Python和超级终端中RS232串行对话中的ASCII到十六进制转换

我正在使用Python通过RS232串行对话从仪器进行数据传输。数据是ASCII格式的。例如，这是我使用Python从仪器中获得的ASCII格式：

'\x02S20390908127F010102F3004000900300000000000000000000000000000300A500000000000202020202020206070505050505050000000707070707070707020700000000000000000000000000000000000000008000800089237A0715047E000000000000000000000000000000000000000000000000000000000000000000000000005350374646303030\x1732'

使用超级终端（Docklight）进行的相同数据传输给了我这个，下面是我在超级终端中得到的十六进制表示：

02 53 31 38 33 38 30 39 30 38 31 32 37 46 30 31 30 31 30 32 46 33 30 30 34 30 30 30 39 30 30 33 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 33 30 30 41 35 30 30 30 30 30 30 30 30 30 30 30 32 30 32 30 32 30 32 30 32 30 32 30 32 30 36 30 37 30 35 30 35 30 35 30 35 30 35 30 35 30 30 30 30 30 30 30 37 30 37 30 37 30 37 30 37 30 37 30 37 30 37 30 32 30 37 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 38 30 30 30 38 30 30 30 38 39 32 33 37 41 30 37 31 35 30 34 37 45 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 35 33 35 30 33 37 34 36 34 36 33 30 33 30 33 30 17 33 38 

---

我的问题是，如何使用Python从我得到的ascii数据中获得相同的十六进制表示。说到这类事情，我真是个笨蛋。提前谢谢。

说明：使用

ord

将每个字符转换为字节，然后格式化为两位十六进制数并将其输入列表。然后使用空格作为分隔符加入列表

>>> ll = ["%02X" % (ord(x)) for x in '\x02S20390908127F010102F3004000900300000000000000000000000000000300A500000000000202020202020206070505050505050000000707070707070707020700000000000000000000000000000000000000008000800089237A0715047E000000000000000000000000000000000000000000000000000000000000000000000000005350374646303030\x1732']
>>> ll           #as a list
['02', '53', '32', '30', '33', '39', '30', '39', '30', '38', '31', '32', '37', '46', '30', '31', '30', '31', '30', '32', '46', '33', '30', '30', '34', '30', '30', '30', '39', '30', '30', '33', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '33', '30', '30', '41', '35', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '32', '30', '32', '30', '32', '30', '32', '30', '32', '30', '32', '30', '32', '30', '36', '30', '37', '30', '35', '30', '35', '30', '35', '30', '35', '30', '35', '30', '35', '30', '30', '30', '30', '30', '30', '30', '37', '30', '37', '30', '37', '30', '37', '30', '37', '30', '37', '30', '37', '30', '37', '30', '32', '30', '37', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '38', '30', '30', '30', '38', '30', '30', '30', '38', '39', '32', '33', '37', '41', '30', '37', '31', '35', '30', '34', '37', '45', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '30', '35', '33', '35', '30', '33', '37', '34', '36', '34', '36', '33', '30', '33', '30', '33', '30', '17', '33', '32']
>>> ' '.join(ll) # as a string
'02 53 32 30 33 39 30 39 30 38 31 32 37 46 30 31 30 31 30 32 46 33 30 30 34 30 30 30 39 30 30 33 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 33 30 30 41 35 30 30 30 30 30 30 30 30 30 30 30 32 30 32 30 32 30 32 30 32 30 32 30 32 30 36 30 37 30 35 30 35 30 35 30 35 30 35 30 35 30 30 30 30 30 30 30 37 30 37 30 37 30 37 30 37 30 37 30 37 30 37 30 32 30 37 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 38 30 30 30 38 30 30 30 38 39 32 33 37 41 30 37 31 35 30 34 37 45 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 35 33 35 30 33 37 34 36 34 36 33 30 33 30 33 30 17 33 32'

---

这似乎是最简单的方法

ascstr = 'x\02S20..."  #your string

#I think you need to get rid of the first 4 characters '\x02' looks like
#it is hex
ascstr = ascstr[4:] 
for c in mystring:
   hexstr = hexstr + hex(ord(c))[2:] + ' '

---

两条被引用的线应该代表相同的信号还是仅仅是相似的信号？它们应该是相同的。我在docklight hyperterminal中也检查了ascii格式的数据，结果是一样的，但我无法让它们匹配。我发现四个不同点：我的开头是

02533230339

，而你的开头是

025331383338

，我的结尾是

32

，而你的结尾是

38

。因为只有四个术语不同，所以很难看出错误是什么。谢谢你的回答，但是你能解释一下这是怎么回事吗？我猜您正在将每个字符转换为其序号，但如何获得我需要的格式。我将如何从这里获得十六进制表示。抱歉，我对所有这些编程都是新手。@AasamTasaddaq抱歉，错过了十六进制要求。。。它现在就在那里：）我在序数上使用了hex（）函数，效果很好。非常感谢。这不会重现OP的结果，但我认为这没关系，因为OP的结果实际上不是相同的信号，不管他说了什么。你不能像那样从字符串中删除随机字符！