selected_combinations = [list(red_ball_combination) + [blue_ball] for red_ball_combination in itertools.combinations(red_balls, 6) if not set([4, 8, 17, 20, 22, 27]).intersection(set(red_ball_combination)) for blue_ball in blue_balls]把这句更改为没有相同的数据才保存

可以使用一个集合来保存已经生成的组合，判断新的组合是否重复即可。修改后的代码如下：

import itertools

red_balls = list(range(1, 34))
blue_balls = list(range(1, 17))
n = 0

selected_combinations = []
generated_combinations = set()

for red_ball_combination in itertools.combinations(red_balls, 6):
    if set([4, 8, 17, 20, 22, 27]).intersection(set(red_ball_combination)):
        continue
    
    for blue_ball in blue_balls:
        combination = list(red_ball_combination) + [blue_ball]
        if tuple(combination) not in generated_combinations:
            selected_combinations.append(combination)
            generated_combinations.add(tuple(combination))

for i, combination in enumerate(selected_combinations):
    hongqiu = combination[:6]
    lanqiu = combination[-1]
    
    max_consecutive = 1
    zdxl = 1
    for j in range(len(hongqiu) - 1):
        if hongqiu[j + 1] == hongqiu[j] + 1:
            zdxl += 1
            if zdxl > max_consecutive:
                max_consecutive = zdxl
        else:
            zdxl = 1
    
    if max_consecutive <= 3:
        n += 1
        print(f"红球: {hongqiu}, 篮球: {lanqiu}, 第{n}组")

这样修改后，可以确保不会有重复的组合出现。