Foundations of Python

134 / 134

Lambda Function and List Comprehension

Not able to play video? Try with vimeo

A lambda function is a small anonymous function and it is defined by the keyword "lambda".

A lambda function can take any number of arguments, but can only have one expression.

A lambda function is defined and used in the same statement.


lambda arguments : expression


Increment a number by 1.

x = lambda a : a + 10

Multiply two arguments return the result:

x = lambda a, b : a * b
print(x(3, 4))

The concept of lambda can be leveraged further by using this inside a normal function. In the below example, a generic function is created which acts a doubler, tripler or quadrupler or any multiplier.

def multiplier(n):
  return lambda a : a * n

doubler = multiplier(2)

tripler= multiplier(3)

# or can do this directly

You can also do operations like filter, map, etc using lambda which can also be using normal functions.

In the example below, we will filter the numbers divisible by 3 using different methods.

Method 1: Using a normal function

foo = [2, 18, 9, 22, 17, 24, 8, 12, 27]

def filter_divisbile_by_n(n, numlist):
    return [num for num in numlist if num%n==0]

divisible_by_3 = filter_divisbile_by_n(3, foo)

print("Method 1 output is:")
print(divisible_by_3, "\n")

Method 2: Using a function as an argument

def is_divisbile_by_n(n, num):
    return num%n==0

def filter_divisbile_by_n(is_divisbile_by_n, n, numlist):
    return [num for num in numlist if is_divisbile_by_n(n, num)]

divisible_by_3 = filter_divisbile_by_n(is_divisbile_by_n, 3, foo)

print("Method 2 output is:")
print(divisible_by_3, "\n")

Method 3: Using a lambda as an argument

def filter_divisbile_by_n(div_func, numlist):
    return [num for num in numlist if div_func(num)]

divisible_by_3 = filter_divisbile_by_n(lambda x: x%3 == 0, foo)

print("Method 3 output is:")
print(divisible_by_3, "\n")

Method 4: Using a filter and normal function

def is_divisbile_by_3(num):
    return num%3==0
divisible_by_3 = filter(is_divisbile_by_3, foo)
print("Method 4 output is:")

Method 5: Using a filter and lambda function

divisible_by_3 = filter(lambda x: x%3 == 0, foo)
print("Method 5 output is:")

Method 6: Using a lambda inside a function

def is_divisbile_by_n(n):
    return lambda num: num%n == 0
divisible_by_3 = filter(is_divisbile_by_n(3), foo)
print("Method 6 output is:")

Using lambda with map In the below example, we double the value of each element.

doubled_list = map(lambda x: x*2, foo) print(list(doubled_list))

Using lambda with reduce In the below example, we sum up all elements. from functools import reduce summed_list = reduce(lambda x, y: x + y, foo) print(summed_list)

Loading comments...