Question

我使用的是我通过Python3 socket图书馆连接的LabVIEW模拟器

import socket
client = socket.socket(socket.AF_INET,socket.SOCK_STREAM)
ip=socket.gethostbyname("127.0.0.1")
port=1234
address=(ip,port)
client.connect(address)
while True:
    data = client.recv(1024)
    print(data)
    client.close()

数据输出为：

B＆＃39;＆安培;＆安培; \ r \ n0108500.00 \ r \ n01410.000000 \ r \ n01420.000000 \ r \ n01430.000000 \ r \ n01440.000000 \ r \ n01450.000000 \ r \ n0146200.000000 \ r \ n0147100.000000 \ r \ n !! \ r \ n＆安培;＆安培; \ r \ n0108500.00 \ r \ n01410.000000 \ r \ n01420.000000 \ r \ n01430.000000 \ r \ n01440 0.000000 \ r \ n01450.000000 \ r \ n0146200.000000 \ r \ n0147100.000000 \ r \ n !! \ r \ n＆＃39;

模拟器输出两个200和100的常量值，我对应\r\n0146200.000000\r\n0147100.000000\ 0146 and 0147是来自模拟器的标签ID。什么是解析这些数据的最佳方法（让我们假设这个问题我只想要tagID和最后两个的值）？我试图输出它：

print(data.decode("utf-8", "strict"))

输出：

＆安培;＆安培; 0108500.00 01410.000000 01420.000000 01430.000000 01440.000000 01450.000000 0146200.000000 0147100.000000 !! ＆安培;＆安培; 0108500.00 01410.000000 01420.000000 01430.000000 01440.000000 01450.000000 0146200.000000 0147100.000000 !!

Answer 1

使用正则表达式提取键和值：

import re

results = [
    dict(re.findall(r"(\d{4})([\d.]+)", string))
    for string in filter(
        bool,  # non-empty
        data.decode("utf8", "replace").split("&&"),
    )
]

您的数据结果：

[{'0108': '500.00',
  '0141': '0.000000',
  '0142': '0.000000',
  '0143': '0.000000',
  '0144': '0.000000',
  '0145': '0.000000',
  '0146': '200.000000',
  '0147': '100.000000'},
 {'0108': '500.00',
  '0141': '0.000000',
  '0142': '0.000000',
  '0143': '0.000000',
  '0144': '0.000000',
  '0145': '0.000000',
  '0146': '200.000000',
  '0147': '100.000000'}]

解码使用Python

1 个答案: