server.py

Question

我正在研究使用Raspberry PI 3 B +的网络视频流解决方案，而低延迟是关键。

我使用的第一种方法是将标准输出从raspivid传递到netcat TCP流中：

a_i

这种方法的延迟很短，我对结果总体上满意。

但是，我需要在客户端进行一些图像处理。我所做的是尝试使用python复制上述方法。我在documentation of the 'picamera' Python module中找到了类似的解决方案：

在树莓上：

# On the Raspberry:
raspivid -w 640 -h 480 --nopreview -t 0 -o - | nc 192.168.64.104 5000

# On the client:
nc -l -p 5000 | mplayer -nolirc -fps 60 -cache 1024 -

在客户端上：

import io
import socket
import struct
import time
import picamera

# Connect a client socket to my_server:8000 (change my_server to the
# hostname of your server)
client_socket = socket.socket()
client_socket.connect(('my_server', 8000))

# Make a file-like object out of the connection
connection = client_socket.makefile('wb')
try:
    camera = picamera.PiCamera()
    camera.resolution = (640, 480)
    # Start a preview and let the camera warm up for 2 seconds
    camera.start_preview()
    time.sleep(2)

    # Note the start time and construct a stream to hold image data
    # temporarily (we could write it directly to connection but in this
    # case we want to find out the size of each capture first to keep
    # our protocol simple)
    start = time.time()
    stream = io.BytesIO()
    for foo in camera.capture_continuous(stream, 'jpeg'):
        # Write the length of the capture to the stream and flush to
        # ensure it actually gets sent
        connection.write(struct.pack('<L', stream.tell()))
        connection.flush()
        # Rewind the stream and send the image data over the wire
        stream.seek(0)
        connection.write(stream.read())
        # If we've been capturing for more than 30 seconds, quit
        if time.time() - start > 30:
            break
        # Reset the stream for the next capture
        stream.seek(0)
        stream.truncate()
    # Write a length of zero to the stream to signal we're done
    connection.write(struct.pack('<L', 0))
finally:
    connection.close()
    client_socket.close()

此方法的延迟要差得多，我正在尝试找出原因。与第一种方法一样，它使用TCP流从内存缓冲区发送帧。

目标只是尽可能快地准备好帧以在客户端上使用OpenCV进行处理。因此，如果有比上述方法更好的方法来实现这一目标，请与我分享。

Answer 1

这主要来自另一篇我现在找不到的帖子。但是我在那里修改了给定的代码。在这一帧上，平均每帧传输时间为0.35秒，与netcat相比仍然很差，但比您提到的顺序捕获代码要好一些。这个也使用套接字，但是代替图片，您处理视频帧：

server.py

import socket
import sys
import cv2
import pickle
import numpy as np
import struct ## new
import time

HOST='ip address'
PORT=8089

s=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
print ('Socket created')

s.bind((HOST,PORT))
print ('Socket bind complete')
s.listen(10)
print ('Socket now listening')

conn,addr=s.accept()

### new
counter=0
data = b''
payload_size = struct.calcsize("<L") 
while True:
    start=time.time()
    while len(data) < payload_size:
        data += conn.recv(8192)
    packed_msg_size = data[:payload_size]
    data = data[payload_size:]
    msg_size = struct.unpack("<L", packed_msg_size)[0]
    while len(data) < msg_size:
        data += conn.recv(8192)
    frame_data = data[:msg_size]
    data = data[msg_size:]
    ###

    frame=pickle.loads(frame_data)

    name='path/to/your/directory'+str(counter)+'.jpg'
    cv2.imwrite(name,frame)
    counter+=1
    end=time.time()
    print("rate is: " ,end-start)

=============

client.py

import cv2
import numpy as np
import socket
import sys
import pickle
import struct ### new code
#cap=cv2.VideoCapture(0)
cap=cv2.VideoWriter()
clientsocket=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
clientsocket.connect(('server ip address',8089))
while True:
    ret,frame=cap.read()

    data = pickle.dumps(frame) ### new code
    clientsocket.sendall(struct.pack("<L", len(data))+data) ### new code

=============

来自Raspberry PI的视频流-Python与raspivid + Netcat

1 个答案:

server.py

client.py