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?

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() …

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.

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

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General Usage

Yifan_Liu February 28, 2021, 8:12pm 3

how to use the regex option?

Safe way to open new file and unlock an existing file to enable deletion, if it exists

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Recommended way to delete user-defined variables

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