13

I'm looking for a way to replace values in Dataframe column with random numbers. They should be different in every row where the substitution was performed.

For example replacing "X" with random numbers drawn from 100:120 range

julia> df = DataFrame(:a=>[1,2,"X","X",5,"X"],)
6×1 DataFrame
│ Row │ a   │
│     │ Any │
├─────┼─────┤
│ 1   │ 1   │
│ 2   │ 2   │
│ 3   │ X   │
│ 4   │ X   │
│ 5   │ 5   │
│ 6   │ X   │

* Replacing X with random values in 100:120 *

julia> df
6×1 DataFrame
│ Row │ a   │
│     │ Any │
├─────┼─────┤
│ 1   │ 1   │
│ 2   │ 2   │
│ 3   │ 103 │
│ 4   │ 110 │
│ 5   │ 5   │
│ 6   │ 116 │

I've tried using replace but rand is evaluated before replace:

julia> replace!(df.a,"X"=> rand(100:120))
julia> df
6×1 DataFrame
│ Row │ a   │
│     │ Any │
├─────┼─────┤
│ 1   │ 1   │
│ 2   │ 2   │
│ 3   │ 115 │
│ 4   │ 115 │
│ 5   │ 5   │
│ 6   │ 115 │
Maciej Fender
  • 311
  • 2
  • 8
  • If you can arrange to have `missing` appear instead of `"X"` it would be more Julian. Then, Przemyslaw's one-liner would be: `replace!(x -> @coalesce(x, rand(100:120)), df.a)` which is more readable. – Dan Getz Dec 07 '22 at 16:18

2 Answers2

12

A one liner could be:

replace!( x-> x=="X" ? rand() : x, df.a)
Przemyslaw Szufel
  • 40,002
  • 3
  • 32
  • 62
6

Alternatively using operation specification syntax:

transform!(df, :a => ByRow(a -> a == "X" ? rand() : a)=> :a)

or you can do:

rand!(view(df.a, df.a .== "X"), Float64)

for another in-place approach (but replace! will likely be more efficient)

Bogumił Kamiński
  • 66,844
  • 3
  • 80
  • 107