<a href="http://www.jianshu.com/p/54870e9541fc">总目录</a>
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
Stop watch
例一：
import time
def time_execution(code):
start = time.clock()
result = eval(code)
run_time = time.clock() - start
return result, run_time
print time_execution("1 + 1")
"""
Console:
(2, 2.5999999999998247e-05) # 写成scientific notation
Process finished with exit code 0
"""
例二：
import time
def time_execution(code):
start = time.clock()
result = eval(code)
run_time = time.clock() - start
return result, run_time
def spin_loop(n):
i = 0
while i < n:
i += 1
print time_execution('spin_loop(100000)')
"""
Console:
(None, 0.0045579999999999996)
Process finished with exit code 0
"""
Better hash functions
# -*- coding: utf-8 -*-
# Define a function, hash_string,
# that takes as inputs a keyword
# (string) and a number of buckets,
# and returns a number representing
# the bucket for that keyword.
def hash_string(keyword,buckets):
r = 0
for char in keyword:
r += ord(char)
return r % buckets
"""
Udacity的做法比较好：
def hash_string(keyword,buckets):
r = 0
for char in keyword:
r = (r + ord(char)) % buckets
# 这样就不会像我的算法一样，可能会加到一个很大的数字
# 不要写成 r += ord(char) % buckets
return r
"""
print hash_string('a',12)
#>>> 1
print hash_string('b',12)
#>>> 2
print hash_string('a',13)
#>>> 6
print hash_string('au',12)
#>>> 10
print hash_string('udacity',12)
#>>> 11
Make empty table
# # Creating an Empty Hash Table
# Define a procedure, make_hashtable,
# that takes as input a number, nbuckets,
# and returns an empty hash table with
# nbuckets empty buckets.
def make_hashtable(nbuckets):
r = []
for i in range(nbuckets):
r.append([])
return r
Finding buckets
# -*- coding: utf-8 -*-
# Define a procedure, hashtable_get_bucket,
# that takes two inputs - a hashtable, and
# a keyword, and returns the bucket where the
# keyword could occur.
def hashtable_get_bucket(htable,keyword):
return htable[hash_string(keyword, len(htable))]
def hash_string(keyword,buckets): # returns hash number
out = 0
for s in keyword:
out = (out + ord(s)) % buckets
return out
def make_hashtable(nbuckets):
table = []
for unused in range(0,nbuckets):
table.append([])
return table
table = [[['Francis', 13], ['Ellis', 11]], [], [['Bill', 17],
['Zoe', 14]], [['Coach', 4]], [['Louis', 29], ['Rochelle', 4], ['Nick', 2]]]
print hashtable_get_bucket(table, "Zoe")
#>>> [['Bill', 17], ['Zoe', 14]]
print hashtable_get_bucket(table, "Brick")
#>>> []
print hashtable_get_bucket(table, "Lilith")
#>>> [['Louis', 29], ['Rochelle', 4], ['Nick', 2]]
Adding keywords
# -*- coding: utf-8 -*-
# Define a procedure,
#
# hashtable_add(htable,key,value)
#
# that adds the key to the hashtable (in
# the correct bucket), with the correct
# value and returns the new hashtable.
#
# (Note that the video question and answer
# do not return the hashtable, but your code
# should do this to pass the test cases.)
def hashtable_add(htable,key,value):
hashtable_get_bucket(htable, key).append([key, value])
return htable
def hashtable_get_bucket(htable,keyword):
return htable[hash_string(keyword,len(htable))]
def hash_string(keyword,buckets):
out = 0
for s in keyword:
out = (out + ord(s)) % buckets
return out
def make_hashtable(nbuckets):
table = []
for unused in range(0,nbuckets):
table.append([])
return table
table = make_hashtable(5)
hashtable_add(table,'Bill', 17)
hashtable_add(table,'Coach', 4)
hashtable_add(table,'Ellis', 11)
hashtable_add(table,'Francis', 13)
hashtable_add(table,'Louis', 29)
hashtable_add(table,'Nick', 2)
hashtable_add(table,'Rochelle', 4)
hashtable_add(table,'Zoe', 14)
print table
#>>> [[['Ellis', 11], ['Francis', 13]], [], [['Bill', 17], ['Zoe', 14]],
#>>> [['Coach', 4]], [['Louis', 29], ['Nick', 2], ['Rochelle', 4]]]
Lookup
# -*- coding: utf-8 -*-
# Define a procedure,
# hashtable_lookup(htable,key)
# that takes two inputs, a hashtable
# and a key (string),
# and returns the value associated
# with that key.
def hashtable_lookup(htable,key):
for item in hashtable_get_bucket(htable, key):
if item[0] == key:
return item[1]
return None
def hashtable_add(htable,key,value):
bucket = hashtable_get_bucket(htable,key)
bucket.append([key,value])
def hashtable_get_bucket(htable,keyword):
return htable[hash_string(keyword,len(htable))]
def hash_string(keyword,buckets):
out = 0
for s in keyword:
out = (out + ord(s)) % buckets
return out
def make_hashtable(nbuckets):
table = []
for unused in range(0,nbuckets):
table.append([])
return table
table = [[['Ellis', 11], ['Francis', 13]], [], [['Bill', 17], ['Zoe', 14]],
[['Coach', 4]], [['Louis', 29], ['Nick', 2], ['Rochelle', 4]]]
print hashtable_lookup(table, 'Francis')
#>>> 13
#print hashtable_lookup(table, 'Louis')
#>>> 29
#print hashtable_lookup(table, 'Zoe')
#>>> 14
Update
# -*- coding: utf-8 -*-
# Define a procedure,
# hashtable_update(htable,key,value)
# that updates the value associated with key. If key is already in the
# table, change the value to the new value. Otherwise, add a new entry
# for the key and value.
# Hint: Use hashtable_lookup as a starting point.
# Make sure that you return the new htable
def hashtable_update(htable,key,value):
# Your code here
if hashtable_lookup(htable,key) == None:
hashtable_add(htable, key, value)
else:
bucket = hashtable_get_bucket(htable,key)
for entry in bucket:
if entry[0] == key:
entry[1] = value
return htable
def hashtable_lookup(htable,key):
bucket = hashtable_get_bucket(htable,key)
for entry in bucket:
if entry[0] == key:
return entry[1]
return None
def hashtable_add(htable,key,value):
bucket = hashtable_get_bucket(htable,key)
bucket.append([key,value])
def hashtable_get_bucket(htable,keyword):
return htable[hash_string(keyword,len(htable))]
def hash_string(keyword,buckets):
out = 0
for s in keyword:
out = (out + ord(s)) % buckets
return out
def make_hashtable(nbuckets):
table = []
for unused in range(0,nbuckets):
table.append([])
return table
table = [[['Ellis', 11], ['Francis', 13]], [], [['Bill', 17], ['Zoe', 14]],
[['Coach', 4]], [['Louis', 29], ['Nick', 2], ['Rochelle', 4]]]
hashtable_update(table, 'Bill', 42)
hashtable_update(table, 'Rochelle', 94)
hashtable_update(table, 'Zed', 68)
print table
#>>> [[['Ellis', 11], ['Francis', 13]], [['Zed', 68]], [['Bill', 42],
#>>> ['Zoe', 14]], [['Coach', 4]], [['Louis', 29], ['Nick', 2],
#>>> ['Rochelle', 94]]]
Is offered
# -*- coding: utf-8 -*-
# Dictionaries of Dictionaries (of Dictionaries)
# The next several questions concern the data structure below for keeping
# track of Udacity's courses (where all of the values are strings):
# { <hexamester>, { <class>: { <property>: <value>, ... },
# ... },
# ... }
#For example,
courses = {
'feb2012': { 'cs101': {'name': 'Building a Search Engine',
'teacher': 'Dave',
'assistant': 'Peter C.'},
'cs373': {'name': 'Programming a Robotic Car',
'teacher': 'Sebastian',
'assistant': 'Andy'}},
'apr2012': { 'cs101': {'name': 'Building a Search Engine',
'teacher': 'Dave',
'assistant': 'Sarah'},
'cs212': {'name': 'The Design of Computer Programs',
'teacher': 'Peter N.',
'assistant': 'Andy',
'prereq': 'cs101'},
'cs253':
{'name': 'Web Application Engineering - Building a Blog',
'teacher': 'Steve',
'prereq': 'cs101'},
'cs262':
{'name': 'Programming Languages - Building a Web Browser',
'teacher': 'Wes',
'assistant': 'Peter C.',
'prereq': 'cs101'},
'cs373': {'name': 'Programming a Robotic Car',
'teacher': 'Sebastian'},
'cs387': {'name': 'Applied Cryptography',
'teacher': 'Dave'}},
'jan2044': { 'cs001': {'name': 'Building a Quantum Holodeck',
'teacher': 'Dorina'},
'cs003': {'name': 'Programming a Robotic Robotics Teacher',
'teacher': 'Jasper'},
}
}
# If you want to loop through the keys in the dictionary,
# you can use the construct below.
# for <key> in <dictionary>:
# <block>
# For example, this procedure returns a list of all the courses offered
# in the given hexamester:
def courses_offered(courses, hexamester):
res = []
for c in courses[hexamester]:
res.append(c)
return res
# You do not need to use this code if you do not want to and may find another,
# simpler method to answer this question, although later ones may require this.
# Define a procedure, is_offered(courses, course, hexamester), that returns
# True if the input course is offered in the input hexamester, and returns
# False otherwise. For example,
#print is_offered(courses, 'cs101', 'apr2012')
#>>> True
#print is_offered(courses, 'cs003', 'apr2012')
#>>> False
# (Note: it is okay if your procedure produces an error if the input
# hexamester is not included in courses.
# For example, is_offered(courses, 'cs101', 'dec2011') can produce an error.)
# However, do not leave any uncommented statements in your code which lead
# to an error as your code will be graded as incorrect.
def is_offered(courses, course, hexamester):
if course in courses[hexamester]:
return True
return False
print is_offered(courses, 'cs101', 'apr2012')
#>>> True
print is_offered(courses, 'cs003', 'apr2012')
#>>> False
print is_offered(courses, 'cs001', 'jan2044')
#>>> True
print is_offered(courses, 'cs253', 'feb2012')
#>>> False
When offered
# -*- coding: utf-8 -*-
# Dictionaries of Dictionaries (of Dictionaries)
# The next several questions concern the data structure below for keeping
# track of Udacity's courses (where all of the values are strings):
# { <hexamester>, { <class>: { <property>: <value>, ... },
# ... },
# ... }
# For example,
courses = {
'feb2012': { 'cs101': {'name': 'Building a Search Engine',
'teacher': 'Dave',
'assistant': 'Peter C.'},
'cs373': {'name': 'Programming a Robotic Car',
'teacher': 'Sebastian',
'assistant': 'Andy'}},
'apr2012': { 'cs101': {'name': 'Building a Search Engine',
'teacher': 'Dave',
'assistant': 'Sarah'},
'cs212': {'name': 'The Design of Computer Programs',
'teacher': 'Peter N.',
'assistant': 'Andy',
'prereq': 'cs101'},
'cs253': {'name': 'Web Application Engineering - Building a Blog',
'teacher': 'Steve',
'prereq': 'cs101'},
'cs262': {'name': 'Programming Languages - Building a Web Browser',
'teacher': 'Wes',
'assistant': 'Peter C.',
'prereq': 'cs101'},
'cs373': {'name': 'Programming a Robotic Car',
'teacher': 'Sebastian'},
'cs387': {'name': 'Applied Cryptography',
'teacher': 'Dave'}},
'jan2044': { 'cs001': {'name': 'Building a Quantum Holodeck',
'teacher': 'Dorina'},
'cs003': {'name': 'Programming a Robotic Robotics Teacher',
'teacher': 'Jasper'},
}
}
# For the following questions, you will find the
# for <key> in <dictionary>:
# <block>
# construct useful. This loops through the key values in the Dictionary. For
# example, this procedure returns a list of all the courses offered in the given
# hexamester:
def courses_offered(courses, hexamester):
res = []
for c in courses[hexamester]:
res.append(c)
return res
# Define a procedure, when_offered(courses, course), that takes a courses data
# structure and a string representing a class, and returns a list of strings
# representing the hexamesters when the input course is offered.
def when_offered(courses,course):
r = []
for hexamester in courses:
for Uclass in courses[hexamester]:
if Uclass == course:
r.append(hexamester)
return r
print when_offered (courses, 'cs101')
#>>> ['apr2012', 'feb2012']
print when_offered(courses, 'bio893')
#>>> []
Involved
# -*- coding: utf-8 -*-
# Dictionaries of Dictionaries (of Dictionaries)
# The next several questions concern the data structure below for keeping
# track of Udacity's courses (where all of the values are strings):
# { <hexamester>, { <class>: { <property>: <value>, ... },
# ... },
# ... }
# For example,
courses = {
'feb2012': { 'cs101': {'name': 'Building a Search Engine',
'teacher': 'Dave',
'assistant': 'Peter C.'},
'cs373': {'name': 'Programming a Robotic Car',
'teacher': 'Sebastian',
'assistant': 'Andy'}},
'apr2012': { 'cs101': {'name': 'Building a Search Engine',
'teacher': 'Dave',
'assistant': 'Sarah'},
'cs212': {'name': 'The Design of Computer Programs',
'teacher': 'Peter N.',
'assistant': 'Andy',
'prereq': 'cs101'},
'cs253':
{'name': 'Web Application Engineering - Building a Blog',
'teacher': 'Steve',
'prereq': 'cs101'},
'cs262':
{'name': 'Programming Languages - Building a Web Browser',
'teacher': 'Wes',
'assistant': 'Peter C.',
'prereq': 'cs101'},
'cs373': {'name': 'Programming a Robotic Car',
'teacher': 'Sebastian'},
'cs387': {'name': 'Applied Cryptography',
'teacher': 'Dave'}},
'jan2044': { 'cs001': {'name': 'Building a Quantum Holodeck',
'teacher': 'Dorina'},
'cs003': {'name': 'Programming a Robotic Robotics Teacher',
'teacher': 'Jasper'},
}
}
# For the following questions, you will find the
# for <key> in <dictionary>:
# <block>
# construct useful. This loops through the key values in the Dictionary. For
# example, this procedure returns a list of all the courses offered in the given
# hexamester:
def courses_offered(courses, hexamester):
res = []
for c in courses[hexamester]:
res.append(c)
return res
# [Double Gold Star] Define a procedure, involved(courses, person), that takes
# as input a courses structure and a person and returns a Dictionary that
# describes all the courses the person is involved in. A person is involved
# in a course if they are a value for any property for the course. The output
# Dictionary should have hexamesters as its keys, and each value should be a
# list of courses that are offered that hexamester (the courses in the list
# can be in any order).
def involved(courses, person):
r = {}
for hexamester in courses:
for Uclass in courses[hexamester]:
for key in courses[hexamester][Uclass]:
if courses[hexamester][Uclass][key] == person:
if hexamester not in r:
r[hexamester] = [Uclass]
else:
r[hexamester].append(Uclass)
return r
# For example:
print involved(courses, 'Dave')
#>>> {'apr2012': ['cs101', 'cs387'], 'feb2012': ['cs101']}
print involved(courses, 'Peter C.')
#>>> {'apr2012': ['cs262'], 'feb2012': ['cs101']}
print involved(courses, 'Dorina')
#>>> {'jan2044': ['cs001']}
print involved(courses,'Peter')
#>>> {}
print involved(courses,'Robotic')
#>>> {}
print involved(courses, '')
#>>> {}
Refactoring
# -*- coding: utf-8 -*-
# 6. In video 28. Update, it was suggested that some of the duplicate code in
# lookup and update could be avoided by a better design. We can do this by
# defining a procedure that finds the entry corresponding to a given key, and
# using that in both lookup and update.
# Here are the original procedures:
def find_entry(bucket, key):
for entry in bucket:
if entry[0] == key:
return entry
return None
def hashtable_update(htable, key, value):
bucket = hashtable_get_bucket(htable, key)
entry = find_entry(bucket, key)
if entry:
entry[1] = value
else:
bucket.append([key, value])
def hashtable_lookup(htable, key):
bucket = hashtable_get_bucket(htable, key)
entry = find_entry(bucket, key)
if entry:
return entry[1]
else:
return None
def make_hashtable(size):
table = []
for unused in range(0, size):
table.append([])
return table
def hash_string(s, size):
h = 0
for c in s:
h = h + ord(c)
return h % size
def hashtable_get_bucket(htable, key):
return htable[hash_string(key, len(htable))]
# Whenever we have duplicate code like the loop that finds the entry in
# hashtable_update and hashtable_lookup, we should think if there is a better way
# to write this that would avoid the duplication. We should be able to rewrite
# these procedures to be shorter by defining a new procedure and rewriting both
# hashtable_update and hashtable_lookup to use that procedure.
# Modify the code for both hashtable_update and hashtable_lookup to have the same
# behavior they have now, but using fewer lines of code in each procedure. You
# should define a new procedure to help with this. Your new version should have
# approximately the same running time as the original version, but neither
# hashtable_update or hashtable_lookup should include any for or while loop, and
# the block of each procedure should be no more than 6 lines long.
# Your procedures should have the same behavior as the originals. For example,
table = make_hashtable(10)
hashtable_update(table, 'Python', 'Monty')
hashtable_update(table, 'CLU', 'Barbara Liskov')
hashtable_update(table, 'JavaScript', 'Brendan Eich')
hashtable_update(table, 'Python', 'Guido van Rossum')
print hashtable_lookup(table, 'Python')
#>>> Guido van Rossum
Memorisation
# [Double Gold Star] Memoization is a way to make code run faster by saving
# previously computed results. Instead of needing to recompute the value of an
# expression, a memoized computation first looks for the value in a cache of
# pre-computed values.
# Define a procedure, cached_execution(cache, proc, proc_input), that takes in
# three inputs: a cache, which is a Dictionary that maps inputs to proc to
# their previously computed values, a procedure, proc, which can be called by
# just writing proc(proc_input), and proc_input which is the input to proc.
# Your procedure should return the value of the proc with input proc_input,
# but should only evaluate it if it has not been previously called.
def cached_execution(cache, proc, proc_input):
if proc_input not in cache:
cache[proc_input] = proc(proc_input)
return cache[proc_input]
# Here is an example showing the desired behavior of cached_execution:
def factorial(n):
print "Running factorial"
result = 1
for i in range(2, n + 1):
result = result * i
return result
cache = {} # start cache as an empty dictionary
### first execution (should print out Running factorial and the result)
print cached_execution(cache, factorial, 50)
print "Second time:"
### second execution (should only print out the result)
print cached_execution(cache, factorial, 50)
# Here is a more interesting example using cached_execution
# (do not worry if you do not understand this, though,
# it will be clearer after Unit 6):
def cached_fibo(n):
if n == 1 or n == 0:
return n
else:
return (cached_execution(cache, cached_fibo, n - 1 )
+ cached_execution(cache, cached_fibo, n - 2 ))
cache = {} # new cache for this procedure
# do not try this at home...at least without a cache!
print cached_execution(cache, cached_fibo,100)
