[Tut] Python One Line FizzBuzz

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Python One Line FizzBuzz

The FizzBuzzproblem is a common exercise posed in code interviews to test your proficiency in writing simple Python code.

Problem: Print all numbers from 1-100 to the shell with three exceptions:

• For each number divisible by three you print
  Code:

  "Fizz"
  ,
  
  
• For each number divisible by five you print
  Code:

  "Buzz"
  , and
  
  
• For each number divisible by three and five you print
  Code:

  "FizzBuzz"
  .
  

Example: The first 15 numbers of the FizzBuzz sequence are the following.

Code:

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz ...

How to write a Python one-liner that solves this problem?

Here’s an interactive overview:

https://repl.it/@finxter/RashMintyConstants?lite=true

Exercise: Do both one-liners produce the same results? Run the code to check!

Let’s dive into those one-liners to gain a deeper understanding and improve your Python skills!

FizzBuzz One-Liner 1: Generator Expression + String Concatenation + Short Circuiting

The following one-liner solves the problem in an elegant way using a fine understanding of more advanced Python features (source).

Code:

print('\n'.join('Fizz' * (i%3==0) + 'Buzz' * (i%5==0) or str(i) for i in range(1,101)))

The one-liner creates a string using the

Code:

join

function with the newline character as a delimiter. Here’s a short explanation of the function:

The

Code:

string.join(iterable)

method concatenates all the string elements in the

Code:

iterable

(such as a list, string, or tuple) and returns the result as a new string. The

Code:

string

on which you call it is the delimiter string—and it separates the individual elements. For example,

Code:

'-'.join(['hello', 'world'])

returns the joined string

Code:

'hello-world'

.

So, what’s the iterable, you pass into the

Code:

join()

function? It’s a generator expression of the form:

Code:

expression for variable in context

. You go over all integer values in the context 1 to 100 using the

Code:

range()

function. So, you obtain the remaining

Code:

expression for i in range(1, 101)

. What’s the expression part?

It consists of three elements:

• Code:

  'Fizz' * (i%3==0)
  — The moduloexpression
  Code:

  i%3==0
  returns
  Code:

  True
  only if the integer
  Code:

  i
  is divisible by 3, otherwise it returns
  Code:

  False
  . So, you multiply the string
  Code:

  'Fizz'
  either with
  Code:

  True
  (=1) or with
  Code:

  False
  (=0). As a result, you obtain the empty string
  Code:

  ''
  in all cases except if the integer
  Code:

  i
  is divisible by 3—in which case you obtain the string
  Code:

  'Fizz'
  .
  
  
• Code:

  'Buzz' * (i%5==0)
  — The moduloexpression
  Code:

  i%5==0
  returns
  Code:

  True
  only if the integer
  Code:

  i
  is divisible by 5, otherwise it returns
  Code:

  False
  . So, you multiply the string
  Code:

  'Buzz'
  either with
  Code:

  True
  (=1) or with
  Code:

  False
  (=0). As a result, you obtain the empty string
  Code:

  ''
  in all cases except if the integer
  Code:

  i
  is divisible by 5—in which case you obtain the string
  Code:

  'Buzz'
  .
  
  
• You use string concatenation to glue together the previously obtained strings. In most cases, this will be the empty string. If
  Code:

  i
  is divisible by 3, you obtain the string
  Code:

  'Fizz'
  . If
  Code:

  i
  is divisible by 5, you obtain the string
  Code:

  'Buzz'
  . And if
  Code:

  i
  is divisible by 3 and 5, you obtain the string
  Code:

  'FizzBuzz'
  .
  
  
• Code:

  or str(i)
  — In the case you obtained a non-empty string in
  Code:

  {'Fizz', 'Buzz', 'FizzBuzz'}
  in the previous step, the
  Code:

  or
  operation simply returns this string. This is called short circuiting—and it’s used in many programming languages such as Python to improve efficiency of logical operations.
  
  
• But if the string is empty, it is interpreted as a logical
  Code:

  False
  . Thus, Python returns the second operand of the
  Code:

  or
  operation. The second operand simply is the string representation of the integer
  Code:

  i
  .
  

A very interesting implementation of the FizzBuzz problem indeed!

FizzBuzz One-Liner 2: Slicing

An alternative is given in the following nice one-liner (source):

Code:

for i in range(1, 101): print('FizzBuzz'[i*i%3*4:8--i**4%5] or i)

Wow—what a short and concise one-liner solution! But how does it work?

• You iterate over all values from
  Code:

  i=1
  to
  Code:

  i=100
  and print a string. So far so good.
  
  
• You use the
  Code:

  or
  operation and slicing to determine the string
  Code:

  'FizzBuzz'[start:end] or i
  generates the output.
  
  
• You use the property of short circuiting in Python: If
  Code:

  'FizzBuzz'[start:end]
  is empty, the integer
  Code:

  i
  is returned, otherwise, the non-empty string is returned.
  
  
• You carve out a substring from
  Code:

  'FizzBuzz'
  using slicingas follows.
  

Slicing is a concept to carve out a substring from a given string. Use slicing notation

Code:

s[start:stop:step]

to access every

Code:

step

-th element starting from index

Code:

start

(included) and ending in index

Code:

stop

(excluded). All three arguments are optional, so you can skip them to use the default values (

Code:

start=0

,

Code:

stop=len(lst)

,

Code:

step=1

). For example, the expression

Code:

s[2:4]

from string

Code:

'hello'

carves out the slice

Code:

'll'

and the expression

Code:

s[:3:2]

carves out the slice

Code:

'hl'

.

In the example, you have the slicing operation

Code:

'FizzBuzz'[i*i%3*4:8--i**4%5]

.

• Code:

  start = i*i%3*4
  — Note that the multiplication
  Code:

  *
  and modulo operation
  Code:

  %
  have the same priority, so they are evaluated from left to right. If integer
  Code:

  i
  is divisible by 3,
  Code:

  i*i
  is also divisible by 3, and the start index is 0. In all other cases, the start index is 4. Thus, the slice either starts with
  Code:

  'Fizz'
  or
  Code:

  'Buzz'
  .
  
  
• Code:

  stop = 8--i**4%5
  — This is 4 in all cases except if the number
  Code:

  i
  is divisible by 5, in which case this is 8.
  

So, there are four cases:

• The number is divisible only by 3:
  Code:

  start=0
  ,
  Code:

  stop=4
  –>
  Code:

  'Fizz'
  
  
• The number is divisible only by 5:
  Code:

  start=4
  ,
  Code:

  stop=8
  –>
  Code:

  'Buzz'
  
  
• The number is divisible by both 3 and 5:
  Code:

  start=0
  ,
  Code:

  stop=8
  –>
  Code:

  'FizzBuzz'
  
  
• The number is divisible by neither 3 nor 5:
  Code:

  start = 4
  ,
  Code:

  stop=4
  –>
  Code:

  ''

Phew! This was a hard nut to crack, wasn’t it?

Python One-Liner 3: Map + Lambda

You can find detailed tutorials on the map and lambda functions here:

• https://blog.finxter.com/daily-python-pu...-function/Mastering the Python Map Function
  
  
• https://blog.finxter.com/a-simple-introd...in-python/Lambda Functions in Python
  

Those two functions can be used to solve the FizzBuzz problem (source):

Code:

print(list(map(lambda i: "Fizz"*(i%3==0)+"Buzz"*(i%5==0) or str(i), range(1,101))))

It’s similar to Method 1 and by now you’re able to figure it out. Think of the different values the integer

Code:

i

can take.

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