Lab 2: HigherOrder Functions, Lambda Expressions
Due by 11:59pm on Wednesday, February 2.
Starter Files
Download lab02.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.
Topics
Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should you get stuck.
Lambda Expressions
Lambda expressions are expressions that evaluate to functions by specifying two things: the parameters and a return expression.
lambda <parameters>: <return expression>
While both lambda
expressions and def
statements create function objects,
there are some notable differences. lambda
expressions work like other
expressions; much like a mathematical expression just evaluates to a number and
does not alter the current environment, a lambda
expression
evaluates to a function without changing the current environment. Let's take a
closer look.
lambda  def  

Type  Expression that evaluates to a value  Statement that alters the environment 
Result of execution  Creates an anonymous lambda function with no intrinsic name.  Creates a function with an intrinsic name and binds it to that name in the current environment. 
Effect on the environment  Evaluating a lambda expression does not create or
modify any variables. 
Executing a def statement both creates a new function
object and binds it to a name in the current environment. 
Usage  A lambda expression can be used anywhere that
expects an expression, such as in an assignment statement or as the
operator or operand to a call expression. 
After executing a def statement, the created
function is bound to a name. You should use this name to refer to the
function anywhere that expects an expression. 
Example 


Currying
We can transform multipleargument functions into a chain of singleargument, higher order functions. For example, we can write a function f(x, y)
as a different
function g(x)(y)
. This is known as currying.
For example, to convert the function add(x, y)
into its curried form:
def curry_add(x):
def add2(y):
return x + y
return add2
Calling curry_add(1)
returns a new function which only performs the addition once the returned function is called with the second addend.
>>> add_one = curry_add(1)
>>> add_one(2)
3
>>> add_one(4)
5
Refer to the textbook for more details about currying.
Required Questions
What Would Python Display?
Important: For all WWPD questions, type
Function
if you believe the answer is<function...>
,Error
if it errors, andNothing
if nothing is displayed.
Q1: WWPD: Lambda the Free
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok q lambda u
As a reminder, the following two lines of code will not display anything in the Python interpreter when executed:>>> x = None >>> x
>>> lambda x: x # A lambda expression with one parameter x
______<function <lambda> at ...>
>>> a = lambda x: x # Assigning the lambda function to the name a
>>> a(5)
______5
>>> (lambda: 3)() # Using a lambda expression as an operator in a call exp.
______3
>>> b = lambda x: lambda: x # Lambdas can return other lambdas!
>>> c = b(88)
>>> c
______<function <lambda> at ...
>>> c()
______88
>>> d = lambda f: f(4) # They can have functions as arguments as well.
>>> def square(x):
... return x * x
>>> d(square)
______16
>>> x = None # remember to review the rules of WWPD given above!
>>> x
>>> lambda x: x
______Function
>>> z = 3
>>> e = lambda x: lambda y: lambda: x + y + z
>>> e(0)(1)()
______4
>>> f = lambda z: x + z
>>> f(3)
______NameError: name 'x' is not defined
>>> higher_order_lambda = lambda f: lambda x: f(x)
>>> g = lambda x: x * x
>>> higher_order_lambda(2)(g) # Which argument belongs to which function call?
______Error
>>> higher_order_lambda(g)(2)
______4
>>> call_thrice = lambda f: lambda x: f(f(f(x)))
>>> call_thrice(lambda y: y + 1)(0)
______3
>>> print_lambda = lambda z: print(z) # When is the return expression of a lambda expression executed?
>>> print_lambda
______Function
>>> one_thousand = print_lambda(1000)
______1000
>>> one_thousand
______# print_lambda returned None, so nothing gets displayed
Q2: WWPD: Higher Order Functions
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok q hofwwpd u
>>> def even(f):
... def odd(x):
... if x < 0:
... return f(x)
... return f(x)
... return odd
>>> steven = lambda x: x
>>> stewart = even(steven)
>>> stewart
______<function ...>
>>> stewart(61)
______61
>>> stewart(4)
______4
>>> def cake():
... print('beets')
... def pie():
... print('sweets')
... return 'cake'
... return pie
>>> chocolate = cake()
______beets
>>> chocolate
______Function
>>> chocolate()
______sweets
'cake'
>>> more_chocolate, more_cake = chocolate(), cake
______sweets
>>> more_chocolate
______'cake'
>>> def snake(x, y):
... if cake == more_cake:
... return chocolate
... else:
... return x + y
>>> snake(10, 20)
______Function
>>> snake(10, 20)()
______30
>>> cake = 'cake'
>>> snake(10, 20)
______30
Parsons Problems
To work on these problems, open the Parsons editor:
python3 parsons
Q3: A Hop, a Skip, and a Jump
Complete hop
, which implements a curried version of the function f(x, y) = y
.
def hop():
"""
Calling hop returns a curried version of the function f(x, y) = y.
>>> hop()(3)(2) # .Case 1
2
>>> hop()(3)(7) # .Case 2
7
>>> hop()(4)(7) # .Case 3
7
"""
"*** YOUR CODE HERE ***"
Q4: Digit Index Factory
Complete the function digit_index_factory
, which takes in two integers k
and num
as input and returns a function. The returned function takes no arguments, and outputs the offset between k
and the rightmost digit of num
. The offset between two numbers is defined to be the number of steps between the two numbers. For example, in 25
, there is an offset of 1
between 2
and 5
.
Note that 0
is considered to have no digits (not even 0
).
def digit_index_factory(num, k):
"""
Returns a function that takes no arguments, and outputs the offset
between k and the rightmost digit of num. If k is not in num, then
the returned function returns 1. Note that 0 is considered to
contain no digits (not even 0).
>>> digit_index_factory(34567, 4)() # .Case 1
3
>>> digit_index_factory(30001, 0)() # .Case 2
1
>>> digit_index_factory(999, 1)() # .Case 3
1
>>> digit_index_factory(1234, 0)() # .Case 4
1
"""
"*** YOUR CODE HERE ***"
Coding Practice
Q5: Lambdas and Currying
Write a function lambda_curry2
that will curry any two argument function
using lambdas.
Your solution to this problem should fit entirely on the return line. You can try first writing a solution without the restriction, and then rewriting it into one line after.
If the syntax check isn't passing: Make sure you've removed the line containing
"***YOUR CODE HERE***"
so that it doesn't get treated as part of the function for the syntax check.
def lambda_curry2(func):
"""
Returns a Curried version of a twoargument function FUNC.
>>> from operator import add, mul, mod
>>> curried_add = lambda_curry2(add)
>>> add_three = curried_add(3)
>>> add_three(5)
8
>>> curried_mul = lambda_curry2(mul)
>>> mul_5 = curried_mul(5)
>>> mul_5(42)
210
>>> lambda_curry2(mod)(123)(10)
3
"""
"*** YOUR CODE HERE ***"
return ______
Use Ok to test your code:
python3 ok q lambda_curry2
Q6: Count van Count
Consider the following implementations of count_factors
and count_primes
:
def count_factors(n):
"""Return the number of positive factors that n has.
>>> count_factors(6)
4 # 1, 2, 3, 6
>>> count_factors(4)
3 # 1, 2, 4
"""
i = 1
count = 0
while i <= n:
if n % i == 0:
count += 1
i += 1
return count
def count_primes(n):
"""Return the number of prime numbers up to and including n.
>>> count_primes(6)
3 # 2, 3, 5
>>> count_primes(13)
6 # 2, 3, 5, 7, 11, 13
"""
i = 1
count = 0
while i <= n:
if is_prime(i):
count += 1
i += 1
return count
def is_prime(n):
return count_factors(n) == 2 # only factors are 1 and n
The implementations look quite similar! Generalize this logic by writing a
function count_cond
, which takes in a twoargument predicate function
condition(n, i)
. count_cond
returns a oneargument function that takes
in n
, which counts all the numbers from 1 to n
that satisfy condition
when called.
def count_cond(condition):
"""Returns a function with one parameter N that counts all the numbers from
1 to N that satisfy the twoargument predicate function Condition, where
the first argument for Condition is N and the second argument is the
number from 1 to N.
>>> count_factors = count_cond(lambda n, i: n % i == 0)
>>> count_factors(2) # 1, 2
2
>>> count_factors(4) # 1, 2, 4
3
>>> count_factors(12) # 1, 2, 3, 4, 6, 12
6
>>> is_prime = lambda n, i: count_factors(i) == 2
>>> count_primes = count_cond(is_prime)
>>> count_primes(2) # 2
1
>>> count_primes(3) # 2, 3
2
>>> count_primes(4) # 2, 3
2
>>> count_primes(5) # 2, 3, 5
3
>>> count_primes(20) # 2, 3, 5, 7, 11, 13, 17, 19
8
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok q count_cond
Submit
Make sure to submit this assignment by running:
python3 ok submit
Optional Questions
Q7: Composite Identity Function
Write a function that takes in two singleargument functions, f
and g
, and
returns another function that has a single parameter x
. The returned
function should return True
if f(g(x))
is equal to g(f(x))
. You can
assume the output of g(x)
is a valid input for f
and vice versa.
Try to use the composer
function defined below for more HOF practice.
def composer(f, g):
"""Return the composition function which given x, computes f(g(x)).
>>> add_one = lambda x: x + 1 # adds one to x
>>> square = lambda x: x**2
>>> a1 = composer(square, add_one) # (x + 1)^2
>>> a1(4)
25
>>> mul_three = lambda x: x * 3 # multiplies 3 to x
>>> a2 = composer(mul_three, a1) # ((x + 1)^2) * 3
>>> a2(4)
75
>>> a2(5)
108
"""
return lambda x: f(g(x))
def composite_identity(f, g):
"""
Return a function with one parameter x that returns True if f(g(x)) is
equal to g(f(x)). You can assume the result of g(x) is a valid input for f
and vice versa.
>>> add_one = lambda x: x + 1 # adds one to x
>>> square = lambda x: x**2
>>> b1 = composite_identity(square, add_one)
>>> b1(0) # (0 + 1)^2 == 0^2 + 1
True
>>> b1(4) # (4 + 1)^2 != 4^2 + 1
False
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok q composite_identity
Q8: I Heard You Liked Functions...
Define a function cycle
that takes in three functions f1
, f2
,
f3
, as arguments. cycle
will return another function that should
take in an integer argument n
and return another function. That
final function should take in an argument x
and cycle through
applying f1
, f2
, and f3
to x
, depending on what n
was. Here's what the final function should do to x
for a few
values of n
:
n = 0
, returnx
n = 1
, applyf1
tox
, or returnf1(x)
n = 2
, applyf1
tox
and thenf2
to the result of that, or returnf2(f1(x))
n = 3
, applyf1
tox
,f2
to the result of applyingf1
, and thenf3
to the result of applyingf2
, orf3(f2(f1(x)))
n = 4
, start the cycle again applyingf1
, thenf2
, thenf3
, thenf1
again, orf1(f3(f2(f1(x))))
 And so forth.
Hint: most of the work goes inside the most nested function.
def cycle(f1, f2, f3):
"""Returns a function that is itself a higherorder function.
>>> def add1(x):
... return x + 1
>>> def times2(x):
... return x * 2
>>> def add3(x):
... return x + 3
>>> my_cycle = cycle(add1, times2, add3)
>>> identity = my_cycle(0)
>>> identity(5)
5
>>> add_one_then_double = my_cycle(2)
>>> add_one_then_double(1)
4
>>> do_all_functions = my_cycle(3)
>>> do_all_functions(2)
9
>>> do_more_than_a_cycle = my_cycle(4)
>>> do_more_than_a_cycle(2)
10
>>> do_two_cycles = my_cycle(6)
>>> do_two_cycles(1)
19
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok q cycle