如果一个列表中的字符串没有包含在另一个字符串中，如何从列表中删除这些字符串并将其放入公式中？

def remove_strings_not_contained(strings):
    formula = []  # 存放删除的字符串的公式
    i = 0
    while i < len(strings):
        current_string = strings[i]
        contained = False
        for other_string in strings:
            if current_string != other_string and current_string in other_string:
                contained = True
                break
        if not contained:
            formula.append(current_string)
            strings.remove(current_string)
        else:
            i += 1
    return formula

# 示例用法
strings = ["abc", "def", "ab", "cd", "ef"]
formula = remove_strings_not_contained(strings)
print("删除的字符串公式:", formula)
print("剩余的字符串列表:", strings)

删除的字符串公式: ['abc', 'def']
剩余的字符串列表: ['ab', 'cd', 'ef']