以尽可能小的数量增加 Python 浮点值 [英] Increment a Python floating point value by the smallest possible amount
问题描述
背景:我使用浮点值作为字典键.
Background: I'm using floating point values as dictionary keys.
偶尔,非常(也许永远不会,但不一定永远不会),会发生碰撞.我想通过尽可能小地增加浮点值来解决这些问题.我该怎么做?
Occasionally, very occasionally (and perhaps never, but not certainly never), there will be collisions. I would like to resolve these by incrementing the floating point value by as small an amount as possible. How can I do this?
在 C 中,我会修改尾数来实现这一点,但我认为这在 Python 中是不可能的.
In C, I would twiddle the bits of the mantissa to achieve this, but I assume that isn't possible in Python.
推荐答案
Python 3.9 及更高版本
从 Python 3.9 开始,于 2020 年 10 月 5 日发布,您可以使用 math.nextafter
函数:
Python 3.9 and above
Starting with Python 3.9, released 2020-10-05, you can use the math.nextafter
function:
math.nextafter(x, y)
math.nextafter(x, y)
返回 x 之后朝向 y 的下一个浮点值.
Return the next floating-point value after x towards y.
如果 x 等于 y,则返回 y.
If x is equal to y, return y.
示例:
math.nextafter(x, math.inf)
上升:向正无穷大.
math.nextafter(x, -math.inf)
下降:向负无穷大.
math.nextafter(x, 0.0)
趋向于零.
math.nextafter(x, math.copysign(math.inf, x))
从零开始.
另见math.ulp()
.
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