Way to give DataFrame columns nicknames?

General Usage
1

I am importing data from Excel into a DataFrame. The column headers are long specific names with spaces and units. They are a pain to work with because every time I need to access that column, I have to type df."Long Specific Name With Spaces (Units)". Is there a good way to give the column a nickname like :x, so I can access it with df.x instead? The method would need to preserve the original headers, so I can transform them back for printing.

I think the correct tool is a dictionary, but I haven’t used them much. I also know that DataFrames.jl uses => for some of its “split-apply-combine” syntax, and I don’t want to accidently interfere with that.

2

How are you importing your data? If using CSV.jl, you can read using the normalizenames argument to get rid of the spaces and add underscores. Something like

 CSV.File(file; normalizenames=true)
3

Yes, I am using CSV.jl. That is good to know, but it is really the length of the name that is the problem.

4

There’s no way to do this afaik, but it seems it would be quite easy to do

long_names = names(df)

short_names = [:x, :y, :z] # whatever you need

and then rename!(df, short_names) to work with the data with nicknames and rename!(df, long_names) again when producing outputs?

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5

Adding to this, you can also provide change individual columns one at a time, for example rename!(df,"Long Name"=>"longname"), or through a Dictionary if you want to change a series of column labels with rename!(df,Dict("Long Name"=>"longname","Other Odd Name"=>"oddname")).

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6

Can you reverse the Dictionary to get the original names back?

7

Or if you are not sure whether the order or the number of columns might change, you can try the following approach:

short_names = string.(range('a', step = 1, length = size(d,2)))
dict1 = Dict(zip(Symbol.(names(d)), short_names))
dict2 = Dict(zip(Symbol.(short_names), names(d)))
rename!(d, dict1)
rename!(d, dict2)

where d is the DataFrame. Note that this might causes problems if you have more than 26 columns.

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8

You can reverse the Dictionary you used to rename everything, for example reverse_dict = Dict(val=>key for (key,val) in dict).

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9

You don’t need a Dict at all, just put the long and short names into two equal-length vectors and broadcast the pair operator in rename! to transparently go back between long and short names for the desired columns:

julia> df = DataFrame(["Long Name $i" => rand(2) for i ∈ 1:5]...)
2×5 DataFrame
 Row │ Long Name 1  Long Name 2  Long Name 3  Long Name 4  Long Name 5 
     │ Float64      Float64      Float64      Float64      Float64     
─────┼─────────────────────────────────────────────────────────────────
   1 │    0.573031    0.0743141     0.187922     0.182269     0.251157
   2 │    0.348444    0.986702      0.620114     0.881127     0.434446

julia> long_names = ["Long Name 1", "Long Name 3"];

julia> short_names = ["SN1", "SN3"]
2-element Vector{String}:
 "SN1"
 "SN3"

julia> rename!(df, long_names .=> short_names)
2×5 DataFrame
 Row │ SN1       Long Name 2  SN3       Long Name 4  Long Name 5 
     │ Float64   Float64      Float64   Float64      Float64     
─────┼───────────────────────────────────────────────────────────
   1 │ 0.573031    0.0743141  0.187922     0.182269     0.251157
   2 │ 0.348444    0.986702   0.620114     0.881127     0.434446

julia> rename!(df, short_names .=> long_names)
2×5 DataFrame
 Row │ Long Name 1  Long Name 2  Long Name 3  Long Name 4  Long Name 5 
     │ Float64      Float64      Float64      Float64      Float64     
─────┼─────────────────────────────────────────────────────────────────
   1 │    0.573031    0.0743141     0.187922     0.182269     0.251157
   2 │    0.348444    0.986702      0.620114     0.881127     0.434446
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