matrix - 如何从向量制作对角矩阵

Question

using LinearAlgebra;
        a = rand(4,1);
        B = diagm(a);
        C  = Diagonal(a);

上面的代码在创建对角矩阵时会导致错误/(无意)。
如果 a = [1 2 3 4]我需要一个矩阵，如:

D = [1 0 0 0;0 2 0 0;0 0 3 0;0 0 0 4].

C = Diagonal(a) 创建
C = [1]
B = 图示(a)；给出错误信息:

Error messages: ERROR: MethodError: no method matchingdiagm(::Matrix{Float64})

You might have used a 2d row vector where a 1d column vector wasrequired. Note the difference between 1d column vector [1,2,3] and 2drow vector [1 2 3]. You can convert to a column vector with the vec()function. Closest candidates are: diagm(::Pair{var"#s832",var"#s831"} where {var"#s832"<:Integer, var"#s831"<:(AbstractVector{T}where T)}...) atC:\buildbot\worker\package_win64\build\usr\share\julia\stdlib\v1.6\LinearAlgebra\src\dense.jl:279diagm(::Integer, ::Integer, ::Pair{var"#s832", var"#s831"} where{var"#s832"<:Integer, var"#s831"<:(AbstractVector{T} where T)}...) atC:\buildbot\worker\package_win64\build\usr\share\julia\stdlib\v1.6\LinearAlgebra\src\dense.jl:280diagm(::AbstractVector{T} where T) atC:\buildbot\worker\package_win64\build\usr\share\julia\stdlib\v1.6\LinearAlgebra\src\dense.jl:329... Stacktrace: [1] top-level scope @ REPL[16]:1

Answer 1

我认为问题是你的a是矩阵。
尝试这个:

a = [1,2,3,4]  # 4-element Vector{Int64}
C = Diagonal(a)
4×4 Diagonal{Int64, Vector{Int64}}:
 1  ⋅  ⋅  ⋅
 ⋅  2  ⋅  ⋅
 ⋅  ⋅  3  ⋅
 ⋅  ⋅  ⋅  4

或者，制作一个真正的对角矩阵:

M = diagm(a)
4×4 Matrix{Int64}:
 1  0  0  0
 0  2  0  0
 0  0  3  0
 0  0  0  4