Question

我下面的Python代码非常慢，是否有可能用Numpy语句完全编写此部分？

getToken = () => {

    axios({
        method: 'get',
        url: '/csrf',
        timeout: 1000,
        headers: 'csrf-token'
    }).then(csrfToken => {
        console.log(csrfToken);
        this.setState({ accessGranted: true })
    }).catch(err => console.log(err));
}

首先，我创建一个带有零（m = len(self.man_list) s = len(self.newnew) self.zeroMatrix = np.zeros((m,s)) for k in range(m): a1 = self.man_list[k][2] b1 = self.man_list[k][0] a2 = self.man_list[k][3] b2 = self.man_list[k][1] for f, part in enumerate(self.extra_list): x1 = self.extra_list[f][0] y1 = self.extra_list[f][2] x2 = self.extra_list[f][1] y2 = self.extra_list[f][3] first = np.array((x1, y1)) second = np.array((x2, y2)) third = np.array((a1, b1)) forth = np.array((a2, b2)) dist1 = np.linalg.norm(first - third) dist2 = np.linalg.norm(second - forth) distance = (dist1 + dist2) self.zeroMatrix[k][f] = distance）的矩阵。

self.zeroMatrix和self.man_list包含直线的起点和终点。例如：

self.extra_list

我得到从第一个列表的每一行到另一列表的每一行的距离，然后将此距离值注册在self.man_list = [ [ [1,2], [3,4] ],...] self.extra_list = [ [ [11,30], [4, 10] ],...]中。

非常感谢您的回答！

Answer 1

您需要对呼叫进行引导：

man_list = np.array(self.man_list)
extra_list = np.array(self.extra_list)

然后创建所需的子矩阵：

first = extra_list[:, None, ::2]
second = extra_list[:, None, 1::2]
third = man_list[None, :, 2::-2]
fourth = man_list[None, :, 3::-2]

现在，在最后一个轴（轴2）上计算范数。

dist1 = np.linalg.norm(first - third, axis=2)
dist2 = np.linalg.norm(second - fourth, axis=2)
distance = (dist1 + dist2)

现在，您应该在distance中拥有所需的矩阵。

numpy语句，而不是双for循环

1 个答案: