Round down to 5 factor value in float value

Question

I want to convert a float value to round down 5 factor value, this means for example 0.05,0.10,0.15.

Suppose I have a value like 9.48, and I want to convert it into 9.45.

I tried with this:

val = 9.48
val - val % 0.05

It returns 9.450000000000001. That is good for me but the problem is when I have 9.60, it converts asn to 9.55.

When the value is in already factor of 5 then it stays as it is.

Darth Kotik · Accepted Answer · 2014-11-17T08:01:41.333

6

You can do this kind of thing

>>> val = 9.68
>>> round(val*20)/20.0
9.6999999999999993

edited Nov 17 '14 at 08:01

answered Nov 17 '14 at 07:42

Darth Kotik

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Better to use round instead of int, result should be 9.7 – Ludovic Viaud Nov 17 '14 at 07:53
thanks, but logic returns round up value. i want round down ,mean i want `9.65` for your example. – i'm PosSible Nov 17 '14 at 08:41
i think int() is working, but why 20? for multiplication and division?? – i'm PosSible Nov 17 '14 at 08:53
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@i'mPosSible `20=1/0.05` – Darth Kotik Nov 17 '14 at 09:24

score 1 · Answer 2 · edited May 23 '17 at 11:49

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This is a result of how floating point numbers are represented -- floats cannot precisely represent 9.60. The closest (64-bit) float to 9.6 is just slightly less 9.6, which is why the operation rounds down. If you need your mathematical operations to be precise then you should be using the decimal module rather than a float.

eg.

val = decimal.Decimal("9.60")
val - val % decimal.Decimal("0.05")

See https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html for an in depth explanation of floating point mathematics.

edited May 23 '17 at 11:49

Community

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answered Nov 17 '14 at 08:03

Dunes

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thanks, Dunes its work but in my side I don't use decimal, there is another solution with float or maths lib. – i'm PosSible Nov 17 '14 at 08:49

Round down to 5 factor value in float value

2 Answers2