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() …

Recommended way to delete user-defined variables

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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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