Python经典面试题与参考答案集锦

 list_1 = [10, 20, 30, 40, 50] # 不能对一个列表同时进行 遍历 和 增删元素 for element in list_1: print(element) if element == 30 or element == 40: list_1.remove(element) print(list_1)

 # 改进 temp_list = list() # 先遍历列表，记录下需要增删的元素 for element in list_1: print(element) if element == 30 or element == 40: temp_list.append(element) # 遍历完列表，再对需要增删的内容进行逐一处理 for element in temp_list: list_1.remove(element) print(list_1) 

 # 题目1 # 按照score的值进行排序。 list_a = [ {"name": "p1", "score": 100}, {"name": "p2", "score": 10}, {"name": "p3", "score": 30}, {"name": "p4", "score": 40}, {"name": "p5", "score": 60}, ] print(sorted(list_a, key=lambda a: a["score"], reverse=False)) 

 # 题目2 # 一行代码 通过filter 和 lambda 函数输出以下列表索引为奇数的对应的元素 list_b = [12, 232, 22, 2, 2, 3, 22, 22, 32] 

 new_list=[x[1]   for  x  in  fliter(lambdy x:x[0]%2==1,enumerate(list_b))] 

 for index, element in enumerate(list_b): print("") if index % 2 == 0 else print(element) 

 loneSum1 = (1, 2, 3) loneSum2 = (3, 2, 3) loneSum3 = (3, 3, 3) counter = int() for element in loneSum1: if loneSum1.count(element) == 1: counter += element print(counter) 

 def main(source_str): reverse_str = source_str[::-1] for i in range(len(source_str)): if reverse_str[:(i+1)] != source_str[:(i+1)]: return source_str[:i] if reverse_str == source_str: return source_str if __name__ == '__main__': print(main(mirrorEnds)) 

 num_list = [1, 3, 5, 7, 9, 11, 13, 15] M = 17 for i in range(len(num_list)): for j in range(i, len(num_list)): for k in range(j, len(num_list)): if num_list[i] + num_list[j] + num_list[k] == M: print((num_list[i], num_list[j], num_list[k])) 

 lass Parent(object): x = 1 class Child1(Parent): pass class Child2(Parent): pass print(Parent.x, Child1.x, Child2.x) Child1.x = 2 print(Parent.x, Child1.x, Child2.x) Parent.x = 3 print(Parent.x, Child1.x, Child2.x) # 提示：print(Child1.__mro__) #答案：1，1，1  1，2，1  3，2，3 

 class Counter(): def __init__(self, count=0): self.count = count def main(): c = Counter() c.name = "zhangsan" times = 0 for i in range(10): increment(c, times) print("count is {0}".format(c.count)) print("times is {0}".format(times)) def increment(c, times): times += 1 c.count += 1 if __name__ == "__main__": main() #答案： # count is 10 # times is 0 

 class Count(object): def __init__(self, count=0): self.count = count def main(): c = Count() n = 1 m(c, n) print("count is {0}".format(c.count)) print("n is {0}".format(n)) def m(c, n): c = Count(5) n = 3 if __name__ == '__main__': main() #答案： # count is 0 # n is 1 

 def outFun(a): def inFun(x): return a * x return inFun flist1 = [outFun(a) for a in range(3)] for func in flist1: print(func(1)) # fist2 = [lambda x:a*x for a in range(3)] flist2 = [lambda x:a*x for a in range(4)] for func in flist2: print(func(2)) # 答案： 0 1 2, 6 6 6 6 

 s = [{1: "a"}, {3: "c"}, {2: "b"}]

 print(sorted(s, key = lambda x: tuple(x.values())[0]))

 print(sorted(s, key = lambda x: list(x.values())[0]))

 sorted(iterable[, cmp[, key[, reverse]]])

 def sor(): li=[{1:"b"},{3:"c"},{2:"a"}] for i in range(len(li)): for j in range(len(li) - i - 1): v1 = [value for value in list(li[j].values())][0] v2 = [value for value in list(li[j+1].values())][0] print(v1,v2) if v1 > v2: li[j],li[j+1] = li[j+1],li[j] return li ret = sor() print(ret) 

 def bin_search(data_set, value): low = 0; high = len(data_set) - 1 while True: mid = (low + high) // 2 if data_set[mid] > value: high = mid - 1 elif data_set[mid] = 0: a -= 1 while data_set[b] == value and b 

 print({key: value for key, value in zip([i for i in a if i % 2 == 0], [i for i in a if i % 2 != 0])})

 def factorial(num): counter = 1 for i in range(1, num+1): counter *= i return counter print(factorial(5)) 

 def search(a): # 省略校验逻辑 # 生成包含 element 和 count 数据的集合 data_set = {(element, count) for element, count in zip(a, [a.count(i) for i in a])} print(data_set) # 转换成列表类型，进行倒序排序 data_list = list(data_set) new_data_list = sorted(data_list, key=lambda x: x[1], reverse=True) # 取出前四名 new_data_list = new_data_list[0:4] # 拼接字符串 return "".join([i[0] for i in new_data_list]) print(search(a)) 

 a = 'hsfbsadgasdgvnhhhadhaskdhwqhcjasd' # 导入Counter模块 from collections import Counter # 创建对象 count = Counter(a) # 获取出现频率最高的四个字符 print(count.most_common(4)) # 输出：[('h', 7), ('s', 5), ('a', 5), ('d', 5)] 

 if other_allocated_amount: data_list.append({ "shop_id": 0, "shop_name": "其他", "destination": 2, "current_storage": 0, "demand_amount": 0, "allocated_amount": other_allocated_amount, "priority": 0, "negative_order": 0 }) # 按优先级排序 data_list = sorted(data_list, key=lambda d: d["priority"], reverse=True) shop_order = {"仓库": -999, "总部": 1, "刘园": 2, "侯台": 3, "咸水沽": 4, "华明": 5, "大港": 6, "杨村": 7, "新立": 8, "大寺": 9, "汉沽": 10, "沧州": 11, "静海": 13, "芦台": 14, "工农村": 15, "唐山": 16, "廊坊": 17, "哈尔滨": 18, "西青道": 19, "双鸭山": 20, "承德": 21, "张胖子": 22, "固安": 23, "燕郊": 24, "胜芳": 25, "蓟县": 26, } data_list = sorted(data_list, key=lambda d: shop_order.get(d["shop_name"], 999))