python division precision

Question

I am aware of floating point being inaccurate and would like to know the best way to get 0.08354 instead of (0.08353999999999999) when I do the following in Python:

d = 8.354/100

print (d)

`round(0.08353999999999999, 6) ` – Padraic Cunningham Oct 21 '16 at 15:13 — Padraic Cunningham, Oct 21 '16 at 15:13

score 6 · Answer 1 · answered Oct 21 '16 at 15:14

6

If you want absolute precision, use Decimals:

>>> import decimal
>>> decimal.Decimal('8.354') / 10
Decimal('0.8354')

answered Oct 21 '16 at 15:14

deceze

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That's not working here, decimal.Decimal(2110.32)/decimal.Decimal(0.24) = Decimal('8793.000000000001007527394847'). Any other solutions? – Deekshant Aug 18 '23 at 19:42
You’re already using floats there. Don’t. Use string literals instead, as shown in my answer. – deceze Aug 19 '23 at 01:32
That worked, decimal.Decimal(str(2110.32))/decimal.Decimal(str(0.24))=Decimal('8793'). Now I am wondering about the tradeoff between 3 conversions (float->str, float->str, str->int) and the accuracy required in my use case. They've gotta add some better methods for this on Python. – Deekshant Aug 20 '23 at 03:25
Why `decimal.Decimal(str(2110.32))`!? Just do `decimal.Decimal('2110.32')`! – deceze Aug 20 '23 at 07:02
Because 2110.32 isn't a constant. It is an input which can be any arbitrary number. – Deekshant Aug 21 '23 at 14:04
Then your case isn’t what you present here. You best write a new question with all the details. – deceze Aug 21 '23 at 21:24

score 4 · Accepted Answer · answered Oct 21 '16 at 15:13

4

Use the builtin round() function:

>>> d = 8.354/100
>>> d
0.08353999999999999
>>> round(d, 6)
0.08354
>>>

answered Oct 21 '16 at 15:13

Christian Dean

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python division precision

2 Answers2