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我想检测前额区域以进行研究。

我尝试将我的 roi 的宽度作为左眼最左边点和右眼最右边点之间的距离。就宽度而言，这似乎可行，但是当涉及高度时，我需要一个适用于所有人的解决方案。请注意，我不希望头发出现在我的投资回报率中。我有一个高度的实现，但它并不适用于所有人。

x = int((shape[21][0]+shape[22][0])/2)
y = int((shape[21][1]+shape[22][1])/2)
w = abs(shape[39][0]-shape[42][0])
h = abs(y-shape[29][1])

y_mid = int((y + (y - h)) / 2)
roi = image[y-h:y_mid,int(x-w/2):int(x+w/2)]

if(roi.shape[0] !=0 and roi.shape[1] != 0):
    roi = imutils.resize(roi, width=250, inter=cv2.INTER_CUBIC)
    cv2.imshow('ROI', roi)

cv2.rectangle(image,(int(x-w/2), int(y-h)), (int(x + w/2), y_mid), (0, 0, 255), 1)
cv2.imshow('Image', image)
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1 回答 1

2

检测额头的代码：

import face_recognition
import cv2
import PIL.Image
import PIL.ImageDraw
import os
#image=cv2.imread(img_path)
unknown_image = face_recognition.load_image_file("16.png")
face_locations = face_recognition.face_locations(unknown_image) # detects all the faces in image
t = len(face_locations)
print(len(face_locations))
print(face_locations)
face_landmarks_list = face_recognition.face_landmarks(unknown_image)
# Drawing rectangles over the faces
pil_image = PIL.Image.fromarray(unknown_image)
for face_location in face_locations:
    #print(face_location)
    top,right,bottom,left =face_location
    draw_shape = PIL.ImageDraw.Draw(pil_image)
    im = PIL.Image.open("16.png")
    #bottom=34
    k = face_landmarks_list[0]['right_eyebrow']
    bottom= face_landmarks_list[0]['right_eyebrow'][0][1]
    for k1 in k :   
        if(bottom>k1[1]):
            bottom=k1[1]
    k = face_landmarks_list[0]['left_eyebrow']
    lbottom= face_landmarks_list[0]['left_eyebrow'][0][1]
    for k1 in k :   
        if(lbottom>k1[1]):
            lbottom=k1[1]
    bottom=min(bottom,lbottom)
    print(bottom)
    im = im.crop((left, top, right, bottom))
    im.save("m2.jpg")    
    draw_shape.rectangle([left, top, right, bottom],outline="blue")
于 2020-08-09T14:36:26.667 回答