Get column names from DataFrame or JuliaDB table

Question

How do I get the column names from a DataFrame object or JuliaDB IndexedTable object? Is this possible?

Reproducible Code:

using JuliaDB
import DataFrames
DF = DataFrames

# CREATES AN EXAMPLE TABLE WITH JULIADB

colnames = [:samples, :A, :B, :C, :D]
primary_key = [:samples]
coltypes = [Int[], Float64[],Float64[],Float64[],Float64[]]
sample_sizes = [100,200,300]
example_values = (1, 0.4, 0.3, 0.2, 0.1)

mytable = table(coltypes..., names=colnames, pkey=primary_key) # initialize empty table

# add some data to table
for i in sample_sizes
    example_values = (i, 0.4, 0.3, 0.2, 0.1)
    table_params = [(col=>val) for (col,val) in zip(colnames, example_values)]

    push!(rows(mytable), (; table_params...)) # add row
    mytable = table(mytable, pkey = primary_key, copy = false) # sort rows by primary key
end
mytable = table(unique(mytable), pkey=primary_key) # remove duplicate rows which don't exist

# MAKES A DATAFRAME FROM JULIADB TABLE

df = DF.DataFrame(mytable)

For instance, given the above code, how would you check with a conditional if there's a column :E in either mytable or df, (for the purposes of adding such a column if it doesn't exist yet)?

Ultimately, I'm looking for the Julia equivalent of the following Python code:

if 'E' in df.columns:
     # ...
else:
     # ...

what about `names(df)`? – Michael K. Borregaard Mar 27 '20 at 11:09 — Michael K. Borregaard, Mar 27 '20 at 11:09
That works for DataFrames. Thank you – bug_spray Mar 27 '20 at 14:44 — bug_spray, Mar 27 '20 at 14:44

score 16 · Accepted Answer · edited Nov 18 '22 at 07:29

16

If df is a data frame you can write:

if "E" in names(df)
...

(in JuliaDB.jl it would be JuliaDB.colnames)

but a faster (in terms of run-time and available for data frames option is:

if hasproperty(df, :E)
...

A bit slower, but useful in other cases is (it also works for JuliaDB.jl but first you have to load Tables.jl and write Tables.columnindex instead):

if columnindex(df, :E) != 0
...

The last example columnindex is probably most complex the way it works is described in its documentation:

help?> columnindex
search: columnindex

  Tables.columnindex(table, name::Symbol)

  Return the column index (1-based) of a column by name in a table with a
  known schema; returns 0 if name doesn't exist in table

  ────────────────────────────────────────────────────────────────────────────

  given names and a Symbol name, compute the index (1-based) of the name in
  names

edited Nov 18 '22 at 07:29

Miles

780
7
19

answered Mar 27 '20 at 11:54

Bogumił Kamiński

66,844
3
80
107

This is excellent, thank you. This solves my problem. That said, though, none of these methods work on a JuliaDB IndexedTable (I'd have to convert it to a DataFrame first). Am I correct in assuming that JuliaDB is still very much a work in progress? – bug_spray Mar 27 '20 at 14:47
Sorry for this - I was writing from my head what should work for Tables.jl :(. I will fix in the answer what you can do. – Bogumił Kamiński Mar 27 '20 at 15:23
I now fixed it. Unfortunately JuliaDB.jl is not fully compliant with Tables.jl interface yet which caused confusion. – Bogumił Kamiński Mar 27 '20 at 15:31

Get column names from DataFrame or JuliaDB table

1 Answers1