为了防止有人也在寻找这个，我目前的实现如下。然而，我会感谢任何建议，将加快这一点或减少内存占用。

s = mat1.multiply(mat2)
s.data[:] = 1.

a1 = mat1.copy()
a1.data[:] = 1.
a1 = (a1 - s).maximum(0)

a2 = mat2.copy()
a2.data[:] = 1.
a2 = (a2 - s).maximum(0)

res = mat1.multiply(a1) + mat2.multiply(a2) + \
      mat1.multiply(s).minimum(mat2.multiply(s))

如果稀疏非零值为正数，则使用maximum的正确 UNION 行为的另一种方法可能是 * 取反并使其为正数 *。

按照你的指导，我明确地用data来做了一些事情。

def sp_min_nz_positive(asp,bsp):  # a and b scipy sparse
    amax = asp.max()
    bmax = bsp.max()
    abmaxplus = max(amax, bmax) # + 1.0 : surprise! not needed.
    # invert the direction, while remaining positive
    arev = asp.copy()
    arev.data[:] = abmaxplus - asp.data[:]
    brev = bsp.copy()
    brev.data[:] = abmaxplus - bsp.data[:]
    out = arev.maximum(brev)  # 
    # revert the direction of these positives
    out.data[:] = abmaxplus - out.data[:]
    return out

由于四舍五入，可能会有不精确之处
还有一个建议是使用稀疏内部函数。一个相当通用的函数是sp.find，它返回 anything 的非零元素。
因此，您也可以尝试使用minimum，它也可以处理负值，如下所示：

import scipy.sparse as sp
def sp_min_union(a, b):
    assert a.shape == b.shape
    assert sp.issparse(a) and sp.issparse(b)
    (ra,ca,_) = sp.find(a)   # over nonzeros only
    (rb,cb,_) = sp.find(b)   # over nonzeros only
    setab = set(zip(ra,ca)).union(zip(rb,cb)) # row-column union-of-nonzero
    r=[]
    c=[]
    v=[]
    for (rr,cc) in setab:
        r.append(rr)
        c.append(cc)
        anz = a[rr,cc]
        bnz = b[rr,cc]
        assert anz!=0 or bnz!=0   # they came from *some* sp.find
        if anz==0: anz = bnz
        #else:
        #    #if bnz==0: anz = anz
        #    #else:      anz=min(anz,bnz)
        # equiv.
        elif bnz!=0: anz=min(anz,bnz)
        v.append(anz)
    # choose what sparse output format you want, many seem
    # constructible as:
    return sp.csr_matrix((v, (r,c)), shape=a.shape)