CODEWITHSUNDEEP
pythonnumpysumrounding
user2237024
user2237024

Reputation: 181

Round a Python list of numbers and maintain their sum

I have a list or an array of decimal numbers in Python. I need to round them to the nearest 2 decimal places as these are monetary amounts. But, I need the overall sum to be maintained, i.e. the sum of the original array rounded to 2 decimal places must be equal to the sum of the rounded elements of the array.

Here's my code so far:

myOriginalList = [27226.94982, 193.0595233, 1764.3094, 12625.8607, 26714.67907, 18970.35388, 12725.41407, 23589.93271, 27948.40386, 23767.83261, 12449.81318]
originalTotal = round(sum(myOriginalList), 2)
# Answer = 187976.61

# Using numpy
myRoundedList = numpy.array(myOriginalList).round(2)
# New Array = [ 27226.95    193.06   1764.31  12625.86  26714.68  18970.35  12725.41 23589.93 27948.4   23767.83  12449.81]

newTotal = myRoundedList.sum()
# Answer = 187976.59

I need an efficient way of amending my new rounded array such that the sum is also 187976.61. The 2 pence difference needs to be applied to items 7 and 6 as these have the greatest difference between the rounded entries and the original entries.

Upvotes: 18

Views: 11829

Answers (5)

AJSV
AJSV

Reputation: 31

I am late, but here is a general solution that works with numpy, which I have found to be ~100 times faster than other implementations proposed on this page:

def round_retain_sum(x):
    x = x*100 # We want 2 decimal precision
    N = np.round(np.sum(x)).astype(int)
    y = x.astype(int)
    M = np.sum(y)
    K = N - M 
    z = y-x 
    if K!=0:
        idx = np.argpartition(z,K)[:K]
        y[idx] += 1     
    return y/100.

EDIT NOTE: Must convert the input list into a NumPy array with np.array(x) as some of the methods/attributes don't work on lists. But it works once converted

Upvotes: 3

Asclepius
Asclepius

Reputation: 63312

As noted in an answer by kettlehell, consider the PyPI package iteround. It does not internally use NumPy, however.

>>> from iteround import saferound
>>> saferound([1.0, 2.1, 3.6], places=0)
[1.0, 2.0, 4.0]

Upvotes: 10

Mark Ransom
Mark Ransom

Reputation: 308206

The first step is to calculate the error between the desired result and the actual sum:

>>> error = originalTotal - sum(myRoundedList)
>>> error
0.01999999996041879

This can be either positive or negative. Since every item in myRoundedList is within 0.005 of the actual value, this error will be less than 0.01 per item of the original array. You can simply divide by 0.01 and round to get the number of items that must be adjusted:

>>> n = int(round(error / 0.01))
>>> n
2

Now all that's left is to select the items that should be adjusted. The optimal results come from adjusting those values that were closest to the boundary in the first place. You can find those by sorting by the difference between the original value and the rounded value.

>>> myNewList = myRoundedList[:]
>>> for _,i in sorted(((myOriginalList[i] - myRoundedList[i], i) for i in range(len(myOriginalList))), reverse=n>0)[:abs(n)]:
    myNewList[i] += math.copysign(0.01, n)

>>> myRoundedList
[27226.95, 193.06, 1764.31, 12625.86, 26714.68, 18970.35, 12725.41, 23589.93, 27948.4, 23767.83, 12449.81]
>>> myNewList
[27226.95, 193.06, 1764.31, 12625.86, 26714.68, 18970.359999999997, 12725.42, 23589.93, 27948.4, 23767.83, 12449.81]
>>> sum(myNewList)
187976.61

Upvotes: 8

Jaime
Jaime

Reputation: 67427

With all the caveats about using floating point numbers:

delta_pence = int(np.rint((originalTotal - np.sum(myRoundedList))*100))
if delta_pence > 0:
    idx = np.argsort(myOriginalList - myRoundedList)[-delta_pence:]
    myRoundedList[idx] += 0.01
elif delta_pence < 0:
    idx = np.argsort(myOriginalList - myRoundedList)[:delta_pence]
    myRoundedList[idx] -= 0.01

>>> myRoundedList.sum()
187976.60999999999

Upvotes: 1

palooh
palooh

Reputation: 368

First of all you shouldn't use floats for storing money (use Decimals instead). But below I provide some quite generic solution - you need to store, accumulate and use the sum of differences in rounding. Some verbose (and not very pythonic ;-) example with your numbers:

# define your accuracy
decimal_positions = 2

numbers = [27226.94982, 193.0595233, 1764.3094, 12625.8607, 26714.67907, 18970.35388, 12725.41407, 23589.93271, 27948.40386, 23767.83261, 12449.81318]
print round(sum(numbers),decimal_positions)
>>> 187976.61

new_numbers = list()
rest = 0.0
for n in numbers:
    new_n = round(n + rest,decimal_positions)
    rest += n - new_n
    new_numbers.append( new_n )

print sum(new_numbers)
>>> 187976.61

Upvotes: 1

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