Python：将特定字符串与字符串组分开

Question

Im working on a GPS tracker. The tracker send some packets. each packet start with start bit(7878) and stop bit(0d0a). Some time i receive group of packets. I want to seperate those group of packets. I use below code to split the two char.

data='78780a1301000400020017e73c0d0a787825260f071a0b050ac70165eaf0089b626021052e09019428139100fa7108000403020004a4810d0a78780a1301000409020016e73c0d0a'
splt_data= [data[i:i+2] for i in range(0, len(data), 2)]

i want like below

78780a1301000400020017e73c0d0a
787825260f071a0b050ac70165eaf0089b626021052e09019428139100fa7108000403020004a4810d0a
78780a1301000409020016e73c0d0a

I can able split those packets if "0a or 0d or 78 or" not inbetween start bit and stop bit. if there is possiblity of getting start bit or stop bit itself how to seperate it. For example below is one of the packet, which contains "0a" inbetween start bit(7878) and stop bit(0d0a).

7878ka130100**0a**00020017e73c0d0a

Thanks

Answer 1

You can use regular expression. re.findall will return all the occurrences that match the pattern:

>>> data = '78780a1301000400020017e73c0d0a787825260f071a0b050ac70165eaf0089b626021052e09019428139100fa7108000403020004a4810d0a78780a1301000409020016e73c0d0a'
>>> import re
>>> re.findall('7878.*?0d0a', data)
['78780a1301000400020017e73c0d0a',
 '787825260f071a0b050ac70165eaf0089b626021052e09019428139100fa7108000403020004a4810d0a',
 '78780a1301000409020016e73c0d0a']

• 7878 , 0d0a matches 7878 , 0d0a literally.
• . matches any character.
• .* matches 0 or more any (.) characters (greedy) . Appending ? will make the match non-greedy.