题目：
# Question 1: Pick One # Define a procedure, pick_one, that takes three inputs: a Boolean # and two other values. If the first input is True, it should return # the second input. If the first input is False, it should return the # third input. # For example, pick_one(True, 37, 'hello') should return 37, and # pick_one(False, 37, 'hello') should return 'hello'. def pick_one(): print pick_one(True, 37, 'hello') #>>> 37 print pick_one(False, 37, 'hello') #>>> hello print pick_one(True, 'red pill', 'blue pill') #>>> red pill print pick_one(False, 'sunny', 'rainy') #>>> rainy我的答案：
def pick_one(Boolean, first, second): return first if Boolean else second视频的答案：
def pick_one(boolean, true_response, false_response): if boolean: return true_response else: return false_response题目：
# Question 2. Triangular Numbers # The triangular numbers are the numbers 1, 3, 6, 10, 15, 21, ... # They are calculated as follows. # 1 # 1 + 2 = 3 # 1 + 2 + 3 = 6 # 1 + 2 + 3 + 4 = 10 # 1 + 2 + 3 + 4 + 5 = 15 # Write a procedure, triangular, that takes as its input a positive # integer n and returns the nth triangular number. def triangular(): print triangular(1) #>>>1 print triangular(3) #>>> 6 print triangular(10) #>>> 55我的答案：
def triangular(n): #if n == 1: return 1 """ i = 0 sum = 0 while i <= n: sum += i i += 1 """ sum = 0 for i in range(1, n+1): sum += i return sum题目：
# Question 4: Remove Tags # When we add our words to the index, we don't really want to include # html tags such as <body>, <head>, <table>, <a href="..."> and so on. # Write a procedure, remove_tags, that takes as input a string and returns # a list of words, in order, with the tags removed. Tags are defined to be # strings surrounded by < >. Words are separated by whitespace or tags. # You may assume the input does not include any unclosed tags, that is, # there will be no '<' without a following '>'. def remove_tags(): print remove_tags('''<h1>Title</h1><p>This is a <a href="http://www.udacity.com">link</a>.<p>''') #>>> ['Title','This','is','a','link','.'] print remove_tags('''<table cellpadding='3'> <tr><td>Hello</td><td>World!</td></tr> </table>''') #>>> ['Hello','World!'] print remove_tags("<hello><goodbye>") #>>> [] print remove_tags("This is plain text.") #>>> ['This', 'is', 'plain', 'text.']我的答案：
def remove_tags(text): result = [] if text.find('<') == -1 and \ text.find('>') == -1: result.append(text) return result ''' for index in range(len(text)): at_kai = False at_bi = False at_space = False result = [] if text[index] == ' ': at_space = True if text[index] == '<': at_kai = True continue if text[index] == '>': at_bi = True at_kai = False if at_bi == True: text = text[index + 1:] result[index] = result[index] + char '''(我的答案很稀烂，没有完成。我想起来了之前的视频看过的代码，只能谈下我的思路：
1. 去掉 < 和 > 之间的文本;
2. 在某些情况下， append 单个字符；【result.append(char)】
3. 在某些情况下，连接单个字符。【result[index] = result[index] + char】
)
视频的答案：
def remove_tags(string): start = string.find('<') while start != -1: end = string.find('>', start) string = string[:start] + " " + string[end + 1:] start = string.find('<') return string.split()题目：
# Question 5: Date Converter # Write a procedure date_converter which takes two inputs. The first is # a dictionary and the second a string. The string is a valid date in # the format month/day/year. The procedure should return # the date written in the form <day> <name of month> <year>. # For example , if the # dictionary is in English, english = {1:"January", 2:"February", 3:"March", 4:"April", 5:"May", 6:"June", 7:"July", 8:"August", 9:"September",10:"October", 11:"November", 12:"December"} # then "5/11/2012" should be converted to "11 May 2012". # If the dictionary is in Swedish swedish = {1:"januari", 2:"februari", 3:"mars", 4:"april", 5:"maj", 6:"juni", 7:"juli", 8:"augusti", 9:"september",10:"oktober", 11:"november", 12:"december"} # then "5/11/2012" should be converted to "11 maj 2012". # Hint: int('12') converts the string '12' to the integer 12. def date_converter(): print date_converter(english, '5/11/2012') #>>> 11 May 2012 print date_converter(english, '5/11/12') #>>> 11 May 12 print date_converter(swedish, '5/11/2012') #>>> 11 maj 2012 print date_converter(swedish, '12/5/1791') #>>> 5 december 1791我的答案：
def date_converter(language, date): first_xie = date.find('/') second_xie = date.find('/', first_xie + 1) return date[first_xie + 1:second_xie] + ' ' + \ language[int(date[:first_xie])] + ' ' + \ date[second_xie + 1:]视频的答案：
def date_converter(language, date): first = date.find('/') month = date[:first] second = date.find('/', first+1) day = date[first + : second]year = date[second + 1 :] year = date[second + 1:] return day + " " + language[int(month)] + " " + year def date_converter(language, date): month, day, year = date.split('/') return day + " " + language[int(month)] + " " + year