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Rounding to 5 coefficient values ​​in floating value

I want to convert a float value to round the value of 5 factors, that means for example 0.05,0.10,0.15

.

Suppose I have a value like this 9.48

and I want to convert it to 9.45

.

I tried with this:

val = 9.48
val - val % 0.05

Returns 9.450000000000001

. This is fine for me, but the problem is when I have 9.60

it converts asn to 9.55

.

When the value already has a factor of 5, it remains as it is.

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2 answers


You can do things like this



>>> val = 9.68
>>> round(val*20)/20.0
9.6999999999999993
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This is the result where floating point numbers are represented - floats cannot represent exactly 9.60

. The closest (64-bit) float is 9.6

slightly smaller 9.6

, so the operation is rounded. If you want your math operations, then you should use a module decimal

, not 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 a detailed explanation of floating point math.

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