along x:',
        length = 50)
        while i < len(qbs):
            # first check if faces are touching
            if (abs(css[j][0] - css[j][3]) == abs(css[j][0] - qbs[i][0])
            and css[j][1] == qbs[i][1]
            and css[j][2] == qbs[i][2]
            # if true up to here then corners are touching
            # below we check if the faces match (same size)
            and abs(css[j][1] - css[j][4]) == abs(qbs[i][1] - qbs[i][4])
            and abs(css[j][2] - css[j][5]) == abs(qbs[i][2] - qbs[i][5])):
                css[j] = [css[j][0],css[j][1],css[j][2],
                qbs[i][3],qbs[i][4],qbs[i][5]]
                qbs = np.delete(qbs, i, axis=0)
                k = 1
            i += 1

        if k == 0:
            css = np.vstack([css, qbs[np.argmin(qbs[:,0])]])
            qbs = np.delete(qbs, np.argmin(qbs[:,0]), axis=0)
            j += 1

    print("")
    return css