Serial Communication ending between Python (OpenCV) and Arduino

Question

I tried to check if my OpenCV code is communicating with Arduino or not.

OpenCV code:

import numpy as np
import cv2
import serial
import time
face_cascade = cv2.CascadeClassifier('haarcascade_frontalface_default.xml')
cap = cv2.VideoCapture(0)
while 1:
    ret, img = cap.read()
    gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
    faces = face_cascade.detectMultiScale(gray, 1.3, 5)

    for (x, y, w, h) in faces:
       cv2.rectangle(img, (x, y), (x + w, y + h), (255, 0, 0), 2)
       detect=x
       print(detect)

       cv2.imshow('img', img)
       k = cv2.waitKey(30) & 0xff

       if 0 < detect < 100:
        ser = serial.Serial("COM1", 19200, timeout=5)
        time.sleep(2)
        ser.write("\x35")
        print "RECIEVED BACK:", repr(ser.read(5000))

       if k == 27:
           break

Arduino code:

int incomingByte = 0;   // for incoming serial data

void setup() {
        Serial.begin(19200);    
}

void loop() {

        // send data only when you receive data:
        if (Serial.available() > 0) {
                // read the incoming byte:
                incomingByte = Serial.read();

                // say what you got:
                Serial.print("I received: ");
                Serial.println(incomingByte, DEC);
        }
}

I get following 'detect' value:

'301
71
RECIEVED BACK: 'I received: 53\r\n'
299
301
301
302
302
301
303
300
306
72'

At detect=71, a signal is sent to Arduino, it returns a value, it works a bit after that, then all communication breaks down and I get the following errors:

Traceback (most recent call last):
  File "C:/Users/khan1/Desktop/python 
  project/tennis_ball_vid/tennis_vid.py", line 40, in <module>
    ser = serial.Serial('COM1', 19200,timeout=5)
  File "C:\Python27\lib\site-packages\serial\serialwin32.py", line 38, in 
 __init__
    SerialBase.__init__(self, *args, **kwargs)
  File "C:\Python27\lib\site-packages\serial\serialutil.py", line 282, in 
 __init__
    self.open()
  File "C:\Python27\lib\site-packages\serial\serialwin32.py", line 66, in 
  open
    raise SerialException("could not open port %r: %r" % (self.portstr, 
    ctypes.WinError()))
serial.serialutil.SerialException: could not open port 'COM1': 
WindowsError(5, 'Access is denied.')

Process finished with exit code 1

Here is my original post: Serial comunication between opencv (python) and arduino

How to keep the communication open?

Please fix your indentation. Where do you send a signal to the Arduino, and where does it get back a value? The line serial.Serial('COM1', 19200,timeout=5) creates a port object but doesn't send or receive anything. — Paul Cornelius
– Paul Cornelius, Commented May 8, 2017 at 20:01
You keep opening the serial but never close it. Why not just open it at start. — gre_gor
– gre_gor, Commented May 8, 2017 at 20:26
I have edited my post. Sorry for the miscommunication. @PaulCornelius — sayem48
– sayem48, Commented May 9, 2017 at 17:40
Seems moving ser port out of while loop did the trick! thanks @gre_gor — sayem48
– sayem48, Commented May 9, 2017 at 17:47

stovfl · Accepted Answer · 2017-05-12 10:18:51Z

1

Comment: I did not understand your question. Can you explain a bit

Your loop should look like this, for instance:

ser = serial.Serial("COM1", 19200, timeout=5)
time.sleep(2)
while True:
    ret, img = cap.read()
    # ... img processing

    for (x, y, w, h) in faces:
        # ... faces processing

       if 0 < detect < 100:
            print('ser.is_open=%s' % ser.is_open() )
            ser.write("\x35")
            print("RECIEVED BACK:", repr(ser.read(5000)) )

Question: How to keep the communication open?

Move this lines of code outside the while ... loop

ser = serial.Serial('COM1', 19200,timeout=5)
time.sleep(6)
print(ser)

edited May 12, 2017 at 10:18

answered May 9, 2017 at 9:19

stovfl

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4 Comments

stovfl Over a year ago

The port keeps open until you call ser.close(). What gives print(ser.is_open) inside if 0<detect<100?

sayem48 Over a year ago

I did not understand your question. Can you explain a bit?

sayem48 Over a year ago

why do I need (ser.is_open)? I am new to this.

stovfl Over a year ago

@sayem48: You don't need it. It's only for tracing your code. It seems we have overlapped comunication. Does it work now?

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