Recommended way to delete user-defined variables

General Usage
1

I currently delete all my user-defined variables like this.

varinfo()
variable1, variable2, variable3 = fill(nothing, 3)

Is there a way to make varinfo() output an array of only user-defined variable names instead of a markdown table with extra things like Base, Core, Main, and InteractiveUtils?

2

I don’t think so. The thing is that Base, Core, etc. are actually “variables” of Main, i.e. Main.Base and Main.Core etc. In contrast, try this: module ABC end; varinfo(ABC).

What you can do is use the regex option of varinfo to filter out Base, Core, etc.

1 Like
3

how to use the regex option?

4

Does this do the job for you?

julia> getvars() = [string(v) for v in sort!(names(Main)) if isdefined(Main, v) && !(v in (:Base, :Main, :Core, :InteractiveUtils, :ans))]
getvars (generic function with 1 method)

julia> getvars()
1-element Array{String,1}:
 "getvars"

julia> x =5
5

julia> getvars()
2-element Array{String,1}:
 "getvars"
 "x"

julia> y = 10
10

julia> getvars()
3-element Array{String,1}:
 "getvars"
 "x"
 "y"
5 Likes
5

This convenient function should a built-in function for Julia.

1 Like