Efficient way of doing linear regression

General Usage Performance
27

what about?

using Random

N = 10000
x = rand(N)
X = [ones(N) x]
y = 10 .+ x .* 0.3

function linreg1(y, X)
    β_hat = (X' * X) \ X' * y
    return(β_hat)
end

function linreg2(y, X)
    β_hat = X \ y
    return(β_hat)
end

using GLM
GLM.fit(LinearModel, X, y, true)

using DataFrames, GLM
data = DataFrame(X = x, Y = y)
lm(@formula(Y ~ X), data)


using BenchmarkTools
@benchmark linreg1(y, X)
@benchmark linreg2(y, X)
@benchmark GLM.fit(LinearModel, X, y, true)
@benchmark lm(@formula(Y ~ X), data)

and the results are

lingerg1:

BenchmarkTools.Trial: 
  memory estimate:  156.78 KiB
  allocs estimate:  8
  --------------
  minimum time:     213.900 μs (0.00% GC)
  median time:      289.351 μs (0.00% GC)
  mean time:        552.780 μs (4.66% GC)
  maximum time:     76.192 ms (0.00% GC)
  --------------
  samples:          8836
  evals/sample:     1

linreg2:

BenchmarkTools.Trial: 
  memory estimate:  303.69 KiB
  allocs estimate:  45
  --------------
  minimum time:     118.300 μs (0.00% GC)
  median time:      170.300 μs (0.00% GC)
  mean time:        344.598 μs (17.20% GC)
  maximum time:     57.911 ms (0.00% GC)
  --------------
  samples:          10000
  evals/sample:     1

fit():

BenchmarkTools.Trial: 
  memory estimate:  470.45 KiB
  allocs estimate:  26
  --------------
  minimum time:     133.700 μs (0.00% GC)
  median time:      189.299 μs (0.00% GC)
  mean time:        428.768 μs (17.15% GC)
  maximum time:     45.171 ms (0.00% GC)
  --------------
  samples:          10000
  evals/sample:     1

lm():

BenchmarkTools.Trial: 
  memory estimate:  1.08 MiB
  allocs estimate:  186
  --------------
  minimum time:     546.600 μs (0.00% GC)
  median time:      839.500 μs (0.00% GC)
  mean time:        1.917 ms (10.94% GC)
  maximum time:     127.190 ms (0.00% GC)
  --------------
  samples:          2588
  evals/sample:     1
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