Any idea why this is not working?
transform(df, All() .=> string; renamecols=false)
Isn't it supposed to apply the string function to all columns and as a result convert them? When adding ByRow it works, but an operation like this should be on entire columns not on each row.
What you describe works as expected. string takes a whole vector and converts it to string (the vector, not its contents). To work on elements of the vector use ByRow, as you have commented, or use broadcasting:
julia> df = DataFrame(x=1:2, y=3:4, z=5:6)
2×3 DataFrame
Row │ x y z
│ Int64 Int64 Int64
─────┼─────────────────────
1 │ 1 3 5
2 │ 2 4 6
julia> transform(df, All() .=> string; renamecols=false)
2×3 DataFrame
Row │ x y z
│ String String String
─────┼────────────────────────
1 │ [1, 2] [3, 4] [5, 6]
2 │ [1, 2] [3, 4] [5, 6]
julia> transform(df, All() .=> ByRow(string); renamecols=false)
2×3 DataFrame
Row │ x y z
│ String String String
─────┼────────────────────────
1 │ 1 3 5
2 │ 2 4 6
julia> string.(df) # broadcasting version
2×3 DataFrame
Row │ x y z
│ String String String
─────┼────────────────────────
1 │ 1 3 5
2 │ 2 4 6
The reason why in All() .=> string you still get a vector is that transform enforces that the number of rows is not changed in the result. Therefore the resulting string is reused. Note that with combine you would get a single row:
julia> combine(df, All() .=> string; renamecols=false)
1×3 DataFrame
Row │ x y z
│ String String String
─────┼────────────────────────
1 │ [1, 2] [3, 4] [5, 6]
To highlight the issue see how string operates on a vector without DataFrames.jl:
julia> string([1, 2])
"[1, 2]"