找到人脸地标之间的欧几里德距离

Question

I've got multiple frames and I've detected the faces in each frame using Retinaface . I would like to keep track of the faces using their landmarks.

To find the similarity between 2 landmarks, I tried to calculate the Eucledian distance :

Input :

landmark_1 = [1828, 911], [1887, 913], [1841, 942], [1832, 974], [1876, 976]
landmark_2 = [1827, 928], [1887, 926], [1848, 963], [1836, 992], [1884, 990]

After referring other links, I wrote the below function, but the values produced are very high :

def euclidean_dist(vector_x, vector_y):
    vector_x, vector_y = np.array(vector_x), np.array(vector_y)
    if len(vector_x) != len(vector_y):
        raise Exception('Vectors must be same dimensions')
    ans = sum((vector_x[dim] - vector_y[dim]) ** 2 for dim in range(len(vector_x)))
    return np.sqrt(np.sum(ans**2))

Output :

euclidean_dist(landmark_1, landmark_2)
>> 1424.9424549784458

(Expecting some smaller value in this case)

I guess the code can only be used for an one dimensional vector, but I'm really stuck here. Any help would be really appreciated.

Answer 1

It looks like you're squaring the answer twice ( ans**2 ). But you can also simplify the function somewhat:

def euclidean_dist(vector_x, vector_y):
    vector_x, vector_y = np.array(vector_x), np.array(vector_y)
    return np.sqrt(np.sum((vector_x - vector_y)**2, axis=-1))

This will automatically raise an exception when the vectors are incompatible shapes.

EDIT: If you use axis=-1 it will sum over the last axis of the array, so you can use a 2-D array of vectors, for example.

Answer 2

You can use linalg.nor too.

def euclidean_dist(vector_x, vector_y):
    distances = np.linalg.norm(np.array(vector_x)-np.array(vector_y), axis=1)
    return distances.tolist()