问如何从Python中的list生成所有可能的排列对？

EN

import itertools
input = [3, 8, 2]
final_list = ["{}-{}".format(*i) for i in itertools.permutations(input, 2)]

['3-8', '3-2', '8-3', '8-2', '2-3', '2-8']

final_list = list(itertools.chain(*[['-'.join(["{}"]*b).format(*i) for i in itertools.permutations(input, b)] for b in range(2, len(input)+1)]))

['3-8', '3-2', '8-3', '8-2', '2-3', '2-8', '3-8-2', '3-2-8', '8-3-2', '8-2-3', '2-3-8', '2-8-3']

import re
def tokenize(s):
   converter = {"-":lambda x, y:x-y, '+':lambda x, y:x+y}
   total = 0
   stack = re.findall('\d+|[\-\+]', s)
   operator = None
   for i in stack:
      if i.isdigit():
         if not operator:
            total += int(i)
         else:
            total = converter[operator](total, int(i))
            operator = None
      else:
          operator = i
   return total

new_list = set(list(itertools.chain(*list(itertools.chain(*[[[''.join([''.join('{}'+i) for i in b]+['{}']).format(*c) for b in itertools.permutations(['-', '+'], len(c)-1)] for c in itertools.permutations(input, x)] for x in range(2, len(input)+1)])))))
final_list = {tokenize(a):a for a in new_list}
new_final_list = [b for a, b in final_list.items()]

['3-2', '8-3', '8-2', '8-3+2', '8-2+3', '8+2', '8+3', '2-8', '3-8', '3+2-8', '2-3']

import itertools

result = ["{}-{}".format(*x) for x in itertools.permutations([3, 8, 2],2)]

print(result)

['3-8', '3-2', '8-3', '8-2', '2-3', '2-8']

from itertools import permutations

input = [3, 8, 2]
output = map('-'.join, permutations(map(str, input), 2))