@BoarGules 是正确的。这是我对这样的字符串进行解码的解决方案:
vt_100_mapping = {
'0x71': '─',
'0x74': '├',
'0x75': '┤',
'0x76': '┴',
'0x77': '┬',
'0x78': '│',
'0x6a': '┘',
'0x6b': '┐',
'0x6c': '┌',
'0x6d': '└',
'0x6e': '┼',
}
from itertools import groupby
def decode_vt_100(iterable, default_set='(B', alt_set='(0'):
for is_escape, group in groupby(iterable, lambda _: _ =='\x1b'):
if is_escape:
continue
characters = ''.join(group)
if characters.startswith(default_set):
yield characters[len(default_set):]
elif characters.startswith(alt_set):
for character in characters[len(alt_set):]:
yield vt_100_mapping[hex(ord(character))]
>>> print(''.join(decode_vt_100("\x1b(0l\x1b(BHeader")))
┌Header