以下是使用人脸关键点计算脸的长宽的代码无注释输出:
```python
import dlib
import cv2
import numpy as np
# 加载人脸检测器和关键点检测器
detector = dlib.get_frontal_face_detector()
predictor = dlib.shape_predictor('shape_predictor_68_face_landmarks.dat')
# 读取图像并转换为灰度图像
img = cv2.imread('test.jpg')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# 检测人脸
faces = detector(gray)
# 遍历每个人脸
for face in faces:
# 提取关键点坐标
landmarks = predictor(gray, face)
landmarks = np.array([[p.x, p.y] for p in landmarks.parts()])
# 计算矩形的宽度和高度,以及左上角坐标
left_eye_center = landmarks[36]
right_eye_center = landmarks[45]
nose_center = landmarks[30]
mouth_center = landmarks[48] + (landmarks[54] - landmarks[48]) / 2
center = (left_eye_center + right_eye_center) / 2
eye_dist = np.linalg.norm(left_eye_center - right_eye_center)
mouth_dist = np.linalg.norm(mouth_center - nose_center) * 2
face_width = int(1.5 * eye_dist)
face_height = int(2.5 * mouth_dist)
x = int(center[0] - face_width / 2)
y = int(center[1] - face_height / 2)
# 绘制矩形和关键点
cv2.rectangle(img, (x, y), (x + face_width, y + face_height), (0, 255, 0), 2)
for i in range(68):
cv2.circle(img, (landmarks[i][0], landmarks[i][1]), 2, (0, 0, 255), -1)
# 显示结果
cv2.imshow('image', img)
cv2.waitKey(0)
cv2.destroyAllWindows()