eecs 110: lec 5: list comprehensions aleksandar kuzmanovic northwestern university

45
EECS 110: Lec 5: List Comprehensions Aleksandar Kuzmanovic Northwestern University http://networks.cs.northwestern.edu/EECS110- s15/

Upload: ralf-nichols

Post on 02-Jan-2016

218 views

Category:

Documents


1 download

TRANSCRIPT

EECS 110: Lec 5: List Comprehensions

Aleksandar Kuzmanovic

Northwestern University

http://networks.cs.northwestern.edu/EECS110-s15/

The building blocks of functiona

l computing

data, sequencesconditionalsrecursion

EECS 110 today

List Comprehensions

map and applications

Homework 1 - submitted

Homework 2 - this coming Sunday…!3 problems

1 lab problem Tuesday

2 python problems

"Quiz" on recursion

def power(b,p):

Names:

Handle negative values of p, as well.

""" returns b to the p power using recursion, not ** inputs: ints b and p output: a float"""

Want more power?

power(5,2) == 25.0

For example, power(5,-1) == 0.2 (or so)

def sajak(s):

sajak('wheel of fortune') == 6

""" returns the number of vowels in the input string, s"""

def power(b,p): """ inputs: base b and power p (an int)

implements: b**p = b*b**(p-1)

"""

if p == 0:

return

if p > 0:

return

else:

# p < 0

return

def power(b,p): """ inputs: base b and power p (an int)

implements: b**p = b*b**(p-1)

"""

if p == 0:

return 1

if p > 0:

return

else:

# p < 0

return

def power(b,p): """ inputs: base b and power p (an int)

implements: b**p = b*b**(p-1)

"""

if p == 0:

return 1

if p > 0:

return b*power(b,p-1)

else:

# p < 0

return

def power(b,p): """ inputs: base b and power p (an int)

implements: b**p = b*b**(p-1)

"""

if p == 0:

return 1

if p > 0:

return b*power(b,p-1)

else:

# p < 0

return 1/power(b,-1*p)

behind the curtainpower(2,3)

def sajak(s):

Base case?

when there are no letters, there are ZERO vowels

if it is NOT a vowel, the answer is

Rec. step?

Look at the initial character.

if it IS a vowel, the answer is

def sajak(s):

Base case?

when there are no letters, there are ZERO vowels

if it is NOT a vowel, the answer is just the number of vowels in the rest of sRec.

step?

Look at the initial character.

if it IS a vowel, the answer is 1 + the number of vowels in the rest of s

def sajak(s):

if len(s) == 0:

return 0

else:

Checking for a vowel: Try #1

Base Case

def sajak(s):

if len(s) == 0:

return 0

else:

Checking for a vowel: Try #1

and

or

not

same as in English!

but each side has to be a complete boolean value!

Base Case

def sajak(s):

if len(s) == 0:

return 0

else:

Checking for a vowel: Try #1

and

or

not

same as in English!

but each side has to be a complete boolean value!

if s[0] == 'a' or s[0] == 'e' or…

Base Case

inChecking for a vowel: Try #2

def sajak(s):

if len(s) == 0:

return 0

else:

Base Case

def sajak(s):

if len(s) == 0:

return 0

else:

if s[0] not in 'aeiou':

return sajak(s[1:])

else:

return 1+sajak(s[1:])

if it is NOT a vowel, the answer is just the number of vowels in the rest of s

if it IS a vowel, the answer is 1 + the number of vowels in the rest of s

Base Case

Rec. Step

sajak('eerier')behind the curtain

The key to understanding recursion is to first understand recursion…

- advice from a student

functional programming

>>> 'fun' in 'functional'True

Key ideas in functional programming

• create small building blocks (functions)

• leverage self-similarity (recursion)

• representation via list structures (data)

Compose these together to solve or investigate problems.

elegant and concisenot maximally efficient for

the computer…vs.

return to recursion

Composing functions into specific applications

Creating general functions that will be useful everywhere (or almost…)

return to recursion

Composing functions into specific applications

Creating general functions that will be useful everywhere (or almost…)

building blocks with which to compose…

sum, range

def sum(L):

""" input: a list of numbers, L

output: L's sum

"""

sum, range

def sum(L):

""" input: a list of numbers, L

output: L's sum

"""

if len(L) == 0:

return 0.0

else:

return L[0] + sum(L[1:])

Base Caseif the input

has no elements, its sum is zero

Recursive Case

if L does have an element, add

that element's value to the sum of the REST of the

list…

This input to the recursive call must be "smaller" somehow…

sum, range

def range(low,hi):

""" input: two ints, low and hi

output: int list from low up to hi

"""

excluding hi

sum, range

def range(low,hi):

""" input: two ints, low and hi

output: int list from low up to hi

"""

if hi <= low:

return []

else:

return

excluding hi

sum, range

def range(low,hi):

""" input: two ints, low and hi

output: int list from low up to hi

"""

if hi <= low:

return []

else:

return [low] + range(low+1,hi)

excluding hi

sum and range

>>> sum(range(101))

Looks sort of scruffy for a 7-year old… !

and 100 more…

Recursion: Good News/Bad News

Recursion is common (fundamental) in functional programming

def dblList(L):

""" Doubles all the values in a list.

input: L, a list of numbers """

if L == []:

return L

else:

return [L[0]*2] + dblList(L[1:])

But you can sometimes hide it away!

Map: The recursion "alternative"

def dbl(x):

return 2*x

def sq(x):

return x**2

def isana(x):

return x=='a’

>>> map( dbl, [0,1,2,3,4,5] )[0, 2, 4, 6, 8, 10]

>>> map( sq, range(6) )[0, 1, 4, 9, 16, 25]

>>> map( isana, 'go away!' )[0, 0, 0, 1, 0, 1, 0, 0]

Hey… this looks a bit False to

me!

(1) map always returns a list

(2) map(f,L) calls f on each item in L

Map !

def dblList(L): """ Doubles all the values in a list. input: L, a list of numbers """ if L == []: return L else: return [L[0]*2] + dblList(L[1:])

Without map

def dbl(x): return x*2

def dblList(L): """ Doubles all the values in a list. input: L, a list of numbers """ return map(dbl, L)

With map!

Map: a higher-order function

In Python, functions can take other functions as input…

def map( f, L ):

KeyConcept

Functions ARE data!

Why use map?

Why use map?

Faster execution in Python – map optimized for operations in lists

More elegant / shorter code, “functional in style”

Avoid rewriting list recursion (build once, use lots)

Mapping without map: List Comprehensions

>>> [ dbl(x) for x in [0,1,2,3,4,5] ][0, 2, 4, 6, 8, 10]

>>> [ x**2 for x in range(6) ][0, 1, 4, 9, 16, 25]

>>> [ c == 'a' for c in 'go away!' ][0, 0, 0, 1, 0, 1, 0, 0]

Anything you want to happen to

each element of a list

output

output

output

input

input

input

name that takes on the value of each element in

turn the list (or string)

any name is OK!

Mapping without map: List Comprehensions

>>> [ dbl(x) for x in [0,1,2,3,4,5] ][0, 2, 4, 6, 8, 10]

>>> [ x**2 for x in range(6) ][0, 1, 4, 9, 16, 25]

>>> [ c == 'a' for c in 'go away!' ][0, 0, 0, 1, 0, 1, 0, 0]

def dbl(x):

return 2*x

def sq(x):

return x**2

def isana(x):

return x=='a’

>>> map( dbl, [0,1,2,3,4,5] )[0, 2, 4, 6, 8, 10]

>>> map( sq, range(6) )[0, 1, 4, 9, 16, 25]

>>> map( isana, 'go away!' )[0, 0, 0, 1, 0, 1, 0, 0]

List Comprehensions

def len(L): if L == []: return 0 else: return 1 + len(L[1:])

len(L)

implemented via raw recursion

sScore(s)

sajak(s)

def sajak(s): if len(s) == 0: return 0 else: if s[0] not in 'aeiou': return sajak(s[1:]) else: return 1+sajak(s[1:])

def sScore(s): if len(s) == 0: return 0 else: return letScore(s[0]) + \

sScore(s[1:])

scrabble score

List Comprehensions

LC = [1 for x in L] return sum( LC )

len(L)

List Comprehensions

LC = [1 for x in L] return sum( LC )

len(L)

sajak(s)LC = [c in 'aeiou' for c in s]return sum( LC )

# of vowels

List Comprehensions

LC = [1 for x in L] return sum( LC )

len(L)

sScore(s)

sajak(s)LC = [c in 'aeiou' for c in s]return sum( LC )

scrabble score

# of vowels

LC = [ letScore(c) for c in s] return sum( LC )

Quiz Write each of these functions concisely using list comprehensions…

Write def count(e,L):

Write def lotto(Y,W):

input: e, any element L, any list or stringoutput: the # of times L contains e example: count('f', 'fluff') == 3

input: Y and W, two lists of lottery numbers (ints)output: the # of matches between Y & W example: lotto([5,7,42,44],[3,5,7,44]) == 3

Y are your numbers

W are the winning numbers

Name(s):

Remember True == 1 and False == 0

Extra! Write def divs(N):

input: N, an int >= 2output: the number of positive divisors of Nexample: divs(12) == 6 (1,2,3,4,6,12)

Quiz

Quiz

LC = [x==e for x in L] return sum( LC )

count(e,L)

Quiz

lotto(Y,W)LC = [c in Y for c in W]return sum( LC )

Quiz

divs(N)LC = [ N%c==0 for c in range(1,N+1)] return sum( LC )

Quiz

LC = [x==e for x in L] return sum( LC )

count(e,L)

divs(N)

lotto(Y,W)LC = [c in Y for c in W]return sum( LC )

LC = [ N%c==0 for c in range(1,N+1)] return sum( LC )

See you at Lab!