# 作 1 :
## Use to use the*dictionary*output directory, you can select the directory to enter , you can go back and exit!
product_list = [[
| Assignment 3 |
a_list = [
Implementation method one:
##a_list = [
Implementation method two
:| ## |
做业五
打印99乘法表
|
#! /usr/bin/python
# Filename : table_9x9.py
print '\n9x9 Table\n'
for i in range(1, 10) :
for j in range(1, i+1) :
print j, 'x', i, '=', j*i, '\t',
# print '%d x %d = %d\t' %(j, i, j*i),
print '\n'
print '\nDone!'
关于该程序的说明：
1. 第一行是特殊注释行，称为组织行，用来告诉GNU/Linux系统应该使用哪个解释器来执行该程序。
2. 第二行至第四行都是一般的注释行，用来说明一些信息的（如文件名，作者，时间等）。
3. 第六行打印一个字符串。
4. 第八行i取值范围为1, 2, 3, 4, 5, 6, 7, 8, 9
5. 第九行j取值范围为1, 2, ..., i
6. 第十行和第十一行效果一样，最后的逗号都是用来取消自动换行的。
7. 第十二行作用是在每个内层for循环结束后换行，即在输出完九九乘法表一行后换行。
8. 第十四行打印一个字符串。
小结：通过这个程序熟悉print的用法，for循环的用法，以及range函数的特点。
|
课堂中提到的方法：
for i in range(1,10):<br> output = ''<br> for j in range(1,i+1):<br> output+= "%s*%s=%2s "%(i,j,i*j)<br> print output
---------------------------------------------------
作业六：
取一个列表中前两位大的数字
<br> #!/usr/bin/python<br><br>NumList = [1,2,3,2,12,3,1,3,21,2,2,3,4111,22,3333,444,111,4,5,777,4222,46,33,45,65555]<br><br>max_num = 0<br>sec_num = 0<br><br>for n in NumList:<br>if max_num sec_num=max_num <br>#获取最大数之前的比较的哪个数肯定是在这之前第二大的<br> max_num = n<br>#但是后面如果再出现比最大的大，比之前比较出料的第二大<br> #小的数时需要再比较下。<br> if n and n > sec_num:<br> sec_num = n<br><br><br>print ('The bigest NUM is : %s' %(max_num))<br>print ('The second one is : %s' %(sec_num))
<br> |
利用单循环来同事去除两个两个数。
作业七
二分查找法（利用递归函数）
num_list = [ |
作业八
利用函数完成用户的注册登录功能
#!/usr/bin/python<br>#encoding: utf-8
import getpass
user_list = {}
def new_user():<br> username = raw_input('please input your new name:')<br> if username in user_list:<br> print '%s is already exits!!' %username<br> else:<br> passwd = getpass.getpass() #已密码不显示的方式输入。<br> user_list[username] = passwd
def old_user():<br> username = raw_input('please input your name:')<br> passwd = raw_input('please input your passwd:')<br> if username in user_list:<br> if user_list[username] == passwd:<br> print 'login successfull!!'<br> else:<br> print 'login fail!!'<br> else:<br> print 'login fail!!!'
CMD = {'n' : new_user , 'o' : old_user}
def main_menu():<br> ppp = '''<br> 新用户注册：n<br> 老用户登录：o<br> 推出：q
'''<br> while True:<br> try:<br> choice = raw_input(ppp)<br> except (KeyboardInterrupt, EOFError): #捕捉ctrl+ D 的错误输出。<br> choice = 'q'
if choice not in 'noq':<br> print 'please input again'<br> continue<br> if choice == 'q':<br> break<br> CMD[choice]() #通过取字典的key，选择来执行函数
if __name__ == '__main__':<br> main_menu()
| 作业九 |
log日志前十排序，其中包括重复的。利用函数简单的html输出
#!/usr/bin/python
#coding:utf8
def openFile(log):
with open(log) as f:
content = f.readlines()
return content
def countLine(fn):
content = fn
count_dict = {}
for line in content:
line = line.split(' ') # string 通过split 转换为list
local_ip,uri,code = line[0],line[6],line[8]
#利用元组来作为key，然后统计数量
count_dict[(local_ip,uri,code)] = count_dict.get((local_ip,uri,code),0) + 1
#这个排序的功能网上copy的，不了解原理
count_dict = sorted(count_dict.items() ,key=lambda item:item[1],reverse=True)
return count_dict
def detailLine(fn):
count_dict = fn
#下面进行重复行计数为一行。
n = 1 #取行计数
num = 0 #用于判断'统计数量'是否出现一样的
# print ('-----IP------------------URI----------CODE---count---')
# for i in count_dict:
# if n no%s' % (i[0][0], i[0][1], i[0][2], i[1], n))
# num = i[1]
# n += 1
# elif i[1] == num:
# print('%-10s%20s%5s%10s' % (i[0][0], i[0][1], i[0][2], i[1]))
with open('count.html','w') as f:
res = "
| IP | URI | CODE | COUNT | %s | %s | %s | %s | '% (i[0][0], i[0][1], i[0][2], i[1]) num = i[1] n += 1 elif i[1] == num: res += '
| %s | %s | %s | %s |
| 作业十， |
生成随机验证码：
方法一：
import string,random
source = string.digits + string.ascio_lowercase
print ("".join(random.sample(source,6)))
方法二：
checkcode = ''
for i in range(6):
current = random.randrange(0,6)
if current != i:
temp = chr(random.randint(65,90))
else:
temp = random.randint(0-9)
checkcode += str(temp)
print(checkcode)
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