python - 尝试解决将字符串转换为列表中的整数时遇到不同的错误

Question

我有以下列表和字符串:

befcodes = ["A1","A2","A3","A4","A5","A6","A7","A8","A9","10","11","12","13","14","15","16","17","18","19","20"]
telegram = "$00;02;A1;00000000*49"

现在，我的代码将字符串中的 A1更改为20次，如下所示:

allbefcodes = []
for i in befcodes:
    dif_tele = telegram.replace("A1", i)
    allbefcodes.append(dif_tele)
print (allbefcodes) 

输出以下列表:

['$00;02;A1;00000000*49', '$00;02;A2;00000000*49', '$00;02;A3;00000000*49', '$00;02;A4;00000000*49', '$00;02;A5;00000000*49', '$00;02;A6;00000000*49', '$00;02;A7;00000000*49', '$00;02;A8;00000000*49', '$00;02;A9;00000000*49', '$00;02;10;00000000*49', '$00;02;11;00000000*49', '$00;02;12;00000000*49', '$00;02;13;00000000*49', '$00;02;14;00000000*49', '$00;02;15;00000000*49', '$00;02;16;00000000*49', '$00;02;17;00000000*49', '$00;02;18;00000000*49', '$00;02;19;00000000*49', '$00;02;20;00000000*49']

现在，我想使用XOR运算符来获取每个电报的二进制校验和(= allbefcodes的每个字符)，我这样做是这样的:

result = functools.reduce(operator.xor,(ord(n) for n in (allbefcodes[0])[1:18]))  
print (f'{result:08b}')

这对于 allbefcodes[0]确实可以正常工作，我得到了输出 01001001。但是我现在想通过循环对 allbefcodes的所有字符执行此操作，在此我会遇到不同的错误.. 这是到目前为止我尝试过的操作:

for x in allbefcodes:
    result = functools.reduce(operator.xor,(ord(n) for n in (allbefcodes[x])[1:18]))  
    print (f'{result:08b}')

错误 TypeError: list indices must be integers or slices, not str发生。我试图这样解决:

for x in allbefcodes:
    allbefcodes = (int(a) for a in allbefcodes)
    result = functools.reduce(operator.xor,(ord(n) for n in (allbefcodes[x])[1:18]))  # "
    print (f'{result:08b}')

但是这里会出现错误 TypeError: 'generator' object is not subscriptable。所以我接下来尝试了这个:

allbefcodes = []
for i in befcodes:
    dif_tele = telegram.replace("A1", i)
    allbefcodes.append(dif_tele)
allbefcodes = (int(a) for a in allbefcodes)     #Tried to solve it with this line
print (allbefcodes)    

for x in allbefcodes:
    result = functools.reduce(operator.xor,(ord(n) for n in (allbefcodes[x])[1:18]))  # "(allbefcodes[0])" statt telegram
    print (f'{result:08b}')

但是现在出现了 ValueError: invalid literal for int() with base 10: '$00;02;A1;00000000*49'错误。进行 (float(int(a)) for a in allbefcodes)不会改变这一点。
我的最后一次尝试是将 (allbefcodes[x])更改为 (allbefcodes[0:])并保留 allbefcodes = (int(a) for a in allbefcodes) ，如下所示:

allbefcodes = []
for i in befcodes:
    dif_tele = telegram.replace("A1", i)
    allbefcodes.append(dif_tele)
print (allbefcodes)    

for x in allbefcodes:
    result = functools.reduce(operator.xor,(ord(n) for n in (allbefcodes[0:])[1:18]))  # "(allbefcodes[0])" statt telegram
    print (f'{result:08b}')

但是现在我得到了 TypeError: ord() expected a character, but string of length 21 found，这使我又回到了整数而不是字符串的问题，我已经尝试解决了……我真的不知道该怎么办了，非常感谢您的帮助!

Answer 1

我想到了!
我只需要像这样用(allbefcodes[0:])替换x:

allbefcodes = []
for i in befehlscodes:
    dif_tele = telegram.replace("A1", i)
    allbefcodes.append(dif_tele)
print (allbefcodes)
   
for x in allbefcodes:
    result = functools.reduce(operator.xor,(ord(n) for n in x[1:18]))