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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

12

$begingroup$

Task

Old mobile phones had the ability to type characters by pressing a number. The letter a could be typed by pressing 2 once. The letter b could be typed by pressing 2 twice.

Given a sequence of numbers, give all possible letter combinations.

For example: The number 23 could give an output ad, ae, af, bd, be, bf, cd, ce, cf

My recursive solution to this problem is given below.

def num_to_char(value):
 if value == 2: return ["a","b","c"]
 if value == 3: return ["d","e","f"]
 if value == 4: return ["g","h","i"]
 if value == 5: return ["j","k","l"]
 if value == 6: return ["m","n","o"]
 if value == 7: return ["p","q","r","s"]
 if value == 8: return ["t","u","v"]
 if value == 9: return ["w","x","y","z"]

def convert_num(number, current_string = ""):
 if number == []:
 print(current_string)
 return 
 get_list = num_to_char(int(number[0]))
 for character in get_list:
 current_string += character
 convert_num(number[1:], current_string)
 current_string = current_string[:-1]

num_to_covert = list("234")
convert_num(num_to_covert)

python python-3.x programming-challenge

share|improve this question

edited Jun 2 at 20:21

200_success

134k21166439

asked Jun 2 at 20:14

EMLEML

682113

$endgroup$

add a comment | 

12

$begingroup$

Task

Old mobile phones had the ability to type characters by pressing a number. The letter a could be typed by pressing 2 once. The letter b could be typed by pressing 2 twice.

Given a sequence of numbers, give all possible letter combinations.

For example: The number 23 could give an output ad, ae, af, bd, be, bf, cd, ce, cf

My recursive solution to this problem is given below.

def num_to_char(value):
 if value == 2: return ["a","b","c"]
 if value == 3: return ["d","e","f"]
 if value == 4: return ["g","h","i"]
 if value == 5: return ["j","k","l"]
 if value == 6: return ["m","n","o"]
 if value == 7: return ["p","q","r","s"]
 if value == 8: return ["t","u","v"]
 if value == 9: return ["w","x","y","z"]

def convert_num(number, current_string = ""):
 if number == []:
 print(current_string)
 return 
 get_list = num_to_char(int(number[0]))
 for character in get_list:
 current_string += character
 convert_num(number[1:], current_string)
 current_string = current_string[:-1]

num_to_covert = list("234")
convert_num(num_to_covert)

python python-3.x programming-challenge

share|improve this question

edited Jun 2 at 20:21

200_success

134k21166439

asked Jun 2 at 20:14

EMLEML

682113

$endgroup$

add a comment | 

$begingroup$

Task

Old mobile phones had the ability to type characters by pressing a number. The letter a could be typed by pressing 2 once. The letter b could be typed by pressing 2 twice.

Given a sequence of numbers, give all possible letter combinations.

For example: The number 23 could give an output ad, ae, af, bd, be, bf, cd, ce, cf

My recursive solution to this problem is given below.

def num_to_char(value):
 if value == 2: return ["a","b","c"]
 if value == 3: return ["d","e","f"]
 if value == 4: return ["g","h","i"]
 if value == 5: return ["j","k","l"]
 if value == 6: return ["m","n","o"]
 if value == 7: return ["p","q","r","s"]
 if value == 8: return ["t","u","v"]
 if value == 9: return ["w","x","y","z"]

def convert_num(number, current_string = ""):
 if number == []:
 print(current_string)
 return 
 get_list = num_to_char(int(number[0]))
 for character in get_list:
 current_string += character
 convert_num(number[1:], current_string)
 current_string = current_string[:-1]

num_to_covert = list("234")
convert_num(num_to_covert)

python python-3.x programming-challenge

share|improve this question

edited Jun 2 at 20:21

200_success

134k21166439

asked Jun 2 at 20:14

EMLEML

682113

$endgroup$

def num_to_char(value):
 if value == 2: return ["a","b","c"]
 if value == 3: return ["d","e","f"]
 if value == 4: return ["g","h","i"]
 if value == 5: return ["j","k","l"]
 if value == 6: return ["m","n","o"]
 if value == 7: return ["p","q","r","s"]
 if value == 8: return ["t","u","v"]
 if value == 9: return ["w","x","y","z"]

def convert_num(number, current_string = ""):
 if number == []:
 print(current_string)
 return 
 get_list = num_to_char(int(number[0]))
 for character in get_list:
 current_string += character
 convert_num(number[1:], current_string)
 current_string = current_string[:-1]

num_to_covert = list("234")
convert_num(num_to_covert)

share|improve this question

edited Jun 2 at 20:21

200_success

134k21166439

asked Jun 2 at 20:14

EMLEML

682113

share|improve this question

edited Jun 2 at 20:21

200_success

134k21166439

asked Jun 2 at 20:14

EMLEML

682113

edited Jun 2 at 20:21

200_success

134k21166439

edited Jun 2 at 20:21

200_success

134k21166439

asked Jun 2 at 20:14

EMLEML

682113

asked Jun 2 at 20:14

EMLEML

682113

1 Answer
1

active

oldest

votes

18

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product

KEYPAD = 
 '2': 'abc', '3': 'def',
 '4': 'ghi', '5': 'jkl', '6': 'mno',
 '7': 'pqrs', '8': 'tuv', '9': 'wxyz',


def convert_num(number):
 letters = [KEYPAD[c] for c in number]
 return [''.join(combo) for combo in product(*letters)]

print(convert_num('234'))

share|improve this answer

answered Jun 2 at 20:40

200_success200_success

134k21166439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant. Sadly I actually had no idea what a cartesian product was so thanks for educating me :)
  $endgroup$
  – EML
  Jun 2 at 20:48
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  In general, any time you want to do some kind of fancy iteration in Python, look at itertools first.
  $endgroup$
  – 200_success
  Jun 2 at 20:55
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  @200_success And if you don't find it in itertools, more_itertools might have it instead (although you need to install it separately).
  $endgroup$
  – Graipher
  Jun 3 at 4:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Beautiful code :)
  $endgroup$
  – JollyJoker
  Jun 3 at 7:51
  

  
  
  
  

add a comment | 

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18

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product

KEYPAD = 
 '2': 'abc', '3': 'def',
 '4': 'ghi', '5': 'jkl', '6': 'mno',
 '7': 'pqrs', '8': 'tuv', '9': 'wxyz',


def convert_num(number):
 letters = [KEYPAD[c] for c in number]
 return [''.join(combo) for combo in product(*letters)]

print(convert_num('234'))

share|improve this answer

answered Jun 2 at 20:40

200_success200_success

134k21166439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant. Sadly I actually had no idea what a cartesian product was so thanks for educating me :)
  $endgroup$
  – EML
  Jun 2 at 20:48
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  In general, any time you want to do some kind of fancy iteration in Python, look at itertools first.
  $endgroup$
  – 200_success
  Jun 2 at 20:55
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  @200_success And if you don't find it in itertools, more_itertools might have it instead (although you need to install it separately).
  $endgroup$
  – Graipher
  Jun 3 at 4:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Beautiful code :)
  $endgroup$
  – JollyJoker
  Jun 3 at 7:51
  

  
  
  
  

add a comment | 

18

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product

KEYPAD = 
 '2': 'abc', '3': 'def',
 '4': 'ghi', '5': 'jkl', '6': 'mno',
 '7': 'pqrs', '8': 'tuv', '9': 'wxyz',


def convert_num(number):
 letters = [KEYPAD[c] for c in number]
 return [''.join(combo) for combo in product(*letters)]

print(convert_num('234'))

share|improve this answer

answered Jun 2 at 20:40

200_success200_success

134k21166439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Brilliant. Sadly I actually had no idea what a cartesian product was so thanks for educating me :)
  $endgroup$
  – EML
  Jun 2 at 20:48
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  In general, any time you want to do some kind of fancy iteration in Python, look at itertools first.
  $endgroup$
  – 200_success
  Jun 2 at 20:55
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  @200_success And if you don't find it in itertools, more_itertools might have it instead (although you need to install it separately).
  $endgroup$
  – Graipher
  Jun 3 at 4:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Beautiful code :)
  $endgroup$
  – JollyJoker
  Jun 3 at 7:51
  

  
  
  
  

add a comment | 

$begingroup$

You're working way too hard:

• 
  itertools.product() produces cartesian products.


• You don't need to convert strings to lists; you can iterate over strings directly.


• Lookups are better done using a dictionary than a chain of if statements.

from itertools import product

KEYPAD = 
 '2': 'abc', '3': 'def',
 '4': 'ghi', '5': 'jkl', '6': 'mno',
 '7': 'pqrs', '8': 'tuv', '9': 'wxyz',


def convert_num(number):
 letters = [KEYPAD[c] for c in number]
 return [''.join(combo) for combo in product(*letters)]

print(convert_num('234'))

share|improve this answer

answered Jun 2 at 20:40

200_success200_success

134k21166439

$endgroup$

from itertools import product

KEYPAD = 
 '2': 'abc', '3': 'def',
 '4': 'ghi', '5': 'jkl', '6': 'mno',
 '7': 'pqrs', '8': 'tuv', '9': 'wxyz',


def convert_num(number):
 letters = [KEYPAD[c] for c in number]
 return [''.join(combo) for combo in product(*letters)]

print(convert_num('234'))

share|improve this answer

answered Jun 2 at 20:40

200_success200_success

134k21166439

answered Jun 2 at 20:40

200_success200_success

134k21166439

answered Jun 2 at 20:40

200_success200_success

134k21166439