How to convert a vector if Arrays into an Array along a new dimension? #1150
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Suppose that we have a vector of 5 Arrays. Also let the shapes of theses Arrays be uniform and known, say (3, 3). let a: Vec<Array2<f64>> = vec![ # five Arrays ];Do we have any constructor method to create a new Array of shape (5, 3, 3) from this vector? |
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Replies: 1 comment 2 replies
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This is an area which could be improved. There are a couple of ways to do this using existing functionality: use ndarray::prelude::*;
fn main() -> Result<(), Box<dyn std::error::Error>> {
let owned: Vec<Array2<f64>> = vec![Array2::zeros([3, 3]), Array2::ones([3, 3])];
// Option 1:
let views: Vec<ArrayView2<'_, f64>> = owned.iter().map(|arr| arr.view()).collect();
let stacked = ndarray::stack(Axis(0), &views)?;
println!("Option 1:\n{stacked}\n");
// Option 2:
let mut stacked: Array3<f64> = Array3::zeros([0, 3, 3]);
for arr in &owned {
stacked.push(Axis(0), arr.view())?;
}
println!("Option 2:\n{stacked}\n");
Ok(())
} |
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Is it possible to have a stack of views along an axis produce a non-owned ArrayView instead of an ArrayBase? |
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This question doesn't seem to be related to the current discussion...? See this issue. The answer is no because each view is allocated randomly in memory and there's no way that this new ArrayView could make sense of it. It must own the data. |
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This is an area which could be improved. There are a couple of ways to do this using existing functionality:
(Playg…