如何使用边界值和step作为浮点值进行'for'循环?

时间:2017-12-18 11:26:41

标签: for-loop rust

我需要实现一个for循环,该循环从一个浮点数转到另一个浮点数,并将该步作为另一个浮点数。

我知道如何用类C语言实现它:

for (float i = -1.0; i < 1.0; i += 0.01) { /* ... */ }

我也知道在Rust中我可以使用step_by指定循环步骤,如果我有边界值并且步长为整数,那么这给了我想要的东西:

#![feature(iterator_step_by)]

fn main() {
    for i in (0..30).step_by(3) {
        println!("Index {}", i);
    }
}

当我使用浮点数时,会导致编译错误:

#![feature(iterator_step_by)]

fn main() {
    for i in (-1.0..1.0).step_by(0.01) {
        println!("Index {}", i);
    }
}

这是编译输出:

error[E0599]: no method named `step_by` found for type `std::ops::Range<{float}>` in the current scope
--> src/main.rs:4:26
  |
4 |     for i in (-1.0..1.0).step_by(0.01) {
  |                          ^^^^^^^
  |
  = note: the method `step_by` exists but the following trait bounds were not satisfied:
          `std::ops::Range<{float}> : std::iter::Iterator`
          `&mut std::ops::Range<{float}> : std::iter::Iterator`

如何在Rust中实现这个循环?

3 个答案:

答案 0 :(得分:7)

如果你还没有,我邀请你阅读Goldberg的What Every Computer Scientist Should Know About Floating-Point Arithmetic

浮点问题在于,您的代码可能正在进行200或201次迭代,具体取决于循环的最后一步是i = 0.99还是i = 0.999999(仍然是{{} 1}}即使非常接近)。

为了避免使用这个步枪,Rust不允许在< 1f32范围内进行迭代。相反,它会强制您使用整体步骤:

f64

答案 1 :(得分:2)

作为一个真正的迭代器:

Playground

/// produces: [ linear_interpol(start, end, i/steps) | i <- 0..steps ]
/// (does NOT include "end")
///
/// linear_interpol(a, b, p) = (1 - p) * a + p * b
pub struct FloatIterator {
    current: u64,
    current_back: u64,
    steps: u64,
    start: f64,
    end: f64,
}

impl FloatIterator {
    pub fn new(start: f64, end: f64, steps: u64) -> Self {
        FloatIterator {
            current: 0,
            current_back: steps,
            steps: steps,
            start: start,
            end: end,
        }
    }

    /// calculates number of steps from (end - start) / step
    pub fn new_with_step(start: f64, end: f64, step: f64) -> Self {
        let steps = ((end - start) / step).abs().round() as u64;
        Self::new(start, end, steps)
    }

    pub fn length(&self) -> u64 {
        self.current_back - self.current
    }

    fn at(&self, pos: u64) -> f64 {
        let f_pos = pos as f64 / self.steps as f64;
        (1. - f_pos) * self.start + f_pos * self.end
    }

    /// panics (in debug) when len doesn't fit in usize
    fn usize_len(&self) -> usize {
        let l = self.length();
        debug_assert!(l <= ::std::usize::MAX as u64);
        l as usize
    }
}

impl Iterator for FloatIterator {
    type Item = f64;

    fn next(&mut self) -> Option<Self::Item> {
        if self.current >= self.current_back {
            return None;
        }
        let result = self.at(self.current);
        self.current += 1;
        Some(result)
    }

    fn size_hint(&self) -> (usize, Option<usize>) {
        let l = self.usize_len();
        (l, Some(l))
    }

    fn count(self) -> usize {
        self.usize_len()
    }
}

impl DoubleEndedIterator for FloatIterator {
    fn next_back(&mut self) -> Option<Self::Item> {
        if self.current >= self.current_back {
            return None;
        }
        self.current_back -= 1;
        let result = self.at(self.current_back);
        Some(result)
    }
}

impl ExactSizeIterator for FloatIterator {
    fn len(&self) -> usize {
        self.usize_len()
    }

    //fn is_empty(&self) -> bool {
    //    self.length() == 0u64
    //}
}

pub fn main() {
    println!(
        "count: {}",
        FloatIterator::new_with_step(-1.0, 1.0, 0.01).count()
    );
    for f in FloatIterator::new_with_step(-1.0, 1.0, 0.01) {
        println!("{}", f);
    }
}

答案 2 :(得分:1)

使用迭代器的另一个答案,但方式略有不同playground

extern crate num;
use num::{Float, FromPrimitive};

fn linspace<T>(start: T, stop: T, nstep: u32) -> Vec<T>
where
    T: Float + FromPrimitive,
{
    let delta: T = (stop - start) / T::from_u32(nstep - 1).expect("out of range");
    return (0..(nstep))
        .map(|i| start + T::from_u32(i).expect("out of range") * delta)
        .collect();
}

fn main() {
    for f in linspace(-1f32, 1f32, 3) {
        println!("{}", f);
    }
}

在夜间,您可以使用conservative impl trait功能来避免Vec分配playground

#![feature(conservative_impl_trait)]

extern crate num;
use num::{Float, FromPrimitive};

fn linspace<T>(start: T, stop: T, nstep: u32) -> impl Iterator<Item = T>
where
    T: Float + FromPrimitive,
{
    let delta: T = (stop - start) / T::from_u32(nstep - 1).expect("out of range");
    return (0..(nstep))
        .map(move |i| start + T::from_u32(i).expect("out of range") * delta);
}

fn main() {
    for f in linspace(-1f32, 1f32, 3) {
        println!("{}", f);
    }
}