'Convert Julia array to dataframe

I have an array X that I'd like to convert to a dataframe. Upon recommendation from the web, I tried converting to a dataframe and get the following error.

julia> y=convert(DataFrame,x) ERROR:converthas no method matching convert(::Type{DataFrame}, ::Array{Float64,2}) in convert at base.jl:13

When I try DataFrame(x), the conversion works but i get a complaint that the conversion is deprecated.

julia> DataFrame(x) WARNING: DataFrame(::Matrix, ::Vector)) is deprecated, use convert(DataFrame, Matrix) instead in DataFrame at /Users/Matthew/.julia/v0.3/DataFrames/src/deprecated.jl:54 (repeats 2 times)

Is there another method I should be aware of to keep my code consistent?

EDIT: Julia 0.3.2, DataFrames 0.5.10 OSX 10.9.5

julia> x=rand(4,4)
4x4 Array{Float64,2}:
 0.467882   0.466358  0.28144   0.0151388
 0.22354    0.358616  0.669564  0.828768
 0.475064   0.187992  0.584741  0.0543435
 0.0592643  0.345138  0.704496  0.844822

julia> convert(DataFrame,x)
ERROR: `convert` has no method matching convert(::Type{DataFrame}, ::Array{Float64,2}) in convert at base.jl:13
arraysdataframejulia


Solution 1:[1]

This works for me:

julia> using DataFrames

julia> x = rand(4, 4)
4x4 Array{Float64,2}:
 0.790912  0.0367989  0.425089  0.670121
 0.243605  0.62487    0.582498  0.302063
 0.785159  0.0083891  0.881153  0.353925
 0.618127  0.827093   0.577815  0.488565

julia> convert(DataFrame, x)
4x4 DataFrame
| Row | x1       | x2        | x3       | x4       |
|-----|----------|-----------|----------|----------|
| 1   | 0.790912 | 0.0367989 | 0.425089 | 0.670121 |
| 2   | 0.243605 | 0.62487   | 0.582498 | 0.302063 |
| 3   | 0.785159 | 0.0083891 | 0.881153 | 0.353925 |
| 4   | 0.618127 | 0.827093  | 0.577815 | 0.488565 |

Are you trying something different?

If that doesn't work try posting a bit more code we can help you better.

Solution 2:[2]

I've been confounded by the same issue a number of times, and eventually realized the issue is often related to the format of the array, and is easily resolved by simply transposing the array prior to conversion.

In short, I recommend: 

julia> convert(DataFrame, x')

Solution 3:[3]

# convert a Matrix{Any} with a header row of col name strings to a DataFrame
# e.g. mat2df(["a" "b" "c"; 1 2 3; 4 5 6])

mat2df(mat) = convert(DataFrame,Dict(mat[1,:],
                     [mat[2:end,i] for i in 1:size(mat,2)]))

# convert a Matrix{Any} (mat) and a list of col name strings (headerstrings) 
# to a DataFrame, e.g. matnms2df([1 2 3;4 5 6], ["a","b","c"])

matnms2df(mat, headerstrs) = convert(DataFrame,
    Dict(zip(headerstrs,[mat[:,i] for i in 1:size(mat,2)])))

Solution 4:[4]

Since this is the first thing that comes up when you google, for more recent versions of DataFrames.jl, you can just use the DataFrame() function now:

julia> x = rand(4,4)
4×4 Matrix{Float64}:
 0.920406  0.738911  0.994401  0.9954
 0.18791   0.845132  0.277577  0.231483
 0.361269  0.918367  0.793115  0.988914
 0.725052  0.962762  0.413111  0.328261

julia> DataFrame(x, :auto)
4×4 DataFrame
 Row ? x1        x2        x3        x4       
     ? Float64   Float64   Float64   Float64  
??????????????????????????????????????????????
   1 ? 0.920406  0.738911  0.994401  0.9954
   2 ? 0.18791   0.845132  0.277577  0.231483
   3 ? 0.361269  0.918367  0.793115  0.988914
   4 ? 0.725052  0.962762  0.413111  0.328261

Solution 5:[5]

So I found this online and honestly felt dumb. 

using CSV
WhatIWant = DataFrame(WhatIHave)

this was adapted from an R guide, but it works so heck

Solution 6:[6]

DataFrame([1 2 3 4; 5 6 7 8; 9 10 11 12], :auto)

This works as per >? DataFrame

Solution 7:[7]

A little late, but with the update to the DataFrame() function, I created a custom function that would take a matrix (e.g. an XLSX imported dataset) and convert it into a DataFrame using the first row as column headers. Saves me a ton of time and, hopefully, it helps you too.

function MatrixToDataFrame(mat)
    DF_mat = DataFrame(
        mat[2:end, 1:end],
        string.(mat[1, 1:end])
    )
    return DF_mat
end

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 spencerlyon2
Solution 2 Baum mit Augen
Solution 3
Solution 4 colebrookson
Solution 5 Josh Glaab
Solution 6 user17029991
Solution 7 Kyle Protho

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