Python_17_Udacity_Evans_Intro to CS_2_How to repeat

总目录

---

课程页面：https://www.udacity.com/course/intro-to-computer-science--cs101
授课教师：Dave Evans https://www.cs.virginia.edu/~evans/
如下内容包含课程笔记和自己的扩展折腾

课堂笔记

find_second

# Define a procedure, find_second, that takes
# two strings as its inputs: a search string
# and a target string. It should return a
# number that is the position of the second
# occurrence of the target string in the
# search string.

def find_second(a, b):
    first = a.find(b)
    second = a.find(b, first+1)
    return second

danton = "De l'audace, encore de l'audace, toujours de l'audace"
print find_second(danton, 'audace')
#>>> 25

twister = "she sells seashells by the seashore"
print find_second(twister,'she')
#>>> 13

is_friend

# Define a procedure, is_friend, that takes
# a string as its input, and returns a
# Boolean indicating if the input string
# is the name of a friend. Assume
# I am friends with everyone whose name
# starts with either 'D' or 'N', but no one
# else. You do not need to check for
# lower case 'd' or 'n'

def is_friend(name):
    return name[0] == "D" or name[0] == "N"

#print is_friend('Diane')
#>>> True

#print is_friend('Ned')
#>>> True

#print is_friend('Moe')
#>>> False

get_next_target

# Modify the get_next_target procedure so that
# if there is a link it behaves as before, but
# if there is no link tag in the input string,
# it returns None, 0.

# Note that None is not a string and so should
# not be enclosed in quotes.

# Also note that your answer will appear in
# parentheses if you print it.

def get_next_target(page):
    start_link = page.find('

【注意】
return之后，script就终止了。参见SO

median

# numbers as its inputs, and returns the median
# of the three numbers.

# Make sure your procedure has a return statement.

def bigger(a,b):
    if a > b:
        return a
    else:
        return b

def biggest(a,b,c):
    return bigger(a,bigger(b,c))

# my code here:
'''
其实比较简单是再写一个求最小值的function，然后求和，依次减去最大最小值就好了。
但是我看udacity既然已经给了我们 求两者、三者最大值的functions，
应该是让我们只用这两个做。也不是不能做。
'''
def median(a, b, c):
    ab_smaller = a + b - bigger(a, b)
    bc_smaller = b + c - bigger(b, c)
    ac_smaller = a + c - bigger(a, c)
    return biggest(ab_smaller, bc_smaller, ac_smaller)

上面可以写简单点一行搞定：

def median(a, b, c):
  return biggest(a + b - bigger(a, b), b + c - bigger(b, c), a + c - bigger(a, c))

Udacity的解法是，先找出biggest，然后用三个if句，if biggest == a, 那么return bigger(b, c). 以此类推。

stamps

# Define a procedure, stamps, which takes as its input a positive integer in
# pence and returns the number of 5p, 2p and 1p stamps (p is pence) required 
# to make up that value. The return value should be a tuple of three numbers 
# (that is, your return statement should be followed by the number of 5p,
# the number of 2p, and the nuber of 1p stamps).
#
# Your answer should use as few total stamps as possible by first using as 
# many 5p stamps as possible, then 2 pence stamps and finally 1p stamps as 
# needed to make up the total.
#
# (No fair for USians to just say use a "Forever" stamp and be done with it!)
#

def stamps(n):
    # Your code here
    five = (n - (n % 5))/5
    two = ((n - five*5) - (n - five*5) % 2) / 2
    one = (n - five*5 - two*2)/1
    return five, two, one

print stamps(8)
#>>> (1, 1, 1)  # one 5p stamp, one 2p stamp and one 1p stamp
print stamps(5)
#>>> (1, 0, 0)  # one 5p stamp, no 2p stamps and no 1p stamps
print stamps(29)
#>>> (5, 2, 0)  # five 5p stamps, two 2p stamps and no 1p stamps
print stamps(0)
#>>> (0, 0, 0) # no 5p stamps, no 2p stamps and no 1p stamps

Superhero Nuisance

# By Sam the Great from forums
# That freaking superhero has been frequenting Udacity
# as his favorite boss battle fight stage. The 'Udacity'
# banner keeps breaking, and money is being wasted on
# repairs. This time, we need you to proceduralize the
# fixing process by building a machine to automatically
# search through debris and return the 'Udacity' banner
# to the company, and be able to similarly fix other goods.

# Write a Python procedure fix_machine to take 2 string inputs
# and returns the 2nd input string as the output if all of its
# characters can be found in the 1st input string and "Give me
# something that's not useless next time." if it's impossible.
# Letters that are present in the 1st input string may be used
# as many times as necessary to create the 2nd string (you
# don't need to keep track of repeat usage).

# NOTE: # If you are experiencing difficulties taking
        # this problem seriously, please refer back to
        # "Superhero flyby", the prequel, in Problem Set 11.

# TOOLS: # if statement
         # while loop
         # string operations
         # Unit 1 Basics

# BONUS: # 
# 5***** #  If you've graduated from CS101,
#  Gold  #  try solving this in one line.
# Stars! #

def fix_machine(debris, product):
    ### WRITE YOUR CODE HERE ###
    r = product
    for i in product:
        if i not in debris:
            r = "Give me something that's not useless next time."
            break
    return r

### TEST CASES ###
print "Test case 1: ", fix_machine('UdaciousUdacitee', 'Udacity') == "Give me something that's not useless next time."
print "Test case 2: ", fix_machine('buy me dat Unicorn', 'Udacity') == 'Udacity'
print "Test case 3: ", fix_machine('AEIOU and sometimes y... c', 'Udacity') == 'Udacity'
print "Test case 4: ", fix_machine('wsx0-=mttrhix', 't-shirt') == 't-shirt'

Days Old

# By Websten from forums
#
# Given your birthday and the current date, calculate your age in days. 
# Account for leap days. 
#
# Assume that the birthday and current date are correct dates (and no 
# time travel). 

def two_digit_date(x):
    if x<10:
        return "0" + str(x)
    else:
        return str(x)

def daysBetweenDates(year1, month1, day1, year2, month2, day2):
    # Your code here.
    # months
    r = 0
    m_day = {1 : 31, 2 : 28, 3: 31, 4 : 30, 5 : 31, 6 : 30, 7 : 31, 8 : 31, 9 : 30, 10 : 31, 11 : 30, 12 : 31}
    # 情况[1]
    if year1 == year2:
        if month1 == month2:
            r = day2 - day1 + 1
            if month1 == 2 and day1 <= 29 and day2 == 29:
                r = r-1
        else:
            for i in range(month1, month2): #算month1到month2-1
                r += m_day[i]
            r = r-day1+1
            r = r+day2
    # 情况[2]
    else:
    # 算第一年,leap year最后算
        for i in range(month1, 13):
            r += m_day[i]
        r = r - day1 + 1
        print "start_year", r
    # 算最后一年,leap year最后算
        for i in range(1, month2):
            r += m_day[i]
        r += day2
        print "end_year", r
    # 算中间年,leap year最后算
        r += 365*(year2-year1-1)
        print "between_year", r

    # 收尾,算闰年
    # 把年份转换成数字,譬如20100101
    # 最终是比较数字大小,因此年份不足4位不用加0,但是月份和日期不足2位要加0
    start_date = int(str(year1) + two_digit_date(month1) + two_digit_date(day1))
    end_date = int(str(year2) + two_digit_date(month2) + two_digit_date(day2))
    print start_date, end_date
    print r
    print range(year1, year2+1)
    leap_days = 0
    for i in range(year1, year2+1):
        if i % 4 == 0:
            leap_date = int(str(i) + "0229")
            if leap_date >= start_date and leap_date <= end_date:
                leap_days += 1
                if i % 100 == 0 and i % 400 != 0:
                    leap_days -= 1 #四年一闰 百年不闰 四百年再闰
                print "add", leap_date
    r += leap_days

    return r-1 #因为是算活了多少天,是跨度,间隔问题,要减一

def test():
    test_cases = [((2012,1,1,2012,2,28), 58), 
                  ((2012,1,1,2012,3,1), 60),
                  ((2011,6,30,2012,6,30), 366),
                  ((2011,1,1,2012,8,8), 585 ),
                  ((1900,1,1,1999,12,31), 36523)]
    for (args, answer) in test_cases:
        result = daysBetweenDates(*args)
        if result != answer:
            print "Test with data:", args, "failed"
        else:
            print "Test case passed!"

test()

脱水版：

def two_digit_date(x):
    if x<10:
        return "0" + str(x)
    else:
        return str(x)

def daysBetweenDates(year1, month1, day1, year2, month2, day2):
    r = 0
    m_day = {1 : 31, 2 : 28, 3: 31, 4 : 30, 5 : 31, 6 : 30, 7 : 31, 8 : 31, 9 : 30, 10 : 31, 11 : 30, 12 : 31}
    if year1 == year2:
        if month1 == month2:
            r = day2 - day1 + 1
            if month1 == 2 and day1 <= 29 and day2 == 29:
                r = r-1
        else:
            for i in range(month1, month2):
                r += m_day[i]
            r = r-day1+1
            r = r+day2
    else:
        for i in range(month1, 13):
            r += m_day[i]
        r = r - day1 + 1

        for i in range(1, month2):
            r += m_day[i]
        r += day2

        r += 365*(year2-year1-1)

    start_date = int(str(year1) + two_digit_date(month1) + two_digit_date(day1))
    end_date = int(str(year2) + two_digit_date(month2) + two_digit_date(day2))

    leap_days = 0
    for i in range(year1, year2+1):
        if i % 4 == 0:
            leap_date = int(str(i) + "0229")
            if leap_date >= start_date and leap_date <= end_date:
                leap_days += 1
                if i % 100 == 0 and i % 400 != 0:
                    leap_days -= 1 
    r += leap_days

    return r-1 

abacus

#########################################################################
#                 10-row School abacus
#                         by
#                      Michael H
#########################################################################
#       Description partially extracted from from wikipedia
#
#  Around the world, abaci have been used in pre-schools and elementary
#
# In Western countries, a bead frame similar to the Russian abacus but
# with straight wires and a vertical frame has been common (see image).
# Helps schools as an aid in teaching the numeral system and arithmetic
#
#         |00000*****   |     row factor 1000000000
#         |00000*****   |     row factor 100000000
#         |00000*****   |     row factor 10000000
#         |00000*****   |     row factor 1000000
#         |00000*****   |     row factor 100000
#         |00000*****   |     row factor 10000
#         |00000*****   |     row factor 1000
#         |00000****   *|     row factor 100     * 1
#         |00000***   **|     row factor 10      * 2
#         |00000**   ***|     row factor 1       * 3
#                                        -----------
#                             Sum                123
#
# Each row represents a different row factor, starting with x1 at the
# bottom, ascending up to x1000000000 at the top row.
######################################################################

# TASK:
# Define a procedure print_abacus(integer) that takes a positive integer
# and prints a visual representation (image) of an abacus setup for a
# given positive integer value.
#
# Ranking
# 1 STAR: solved the problem!
# 2 STARS: 6 < lines <= 9
# 3 STARS: 3 < lines <= 6
# 4 STARS: 0 < lines <= 3

def print_abacus(value): 
       value = "0"*(10-len(str(value))) + str(value)
       abacus_line = "00000*****"
       for i in value:
           digit = int(i)
           print "|" + abacus_line[:10-digit] + "   " + abacus_line[10-digit:] + "|"


###  TEST CASES
print "Abacus showing 0:"
print_abacus(0)
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
print "Abacus showing 12345678:"
print_abacus(12345678)
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000****   *|
#>>>|00000***   **|
#>>>|00000**   ***|
#>>>|00000*   ****|
#>>>|00000   *****|
#>>>|0000   0*****|
#>>>|000   00*****|
#>>>|00   000*****|
print "Abacus showing 1337:"
print_abacus(1337)
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000*****   |
#>>>|00000****   *|
#>>>|00000**   ***|
#>>>|00000**   ***|
#>>>|000   00*****|

Leap year baby

# By Ashwath from forums

# A leap year baby is a baby born on Feb 29, which occurs only on a leap year.

# Define a procedure is_leap_baby that takes 3 inputs: day, month and year
# and returns True if the date is a leap day (Feb 29 in a valid leap year)
# and False otherwise.

# A year that is a multiple of 4 is a leap year unless the year is
# divisible by 100 but not a multiple of 400 (so, 1900 is not a leap
# year but 2000 and 2004 are).

def is_leap_baby(day,month,year):
    # Write your code after this line.
    if year % 4 == 0 and month == 2 and day == 29:
        r = True
        if year % 100 == 0 and year % 400 != 0:
            r = False
    else:
        r = False
    return r


# The function 'output' prints one of two statements based on whether 
# the is_leap_baby function returned True or False.

def output(status,name):
    if status:
        print "%s is one of an extremely rare species. He is a leap year baby!" % name
    else:
        print "There's nothing special about %s's birthday. He is not a leap year baby!" % name

# Test Cases

output(is_leap_baby(29, 2, 1996), 'Calvin')
#>>>Calvin is one of an extremely rare species. He is a leap year baby!

output(is_leap_baby(19, 6, 1978), 'Garfield')
#>>>There's nothing special about Garfield's birthday. He is not a leap year baby!

output(is_leap_baby(29, 2, 2000), 'Hobbes')
#>>>Hobbes is one of an extremely rare species. He is a leap year baby!

output(is_leap_baby(29, 2, 1900), 'Charlie Brown')
#>>>There's nothing special about Charlie Brown's birthday. He is not a leap year baby!

output(is_leap_baby(28, 2, 1976), 'Odie')
#>>>There's nothing special about Odie's birthday. He is not a leap year baby!