Manipulating data in DataFrame: how to calculate the square of a column

user_9381746 0
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I would like to calculate the square of a column A 1,2,3,4, process it with other calculation store it in column C

using CSV, DataFrames
df = DataFrame(A = 1:4, B = ["M", "F", "F", "M"])
df.C = ((((df.A./2).^2).*3.14)./1000)

Is there an easier way to write it?

calculation julia dataframe
2 Answers

I am not sure how much shorter you would want the formula to be, but you can write:

df.C = @. (df.A / 2) ^ 2 * 3.14 / 1000

to avoid having to write . everywhere.

Or you can use transform!, but it is not shorter (its benefit is that you can uset it in a processing pipeline, e.g. using Pipe.jl):

transform!(df, :A => ByRow(a -> (a / 2) ^ 2 * 3.14 / 1000) => :C)
user_12695670

Try this:

df.D = .5df.A .^2 * 0.00314

Explanation:

  • not so many parentheses needed
  • multiplying scalar by vector is here as good as the vectorization for short vectors (up two something like 100 elements)

A simple benchmark using BenchmarkTools:

julia> @btime $df.E = .5*$df.A .^2 * 0.00314;
  592.085 ns (9 allocations: 496 bytes)

julia> @btime $df.F = @. ($df.A / 2) ^ 2 * 0.00314;
  875.490 ns (11 allocations: 448 bytes)

The fastest is however a longer version where you provide the type information @. (df.A::Vector{Int} / 2) ^ 2 * 0.00314 (again this matters rather for short DataFrames and note that here the Z column must exist so we create it here):

julia> @btime begin $df.Z = Vector{Float64}(undef, nrow(df));@. $df.Z = ($df.A::Vector{Int} / 2.0) ^ 2.0 * 0.00314; end;
  162.564 ns (3 allocations: 208 bytes)
user_99577100
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