-1

我想要做的是，一旦一个字符在 Python 中的字符串中出现多次，程序就会中断。

这是我得到的输出：

b'30,22,37,293,29,26,55,30D\r\n$GPGSV,3,1,11,01,04,250,,03,21,309,24,08,15,200,,10,20,157,*7B \r\n$GPGSV,3,2,11,14,49,062,37,16' b'8,W,001328.00,A,A*70\r\n9,26,55,359,30*75\r\n $GPGSV,3,3,11,27,40,170,38,31,20,022,14,32,42,094,35*4C\r\n$GPGLL,0226.72251,N,0' b'A*2E\r\n $GPGGA,001329.00,A,A*70\r\n$GPRMC,001329.00,A,0226.72248,N,07635.92031,W,0.008,,011217,,,A*64\r\n$GPVTG,,T,, M,0.008,N,'b'2.25,1.08,1.98*0D\r\n$GPG1329.00,0226.72248,N,07635.92031,W,1,08,1.08,1769.1,M,9.8,M,,*4C \r\n$GPGSA,A,3,16,32,26,31,14,22,27'

所以我想在第五个“b”出现后停下来。

4

2 回答 2

0

您可以尝试正则表达式并捕获 group()，

您是否正在寻找这样的东西：

import re
pattern=r"b'[0-9]+"

data="""b'30,22,37,293,29,26,55,30D\r\n$GPGSV,3,1,11,01,04,250,,03,21,309,24,08,15,200,,10,20,157,*7B\r\n$GPGSV,3,2,11,14,49,062,37,16'
b'8,W,001328.00,A,A*70\r\n9,26,55,359,30*75\r\n$GPGSV,3,3,11,27,40,170,38,31,20,022,14,32,42,094,35*4C\r\n$GPGLL,0226.72251,N,0'
b'A*2E\r\n$GPGGA,001329.00,A,A*70\r\n$GPRMC,001329.00,A,0226.72248,N,07635.92031,W,0.008,,011217,,,A*64\r\n$GPVTG,,T,,M,0.008,N,'
b'2.25,1.08,1.98*0D\r\n$GPG1329.00,0226.72248,N,07635.92031,W,1,08,1.08,1769.1,M,9.8,M,,*4C\r\n$GPGSA,A,3,16,32,26,31,14,22,27'"""

all_items=[]
track=[]
for i in data.split(','):
    pr=re.search(pattern,i)
    if len(track)==5:
        break
    else:
        if pr!=None:
            track.append(pr.group())
            all_items.append(pr.group())
        else:
            all_items.append(i)

print(all_items)

如果出现第二个“b”，让我们尝试代替 5 ：

all_items=[]
track=[]
for i in data.split(','):
    pr=re.search(pattern,i)
    if len(track)==2:
        break
    else:
        if pr!=None:
            track.append(pr.group())
            all_items.append(pr.group())
        else:
            all_items.append(i)

print(all_items)

输出：

["b'30", '22', '37', '293', '29', '26', '55', '30D\r\n$GPGSV', '3', '1', '11', '01', '04', '250', '', '03', '21', '309', '24', '08', '15', '200', '', '10', '20', '157', '*7B\r\n$GPGSV', '3', '2', '11', '14', '49', '062', '37', "b'8"]

您可以在那里替换 5 而不是 3。

于 2017-12-01T08:19:48.857 回答
0 
import re
from machine import UART
uart = UART(2, 9600)

pattern=r"b'[0-9]+"

while True:
  if uart.any():
    data=uart.read()
    all_items=[]
    track=[]
    for i in data.split(','):
        pr=re.search(pattern,i)
        if len(track)==5:
            break
        else:
            if pr!=None:
                track.append(pr.group())
                all_items.append(pr.group())
            else:
                all_items.append(i)

    print(all_items)



this would be the code apllied to my project, but I get this error:


Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<string>", line 12, in <module>`enter code here`
TypeError: can't convert 'str' object to bytes implicitly
于 2017-12-04T21:32:11.180 回答