Factoring input into listed frequencies

[blahblahblah123456789]

Details are here:

https://stackoverflow.com/q/53291892/10572106

help plz

[koolinka]

You could just use integers and then divide by 100 for the result.

[ichabod801]

You need a loop, starting with the amount and the coin counts (a list like [0, 0, 0, ..., 0]). Make it a while loop (while amount:). Each time through the loop, use the if clauses to figure out the largest coin that can still be used (the coin value is less than amount). Once you find that value, add one to the count for that coin, and subtract that coin value from amount.

There's a simpler way to do this without the if clauses, but let's see if you can get the if clauses working first.

[blahblahblah123456789]

You need a loop, starting with the amount and the coin counts (a list like [0, 0, 0, ..., 0]). Make it a while loop (while amount:). Each time through the loop, use the if clauses to figure out the largest coin that can still be used (the coin value is less than amount). Once you find that value, add one to the count for that coin, and subtract that coin value from amount.

There's a simpler way to do this without the if clauses, but let's see if you can get the if clauses working first.

I fixed my code according to a reply I received. Would the following be more efficient?

def pay_with_coins( amount ):
    all_coins = [2.00, 1.00, 0.50, 0.20, 0.10, 0.05, 0.02, 0.01]
    
    change = []
    for each_coin in all_coins:
        num = (int)(amount / each_coin)
    change.append(num)
    amount = round(amount - (num * each_coin), 2)
    
    return change

print(pay_with_coins(5.18))
#^^This just prints 518 instead of change how I want it formatted

[DeaD_EyE]

In line number 6 you've an error:
num = (int)(amount / each_coin)
this should be num = int(amount / each_coin)).

First I did not understand why it did not thrown an error. But the explanation is very easy.
(int) returns the built-in function int. The parenthesis after int, are calling the returned int function.
If you have written (int,), it makes a tuple with the int function as a single element.

In line 7 you append num to change, but outside of the loop. You get only the last value.
In line 8 you set amount to the new amount. But this is also outside of the loop. So, only the last value will be evaluated, but it's not used inside the loop. Line 7 and 8 have to be indented.

A corrected version:

def pay_with_coins( amount ):
    all_coins = [2.00, 1.00, 0.50, 0.20, 0.10, 0.05, 0.02, 0.01]
    change = []
    for each_coin in all_coins:
        num = (int(amount / each_coin))
        change.append(num)
        amount = round(amount - (num * each_coin), 2)
    return change

I made a little different implementation.
Instead of using float, I use int for the smallest currency unit. The cause: 0.1 + 0.2 == 0.30000000000000004.
The function divmod does what you want to have. The function divmod returns the result and the rest.
If you still want to work with floats, you should look into the module decimal.

print(divmod(10, 3))

Output:
(3,1)

10 divided by 3 == 3 + rest 1

def pay_with_coins(amount):
    amount = int(amount * 100)
    # convert amount into the smallest currency unit (cent)
    all_coins_cent = [200, 100, 50, 20, 10, 5, 2, 1]
    # values as integers in the smallest currency unit
    change = []
    for each_coin in all_coins_cent:
        factor, rest = divmod(amount, each_coin)
        change.append(factor)
        amount = rest
        # print(rest)
    return  {'change': change, 'rest': rest}

print(pay_with_coins(5.18))

Additionally I return a dict, just to see if there is any rest left.
This should not happen, because we have also 1 cent. If you remove this coin,
the rest will be one cent.
If you remove this coin,

[koolinka]

Btw, you could also use ${:0,.2f}.format(pay_with_coins). This makes pay_with_coins print out in this type of format $80,000.00. Just a suggestion to make it look nice