Python3: List Comprehensions—Basics

List comprehensions are a compact, Pythonic way of writing for loops. In this series, we will learn how to write list comprehensions through a series of simple looping exercises. The solutions provided are not comprehensive (pun unintended), so take the time to come up with your own solutions or improve on the ones shown!

Basic iterations

Let’s take a look at the parts of a generic for loop:

for {element} in {list}:
    {action}

A list comprehension has the same parts, just rearranged:

[{action} for {element} in {list}]

Exercise 1:

'''
Determine whether the integers in a list are even (True) or odd (False).
'''

[1, 2, 3, 4]
>>>[False, True, False, True]

Essentially, we want to return the following list:

>>>[(1 % 2 == 0), (2 % 2 == 0), (3 % 2 == 0), (4 % 2 == 0)]

This can be achieved using a for loop:

even_list = []
for num in [1, 2, 3, 4]:
    even_list.append(num % 2 == 0)
print(even_list)

However, we can use list comprehension to write more compact code:

even_list = [(num % 2 == 0) for num in [1, 2, 3, 4]]
print(even_list)

If you’re still confused, see if this helps:

[num for num in [1, 2, 3, 4]]
>>>[1, 2, 3, 4]

[(num % 2 == 0) for num in [1, 2, 3, 4]]
>>>[(1 % 2 == 0), (2 % 2 == 0), (3 % 2 == 0), (4 % 2 == 0)]
>>>[False, True, False, True]

Exercise 2:

'''
Convert all month names to their mmm abbreviations.
'''

months = [
  'January',
  'February',
  'March',
  'April',
  'May',
  'June',
  'July',
  'August',
  'September',
  'October',
  'November',
  'December'
  ]
>>>['jan', 'feb', 'mar', 'apr', 'may', 'jun', 'jul', 'aug', 'sep', 'oct', 'nov', 'dec']

for loop solution:

mmm = []
for month in months:
    mmm.append(month[:3].lower())
print(mmm)

list comprehension solution:

mmm = [month[:3].lower() for month in months]
print(mmm)

Conditional iterations

using if statements

If we only have an if conditional, it follows the {list} section in a list comprehension.

[{action} for {element} in {list} if {condition}]

##

'''
Define a function that replaces every letter of a string with its position in the alphabet.
(source: codewars)
'''

alphabet_position('hello 123')
>>>"8 5 12 12 15"

(one of many) for loop solutions:

def alphabet_position(text):
    az = 'abcdefghijklmnopqrstuvwxyz'
    result = []
    for c in text.lower():
        if c.isalpha():
            result.append(str(az.index(c) + 1))
    return ' '.join(result)

the corresponding list comprehension solution:

def alphabet_position(text):
    az = 'abcdefghijklmnopqrstuvwxyz'
    return ' '.join([str(az.index(c) + 1) for c in text.lower() if c.isalpha()])

using if-else statements

However, if we have an if-else conditional, it follows the {action} section in a list comprehension.

[{action1} if {condition1} else {action2} for {element} in {list}]

'''
Write a function that scores a list of cards in Blackjack. Return the highest score of cards that is less than or equal to 21 or the lowest score of cards that is greater than 21.
(source: codewars)
'''

score_hand(['2', '8', 'J'])
>>>20

score_hand(['A', 'A'])
>>>12

score_hand(['A', 'J', '2'])
>>>23

(one of many) for loop solutions:

def score_hand(cards):
    score = 0
    aces = cards.count('A')
    score += aces
    for card in cards:
        if card in 'JKQ':
            score += 10
        elif card.isnumeric():
            score += int(card)
    if (21 - score >= 10) and (aces > 0):
        score += 10
    return score

the corresponding list comprehension solution:

def score_hand(cards):
    aces = cards.count('A')
    score = sum([10 if c in 'JQK' else int(c) for c in cards]) + aces
    if (21 - score >= 10) and (aces > 0):
        score += 10
    return score

using multiple if-else iterations part 2

We can even incorporate multiple conditions into a list comprehension.

[{action1} if {condition1} else {action2} if {condition2} else {action3} for {element} in {list}]

Exercise 6: Nested iterations