Python basics and data type exercises

# One, the basic problem.
# 1, describe the variable naming specification (3 points)
‘‘‘
Variable name cannot be a keyword in python
Variable names can be combinations of alphanumeric underscores
Variable names cannot start with a number
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# 2, byte and bit relationship. (2 points)
# 1 bytes byte =8 bit bit
# 3, ' elder brother ' when using utf-8 encoding, what is the number of bits and bytes? What is the number of bits and bytes when using GBK encoding? (2 points)
# s= ' two elder brother '
# res=s.encode (' utf-8 ') #6个字节 * 8 bit bit = 48 positions
# res1=s.encode (' GBK ') #4个字节 * 8 bit bit = 32 positions
# Print (RES)
# Print (RES1)
# 4, the 12 functions of the dictation string, and describes its role. (12 points)
#
# 5, number, string, list, meta-ancestor, dictionary corresponding to the Boolean value of the false what is the difference? (5 points)
‘‘‘
Number: 0 string: ' list: [] tuple: () dictionary: {}
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# 6, the writing Python2 is different from the three in the Python3. (3 points)
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Python2
Xrange
Input
Print printing without parentheses
Python3
Range
Raw_input
Print ()
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# 7, write code, such as the following table, the use of slicing to achieve each function (one point per question, a total of 4 points)
#li = [1,3,2, ' A ', 4, ' B ', 5, ' C ']
# 1) Create a new list by slicing the li list l3,l3 = [' 1,2,4,5]
# S1=li[0::2]
# print (S1)
# 2) Create a new list by slicing the li list l4,l4 = [3, ' A ', ' B ']
# S2=li[1:-1:2]
# Print (S2)
# 3) Create a new list by slicing the li list l5,l5 = [' C ']
# S3=li[-1]
# Print (S3)
# 4) Create a new list by slicing the li list l6,l6 = [' B ', ' A ', 3]
# S4=li[5:0:-2]
# Print (S4)
# 8, combining nested questions.
# A, write code, like the following table, according to the requirements to achieve each function (3 points per question, write a method to get 1 points, write out two methods 3 points. A total of 9 points for this question)
# (each is a single line of code implementation)
#lis = [[' K ', [' qwe ', 20,{' K1 ': [' TT ', 3, ' 1 ']},89], ' AB ']
# 1) Turn the ' TT ' in the list Lis into uppercase (in two ways).
# lis[0][1][2][' K1 ']=lis[0][1][2][' K1 ']= ' TT '
# Print (LIS)
# lis[0][1][2][' K1 '][0]=lis[0][1][2][' K1 '][0].upper ()
# Print (LIS)
# 2) Change the number 3 in the list to the string ' 100 ' (in two ways).
# lis[0][1][2][' K1 '][1]= (str (lis[0][1][2][' K1 '][1]+97))
# Print (LIS)
# lis[0][1][2][' K1 '][1]=lis[0][1][2][' k1 '][1]= ' 100 '
# Print (LIS)
# 3) Change the string ' 1 ' in the list to the number 101 (in two ways).
# lis[0][1][2][' K1 '][2]=lis[0][1][2][' K1 '][2]+ ' 01 '
# Print (LIS)
# lis[0][1][2][' K1 '][2]=str (int (lis[0][1][2][' K1 '][2]) +100)
# Print (LIS)
# b, write code, have the following dictionary, as required to achieve each function (5 points)
#dic = {' K1 ': ' v1 ', ' K2 ': [' Alex ', ' SB '], (1,2,3,4,5): {' K3 ': [' 2 ', ', ' Wer ']}}
# 1) Add an element ' 23 ' to the last face of the value corresponding to ' K2 '.
# dic[' K2 '].append (' 23 ')
# Print (DIC)
# 2) Insert an element ' a ' in the first position of the value corresponding to ' K2 '.
# dic[' K2 '].insert (0, ' a ')
# Print (DIC)
# 3) Add (1,2,3,4,5) The corresponding value of a key value to ' K4 ', ' v4 '.
# dic[(1,2,3,4,5)].setdefault (' K4 ', ' v4 ')
# Print (DIC)
# 4) Add the corresponding value (1,2,3,4,5) to a key-value pair (all-in-one), ' OK '.
# dic[(1,2,3,4,5)].setdefault ((+), ' OK ')
# Print (DIC)
# 5) Change the ' wer ' of the value corresponding to ' K3 ' to ' QQ '.
# dic[(1,2,3,4,5)][2]= ' QQ '
# Print (DIC)
# 9, conversion problem (4 points).
#
# How is the conversion between int and STR, and what is the result of the transformation? Is there a condition?
# Str->int Int (' 5 ')
#int->str Str (5)
# How is the conversion between INT and bool, and what is the result of the transformation? Is there a condition?
# int:0 1
# 0-->false
# 1-->true
# How are the conversions between STR and bool, and what is the result of the conversion? Is there a condition?
# empty string is False '-->fales
# non-empty string is True ' Hello '-->true
# can STR and list be converted? How to convert?
#str List Str.split ()
# list-->str '. Join (list)
# 10, achieve the following results (5 points).
# 1) There is a list of li = [' Alex ', ' Wusir ', ' rain '] construct a string by manipulating the list s= ' Alexwusirrain '

# s= '. Join (LI)
# print (s)
# 2) There is a list of li = [' Alex ', ' Wusir ', ' rain '] construct a string by manipulating the list s= ' Alex*wusir*rain '
# s1= ' * '. Join (LI)
# print (S1)
# 3) There is a string s = ' Alexwusirlex ', constructed by manipulating the string to construct a list li = [' A ', ' Exwusirlex ']
# s = ' Alexwusirlex '
# s1=s.split (' l ', 1)
# print (S1)

# 4) There is a string s = ' Alex Wusir ', construct a list by manipulating the string li = [' Alex ', ' Wusir ']
# s= ' Alex Wusir '
# Print (S.split ())

# 5) There is a string s = ' Alex ' constructs a string by manipulating the string S1 = ' a_l_e_x '
# s= ' Alex '
# s1= ' _ '. Join (s)
# print (S1)

# 11, respectively, using the while loop, and the for loop prints the results of the 1-2+3-4+5.......+99. (10 points)
# count=0
# for I in range (1,100):
# if I% 2 = = 0:
# count-=i
# Else:
# count+=i
#print (count)

# 12, use range to print 100,99,98, .... 1,0 (2 min)
# for I in Range (100,-1,-1):
# Print (i)

# 13, calculate the number of the index in user input is odd and the corresponding element is number (no number is 0) (6 points)
# count=0
# info=input (' Please enter content: '). Strip ()
# for I in range (len (info)):
# if I% 2 = = 1 and info[i].isdigit ():
# count+=1
# Print (count)

# 14, Supplemental code (continue writing under existing code): (6 points)
# with the following value li= [11,22,33,44,55,77,88,99,90], all values greater than 66 are saved to the first key in the dictionary, and values less than 66 are saved to the value of the second key.
# li = [11,22,33,44,55,77,88,99,90]
# result = {}
# for row in Li:
# if row > 66:
# if ' K1 ' not in result:
# result[' K1 ']=[]
# Else:
# result[' K1 '].append (row)
# Else:
# if ' K2 ' not in result:
# result[' K2 ']=[]
# Else:
# result[' K2 '].append (Row)
# Print (Result)
#方法二
# Result.setdefault (' K1 ', [])
# Result.setdefault (' K2 ', [])
# for I in Li:
# If I > 66:
# result[' K1 '].append (i)
# Else:
# result[' K2 '].append (i)
# Print (Result)
# 15, find the elements in the list Li, remove the spaces for each element, and find all the elements that start with ' a ' or ' a ' and end With ' C ', add to a new list, and then loop through the new list. (6 points)
# li = [' Taibai ', ' alexc ', ' AbC ', ' Egon ', ' Ritian ', ' wusir ', ' AQC ']
# new=[]
# for I in Li:
# S=i.strip ()
# if s[0].upper () = = ' A ' and s[-1]== ' C ':
# new.append (i)
# for X in New:
# Print (x)

# 16, Implementing an Integer addition Calculator: (6 points)
# such as: content = input (' Please enter content: ') # If user input: 5+8+7 .... (add at least two numbers), then split and then calculate, adding the final result to this dictionary (replace none):
# dic={' final calculation result ': None}.
# info=input (' >>> '). Strip ()
# dic={' final result ': None}
# info_list = info.split (' + ')
# num=0
# for I in Info_list:
# I=int (i)
# num+=i
# dic[' The final result ']=num
# Print (DIC)

# 17, write the program: simulate the company HR input employee account password procedures. (10 points)
# 1), the employee's account password is stored in this data type:
# 2) Illegal character Template: board =/[' Zhang San ', ' Li Xiaoxi ', ' King of two leper ']
# 3) HR input user name, password (sustainable input, if you want to terminate the program, then enter the user name when entering Q or Q exit program), in the HR input user name, detect if this user name has board inside the illegal characters,
# If there are illegal characters, then replace the illegal characters with the same number of * (such as Wang er), and then add to the user_list, if there is no illegal characters, then added directly to User_list, each time the addition of the successful, the new user name, the password was added.

# user_list = [
# {' username ': ' Barry ', ' Password ': ' 1234 '},
# {' username ': ' Alex ', ' Password ': ' Asdf '},
# ]
# board = [' Zhang San ', ' Li Xiaoxi ', ' King two leper ']
# while 1:
# username=input (' username '). Strip ()
# if username.upper () = = ' Q ': Break
# password=input (' password '). Strip ()
# for I in board:
# If I in Username:
# username=username.replace (i, ' * ' *len (i))
# user_list.append ({' username ': username, ' password ':p assword})
# print ({' username ': username, ' password ':p assword})
# Print (user_list)

Python basics and data type exercises