将字符串转换为十六进制

Question

我有一个字符串文件，但其中的字符串表示十六进制值。例如，我的文件中有这样的字符串：

1091    A3B7    56FF    ...

我不想将它们用作字符串，而是用作十六进制值；然后将十六进制转换为int。

例如：

1091(in string)---> 1091(in hexa)---> 4241 # The int value of 1091 in hexa

所以我上网看了看。我尝试了很多不同的方法，比如：

1. https://stackoverflow.com/questions/12214801/print-a-string-as-hex-bytes
2. https://stackoverflow.com/questions/209513/convert-hex-string-to-int-in-python

但没有任何东西与我所需要的完全相符，或者根本不起作用。

这是我代码的一部分：

t = False
i = 0
while t != True and h != True or i <=100: # My way to look each string of my file
    file = row[1]
    read_byte = file[i]

    if read_byte == 'V': #V is the letter in my file which says that it s the data that I want then there is 2 kinds of channel 01 and 02 that interest me

        i=i+1
        a=i+2
        read_bytechannel = file[i:a]  #read 2 strings because the channel can be '01' or '02'

        if read_bytechannel == '01':
            print(read_bytechannel)
            i=i+1
            a=i+4
            read_bytetemp = file[i:a] # Reading 4 strings because the value is the int value of the two hexa.
            realintvalue= # (?????????convert the read_bytetemp string value into an hexa value, then into an int from the hexa)

            tempfinal = realintvalue/100 # I have to divide by 100 the int value to get the real temperature
            t = True # This condition just tell me that I already know the temporary

    i = i+1

这是我想读的那种文件：

@
I01010100B00725030178
V01109103
I02020100B00725030148
V0215AA5C
$
@

Answer 1

>>> int('1091', 16)
4241

int的第二个参数是解释第一个参数的基础。

Answer 2

您可以尝试这样的方法，带基数为16的int：

>>> my_string = "1091 A3B7 56FF"
>>> map(lambda x:int(x,16), my_string.split())
[4241, 41911, 22271]

或者你可以两者兼得：

>>> map(lambda x:[x,int(x,16)], my_string.split())
[['1091', 4241], ['A3B7', 41911], ['56FF', 22271]]

如果您对lambda和map不满意，可以使用列表理解。

>>> [[x,int(x,16)] for x in my_string.split()]
[['1091', 4241], ['A3B7', 41911], ['56FF', 22271]]