CODEWITHSUNDEEP
julia
hooman
hooman

Reputation: 35

What ways exist to create an array with given dimensions from a given sequence in Julia?

I'm new to Julia and I could not find any useful information on the following: I would like to create an array of given dimensions and fill it with a given sequence.

m,n = 10,10 # dimensions
i = 1:100 # sequence

I've tried to use collect, but this gives me a single column array. I have also tried it the Julia way 

[? for i in 1:m, j in 1:n]

but I don't know what I could insert for ?.

Upvotes: 1

Views: 81

Answers (1)

carstenbauer
carstenbauer

Reputation: 10127

The easiest way is reshape(i, m,n) (potentially together with a collect if you really need an Array{Int64,2}):

julia> reshape(i,m,n)
10×10 reshape(::UnitRange{Int64}, 10, 10) with eltype Int64:
  1  11  21  31  41  51  61  71  81   91
  2  12  22  32  42  52  62  72  82   92
  3  13  23  33  43  53  63  73  83   93
  4  14  24  34  44  54  64  74  84   94
  5  15  25  35  45  55  65  75  85   95
  6  16  26  36  46  56  66  76  86   96
  7  17  27  37  47  57  67  77  87   97
  8  18  28  38  48  58  68  78  88   98
  9  19  29  39  49  59  69  79  89   99
 10  20  30  40  50  60  70  80  90  100

julia> collect(ans)
10×10 Array{Int64,2}:
  1  11  21  31  41  51  61  71  81   91
  2  12  22  32  42  52  62  72  82   92
  3  13  23  33  43  53  63  73  83   93
  4  14  24  34  44  54  64  74  84   94
  5  15  25  35  45  55  65  75  85   95
  6  16  26  36  46  56  66  76  86   96
  7  17  27  37  47  57  67  77  87   97
  8  18  28  38  48  58  68  78  88   98
  9  19  29  39  49  59  69  79  89   99
 10  20  30  40  50  60  70  80  90  100

To answer your question what to put as ? in the array comprehension approach, you must convert the cartesian index to a linear index, for example like so:

julia> [i[LinearIndices((m,n))[p,q]] for p in 1:m, q in 1:n]
10×10 Array{Int64,2}:
  1  11  21  31  41  51  61  71  81   91
  2  12  22  32  42  52  62  72  82   92
  3  13  23  33  43  53  63  73  83   93
  4  14  24  34  44  54  64  74  84   94
  5  15  25  35  45  55  65  75  85   95
  6  16  26  36  46  56  66  76  86   96
  7  17  27  37  47  57  67  77  87   97
  8  18  28  38  48  58  68  78  88   98
  9  19  29  39  49  59  69  79  89   99
 10  20  30  40  50  60  70  80  90  100

Of course, you can also calculate the linear index yourself, [i[(q-1)*m + p] for p in 1:m, q in 1:n].

Alternatively, you can preallocate the array and fill it in a linear fashion:

julia> result = Matrix{Int64}(undef, m,n);

julia> result[:] .= i;

julia> result
10×10 Array{Int64,2}:
  1  11  21  31  41  51  61  71  81   91
  2  12  22  32  42  52  62  72  82   92
  3  13  23  33  43  53  63  73  83   93
  4  14  24  34  44  54  64  74  84   94
  5  15  25  35  45  55  65  75  85   95
  6  16  26  36  46  56  66  76  86   96
  7  17  27  37  47  57  67  77  87   97
  8  18  28  38  48  58  68  78  88   98
  9  19  29  39  49  59  69  79  89   99
 10  20  30  40  50  60  70  80  90  100

which is basically equivalent to the naive, explicit solution

julia> result = Matrix{Int64}(undef, m,n);

julia> for k in eachindex(i) result[k] = i[k] end

julia> result
10×10 Array{Int64,2}:
  1  11  21  31  41  51  61  71  81   91
  2  12  22  32  42  52  62  72  82   92
  3  13  23  33  43  53  63  73  83   93
  4  14  24  34  44  54  64  74  84   94
  5  15  25  35  45  55  65  75  85   95
  6  16  26  36  46  56  66  76  86   96
  7  17  27  37  47  57  67  77  87   97
  8  18  28  38  48  58  68  78  88   98
  9  19  29  39  49  59  69  79  89   99
 10  20  30  40  50  60  70  80  90  100

Upvotes: 2

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