Julia中的下三角矩阵 [英] Lower triangular matrix in julia

问题描述

我的列数等于行数.并且对角线等于零.如何构建这个矩阵?

I have the number of columns equals the number of rows. And the the diagonal is is equal to zero. How can I build this matrix?

#mat
#     [,1] [,2] [,3] [,4]
#[1,]   0   NA   NA   NA
#[2,]    1   0   NA   NA
#[3,]    2    4   0   NA
#[4,]    3    5    6   0

我试过了

x=rand(4,4)
4x4 Array{Float64,2}:
 0.60064   0.917443  0.561744   0.135717 
 0.106728  0.72391   0.0894174  0.0656103
 0.410262  0.953857  0.844697   0.0375045
 0.476771  0.778106  0.469514   0.398846 

c=LowerTriangular(x)

4x4 LowerTriangular{Float64,Array{Float64,2}}:
 0.60064   0.0       0.0       0.0     
 0.106728  0.72391   0.0       0.0     
 0.410262  0.953857  0.844697  0.0     
 0.476771  0.778106  0.469514  0.398846

但我正在寻找这样的东西

but l'm looking for something like this

c=LowerTriangular(x)
4x4 LowerTriangular{Float64,Array{Float64,2}}:
 0.0   NA      NA       NA    
 0.106728  0.0    NA      NA    
 0.410262  0.953857  0.0 NA     
 0.476771  0.778106  0.469514  0

对角线应该为零.

推荐答案

这里的灵感来自 Stefan Karpinski 在 Julia 用户的 列表:

Here is something taking inspiration from Code by Stefan Karpinski on the Julia User's list:

function vec2ltri_alt{T}(v::AbstractVector{T}, z::T=zero(T))
    n = length(v)
    v1 = vcat(0,v)
    s = round(Int,(sqrt(8n+1)-1)/2)
    s*(s+1)/2 == n || error("vec2utri: length of vector is not triangular")
    s+=1
    [ i>j ? v1[round(Int, j*(j-1)/2+i)] : (i == j ? z : NaN) for i=1:s, j=1:s ]
end

julia> vec2ltri_alt(collect(1:6))
4x4 Array{Any,2}:
 0  NaN  NaN  NaN
 1    0  NaN  NaN
 2    3    0  NaN
 3    4    6    0

注意:如果需要，请查看官方文档关于三元运算符，以便更清楚地了解 发生了什么?... : 语法在这里.

Note: If desired, check out the official documentation on the ternary operator for a bit more clarity on what is going on with the ? ... : syntax here.

对于那些寻找更标准"的对角矩阵解决方案的人:

这是一个创建更标准解决方案的版本:

Here is a version that creates a more standard solution:

function vec2ltri{T}(v::AbstractVector{T}, z::T=zero(T))
    n = length(v)
    s = round(Int,(sqrt(8n+1)-1)/2)
    s*(s+1)/2 == n || error("vec2utri: length of vector is not triangular")
    [ i>=j ? v[round(Int, j*(j-1)/2+i)] : z for i=1:s, j=1:s ]
end

a = vec2ltri(collect(1:6))

julia> a = vec2ltri(collect(1:6))
3x3 Array{Int64,2}:
 1  0  0
 2  3  0
 3  4  6

julia> istril(a)  ## verify matrix is lower triangular
true

如果您想要上三角: 而不是下三角，只需将 i<=j 更改为 i>=j.

If you want upper triangular: instead of lower, just change the i<=j to i>=j.

其他随机工具注意还有像 tril!(a) 这样的函数，它将给定矩阵原地转换为下三角形，用零替换主对角线以上的所有内容.有关此功能的更多信息，请参阅 Julia 文档作为各种其他相关工具.

Other random tools Note also functions like tril!(a) which will convert in place a given matrix to lower triangular, replacing everything above the main diagonal with zeros. See the Julia documentation for more info on this function, as well as various other related tools.

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