如何转换 Vec<ndarray::Array1> 的集合成一个Array2? [英] How to convert a collection of Vec<ndarray::Array1> into an Array2?

问题描述

我正在尝试使用 ndarray crate 从一维数组的 Vec 创建一个二维数组.在当前的实现中，我将 Vec> 作为一维数组的集合，我很难弄清楚如何将其转换为 Array2;.我已经在 Vec> 上尝试了 from_vec() 但它产生了 Array1>.我想过使用 stack! 宏，但我不确定如何在上面的 Vec 上调用它.我使用的是 ndarray 0.12.1 和 Rust 1.31.0.

I'm trying to create a 2D array from a Vec of 1D arrays using the ndarray crate. In the current implementation, I have Vec<Array1<u32>> as the collection of 1D arrays, and I'm having a hard time figuring out how to convert it to Array2<u32>. I've tried from_vec() on Vec<Array1<u32>> but it yielded Array1<Array1<u32>>. I thought of using the stack! macro, but I'm not sure how to call it on the above Vec. I'm using ndarray 0.12.1 and Rust 1.31.0.

推荐答案

我对 ndarray 不是很熟悉，但看起来你必须将数据展平作为中间步骤然后重建从那.迭代器可能会更有效，但我没有看到从迭代器构建的方法也可以让您指定形状.

I'm not hugely familiar with ndarray, but it looks like you have to flatten the data as an intermediate step and then rebuild from that. An iterator would probably have been more efficient but I don't see a method to build from an iterator that also lets you specify a shape.

这可能不是最性能的方法，但它至少有效:

It likely isn't the most performant way to to this, but it does at least work:

fn to_array2<T: Copy>(source: &[Array1<T>]) -> Result<Array2<T>, impl std::error::Error> {
    let width = source.len();
    let flattened: Array1<T> = source.into_iter().flat_map(|row| row.to_vec()).collect();
    let height = flattened.len() / width;
    flattened.into_shape((width, height))
}

请注意，如果源数组的长度不同，它可能会失败.该解决方案不是 100% 稳健的，因为如果一个阵列较小但由另一个更长的阵列补偿，它不会失败.可能值得在那里添加一个检查来防止这种情况发生，但我会把它留给你.

Note that it can fail if the source arrays has different lengths. This solution is not 100% robust because it won't fail if one array is smaller but compensated by another array being longer. It is probably worth adding a check in there to prevent that, but I'll leave that to you.

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