如何将一个mxn矩阵转换成具有相同行数的向量?

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英文:

How to convert a mxn matrix into a vector with same rows?

问题

I'm trying to convert a Matrix into SVector in Julia. Following is a sample of how to create a matrix:

p=randn(Point3,100)
p_mat=collect(reshape(reduce(vcat,p),3,length(p))')

Here is the matrix.

I've one method but it's not very optimal:

[SVector{size(p_mat,2), Real}(p_mat[i,:]) for i in 1:size(p_mat,1)]

I'm trying to avoid the for loop and convert matrix into SVector. Is there another possible way?

英文:

I'm trying to convert a Matrix into SVector in Julia. Following is a sample of how to create a matrix

p=randn(Point3,100)
p_mat=collect(reshape(reduce(vcat,p),3,length(p))')

如何将一个mxn矩阵转换成具有相同行数的向量?
Here is the matrix.
如何将一个mxn矩阵转换成具有相同行数的向量?

I've one method but its not very optimal.

[SVector{size(p_mat,2), Real}(p_mat[i,:]) for i in 1:size(p_mat,1)]

I'm trying to avoid the for loop and convert matrix into SVector.
Is there another possible way?

答案1

得分: 2

我会假设Point3只是SVector{3, Float64}的另一个名称。然后,你可以通过以下方式从矩阵的行创建一个向量:Point3.(eachrow(p_mat)),代码如下:

using StaticArrays
const Point3 = SVector{3, Float64}

p = randn(Point3, 100)
100-element Vector{SVector{3, Float64}}:
 [-0.643411054127986, -1.3146948793707078, 0.8560043466425089]
 [-1.2014944236615475, -0.16332265198198045, 0.7899287791641353]
 [-0.4468974734632806, -2.115137061106782, 1.2797367438232168]
 ...

p_mat = [v[i] for v in p, i in 1:3]
100×3 Matrix{Float64}:
 -0.643411   -1.31469     0.856004
 -1.20149    -0.163323    0.789929
 -0.446897   -2.11514     1.27974
 ...

p_vec = Point3.(eachrow(p_mat)) # 与p相同
100-element Vector{SVector{3, Float64}}:
 [-0.643411054127986, -1.3146948793707078, 0.8560043466425089]
 [-1.2014944236615475, -0.16332265198198045, 0.7899287791641353]
 [-0.4468974734632806, -2.115137061106782, 1.2797367438232168]
 ...
英文:

I will assume Point3 is just another name for SVector{3, Float64}. Then, you can create a Vector from the matrix rows by Point3.(eachrow(p_mat)) as follows:

using StaticArrays
const Point3 = SVector{3, Float64}

p = randn(Point3, 100)
100-element Vector{SVector{3, Float64}}:
 [-0.643411054127986, -1.3146948793707078, 0.8560043466425089]
 [-1.2014944236615475, -0.16332265198198045, 0.7899287791641353]
 [-0.4468974734632806, -2.115137061106782, 1.2797367438232168]
 ...

p_mat = [v[i] for v in p, i in 1:3]
100×3 Matrix{Float64}:
 -0.643411   -1.31469     0.856004
 -1.20149    -0.163323    0.789929
 -0.446897   -2.11514     1.27974
 ...

p_vec = Point3.(eachrow(p_mat)) # the same as p
100-element Vector{SVector{3, Float64}}:
 [-0.643411054127986, -1.3146948793707078, 0.8560043466425089]
 [-1.2014944236615475, -0.16332265198198045, 0.7899287791641353]
 [-0.4468974734632806, -2.115137061106782, 1.2797367438232168]
 ...


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  • 本文由 发表于 2023年5月26日 08:46:59
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