Difference between Python 3.7 math.remainder and %(modulo operator)

Question

From What’s New In Python 3.7 we can see that there is new `math.remainder`. It says Return the IEEE 754-style remainder of x with respect to y. For finite x and finite nonzero y, this is the difference `x - n*y`, where n is the closest integer to the exact value of the quotient `x / y`. If `x / y` is exactly halfway between two consecutive integers, the nearest even integer is used for `n`. The remainder `r = remainder(x, y)` thus always satisfies `abs(r) <= 0.5 * abs(y)`. Special cases follow IEEE 754: in particular, `remainder(x, math.inf)` is x for any finite x, and `remainder(x, 0)` and `remainder(math.inf, x)` raise `ValueError` for any non-NaN x. If the result of the remainder operation is zero, that zero will have the same sign as x. On platforms using IEEE 754 binary floating-point, the result of this operation is always exactly representable: no rounding error is introduced. But we also remember that there is `%` symbol which is remainder of `x / y` We also see that there is…

Accepted Answer

Return the IEEE 754-style remainder of x with respect to y. For finite x and finite nonzero y, this is the difference `x - n*y`, where n is the closest integer to the exact value of the quotient `x / y`. If `x / y` is exactly halfway between two consecutive integers, the nearest even integer is used for n. The remainder `r = remainder(x, y)` thus always satisfies `abs(r) <= 0.5 * abs(y)`.

for the modulo this is `m = x - n*y` where `n` is the `floor(x/y)`, so `0 <= m < y` instead of `abs(r) <= 0.5 * abs(y)` for the remainder.

so

```
modulo(2.7, 1) = 0.7
remainder(2.7, 1) = -0.3
```

Difference between Python 3.7 math.remainder and %(modulo operator)

Solution

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