Question

我有一个S19文件，如下所示：

S0030000FC
S30D0003C0000F0000000000000020
S3FD00000000782EFF1FB58E00003D2B00003D2B00003D2B00003D2B00003D2B00003D
S3ED000000F83D2B00003D2B00003D2B00003D2B00003D2B00003D2B00003D2B00003D
S31500000400FFFFFFFFFFFFFFFFFFFFFFFF7EF9FFFF7D
S3FD0000041010B5DFF828000468012147F22C10C4F20300016047F22010C4F2030000
S70500008EB4B8

我想分开前两个字符以及接下来的两个字符，依此类推......我希望它看起来如下所示（每行最后两个字符也要分开）：

S0, 03, 0000, FC
S3, 0D, 0003C000, 0F00000000000000, 20
S3, FD, 00000000, 782EFF1FB58E00003D2B00003D2B00003D2B00003D2B00003D2B0000, 3D
S3, ED, 000000F8, 3D2B00003D2B00003D2B00003D2B00003D2B00003D2B00003D2B0000, 3D
S3, 15, 00000400, FFFFFFFFFFFFFFFFFFFFFFFF7EF9FFFF, 7D
S3, FD, 00000410, 10B5DFF828000468012147F22C10C4F20300016047F22010C4F20300, 00
S7, 05, 00008EB4, B8

我怎样才能在Python中执行此操作？我有这样的事情：

 #!/usr/bin/python
 import string,os,sys,re,fileinput
 print "hi"
 inputfile = "k60.S19"
 outputfile = "k60_out.S19"

 # open the source file and read it
 fh = file(inputfile, 'r')
 subject = fh.read()
 fh.close()

 # create the pattern object. Note the "r". In case you're unfamiliar with Python
 # this is to set the string as raw so we don't have to escape our escape characters

 pattern2 = re.compile(r'S3')
 pattern3 = re.compile(r'S7')
 pattern1 = re.compile(r'S0')
 # do the replace
 result1 = pattern1.sub("S0, ", subject)
 result2 = pattern2.sub("S3, ", subject)
 result3 = pattern3.sub("S7, ", subject)

 # write the file
 f_out = file(outputfile, 'w')

 f_out.write(result1)
 f_out.write(result2)
 f_out.write(result3)
 f_out.close()

 #EoF

但是我不喜欢!!有人可以帮我解决如何使用正确的正则表达式吗？

Answer 1

尝试打包bincopy，也许你需要它。

bincopy - 将字符串解释为压缩二进制数据

管理传输二进制信息的各种文件格式（Motorola S-Record，Intel HEX和二进制文件）。

import bincopy
f = bincopy.BinFile()
f.add_srec_file("path/to/your/s19/flie.s19")
f.as_binary() # print s19 as binary

或者您可以轻松地使用open（）作为文件：

with open("path/to/your/s19/flie.s19") as s19:
    for line in s19:
        type = line[0:2]
        count = line[2:4]
        adress = line[4:12]
        data = line[12:-2]
        crc = line[-2:]
        print type + ", "+ count + ", " + adress + ", " + data + ", " + crc + "\n"

希望它有所帮助。 Motorola S-record file format

Answer 2

您可以使用回调函数替换re.sub：

#!/usr/bin/python
import re

data = r'''S0030000FC
S30D0003C0000F0000000000000020
S3FD00000000782EFF1FB58E00003D2B00003D2B00003D2B00003D2B00003D2B00003D
S3ED000000F83D2B00003D2B00003D2B00003D2B00003D2B00003D2B00003D2B00003D
S31500000400FFFFFFFFFFFFFFFFFFFFFFFF7EF9FFFF7D
S3FD0000041010B5DFF828000468012147F22C10C4F20300016047F22010C4F2030000
S70500008EB4B8'''

pattern = re.compile(r'^(..)(..)((?:.{4}){1,2})(.*)(?=..)', re.M)

def repl(m):
    repstr = ''
    for g in m.groups():
        if (g):
            repstr += g + ', '
    return repstr

print re.sub(pattern, repl, data)

然而，正如Mark Setchell所注意到的那样，切片可能是一种很好的方法。

Answer 3

我知道你在考虑使用Python和正则表达式，但这是为awk做的，以下内容可能会帮助你找到使用切片的方法：

awk '{r=length($0);print substr($0,1,2),substr($0,3,2),substr($0,5,8),substr($0,13,r-14),substr($0,r-1)}' OFS=, k60.s19

那说＆＃34; 获取变量r中行的长度，然后打印前两个字符，接下来的两个字符，接下来的8个字符等等......然后使用逗号作为字段分隔符＆＃34;。

EDITED

以下是一些可以帮助您入门的提示......

如果你想避免印刷第1行，你可以

awk 'FNR==1{next}  ...rest of awk script above ... '

如果您只想处理超过40个字符的行，可以执行

awk 'length($0)>40 {print}' yourfile

如果您只想处理第二个字段为＆＃34; xx＆＃34;的行，您可以

awk '$2 ~ "xx" {print}' yourfile

如何操作SREC文件

3 个答案: