python，字符串到十六进制数组

Question

快速提问，我想要这个（当然是Python3）：

frame = "AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA"

为此：

frame = [0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA, 0x5, 0x5, 0x5, 0xA, 0xA, 0xA]

Answer 1

我们可以试试这个：

frame = "AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA"
[int(character, 16) for character in frame]
# [10, 10, 10, 5, 5, 5, 10, 10, 10, 5, ... ]

正如@Reti43在a comment中所指出的那样，像0xA这样的十六进制值对应于像10这样的数字，这就是结果列表有整数的原因

如果OP需要十六进制，则可以使用以下字符串表示法。然后可以将该字符串传递给int以获得相应的整数

frame = "AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA555AAA"
[f"0x{character}" for character in frame]
# ['0xA', '0xA', '0xA', '0x5', '0x5', '0x5', '0xA', '0xA', '0xA', '0x5', ...]