numpy.ufunc.identity的用法

Question

I wonder what line r = ufct.identity does inside function ufunc_reduce . Does it just initialize the operation exactly like r = 0 ?

a = np.array([2,3,4,5])
b = np.array([8,5,4])
c = np.array([5,4,6,8,3])

def ufunc_reduce(ufct, *vectors):
    vs = np.ix_(*vectors)
    r = ufct.identity
    for v in vs:
        r = ufct(r,v)
    return r

ufunc_reduce(np.add,a,b,c)

Answer 1

Yes, for the ufunc np.add , ufct.identity is 0 . But for another function it may be something else. For instance, np.multiply.identity is 1 .

The ufunc_reduce function you show doesn't know in advance what function it will be given, so it can't be hard-coded to use either 0 or 1 . It gets the proper value by checking the identity attribute of the ufunc instead.