1.如何实现对python列表去重，并保持原来顺序
li = [1,2,5,3,1,6,3,8,0,3,2,4] l = [] for i in li: if i not in l: l.append(i) print(l)
li = [1,2,5,3,1,6,3,8,0,3,2,4] l = list(set(li)) l.sort(key=lambda i:li.index(i)) print(l) # 先去重，再按照li的索引进行排序
li = [1,2,5,3,1,6,3,8,0,3,2,4] from collections import OrderedDict d = OrderedDict d = d.fromkeys(li) print(list(d.keys())) # 用有序字典进行去重排序
2.请给出二分查找的python示例代码
def search(li,i,start=0,end=None): if end == None:end = len(li) - 1 if start <= end: mid = (end - start) // 2 + start if i < li[mid]:end = mid - 1 elif i > li[mid]:start = mid + 1 else:return '在该列表的第%s个数'%(mid + 1) return search(li,i,start,end) else:return '该列表没有这个数' l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88] print(search(l,42))
def select(l,aim,start = 0,end = None): if end == None: end = len(l) if start > end: return '列表没有该元素' mid = (end - start) // 2 + start if aim > l[mid]: return select(l,aim,mid + 1,end) elif aim < l[mid]: return select(l,aim,start,mid - 1) else: return aim,mid l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88] print(select(l,42))
3.现有两元组(('a','b')),(('c','d')),请使用匿名函数生成列表[{'a':'c'},{'b':'d'}]
#方法一： print(list(map(lambda t:{t[0]:t[1]},zip(('a','b'),('c','d'))))) #方法二： test = lambda t1,t2:[{i:j} for i,j in zip(t1,t2)] print(test(('a','b'),('c','d')))
4.用python实现'九九乘法表'（用两种不同的方式实现）
#方法一： for j in range(1,10): i = 1 l = [] while i <= j: sum = i * j l.append('%d * %d = %d'%(i,j,sum)) i += 1 print(' '.join(l)) #方法二： j = 1 while j < 10: i = 1 s = '' while i <= j: sum = i * j s += '{} * {} = {} '.format(i,j,sum) i += 1 print(s) j += 1
5.三级菜单（可显示下级菜单，按q退出，按b返回）
menu = { '北京': { '海淀': { '五道口': { 'soho': {}, '网易': {}, 'google': {} }, '中关村': { '爱奇艺': {}, '汽车之家': {}, 'youku': {}, }, '上地': { '百度': {}, }, }, '昌平': { '沙河': { '老男孩': {}, '北航': {}, }, '天通苑': {}, '回龙观': {}, }, '朝阳': {}, '东城': {}, }, '上海': { '闵行': { "人民广场": { '炸鸡店': {} } }
l = [menu] while True: ret = l[-1] #取列表最后一个元素 for k in ret: print(k) choice = input('请选择地点（按B/b返回，按Q/q退出）：') if choice in ret and ret[choice]: l.append(ret[choice]) #添加到列表末尾 elif choice.upper() == 'B' and len(l) > 1: l.pop() #删除列表末尾的元素 elif choice.upper() == 'Q': break
def menu_tree(menu): while True: for k in menu: print(k) choice = input('请选择地点（按B/b返回，按Q/q退出）：') if choice.upper() == 'B' or choice.upper() == 'Q': return choice elif choice in menu and menu[choice]: ret = menu_tree(menu[choice]) if ret == 'Q': return 'q' menu_tree(menu)
6.哥德巴赫偶数拆分质数，请拆分100-120之间的哥德巴赫数。
def prime_num(n): m = n ** 0.5 i = 2 while i < m+1: if n % i == 0: return False i += 1 return True for i in range(100,121): a = 2 while a <= i/2: b = i - a ret1 = prime_num(a) ret2 = prime_num(b) if ret1 == True and ret2 == True: print('%d = %d + %d'%(i,a,b)) a += 1
7.设计一个计算器，计算'1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )'的结果
import re def main(express): express = express.replace(' ','') #对表达式去空格 while True: express_bracket = re.search('([^(]+?)',express) #循环寻找带括号的式子 if express_bracket: express_bracket = express_bracket.group() result = real_express(express_bracket) #调用real_express函数 express = express.replace(express_bracket,result,1) #把式子替换成计算结果 else:break return real_express(express) #调用real_express函数 def real_express(express): while True: real_exp = re.search('d+.?d*[*/]-?d+.?d*',express) #循环寻找乘法/除法式子 if real_exp: real_exp = real_exp.group() final_num = mul_div(real_exp) #调用mul_div函数 express = express.replace(real_exp,final_num,1) #把式子替换成计算结果 else:break express = deal_symbol(express) #对式子处理符号 num_l = re.findall('-?d+.?d*',express) #匹配式子的所有数字 sum = 0 for num in num_l: sum += float(num) #将所有数字相加 return str(sum) def mul_div(express): if '*' in express: #判断是否是乘法式子 a,b = express.split('*') return str(float(a) * float(b)) if '/' in express: #判断是否是除法式子 a,b = express.split('/') return str(float(a) / float(b)) def deal_symbol(express): #处理符号 if '--' in express:express = express.replace('--','+') if '-+' in express:express = express.replace('-+','-') if '+-' in express:express = express.replace('+-','-') if '++' in express:express = express.replace('++','+') return express express = '1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )' print(main(express))
import re def math(dic): x,y = float(dic['x']),float(dic['y']) if dic['mark'] == '*':return x * y elif dic['mark'] == '/':return x / y elif dic['mark'] == '+':return x + y else:return x - y def suansu(re_str): ret = re.search('(?P<x>d+.?d*)(?P<mark>[*/])(?P<y>-?d+.?d*)',re_str) try: while True: re_str = re_str.replace(ret.group(),str(math(ret.groupdict()))) if '++' in re_str:re_str = re_str.replace('++','+') if '--' in re_str:re_str = re_str.replace('--','+') ret = re.search('(?P<x>d+.?d*)(?P<mark>[*/])(?P<y>-?d+.?d*)', re_str) except AttributeError:pass ret = re.search('(?P<x>-?d+.?d*)(?P<mark>[+-])(?P<y>-?d+.?d*)',re_str) try: while True: re_str = re_str.replace(ret.group(),str(math(ret.groupdict()))) ret = re.search('(?P<x>-?d+.?d*)(?P<mark>[+-])(?P<y>-?d+.?d*)', re_str) except AttributeError:return re_str def main(user_inp): while True: if not re.search('([^(]+?)',user_inp): return suansu(user_inp) else: for i in re.findall('([^(]+?)',user_inp): user_inp = user_inp.replace(i,suansu(i.replace('(','').replace(')',''))) while True:print(main(input('>>:').replace(' ','')))
1 import re 2 def math(x, y, z): 3 if y in ['*', '/', '+', '-']: return eval(f'{x} {y} {z}') 4 def deal_express(express, reg, flag=''): 5 while True: 6 ret = re.search(f'({flag}d+.?d*)({reg})(-?d+.?d*)', express) 7 if ret: express = express.replace(ret.group(), str(math(ret.group(1), ret.group(2), ret.group(3)))) 8 else: return express 9 def count(re_str): 10 return deal_express(deal_express(re_str, '[*/]'), '[+-]', '-?') 11 def main(user_inp): 12 while True: 13 ret = re.search('([^(]+?)', user_inp) 14 if ret: user_inp = user_inp.replace(ret.group(), count(ret.group())[1: -1], 1) 15 else: return count(user_inp) 16 while True: print(main(input('>>:').replace(' ', '')))
8.请用python函数实现n次登陆功能
def login(n,i = 1): name = input('请输入用户名：').strip() pwd = input('请输入密码：').strip() if name == 'alex' and pwd == '123':return '登陆成功' elif i == n:return '您的账户已经被锁定' else: print('登陆失败，你还有%s次机会'%(n - i)) i += 1 return login(n,i) print(login(3))
9.有两个序列a,b，大小都为n,序列元素的值任意整形数，无序；要求：通过交换a,b中的元素，使[序列a元素的和]与[序列b元素的和]之间的差最小。
a = [1,3,5,8,3,24,53,11,34,56] b = [6,51,32,45,65,77,88,99,34,44] x = y = 0 min = abs(sum(a) - sum(b)) while x < len(a): while y < len(b): a[x],b[y] = b[y],a[x] tmp = abs(sum(a) - sum(b)) if min > tmp:min = tmp else:a[x],b[y] = b[y],a[x] y += 1 x += 1 y = 0 print(min)
def change(a,b,x=0): y = 0 num = abs(sum(a) - sum(b)) while y < len(b): a[x],b[y] = b[y],a[x] tmp = abs(sum(a) - sum(b)) if num > tmp:num = tmp else:a[x],b[y] = b[y],a[x] y += 1 x += 1 if x < len(a):return change(a,b,x) return a,b,num
def change(a,b): x = y = 0 num = abs(sum(a) - sum(b)) while x < len(a): while y < len(b): a[x],b[y] = b[y],a[x] tmp = abs(sum(a) - sum(b)) if num > tmp:num = tmp else:a[x],b[y] = b[y],a[x] y += 1 y = 0 x += 1 return a,b,num
10.一个列表含有相同键的字典合并
题目方法一方法二方法三11.给两个字符串s和t，判断t是否为s的重新排列后组成的单词，s='anagram'，t='nagaram'，return True；s='rat'，t='car'，return False；
def check(s, t): li = [k for k in s] try: for w in t: li.remove(w) except Exception as exc: print(exc) return False if len(li) == 0: return True else: return False
def check(s, t): dic1, dic2 = {}, {} for k in s: if k in dic1: dic1[k] += 1 else: dic1 = {k: 0} for w in t: if w in dic2: dic2[w] += 1 else: dic2 = {w: 0} if dic1 == dic2: return True else: return False
12.给定一个m*n的二维列表，查找一个数是否存在。列表有下列特性：每一行的列表从左到右已经排序好；每一行第一个数比上一行最后一个数大；[[1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
def search(li, k): if k > li[-1][-1]: return False for item in li: if k == item[-1]: return True elif k < item[-1]: if k in item: return True else: return False return False
13.给定一个列表和一个整数，设计算法找到两个数的下标，使得两个数之和为给定的整数。例:列表[1,2,5,4]与目标整数3，1+2=3，结果为(0, 1).
def find(li, k): ret = [] for i in range(len(li)): for j in range(i + 1, len(li)): if li[i] + li[j] == k: ret.append((i, j)) return ret