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Split coins into combinations of different denominations

Split coins into combinations of different denominations

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6

$begingroup$

I have 3 types of coins: Gold, Silver, Copper.

1 silver = 100 copper.

1 gold = 100 silver.

My input is always in coppers, and I want to be able to make it a bit more readable. So far my code is:

def api_wallet_translate_gold(value):
 """Translate a value into string of money"""
 if value >= 10000: # Gold
 return ("0 gold, 1 silver and 2 copper."
 .format(str(value)[:-4], str(value)[-4:-2], str(value)[-2:]))
 elif value >= 100: # Silver
 return "0 silver and 1 copper.".format(str(value)[-4:-2], str(value)[-2:])
 else: # Copper
 return "0 copper.".format(str(value)[-2:])

It works, but I am wondering how could it be improved. I think there was a way to format it like xx:2:2 or something but I can't remember how to do it.

Note: We never know how many gold digits we have, it could be 999999 to 1

python python-3.x formatting

share|improve this question

edited 6 hours ago

200_success

131k17157422

asked 6 hours ago

SaelythSaelyth

1553

$endgroup$

add a comment | 

6

$begingroup$

I have 3 types of coins: Gold, Silver, Copper.

1 silver = 100 copper.

1 gold = 100 silver.

My input is always in coppers, and I want to be able to make it a bit more readable. So far my code is:

def api_wallet_translate_gold(value):
 """Translate a value into string of money"""
 if value >= 10000: # Gold
 return ("0 gold, 1 silver and 2 copper."
 .format(str(value)[:-4], str(value)[-4:-2], str(value)[-2:]))
 elif value >= 100: # Silver
 return "0 silver and 1 copper.".format(str(value)[-4:-2], str(value)[-2:])
 else: # Copper
 return "0 copper.".format(str(value)[-2:])

It works, but I am wondering how could it be improved. I think there was a way to format it like xx:2:2 or something but I can't remember how to do it.

Note: We never know how many gold digits we have, it could be 999999 to 1

python python-3.x formatting

share|improve this question

edited 6 hours ago

200_success

131k17157422

asked 6 hours ago

SaelythSaelyth

1553

$endgroup$

add a comment | 

$begingroup$

I have 3 types of coins: Gold, Silver, Copper.

1 silver = 100 copper.

1 gold = 100 silver.

My input is always in coppers, and I want to be able to make it a bit more readable. So far my code is:

def api_wallet_translate_gold(value):
 """Translate a value into string of money"""
 if value >= 10000: # Gold
 return ("0 gold, 1 silver and 2 copper."
 .format(str(value)[:-4], str(value)[-4:-2], str(value)[-2:]))
 elif value >= 100: # Silver
 return "0 silver and 1 copper.".format(str(value)[-4:-2], str(value)[-2:])
 else: # Copper
 return "0 copper.".format(str(value)[-2:])

It works, but I am wondering how could it be improved. I think there was a way to format it like xx:2:2 or something but I can't remember how to do it.

Note: We never know how many gold digits we have, it could be 999999 to 1

python python-3.x formatting

share|improve this question

edited 6 hours ago

200_success

131k17157422

asked 6 hours ago

SaelythSaelyth

1553

$endgroup$

def api_wallet_translate_gold(value):
 """Translate a value into string of money"""
 if value >= 10000: # Gold
 return ("0 gold, 1 silver and 2 copper."
 .format(str(value)[:-4], str(value)[-4:-2], str(value)[-2:]))
 elif value >= 100: # Silver
 return "0 silver and 1 copper.".format(str(value)[-4:-2], str(value)[-2:])
 else: # Copper
 return "0 copper.".format(str(value)[-2:])

share|improve this question

edited 6 hours ago

200_success

131k17157422

asked 6 hours ago

SaelythSaelyth

1553

share|improve this question

edited 6 hours ago

200_success

131k17157422

asked 6 hours ago

SaelythSaelyth

1553

edited 6 hours ago

200_success

131k17157422

edited 6 hours ago

200_success

131k17157422

asked 6 hours ago

SaelythSaelyth

1553

asked 6 hours ago

SaelythSaelyth

1553

1 Answer
1

active

oldest

votes

8

$begingroup$

It may less fragile if you deal with the numbers directly rather than converting to strings. It will also be cleaner code.

You could start with your values in a list sorted highest to lowest. Then in your function you can find the next-largest value and remained with divmod(). After than it's a matter of deciding how you want to format the resulting dict:

coins = [
 ("gold", 100 * 100),
 ("silver", 100), 
 ("copper", 1)
]

def translate_coins(value, coins):
 res = 
 for coin, v in coins:
 res[coin], value = divmod(value, v)
 return res 

translate_coins(1013323, coins)

Result:

'gold': 101, 'silver': 33, 'copper': 23

share|improve this answer

edited 6 hours ago

answered 6 hours ago

MarkMMarkM

28316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is almost perfect @MarkM, though the def translate_coins(value, coins): line combined with res[coin], value = divmod(value, v), could have the values (one of'em) renamed for easier reading... Neat trick with the dictionary assignment there... One question too, why not use a dict for coins instead of a list of tuples?... Then for coin, quantity in coins.items(): could be used with similar effect and one less object within another. That all said I think your answer is solid, just had a few nits to be picked that I could see.
  $endgroup$
  – S0AndS0
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the comment @S0AndS0 those a re great suggestions. I didn't use a dictionary for the coin values because this depends on doing the division in order from highest to lowest. It's only recently that you can count on the order of python dictionaries.
  $endgroup$
  – MarkM
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  After further testing I see what you're doing, clever @MarkM, really clever there with reassigning value... though I don't envy debugging such mutations, it does totally make sense in this context to mutate... I also now see your wisdom in using a list of tuples, and retract my previous question in regards to using a dict as input to the translate_coins function; that would have made code far hairier than needed... Consider me impressed, eleven lines of code and you've taught me plenty new perversions with Python.
  $endgroup$
  – S0AndS0
  4 hours ago
  

  
  
  
  

add a comment | 

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8

$begingroup$

It may less fragile if you deal with the numbers directly rather than converting to strings. It will also be cleaner code.

You could start with your values in a list sorted highest to lowest. Then in your function you can find the next-largest value and remained with divmod(). After than it's a matter of deciding how you want to format the resulting dict:

coins = [
 ("gold", 100 * 100),
 ("silver", 100), 
 ("copper", 1)
]

def translate_coins(value, coins):
 res = 
 for coin, v in coins:
 res[coin], value = divmod(value, v)
 return res 

translate_coins(1013323, coins)

Result:

'gold': 101, 'silver': 33, 'copper': 23

share|improve this answer

edited 6 hours ago

answered 6 hours ago

MarkMMarkM

28316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is almost perfect @MarkM, though the def translate_coins(value, coins): line combined with res[coin], value = divmod(value, v), could have the values (one of'em) renamed for easier reading... Neat trick with the dictionary assignment there... One question too, why not use a dict for coins instead of a list of tuples?... Then for coin, quantity in coins.items(): could be used with similar effect and one less object within another. That all said I think your answer is solid, just had a few nits to be picked that I could see.
  $endgroup$
  – S0AndS0
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the comment @S0AndS0 those a re great suggestions. I didn't use a dictionary for the coin values because this depends on doing the division in order from highest to lowest. It's only recently that you can count on the order of python dictionaries.
  $endgroup$
  – MarkM
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  After further testing I see what you're doing, clever @MarkM, really clever there with reassigning value... though I don't envy debugging such mutations, it does totally make sense in this context to mutate... I also now see your wisdom in using a list of tuples, and retract my previous question in regards to using a dict as input to the translate_coins function; that would have made code far hairier than needed... Consider me impressed, eleven lines of code and you've taught me plenty new perversions with Python.
  $endgroup$
  – S0AndS0
  4 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

It may less fragile if you deal with the numbers directly rather than converting to strings. It will also be cleaner code.

You could start with your values in a list sorted highest to lowest. Then in your function you can find the next-largest value and remained with divmod(). After than it's a matter of deciding how you want to format the resulting dict:

coins = [
 ("gold", 100 * 100),
 ("silver", 100), 
 ("copper", 1)
]

def translate_coins(value, coins):
 res = 
 for coin, v in coins:
 res[coin], value = divmod(value, v)
 return res 

translate_coins(1013323, coins)

Result:

'gold': 101, 'silver': 33, 'copper': 23

share|improve this answer

edited 6 hours ago

answered 6 hours ago

MarkMMarkM

28316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is almost perfect @MarkM, though the def translate_coins(value, coins): line combined with res[coin], value = divmod(value, v), could have the values (one of'em) renamed for easier reading... Neat trick with the dictionary assignment there... One question too, why not use a dict for coins instead of a list of tuples?... Then for coin, quantity in coins.items(): could be used with similar effect and one less object within another. That all said I think your answer is solid, just had a few nits to be picked that I could see.
  $endgroup$
  – S0AndS0
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the comment @S0AndS0 those a re great suggestions. I didn't use a dictionary for the coin values because this depends on doing the division in order from highest to lowest. It's only recently that you can count on the order of python dictionaries.
  $endgroup$
  – MarkM
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  After further testing I see what you're doing, clever @MarkM, really clever there with reassigning value... though I don't envy debugging such mutations, it does totally make sense in this context to mutate... I also now see your wisdom in using a list of tuples, and retract my previous question in regards to using a dict as input to the translate_coins function; that would have made code far hairier than needed... Consider me impressed, eleven lines of code and you've taught me plenty new perversions with Python.
  $endgroup$
  – S0AndS0
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It may less fragile if you deal with the numbers directly rather than converting to strings. It will also be cleaner code.

You could start with your values in a list sorted highest to lowest. Then in your function you can find the next-largest value and remained with divmod(). After than it's a matter of deciding how you want to format the resulting dict:

coins = [
 ("gold", 100 * 100),
 ("silver", 100), 
 ("copper", 1)
]

def translate_coins(value, coins):
 res = 
 for coin, v in coins:
 res[coin], value = divmod(value, v)
 return res 

translate_coins(1013323, coins)

Result:

'gold': 101, 'silver': 33, 'copper': 23

share|improve this answer

edited 6 hours ago

answered 6 hours ago

MarkMMarkM

28316

$endgroup$

coins = [
 ("gold", 100 * 100),
 ("silver", 100), 
 ("copper", 1)
]

def translate_coins(value, coins):
 res = 
 for coin, v in coins:
 res[coin], value = divmod(value, v)
 return res 

translate_coins(1013323, coins)

'gold': 101, 'silver': 33, 'copper': 23

share|improve this answer

edited 6 hours ago

answered 6 hours ago

MarkMMarkM

28316

answered 6 hours ago

MarkMMarkM

28316

answered 6 hours ago

MarkMMarkM

28316