1.使用格式化输出的三种方式实现以下输出(name换成自己的名字,既得修改身高体重,不要厚颜无耻)
name = 'ABDMLBM'
height = 175
weight = 140
# "My name is 'Nick', my height is 180, my weight is 140"
print('My name is %s,my height is %s,my weight is %s'%(name,height,weight))
print('My name is {},my height is {},my weight is {}'.format(name,height,weight))
print(f'My name is {name},my height is {height},my weight is {weight}')
2.输入姑娘的年龄后,进行以下判断:
-
如果姑娘小于18岁,打印“不接受未成年”
-
如果姑娘大于18岁小于25岁,打印“心动表白”
-
如果姑娘大于25岁小于45岁,打印“阿姨好”
-
如果姑娘大于45岁,打印“奶奶好”
while True:
girl_age =int( input('输入美女年龄'))
if girl_age < 18:
print('不接受未成年')
elif girl_age >= 18 and girl_age < 25:
print('心动表白')
elif girl_age >= 25 and girl_age < 45:
print('阿姨好')
else :
print('奶奶好')
3.预习while循环,打印1-100之间的偶数:
i = 1
while i < 101:
oi = i % 2
if oi == 0:
print(i)
i += 1
4.通过预习写一个猜年龄游戏,需求:给定一个标准年龄,用户通过输入年龄判断年龄是否等于标准年龄,如果等于——打印猜对了;如果小于——打印猜小了;如果大于——打印猜大了,增加用户输入年龄功能,并可以参考while循环博客,为应用程序添加循环。
预习while循环,猜年龄游戏升级版,有以下三点要求:
-
允许用户最多尝试3次
-
每尝试3次后,如果还没猜对,就问用户是否还想继续玩,如果回答Y或y, 就继续让其猜3次,以此往复,如果回答N或n,就退出程序
-
如果猜对了,就直接退出
age = 25
count = 0
while count < 4:
user_age = int(input('请输入你的年龄:'))
if user_age == age:
print('你猜对了')
break
elif user_age > age:
print('你猜大了')
else:
print('你猜小了')
count += 1
if count != 3 :
continue
else:
print('你已经猜了三次,答"Y"或者"y"还想再玩,答"N"或者"n"退出')
user_player = input('请输入:')
if user_player == "Y" or user_player == "y":
count = 0
else:
break
5.统计s = 'hello alex alex say hello sb sb'中每个单词的个数
s = 'hello alex alex say hello sb sb'
l=s.split()
print(l)
dic = {}
for item in l:
if item in dic:
dic[item]=dic[item]+1
else:
dic[item]=1
print(dic)
6.统计一篇英文文章内每个单词出现频率,并返回出现频率最高的前10个单词及其出现次数
from collections import Counter
import re
with open('a.txt', 'r', encoding='utf-8') as f:
txt = f.read()
c = Counter(re.split('\W+',txt)) #取出每个单词出现的个数
print(c)
ret = c.most_common(10) #取出频率最高的前10个
print(ret)
7.冒泡排序
def mao_pao(li):
for i in range(len(li)):
for j in range(len(li)):
if li[i] < li[j]:
li[i],li[j] = li[j] ,li[i]
import random
li = list(range(10))
random.shuffle(li)
print(li)
mao_pao(li)
print(li)
8.删除列表中的重复元素
#方式一
li = [1,5,5,4,12,3,1,5]
print(list(set(l)))
#方式二
li = [1,5,5,4,12,3,1,5]
def func(li):
l = []
for i in li:
if i not in l:
l.append(i)
return l
print(func(li))
9.二分查找
#方式一:递归版
li = [1,2,3,4,5,6,7,8,9,10]
def erfen(li,aim ,start=0 ,end=len(li)-1):
if start <= end:
mid = (start+end)//2
if li[mid] >aim : #如果中间的值比目标值大,就从左边找
return erfen(li,aim,start,mid-1)
elif li[mid]<aim : #从右边找
return erfen(li,aim,mid+1,end)
else:
return mid
ret = erfen(li,10)
print(ret)
#方式二:循环版
li = [1,2,3,4,5,6,7,8,9,10]
def erfen(li,aim,start=0,end=len(li)-1):
while start <= end:
mid = (start+end) //2
if li[mid] < aim:
start = li[mid] +1
elif li[mid] >aim:
end = li[mid] -1
else:
return mid
print(erfen(li,10))
10、写出下面代码的输出结果
def f(x,l=[]):
for i in range(x):
l.append(i*i)
print(l)
f(2) #[0,1]
f(3,[3,2,1]) #[3,2,1,0 1 4]
f(3) #[0,1,0,1,4]
11. 实现字符串反转 输入str="string"输出'gnirts'
# 方式一
def str_reverse(str):
return str[::-1] #从前到后步长为-1
print(str_reverse('string'))
# 方式二
def str_reverse2(str):
l = list(str)
l.reverse()
new_str = ''.join(l)
return new_str
print(str_reverse2('string'))