Question

我有以下代码

np.array([points[label==k].mean(axis = 0) for k in range(self.k)])

Points是一个n x d数组，label是一个1 x n数组，其值最大为k，k为一个数字。

我的目标是删除轴参数并仍然得到相同的结果，并且还要对数组部件标签== k进行索引，我想重写。

你们中有人有这样做的方法吗？

Answer 1

我猜您正在寻求矢量化解决方案。这是一个matrix-multiplication-

def matmul(points, label):
    k = label.max()+1
    mask = label == np.arange(k)[:,None]
    out = mask.dot(points)/mask.sum(1,keepdims=True)
    return out

这里还有np.add.reduceat-

def add_reduceat(points, label):
    k = label.max()+1
    sidx = label.argsort()
    ps = points[sidx]
    ls = label[sidx]

    cutidx = np.flatnonzero(np.r_[True,ls[:-1] != ls[1:],True])
    lens = np.diff(cutidx)
    out = np.full((k,points.shape[1]),np.nan)

    idx_rows = ls[cutidx[:-1]]
    mean_vals = np.add.reduceat(ps,cutidx[:-1],axis=0)/lens[:,None]
    out[idx_rows] = mean_vals
    return out

样品运行-

In [220]: n,d,k = 10000,100,100
     ...: np.random.seed(0)
     ...: points = np.random.rand(n,d)
     ...: label = np.random.randint(0,k,(n))

In [221]: out0 = np.array([points[label==k_i].mean(axis = 0) for k_i in range(k)])

In [222]: np.allclose(matmul(points, label),out0)
Out[222]: True

In [223]: np.allclose(add_reduceat(points, label),out0)
Out[223]: True

如何在numpy

1 个答案: