How to use filter, map, and reduce in Python 3

The functionality of map and filter was intentionally changed to return iterators, and reduce was removed from being a built-in and placed in functools.reduce.

So, for filter and map, you can wrap them with list() to see the results like you did before.

>>> def f(x): return x % 2 != 0 and x % 3 != 0
...
>>> list(filter(f, range(2, 25)))
[5, 7, 11, 13, 17, 19, 23]
>>> def cube(x): return x*x*x
...
>>> list(map(cube, range(1, 11)))
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]
>>> import functools
>>> def add(x,y): return x+y
...
>>> functools.reduce(add, range(1, 11))
55
>>>

The recommendation now is that you replace your usage of map and filter with generators expressions or list comprehensions. Example:

>>> def f(x): return x % 2 != 0 and x % 3 != 0
...
>>> [i for i in range(2, 25) if f(i)]
[5, 7, 11, 13, 17, 19, 23]
>>> def cube(x): return x*x*x
...
>>> [cube(i) for i in range(1, 11)]
[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]
>>>

They say that for loops are 99 percent of the time easier to read than reduce, but I’d just stick with functools.reduce.

Edit: The 99 percent figure is pulled directly from the What’s New In Python 3.0 page authored by Guido van Rossum.

As an addendum to the other answers, this sounds like a fine use-case for a context manager that will re-map the names of these functions to ones which return a list and introduce reduce in the global namespace.

A quick implementation might look like this:

from contextlib import contextmanager    

@contextmanager
def noiters(*funcs):
    if not funcs: 
        funcs = [map, filter, zip] # etc
    from functools import reduce
    globals()[reduce.__name__] = reduce
    for func in funcs:
        globals()[func.__name__] = lambda *ar, func = func, **kwar: list(func(*ar, **kwar))
    try:
        yield
    finally:
        del globals()[reduce.__name__]
        for func in funcs: globals()[func.__name__] = func

With a usage that looks like this:

with noiters(map):
    from operator import add
    print(reduce(add, range(1, 20)))
    print(map(int, ['1', '2']))

Which prints:

190
[1, 2]

Just my 2 cents 🙂

Since the reduce method has been removed from the built in function from Python3, don’t forget to import the functools in your code. Please look at the code snippet below.

import functools
my_list = [10,15,20,25,35]
sum_numbers = functools.reduce(lambda x ,y : x+y , my_list)
print(sum_numbers)

Here are the examples of Filter, map and reduce functions.

numbers = [10,11,12,22,34,43,54,34,67,87,88,98,99,87,44,66]

//Filter

oddNumbers = list(filter(lambda x: x%2 != 0, numbers))

print(oddNumbers)

//Map

multiplyOf2 = list(map(lambda x: x*2, numbers))

print(multiplyOf2)

//Reduce

The reduce function, since it is not commonly used, was removed from the built-in functions in Python 3. It is still available in the functools module, so you can do:

from functools import reduce

sumOfNumbers = reduce(lambda x,y: x+y, numbers)

print(sumOfNumbers)

One of the advantages of map, filter and reduce is how legible they become when you “chain” them together to do something complex. However, the built-in syntax isn’t legible and is all “backwards”. So, I suggest using the PyFunctional package (https://pypi.org/project/PyFunctional/).
Here’s a comparison of the two:

flight_destinations_dict = {'NY': {'London', 'Rome'}, 'Berlin': {'NY'}}

PyFunctional version

Very legible syntax. You can say:

“I have a sequence of flight destinations. Out of which I want to get
the dict key if city is in the dict values. Finally, filter out the
empty lists I created in the process.”

from functional import seq  # PyFunctional package to allow easier syntax

def find_return_flights_PYFUNCTIONAL_SYNTAX(city, flight_destinations_dict):
    return seq(flight_destinations_dict.items()) 
        .map(lambda x: x[0] if city in x[1] else []) 
        .filter(lambda x: x != []) 

Default Python version

It’s all backwards. You need to say:

“OK, so, there’s a list. I want to filter empty lists out of it. Why?
Because I first got the dict key if the city was in the dict values.
Oh, the list I’m doing this to is flight_destinations_dict.”

def find_return_flights_DEFAULT_SYNTAX(city, flight_destinations_dict):
    return list(
        filter(lambda x: x != [],
               map(lambda x: x[0] if city in x[1] else [], flight_destinations_dict.items())
               )
    )

from functools import reduce

def f(x):
    return x % 2 != 0 and x % 3 != 0

print(*filter(f, range(2, 25)))
#[5, 7, 11, 13, 17, 19, 23]

def cube(x):
    return x**3
print(*map(cube, range(1, 11)))
#[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]

def add(x,y):
    return x+y

reduce(add, range(1, 11))
#55

It works as is. To get the output of map use * or list

Lambda

Try to understand the difference between a normal def defined function and lambda function. This is a program that returns the cube of a given value:

# Python code to illustrate cube of a number 
# showing difference between def() and lambda(). 
def cube(y): 
    return y*y*y 
  
lambda_cube = lambda y: y*y*y 
  
# using the normally 
# defined function 
print(cube(5)) 
  
# using the lamda function 
print(lambda_cube(5)) 

output:

125
125

Without using Lambda:

• Here, both of them return the cube of a given number. But, while using def, we needed to define a function with a name cube and needed to pass a value to it. After execution, we also needed to return the result from where the function was called using the return keyword.

Using Lambda:

• Lambda definition does not include a “return” statement, it always contains an expression that is returned. We can also put a lambda definition anywhere a function is expected, and we don’t have to assign it to a variable at all. This is the simplicity of lambda functions.

---

Lambda functions can be used along with built-in functions like filter(), map() and reduce().

lambda() with filter()

The filter() function in Python takes in a function and a list as arguments. This offers an elegant way to filter out all the elements of a sequence “sequence”, for which the function returns True.

my_list = [1, 5, 4, 6, 8, 11, 3, 12]

new_list = list(filter(lambda x: (x%2 == 0) , my_list))

print(new_list)


ages = [13, 90, 17, 59, 21, 60, 5]

adults = list(filter(lambda age: age>18, ages)) 
  
print(adults) # above 18 yrs 

output:

[4, 6, 8, 12]
[90, 59, 21, 60]

lambda() with map()

The map() function in Python takes in a function and a list as an argument. The function is called with a lambda function and a list and a new list is returned which contains all the lambda modified items returned by that function for each item.

my_list = [1, 5, 4, 6, 8, 11, 3, 12]

new_list = list(map(lambda x: x * 2 , my_list))

print(new_list)


cities = ['novi sad', 'ljubljana', 'london', 'new york', 'paris'] 
  
# change all city names 
# to upper case and return the same 
uppered_cities = list(map(lambda city: str.upper(city), cities)) 
  
print(uppered_cities)

output:

[2, 10, 8, 12, 16, 22, 6, 24]
['NOVI SAD', 'LJUBLJANA', 'LONDON', 'NEW YORK', 'PARIS']

reduce

reduce() works differently than map() and filter(). It does not return a new list based on the function and iterable we’ve passed. Instead, it returns a single value.

Also, in Python 3 reduce() isn’t a built-in function anymore, and it can be found in the functools module.

The syntax is:

reduce(function, sequence[, initial])

reduce() works by calling the function we passed for the first two items in the sequence. The result returned by the function is used in another call to function alongside with the next (third in this case), element.

The optional argument initial is used, when present, at the beginning of this "loop" with the first element in the first call to function. In a way, the initial element is the 0th element, before the first one, when provided.

lambda() with reduce()

The reduce() function in Python takes in a function and a list as an argument. The function is called with a lambda function and an iterable and a new reduced result is returned. This performs a repetitive operation over the pairs of the iterable.

from functools import reduce

my_list = [1, 1, 2, 3, 5, 8, 13, 21, 34] 

sum = reduce((lambda x, y: x + y), my_list) 

print(sum) # sum of a list
print("With an initial value: " + str(reduce(lambda x, y: x + y, my_list, 100)))

88
With an initial value: 188

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These functions are convenience functions. They are there so you can avoid writing more cumbersome code, but avoid using both them and lambda expressions too much, because "you can", as it can often lead to illegible code that’s hard to maintain. Use them only when it’s absolutely clear what’s going on as soon as you look at the function or lambda expression.