return mag, deg F1 = getVector(100, 30) # 大小 100N, 方向 30˚ F2 = getVector(120, 60) # 大小 120N, 方向 60˚ Fsum = F1 + F2 magn, angle = getMagDeg(Fsum) print("大小: ", magn) print("方向: ", angle) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14...
A-1, 1)dihedral = np.einsum('ijk,ijk->ij', vec_01, vec_12)[...,None]# (B, A-3, 1), (B, A-2, 1), (B, A-1, 1)returnnp.arccos(dihedral[:, :-2, :]), angle, disdefget
下面是围绕z轴旋转的示例。 defrotate_z(points,angle):# 将角度转换为弧度theta=np.radians(angle)cos_theta=np.cos(theta)sin_theta=np.sin(theta)# 构造z轴旋转矩阵rotation_matrix=np.array([[cos_theta,-sin_theta,0],[sin_theta,cos_theta,0],[0,0,1]])# 旋转操作rotated_points=points.dot(r...
Win1 = vectorSumPbMatrix[0][n]ifnumpy.abs(Win1) >0.:forminrange(0,16) : vectorSumPbMatrix[m][n] = vectorSumPbMatrix[m][n] * numpy.conj(Win1)# ... don't bother normalizing, as this happens at the very end anywayprint"\nafter rotation:"printnumpy.array_str( numpy.angle(vecto...
arccos(vector_angle_cos(u, v)) elif direction is True: return np.arctan2(v[1], v[0]) - np.arctan2(u[1], u[0]) elif direction is False: return np.arctan2(u[1], u[0]) - np.arctan2(v[1], v[0]) else: axis1 = normalize(u) axis2 = normalize(np.cross(u, v)) ...
v = matrix(v1).T# make it a column vectorv => matrix([[0], [1], [2], [3], [4]]) M * M => matrix([[300,310,320,330,340], [1300,1360,1420,1480,1540], [2300,2410,2520,2630,2740], [3300,3460,3620,3780,3940], ...
import numpy as npfrom PIL import Image # 图像缩放、旋转、翻转、裁剪img = np.array(Image.open('image.png'))img_resize = np.array(Image.fromarray(img).resize((width, height)))img_rotate = np.array(Image.fromarray(img).rotate(angle))img_flip = np.array(Image.fromarray(img).transpose(...
angle = np.arccos(normal_vector[0] / norm_length) 在这个例子中,我们计算了法向量与x轴的夹角。如果希望计算法向量与其他轴的夹角,只需要相应地修改代码。 我们可以将弧度转换为角度,以便更直观地理解夹角的大小。可以使用以下代码实现: degree = np.degrees(angle) 在这个例子中,我们将弧度转换为角度。如果...
sin(angle) print(sin_value) # Output: 1.0 其他类似概念 其他三角函数如 cos, tan, arcsin, arccos, arctan 等。 详细区别 numpy.cos 计算余弦值 numpy.tan 计算正切值 官方链接 numpy.org/doc/stable/re (其余函数依此类推,每个函数的说明按照相同的格式展示,简要介绍存在的特殊意义、原理、使用场景、...
half a barb. Mag should be the magnitude of a vector (ie. >= 0). This returns a tuple of: (*number of flags*, *number of barbs*, *half_flag*, *empty_flag*) *half_flag* is a boolean whether half of a barb is needed, ...