问在python列表中筛选出重复的字符串算术操作的最佳方法？

EN

import re

eqs =  ['2*3=6', '3*2=6', '1+10/2=6', '2/1*3=6', '2*3/1=6', '3/1*2=6', '3*2/1=6', '10/2+1=6']

by_digits_used = {}

remaining_digits = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

for eq in eqs:
    used_digits = [int(f) for f in re.findall(r'\d*', eq) if f != ""]  # Find all digits in the equations
    used_digits.sort()                                                 # Ensure they're always in the same order
    used_digits = tuple(used_digits)                                   # Convert to tuple

    # Remove digits that were used
    for digit in used_digits:
        if digit in remaining_digits:
            remaining_digits.remove(digit)

    # Store equation by the digits used
    by_digits_used[used_digits] = by_digits_used.get(used_digits, []) + [eq]

print(by_digits_used)
print(remaining_digits)

# Equations by digits used
{(2, 3, 6): ['2*3=6', '3*2=6'], (1, 2, 6, 10): ['1+10/2=6', '10/2+1=6'], (1, 2, 3, 6): ['2/1*3=6', '2*3/1=6', '3/1*2=6', '3*2/1=6']}
# Unused digits
{0, 4, 5, 7, 8, 9}

import re
from itertools import chain

ls = ['2*3=6', '3*2=6', '1+10/2=6.0', '2/1*3=6', '2*3/1=6', '3/1*2=6', '3*2/1=6', '10/2+1=6']

exp = '10/2+1=6'

sorted_nums = [sorted([nums for nums  in re.findall("(\d+)", exp )]) for exp in ls]

unused_nums = set(map(str, range(1,10)))-set(chain(*sorted_nums))

nums_in_exps = {}
for nums, exp in zip(sorted_nums, ls):
    key = "-".join(nums)
    if key not in nums_in_exps:
        nums_in_exps[key] = []

    nums_in_exps[key] += [exp]

print("expression with nums")
print(nums_in_exps)

print("unused numbers")
print(unused_nums)

expression with nums
{'2-3-6': ['2*3=6', '3*2=6'], '0-1-10-2-6': ['1+10/2=6.0'], '1-2-3-6': ['2/1*3=6', '2*3/1=6', '3/1*2=6', '3*2/1=6'], '1-10-2-6': ['10/2+1=6']}
unused numbers
{'8', '4', '9', '7', '5'}

import re
import string

s = ['2*3=6', '3*2=6', '1+10/2=6.0', '2/1*3=6', '2*3/1=6', '3/1*2=6', '3*2/1=6', '10/2+1=6']
s = [i.replace('.0', '') for i in s]  # remove floats

d = {frozenset(re.split('[\+\-\*\/=]', i)): i for i in s}
print('Unique equations: ' , *d.values())
print('Unused numbers: ', set(string.digits) - frozenset.union(*d.keys()))

Unique equations:  3*2=6 10/2+1=6 3*2/1=6
Unused numbers:  {'4', '0', '5', '9', '7', '8'}