Rounding a number in Python but keeping ending zeros

Question:

I’ve been working on a script that takes data from an Excel spreadsheet, rounds the numbers, and removes the decimal point, for example, 2606.89579999999 becomes 26069. However, I need the number to round to two decimal places even if there would be a trailing zero, so 2606.89579999999 should become 260690.

I currently have it so i takes the data from the cell in Excel, and rounds it to two decimal places (i = round(i, 2)) which gives me the single decimal point in the above example.

I’ve tried figuring out how to get this to work with Decimal, but I can’t seem to get it working.

All other numbers that get rounded, if the rounded value doesn’t end in ‘0’, work fine with round(i, 2), but if the numbers just so happen to end in *.x0, that 0 gets dropped off and messes with the data.

Asked By: Seth Koberg

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Answers:

As you are talking about trailing zeros, this is a question about representation as string,
you can use

>>> "%.2f" % round(2606.89579999999, 2)
'2606.90'

Or use modern style with format function:

>>> '{:.2f}'.format(round(2606.89579999999, 2))
'2606.90'

and remove point with replace or translate (_ refers to result of previous command in python console):

>>> _.translate(None, '.')
'260690'

Note that rounding is not needed here, as .2f format applyies the same rounding:

>>> "%.2f" % 2606.89579999999
'2606.90'

But as you mentioned excel, you probably would opt to roll your own rounding function, or use decimal, as float.round can lead to strange results due to float representation:

>>> round(2.675, 2)
2.67
>>> round(2606.89579999999, 2)
2606.89

With decimal use quantize:

>>> from decimal import *
>>> x = Decimal('2606.8950000000001')
# Decimal('2606.8950000000001')
>>> '{}'.format(x.quantize(Decimal('.01'), rounding=ROUND_HALF_EVEN))
'2606.90'

That, for your original task, becomes:

>>> x = Decimal('2606.8950000000001')
>>> int((x*100).quantize(1, rounding=ROUND_HALF_EVEN))
260690

And the reason of strange rounding comes to the front with Decimal:

>>> x = Decimal(2606.8950000000001)
# Decimal('2606.89499999999998181010596454143524169921875') # internal float repr
Answered By: alko
>>> '{:.2f}'.format(2606.89579999999).replace('.', '')
'260690'
Answered By: Maciej Gol
>>> int (round (2606.89579999999,2)*100)
260690
Answered By: user2437648

As of Python 3.6, you can also use an f-string to inline format the number. In this case, the desired format is floating point with 2 decimal places so you would use .2f as the format specifier:

x = 2606.89579999999
x = round(x, 2)      # not strictly necessary as format will round for you
print(f'{x:.2f}')

Output:

2606.90
Answered By: Nick

In case you want to dynamically change the number of decimal places, you can use the modification below:

def formatNumber(n, digits):
    formatter = '{:.' + '{}'.format(digits) + 'f}'
    x = round(n, digits)
    return formatter.format(x)

x = 2606.89579999999
digits = 2

formatted_number = formatNumber(x, digits)

That way, all you have to do, it to change the digits variable. This will return:

2606.90
Answered By: Vasilis G.

I the answers that I saw here did not satisfy me.
So here my solution:

def round_with_padding(value, round_digits):
    return format(round(value,round_digits), "."+str(round_digits)+"f")

Hope you like it

Answered By: ChaosPredictor