ARDUINO / PYTHON / SERIAL

Question

我需要一点帮助才能使我的代码正常工作。 以下将是 arduino 和 python 代码。

我需要发生的是，当我戴上脸时，绿色 LED 亮起，表示它已释放，如果出现未注册的脸。 红色 LED 将亮起。 但我认为我的代码不正确。

PYTHON 代码

import face_recognition as fr
import os
import serial

porta_serial = serial.Serial('COM3',9600)

encoders = []
nomes = []

def criarEncoders():
    lista = os.listdir('Pessoas')
    for arquivo in lista:
        imAtual = fr.load_image_file(f'Pessoas/{arquivo}')
        imAtual = cv2.cvtColor(imAtual,cv2.COLOR_BGR2RGB)
        encoders.append(fr.face_encodings(imAtual)[0])
        nomes.append(os.path.splitext(arquivo)[0])

def compararWebcam():
    video = cv2.VideoCapture(0)

    while True:
        check,img = video.read()
        imgP = cv2.resize(img, (0, 0), None, 0.25, 0.25)

        imgP = cv2.cvtColor(imgP,cv2.COLOR_BGR2RGB)

        try:
            faceLoc = fr.face_locations(imgP)[0]
        except:
            faceLoc = []

        if faceLoc:
            y1,x2,y2,x1 = faceLoc
            y1, x2, y2, x1 = y1*4, x2*4, y2*4, x1*4
            cv2.rectangle(img,(x1,y1),(x2,y2),(0,255,0),2)  # criando o retângulo em volta da face na webcam com a cor verde
            porta_serial.write(b'o')
            encodeImg = fr.face_encodings(imgP)[0]

            for id,enc in enumerate(encoders):             # loop para comparar imagem com o banco de dados
                comp = fr.compare_faces([encodeImg],enc)
                if comp[0]:
                    cv2.rectangle(img,(x1,y2-35),(x2,y2),(0,255,0),-1)
                    cv2.putText(img,nomes[id],(x1+6,y2-6),cv2.FONT_HERSHEY_COMPLEX,1,(255,255,255),2)
                    porta_serial.write(b'f')
                    print(nomes[id])

        cv2.imshow('Webcam', img)
        cv2.waitKey(1)

criarEncoders()
compararWebcam()

ARDUINO 代码

int led_liberado = 8;
int led_nao_liberado = 7;
char var;

void setup() {
  Serial.begin(9600);
  pinMode(led_liberado, OUTPUT);
  pinMode(led_nao_liberado, OUTPUT);
  digitalWrite(led_nao_liberado, HIGH);
}

void loop() {
  while (Serial.available() > 0) {
    if (var=='o') {
      digitalWrite(led_liberado, HIGH);
      digitalWrite(led_nao_liberado, LOW);
    } else if(var=='f') {
      digitalWrite(led_nao_liberado,HIGH);
      digitalWrite(led_liberado, LOW);
    } else if (var=='q') {
      digitalWrite(led_nao_liberado, HIGH);
      digitalWrite(led_liberado, LOW);
    }
  }
}

Answer 1

我敢冒险这与这条线有关。

y1,x2,y2,x1 = faceLoc
y1, x2, y2, x1 = y1*4, x2*4, y2*4, x1*4
#If (x1,y1),(x2,y2) indicate some coordinate of where a face is which seems like it, multiplying them by a scalar loses that information?
#Consider your ROI is at (2,2) and your function grids it at (1,1),(3,3).
#Your code would then change the rectangle to (4,4),(12,12) 
#which doesnt even contain your point of interest. Meaning...no face? 

当然这只是推测。 您可能仍需要调试代码的各个方面。 1) Arduino 命令 2) Python 到 Arduino 接口 3) 人脸识别 ZC1C425268E68385D14AB5074C17A 的东西9