Python MIT OOPS
[b]
L3:
Italian program
commands
assignment
input/output
condition
loop
print x, 'hello world'
Defensive programming
@ToDo: To test the code List: mylist = [] mylist = mylist + [i] #Here the '+' is overridden already.
mytuple = () mytuple = mytuple + (i) # Does it really work? To be checked.
Answer: Both can be.
[code="python"]mytuple = ('abc')
mytuple = mytuple + ('de')
print mytuple #Show 'abcde' instead of ('abc','de')
[code="python"]mytuple = ('abc','cde')
[code="python"]mytuple += ('fg') #Cause error. Last example because, the tuple convert to string, and the () is ignored.
[img]/admin/blogs/images/callouts/1.png" border="0" alt="1" width="12" height="12[/img]
You can't add elements to a tuple. Tuples have no append
or extend
method.
[img]/admin/blogs/images/callouts/2.png" border="0" alt="2" width="12" height="12[/img]
You can't remove elements from a tuple. Tuples have no remove
or pop
method.
[img]/admin/blogs/images/callouts/3.png" border="0" alt="3" width="12" height="12[/img]
You can't find elements in a tuple. Tuples have no index
method.
[img]/admin/blogs/images/callouts/4.png" border="0" alt="4" width="12" height="12[/img]
You can, however, use in
to see if an element exists in the tuple.
L4:
@Todo: To simulate
return None , no value assign to variable, but when use '==' to check, it return 'true'. Too suprised
[code="python"]def returnNone():
return None
result = returnNone()
print result # show None # In shell mode, it would be nothing
print (result == None) #show true
@Todo: 20heads, 56 legs, how many chicken and pigs
Done, looping all possible values
@Todo: to verify it return [None,None] return [a,b,c] #return more than 1 value a,b,c = return [a,b,c] #assign to more than 1 variable
Done, actually, it return the tuple or list
@Todo: check Fibonacci number Use recursive method to calculate the numbers
Done, remember to handle the 0 & 1 number.
@Todo:Palindrome, to be write program to test
LS5:
@Todo: 2**1000
In shell prompt, it would large number with 'L'. When use print, auto conver and remove L
@Todo
: x=0.1, print x IEEE 754 floating point @Todo 64bits?? modem computer 64bit? what's that? 1 bit negative/positive 11 bit exponent 52 bit number (mantissa) 17 decimal number @Todo: for i in range(10) s+=0.1 actual result 0.999999999999989, when print, it auto round it.
successive approxmaition (极限法)
bisection method (二分法)
@Todo , try assert in sqrt method
Done. assert (condition), 'not match condition , error and terminate'
@Todo: find out sqrt To solve decimal fraction issue, to tune the program for the espion.
Done, use x*x - y 0):
ans *= a
b -= 1
return ans
Use big O notation
汉诺塔
@Todo: to simulate the twoner
Done
[code="python"]def move (element, from, target)
print 'move',element, 'from', from, 'target', target
def tower(i, from, target, tmp)
if (i == 1):
move(i,from,target)
elsif
#move n tower to tmp
tower(i-1,from,tmp,target)
#move last one to target
move(i,from,target)
#move n-1 tower to target
twoer(i-1,tmp,target,from)
Complexity calculation as below
t(n) = t(n-2) + t(n-2) + t(n-2) + t(n-2)
t(n) = 2^k ( t(n-k) )
t(n) = 2^(n-1) t(1)
4种算法
@Todo: compare the 4 different alg
@Todo: write Binary search, then compare the one by one search
[code="python"]bsearch(num, list, first, last):
i= ( first + last ) / 2
if list[i] == num:
return true, i
elif last - first == 1:
return false
elif list[i] num:
return bsearch (num, list, first, i)
#Complexity calculation
t(n) = t(n/2)
t(n) = t(n/2/2)
t(n) = t(n/2/2/2)
t(n) = t(n/2^k)
2^k= n
k = log2(n)
t(n) = O(
@Todo: check link list ,constant list
LinkList: use 2 cell, one for value, the other for next element memory address, when add new one, just change one element, if you find one, you need to iternative some element, next next to skip skip ...
Depends on what element on your hand, if next element is far away, it cost more time.
Constant List: fixed space of each cell value, if add new element, it would reconstrut list
.
Access each element cost same time. You can calculate each skip steps
L9
:
linear
list, python, list of list
use box to save the pointer, each pointer point to a special area contains value.
Before search, sorted or not?
Linear search
N times
@Todo: check linklist, how to store object of object
Sorted + Search
nlog2(n), + log2(n)
If only search 1 time, use linear search
ksearch times
@Use selection sorting method with binary search, found out how many element that worth to sort before search
@Todo: Bubble sorting & selection sorting
Sorting method
Swap the small one
Selection sorting
Bubble section
Done
@Todo, dynamic function.
val=int
val(x)
@Todo list have sort function, to be try to found out how fast of it.
@Todo try assert
raise 'exception'
assert precondition
LS11
function debugging
performance debugging
LS11
Defensive programming
Both validation & debugging
Test: examine input/output
Unit Testing: validate each piece of program indivually, functions classes
Integration testing: PUt whole program togather, see does the program work?
Rush to integration test, but it's big mistake.
Test each part of unit
Easier to test & debug small thing than big thing
Why testing is always challenge
2 best way
Print Statement
Reading
Be systemmatic
@Todo clone list
list = otherlist[:]
different
list = otherlist
LS12[/b]
code should not always grow
Add more code to fix bug
Big Packing
Shortest path
Travles sales person
Sequence alignment
Knapsack problem
Greedy algor
locall optimal decision do not always lead to global opt.
@Todo: Knapsack problem
@Todo: To check optimal fibionic sequence because overlapping problem
overlapping subproblem
optimal structure
Optimal fibionic sequence
fibnoic, optimal to save the value
[code="python"]memres = {0:0,1:1} #init first elements
def getfib(num=10):
if num in memres: #check key exist:
return memres[num]
memres[num] = getfib(num-1) + getfib(num-2)
return memres[num]
#complexity, O(n) if not save value, O(2^n)
Decision tree
w=5,3,2
V=9,7,8
Each n
2,5,0
index,weight still available,value
2,5,0
1,5,0
0,5,0
@todo, write a decision tree daigram and code
@Todo:use try and raise the exception to check whether list element existing
[code="python"]num=10
assert num>7, 'error' #Assume the exper is correct, otherwise prompt error msg.
decision tree @to check
1. Depth first , left first
2. Back check
LS14:
Binary decision tree
Size of the problem? Size of the solution?
@Todo: problem? Solution?
@Knapsack problem, constraint volumn also
@Todo: what's pseudo polynomic?
LS15
Shallow equality ( point to same reference)
Deep equality
pass keyword
__init__(self,x,y)
__str__
__cmp__
__eq__ @tocheck this overloading use ==
dir(class), you know how many methods
dir(1)
dir('1')
LS16
class template for data type
data hidding, encapution
class Person(object) #Extend from object
if type (xxx) == QQ
@Todo type to judge class
LS17
class Location(object)
def __init_(self,x,y)
x = float(x)
y = y...
def move(xc,yc)
return Location(x+xc,y+yc)
def get Coords
return x,y
def getDis(other)
cx,cy = getCoords(other)
xd = x -xc
yd = y-yc
return math.sqrt(xd**2,yd**2)
class CompassPt(obj)
possibles = {n,w,w,..}
def init(pt)
if pt in possibles, self.pt = pt
else: raise error
def move (dist)
if self.pt == n, return (0,dist)
elif pt == S, return (0,-dist)
w (dist,0)
e (0,dist)
else: raise error
class field(obj)
def init (drunk, loc)
self.d = d
self.l = l
def move (self, cp , dist)
oldLoc = self.d
xc,yc = cp.move(dist)
self.loc = oldLoc.move(xc,yc)
def getLoc(self)
return self.loc
def getDrunk(self)
return self.drunk
class Drunk(obj)
def init(self,name)
def move(self, field, time = 1)
if field.getDrunk() == self:
raise error
for i in range(time)
pt = CompressPt(random.choice(CompressPt.possibles))
field.move(pt,1)
def performTrial(time,f):
start = f.getLoc()
distance = [0,0]
for t in range(1,time+1):
f.getDrunk().move(f)
newLoc = f.getLoc()
distance = newLoc.getDist(start)
distance.append(distance)
return distance
drunk = Drunk(hello world)
for i in range(j):
f = Field(drunk, Location(0,0))
distance = performTrial(500,f)
pylab.plot(distances)
pylab.title('hello')
pylab.xlabel(ok)
pylab.ylabel(okok)
pylab.show