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

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

Topic		Replies	Views
Clearing a variable? New to Julia	7	4346	December 24, 2019
List existing global variables, e.g. via varinfo() General Usage interactiveutils	1	453	February 24, 2023
How do I list the variables I've created? New to Julia pluto	3	661	August 8, 2021
Destructuring NamedTuples with Regular Expressions General Usage destructuring	6	331	October 10, 2022
Delete all variables - Pluto notebook - Julia General Usage question , pluto	6	483	January 22, 2023

Recommended way to delete user-defined variables

Related topics