Study python of day2

today,we will study python's list / dictionary opt
dict detail informations's reference 5.Data Strunctures :https://docs.python.org/2.7/library/stdtypes.html#typesmapping

1.using list as stacks:==>>last-in, first-out”,we need used the collections.deque
stack = [1,2,3,4,5,6,7]
stack.append(8)
stack.append(9)
stack
Out[75]: [1, 2, 3, 4, 5, 6, 7, 8, 9]
stack.pop()
Out[76]: 9
stack.pop()
Out[77]: 8
stack.pop()
Out[80]: 7
stack.pop()
Out[81]: 6
stack
Out[82]: [1, 2, 3, 4, 5]

2.using lists as queues:===>>first-in, first-out
stack
Out[83]: [1, 2, 3, 4, 5]
from collections import deque
stack.append(6)
stack
Out[86]: [1, 2, 3, 4, 5, 6]
stack = deque(stack)
stack.popleft()
Out[90]: 1
stack.popleft()
Out[91]: 2
stack
Out[92]: deque([3, 4, 5, 6])

3.coding informations:
# -*- coding: utf-8 -*-
"""
Created on Mon Sep 26 12:36:37 2016
Using the while function to achieve the following functions
1.sum for 1..100
2.get all odd and even from 1..100
3.sum for 1-2+3-4..100
4.The 100 random integers were placed in the dictionary of dic key k1(values >50) and k2(values<50)
@author: trsenzhang
"""

import random

#sum(1..100)=5050
def getSum():
num = 0
result = 0
while num < 101:
result += num
num += 1
return result

#get 1…100 o
def getOddEven(type):
l1 = []
l2 = []
num = 1
while num < 101:
if (num % 2) == 0:
l1.append(num)
else:
l2.append(num)
num += 1
if type == '1':
return l2
elif type == '0':
return l1
else:
return '结果异常'

#sum 1-2+3-4+5-6….99
def getSum1():
num = 0
result = 0
flag = 1
while num < 101:
if (num % 2) == 0:
flag = -1
else:
flag = 1
result = result +num*flag
num += 1
return result

#produce 100 random integer
def ProduceRange():
l_num = []
for i in range(1,100):
num = random.randint(1,100)
l_num.append(num)
i += 1
return l_num

#if l_num > 50 : put l_num into the k1 for dict else put l_num into k2 for  dict
def getDict():
l_num = ProduceRange()
print "the list' number is : %s " %l_num
dic = {}
print str.center('The Result',80,'#')
for i in l_num:
if i > 50:
try :
if dic.has_key('k1'):
dic[‘k1’].append(i,)
else:
dic[‘k1’] = [i,]
except:
return '%s values have problem' % i
else:
try :
if dic.has_key('k2'):
dic[‘k2’].append(i,)
else:
dic[‘k2’] = [i,]
except:
return '%s values have problem' % i
return dic

if __name__ == '__main__':
result = getSum()
print result

result1 = getOddEven('1')
print 'even is  %s' % result1

result2 = getOddEven('0')
print 'odd is %s' % result2

result3 = getSum1()
print result3

result5 = getDict()
print result5

Study python of day1

```"""
require:user login and verify user information on login user information file. If not ,after the three
attempt to lock the user.if enter username on black user information file,then printing the "user locked".
define login user  and black user information file.
@trsenzhang
"""

import os
import sys

#define variable
account_file = 'D:/study_python/day1/match.txt'
account_name = 'match.txt'
lock_file = 'D:/study_python/day1/lock.txt'
lock_name='lock.txt'

def file_exists(a,b):
if os.path.exists(a):
print("The  file %s is exist" %b)
else:
account_name = open(b,"w")
account_name.close()

with open(lock_file,'a') as deny_f:

def main():

#determine the acount is or not locked
file_exists(account_file, account_name)
file_exists(lock_file, lock_name)

#define variable
retry_count = 0
retry_limit = 3

while retry_count < retry_limit:
with open(lock_file,'r') as lock_f:
if len(line) == 0:
continue
sys.exit()
continue

with open(account_file,'r') as account_f:
flag = False
user,pd = line.strip().split()
if username == user and pwd == pd:
print("you are success!!")
flag = True
break
if flag == False:
if retry_count <2: