如何在julia中启动列表元素列表?

Cho*_*ang 3 arrays julia

我想在朱莉娅列出一份清单.它应该是这样的:

a = [1,"char", [1, 2, 3]]
a[3]
# ouput should be [1,2,3]
Run Code Online (Sandbox Code Playgroud)

但是,朱莉娅会自动进行连接,所以它最终会被连接起来 [1, "char", 1, 2, 3]

如何在julia中执行此类操作而不使用其他值启动列表,然后为其分配列表,如:

a = [1, "char", 3]
a[3] = [1, 2, 3]
Run Code Online (Sandbox Code Playgroud)

Rez*_*lan 6

julia> a = Any[1,"char", [1, 2, 3]]
3-element Array{Any,1}:
 1
  "char"
  [1,2,3]
Run Code Online (Sandbox Code Playgroud)

数组连接的样式已经改变,现在对concat数组,正确的语法是:a = [1;"char"; [1, 2, 3]],所以[1;"char"; [1, 2, 3]]==Any[1;"char"; [1, 2, 3]] # => true但是当Julia [1,"char", [1, 2, 3]]默认面对时,它会尝试连接可能的元素类型.