把以下matlab代码转换为python代码（clear;clc; global_coords = [10, 10; 20, 20; 40, 20; 30, 10];%%%%测量时候桥梁坐标 juli_qiaoche =[2,2,2,2];%%%%测量时候车辆中心距离距离桥梁距离 % 转换为局部坐标 point_bridge1 = zuobiao_qiao( global_coords,juli_qiaoche )）

以下是将 MATLAB 代码转换为 Python 代码后的结果：

import numpy as np

def zuobiao_qiao(global_coords, juli_qiaoche):
    local_coords = np.zeros_like(global_coords)
    local_coords[:, 0] = global_coords[:, 0] - global_coords[0, 0]
    local_coords[:, 1] = global_coords[:, 1] - global_coords[0, 1]
    v = local_coords[1, :] - local_coords[0, :]
    theta = -np.arctan2(v[1], v[0])
    R = np.array([[np.cos(theta), -np.sin(theta)], [np.sin(theta), np.cos(theta)]])
    rotated_point = np.dot(local_coords, R)
    point_bridge1 = np.zeros((4, 2))
    point_bridge1[:, 0] = rotated_point[:, 0]
    point_bridge1[0, 1] = rotated_point[0, 1] - juli_qiaoche[0]
    point_bridge1[1, 1] = rotated_point[1, 1] + juli_qiaoche[1]
    point_bridge1[2, 1] = rotated_point[2, 1] + juli_qiaoche[2]
    point_bridge1[3, 1] = rotated_point[3, 1] - juli_qiaoche[3]
    point_bridge1[:, 1] = point_bridge1[:, 1] - point_bridge1[0, 1]
    return point_bridge1

global_coords = np.array([[10, 10], [20, 20], [40, 20], [30, 10]])
juli_qiaoche = np.array([2, 2, 2, 2])
point_bridge1 = zuobiao_qiao(global_coords, juli_qiaoche)

print(point_bridge1)

请注意，在 Python 中，您需要从 NumPy 库中导入 `numpy`，并且在调用函数时需要将输入的数组转换为 NumPy 数组。此外，Matlab 中的 `clear` 和 `clc` 不需要在 Python 中使用。最后，您可以使用 `print(point_bridge1)` 打印输出 `point_bridge1`。