python day04 作业答案
1.
1)
li=['alex','WuSir','ritian','barry','wenzhou']
print(len(li))
2)
li=['alex','WuSir','ritian','barry','wenzhou']
li.append('seven')
print(li)
3)
li=['alex','WuSir','ritian','barry','wenzhou']
li.insert(1,'Tony')
print(li)
4)
li=['alex','WuSir','ritian','barry','wenzhou']
li[2]='Kelly'
print(li)
5)
li=['alex','WuSir','ritian','barry','wenzhou']
li2=[1,'a',3,4,'heart']
li.extend(li2)
print(li)
6)
li=['alex','WuSir','ritian','barry','wenzhou']
li2='qwert'
li.extend(li2)
print(li)
7)
li=['alex','WuSir','ritian','barry','wenzhou']
li.remove('alex')
print(li)
8)
li=['alex','WuSir','ritian','barry','wenzhou']
a=li.pop(2)
print(li)
print(a)
9)
li=['alex','WuSir','ritian','barry','wenzhou']
del li[2:4]
print(li)
10)
li=['alex','WuSir','ritian','barry','wenzhou']
a=[]
li2=li[::-1]
a.extend(li2)
print(a)
11)
li=['alex','WuSir','ritian','barry','wenzhou']
a=li.count('alex')
print(a)
2.
1)
li=[1,3,2,'a',4,'b',5,'c']
l1=li[0:3]
print(l1)
2)
li=[1,3,2,'a',4,'b',5,'c']
l1=li[3:6]
print(l1)
3)
li=[1,3,2,'a',4,'b',5,'c']
l1=li[0::2]
print(l1)
4)
li=[1,3,2,'a',4,'b',5,'c']
l1=li[1:-2:2]
print(l1)
5)
li=[1,3,2,'a',4,'b',5,'c']
l1=li[-1]
print(l1)
6)
li=[1,3,2,'a',4,'b',5,'c']
l1=li[-3:0:-2]
print(l1)
3.
1)
li=[2,3,'k',['qwe',20,['k1',['tt',3,'']],89],'ab','adv']
li[3][2][1][0]='TT'
lis[3][2][1][0] = lis[3][2][1][0].replace("t", "T")
print(li)
2)
li=[2,3,'k',['qwe',20,['k1',['tt',3,'']],89],'ab','adv']
li[3][2][1][1]=100
li[1]=100
li[3][2][1][1] = str(li[3][2][1][1]+97)
print(li)
3)
li=[2,3,'k',['qwe',20,['k1',['tt',3,'']],89],'ab','adv']
li[3][2][1][2]=101
li[3][2][1][1] = str(int(li[3][2][1][2])+100)
print(li)
4.
1)
li=['alex','eric','rain']
li1=''
for i in li:
li1=li1+i+'_'
print(li1[0:-1])
5.
li = ["alex", "WuSir", "ritian", "barry", "wenzhou", ""]
for i in range(0,len(li)):
print(i)
6.
lst=[]
for i in range(0,101):
if i %2 ==0:
lst.append(str(i))
print(lst)
7.
lst=[]
for i in range(0,51):
if i %3 ==0:
lst.append(str(i))
print(lst)
8.
for i in range(100,0,-1):
print(i)
9.
a=[]
b=[]
for i in range(100,9,-1):
if i %2==0:
a.append(i)
for num in a:
if num %4 ==0:
b.append(num)
print(b)
10.
a=[]
b=[]
for i in range(1,31):
a.append(i)
for j in a :
if j %3==0:
j='*'
b.append(j)
print(b)
11.
a=[]
li=['TaiBai,','xC','AbC','egon','TiAn','WuSir','aqc']
for i in li :
i.replace(' ','')
if (i.startswith('a') or i.startswith('A')) and i.endswith('c'):
a.append(i)
print(a)
12.
a=input('输入:')
b=''
c=[]
li=['苍老师','东京热','武藤兰','波多野结衣']
for j in li :
if str(j) in a :
a=a.replace(str(j),('*'*len(j)))
c.append(a)
print(c)
13.
a=[]
for i in li :
if type(i)!=list:
a.append(i)
else:
a.extend(i)
for j in a :
print(j)
14.
d=0
c=[]
while 1 :
a=input('输入:')
if a.lower() !='q':
e=a.split('_')
c.append(a)
d=d+int(e[1])
else:
break
print(d/len(c))
python day04 作业答案的更多相关文章
- python day10作业答案
2.def func(*args): sum = 0 for i in args: sum=sum+int(i) return sum a=func(2,3,9,6,8) print(a) 3. a= ...
- python day09作业答案
2. def lst(input): lst2=[] count=0 for i in range(0,len(input)): if i %2!=0: lst2.append(input[i]) r ...
- python day08作业答案
1. a f=open('11.txt','r',encoding='utf-8') a=f.read() print(a) f.flush() f.close() b. f=open('11.txt ...
- python day07作业答案
1. sum=0 a=input() for i in a: sum=sum+int(i)**3 if sum==int(a): print('水仙数') 2. lst=[100,2,6,9,1,10 ...
- python day06 作业答案
1. count=1 while count<11: fen=input('请第{}个评委打分' .format( count)) if int(fen) >5 and int(fen) ...
- python day05 作业答案
1. b.不可以 c.tu=("alex",[11,22,{"k1":"v1","k2":["age" ...
- python day02 作业答案
1. (1).false (2).false 2. (1).8 (2).4 3. (1).6 (2).3 (3).false (4).3 (5).true (6).true (7) ...
- python day04作业
- 笔试 - 高德软件有限公司python问题 和 答案
高德软件有限公司python问题 和 答案 本文地址: http://blog.csdn.net/caroline_wendy/article/details/25230835 by Spike 20 ...
随机推荐
- ACM-选人问题(救济金发放)
n(n<20)个人站成一圈,逆时针编号为1-n.有两个官员,A从1开始逆时针数,B从n开 始顺时针数.在每一轮中,官员A数k个就停下来,官员B数m个就停下来(注意有可能两个 官员停在同一个人上) ...
- python-flask-SQLAlchemy-Utils组件
SQLAlchemy-Utils,提供choice功能 定义: # pip3 install sqlalchemy-utils from sqlalchemy_utils import ChoiceT ...
- 牛客练习赛30-A/C
链接:https://ac.nowcoder.com/acm/contest/216/A来源:牛客网 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 32768K,其他语言65536K ...
- ATOM常用插件推荐
转载:http://blog.csdn.net/qq_30100043/article/details/53558381 ATOM常用插件推荐 simplified-chinese-menu ATOM ...
- virtualbox 中centOS在不能ssh
这个重要跟虚拟机的网络设置有关系.废话不多说. 针对一个网卡的形式.可以如下进行配置 1.网络-- 连接方式还选择“网络地址转换(NAT)” 其他不变,展开高级,设置端口转发 主机IP设为本机IP, ...
- 22. Generate Parentheses C++回溯法
把左右括号剩余的次数记录下来,传入回溯函数. 判断是否得到结果的条件就是剩余括号数是否都为零. 注意判断左括号是否剩余时,加上left>0的判断条件!否则会memory limited erro ...
- [LeetCode] Network Delay Time 网络延迟时间——最短路算法 Bellman-Ford(DP) 和 dijkstra(本质上就是BFS的迭代变种)
There are N network nodes, labelled 1 to N. Given times, a list of travel times as directed edges ti ...
- Apache隐藏版本号教程(CentOS)
1 找到Apache配置文件/etc/httpd/conf/httpd.conf 2 给该文件添加写权限:chmod u+w httpd.conf 3 打开该文件找到ServerTokens字段将其值 ...
- nginx源码安装教程(CentOS)
1.说明 官方源码安装说明:http://nginx.org/en/docs/configure.html 源码包下载地址:http://nginx.org/en/download.html 版本说明 ...
- vue 添加vux
1.命令添加vux npm install vux --save 2.在build/webpack.base.conf.js中配置 const vuxLoader = require('vux-loa ...