1. 练习：选一个
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_response2. 练习：三角数字
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 sum3. 练习：线性时间
4. 练习：删除标签
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()5. 练习：日期转换器
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)] + " " + year6. 练习：终止
题目：
# Question 7: Find and Replace
# For this question you need to define two procedures:
# make_converter(match, replacement)
# Takes as input two strings and returns a converter. It doesn't have
# to make a specific type of thing. It can
# return anything you would find useful in apply_converter.
# apply_converter(converter, string)
# Takes as input a converter (produced by make_converter), and
# a string, and returns the result of applying the converter to the
# input string. This replaces all occurrences of the match used to
# build the converter, with the replacement. It keeps doing
# replacements until there are no more opportunities for replacements.
def make_converter(match, replacement):
def apply_converter(converter, string):
# For example,
c1 = make_converter('aa', 'a')
print apply_converter(c1, 'aaaa')
#>>> a
c = make_converter('aba', 'b')
print apply_converter(c, 'aaaaaabaaaaa')
#>>> ab
# Note that this process is not guaranteed to terminate for all inputs
# (for example, apply_converter(make_converter('a', 'aa'), 'a') would
# run forever).(我没有思路，看不懂描述。)
视频的答案：
def make_converter(math, replacement):
return [match, replacement]
def apply_converter(converter, string):
previous = None
while previos != string:
position = string
position = string.find(converter[0])
if position != -1:
string = string[:position] + converter [1] + string[position + len(converter[0]):]
return string7. 练习：查找替换
8. 练习：最长重复
题目：
# Question 8: Longest Repetition
# Define a procedure, longest_repetition, that takes as input a
# list, and returns the element in the list that has the most
# consecutive repetitions. If there are multiple elements that
# have the same number of longest repetitions, the result should
# be the one that appears first. If the input list is empty,
# it should return None.
def longest_repetition():
#For example,
print longest_repetition([1, 2, 2, 3, 3, 3, 2, 2, 1])
# 3
print longest_repetition(['a', 'b', 'b', 'b', 'c', 'd', 'd', 'd'])
# b
print longest_repetition([1,2,3,4,5])
# 1
print longest_repetition([])
# None(我的答案并不完整。。。
def longest_repetition(alist):
if alist == []:
return None
sum = 1
for index in range(len(alist)):
if alist[index+1] == alist[index]:
sum += 1
else:
sum =)
视频的答案：
9. 练习：深度翻转
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原文链接：《计算机科学导论》学习笔记（27） - 课程 27: 累积实践问题，转载请注明来源！