How to have multidimensional array with different length in Julia
up vote
1
down vote
favorite
I need to make a sequence of an array with different length by reading a dataset. I need to call each of them in a loop so probably I need some sort of indexing in order to call them. For example, how can I create the following sequence:
P[1]=[1 2 3 4]
P[2]=[1 4]
P[3]=[8 9 0 0 5 6]
.
.
.
arrays julia-lang
edited Nov 10 at 22:03
marc_s
569k12811001250
asked Nov 10 at 21:59
Reabo
114
add a comment |
up vote
1
down vote
favorite
I need to make a sequence of an array with different length by reading a dataset. I need to call each of them in a loop so probably I need some sort of indexing in order to call them. For example, how can I create the following sequence:
P[1]=[1 2 3 4]
P[2]=[1 4]
P[3]=[8 9 0 0 5 6]
.
.
.
arrays julia-lang
edited Nov 10 at 22:03
marc_s
569k12811001250
asked Nov 10 at 21:59
Reabo
114
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to make a sequence of an array with different length by reading a dataset. I need to call each of them in a loop so probably I need some sort of indexing in order to call them. For example, how can I create the following sequence:
P[1]=[1 2 3 4]
P[2]=[1 4]
P[3]=[8 9 0 0 5 6]
.
.
.
arrays julia-lang
edited Nov 10 at 22:03
marc_s
569k12811001250
asked Nov 10 at 21:59
Reabo
114
I need to make a sequence of an array with different length by reading a dataset. I need to call each of them in a loop so probably I need some sort of indexing in order to call them. For example, how can I create the following sequence:
P[1]=[1 2 3 4]
P[2]=[1 4]
P[3]=[8 9 0 0 5 6]
.
.
.
arrays julia-lang
arrays julia-lang
edited Nov 10 at 22:03
marc_s
569k12811001250
asked Nov 10 at 21:59
Reabo
114
edited Nov 10 at 22:03
marc_s
569k12811001250
asked Nov 10 at 21:59
Reabo
114
edited Nov 10 at 22:03
marc_s
569k12811001250
edited Nov 10 at 22:03
marc_s
569k12811001250
edited Nov 10 at 22:03
marc_s
569k12811001250
569k12811001250
asked Nov 10 at 21:59
Reabo
114
asked Nov 10 at 21:59
Reabo
114
asked Nov 10 at 21:59
Reabo
114
114
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
Here it is:
julia> P = VectorVectorInt64([[1,2,3,4],[1,4],[8,9,0,0,5,6]])
3-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
julia> P[1]
4-element ArrayInt64,1:
1
2
3
4
julia> P[2]
2-element ArrayInt64,1:
1
4
julia> P[3]
6-element ArrayInt64,1:
8
9
0
0
5
6
If you want to add a new element use push!():
julia> push!(P,[7,8,9])
4-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
[7, 8, 9]
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
Thanks, but my problem is that this sequence is very long and I cannot do the first line of your code manually.
– Reabo
Nov 10 at 22:12
I found this and seems it works, thanks V=VectorVectorFloat64(100)
– Reabo
Nov 10 at 22:18
no problem - I edited my answer. You can create an emptyArraybyP = VectorVectorInt64()and then justpush!to it.
– Przemyslaw Szufel
Nov 10 at 22:18
if you know theArraysizeVectorVectorInt64(undef, N)and then usingP[n]=...is of course better (note theundef)
– Przemyslaw Szufel
Nov 10 at 22:21
1
Initialize empty:P = VectorInt. If you already know the entries:P = [[1,2,3,4],[1,4],[8,9,0,0,5,6]]. No need to specify the type in front of the literal.
– DNF
Nov 11 at 15:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here it is:
julia> P = VectorVectorInt64([[1,2,3,4],[1,4],[8,9,0,0,5,6]])
3-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
julia> P[1]
4-element ArrayInt64,1:
1
2
3
4
julia> P[2]
2-element ArrayInt64,1:
1
4
julia> P[3]
6-element ArrayInt64,1:
8
9
0
0
5
6
If you want to add a new element use push!():
julia> push!(P,[7,8,9])
4-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
[7, 8, 9]
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
Thanks, but my problem is that this sequence is very long and I cannot do the first line of your code manually.
– Reabo
Nov 10 at 22:12
I found this and seems it works, thanks V=VectorVectorFloat64(100)
– Reabo
Nov 10 at 22:18
no problem - I edited my answer. You can create an emptyArraybyP = VectorVectorInt64()and then justpush!to it.
– Przemyslaw Szufel
Nov 10 at 22:18
if you know theArraysizeVectorVectorInt64(undef, N)and then usingP[n]=...is of course better (note theundef)
– Przemyslaw Szufel
Nov 10 at 22:21
1
Initialize empty:P = VectorInt. If you already know the entries:P = [[1,2,3,4],[1,4],[8,9,0,0,5,6]]. No need to specify the type in front of the literal.
– DNF
Nov 11 at 15:56
add a comment |
up vote
2
down vote
Here it is:
julia> P = VectorVectorInt64([[1,2,3,4],[1,4],[8,9,0,0,5,6]])
3-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
julia> P[1]
4-element ArrayInt64,1:
1
2
3
4
julia> P[2]
2-element ArrayInt64,1:
1
4
julia> P[3]
6-element ArrayInt64,1:
8
9
0
0
5
6
If you want to add a new element use push!():
julia> push!(P,[7,8,9])
4-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
[7, 8, 9]
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
Thanks, but my problem is that this sequence is very long and I cannot do the first line of your code manually.
– Reabo
Nov 10 at 22:12
I found this and seems it works, thanks V=VectorVectorFloat64(100)
– Reabo
Nov 10 at 22:18
no problem - I edited my answer. You can create an emptyArraybyP = VectorVectorInt64()and then justpush!to it.
– Przemyslaw Szufel
Nov 10 at 22:18
if you know theArraysizeVectorVectorInt64(undef, N)and then usingP[n]=...is of course better (note theundef)
– Przemyslaw Szufel
Nov 10 at 22:21
1
Initialize empty:P = VectorInt. If you already know the entries:P = [[1,2,3,4],[1,4],[8,9,0,0,5,6]]. No need to specify the type in front of the literal.
– DNF
Nov 11 at 15:56
add a comment |
up vote
2
down vote
up vote
2
down vote
Here it is:
julia> P = VectorVectorInt64([[1,2,3,4],[1,4],[8,9,0,0,5,6]])
3-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
julia> P[1]
4-element ArrayInt64,1:
1
2
3
4
julia> P[2]
2-element ArrayInt64,1:
1
4
julia> P[3]
6-element ArrayInt64,1:
8
9
0
0
5
6
If you want to add a new element use push!():
julia> push!(P,[7,8,9])
4-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
[7, 8, 9]
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
Here it is:
julia> P = VectorVectorInt64([[1,2,3,4],[1,4],[8,9,0,0,5,6]])
3-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
julia> P[1]
4-element ArrayInt64,1:
1
2
3
4
julia> P[2]
2-element ArrayInt64,1:
1
4
julia> P[3]
6-element ArrayInt64,1:
8
9
0
0
5
6
If you want to add a new element use push!():
julia> push!(P,[7,8,9])
4-element ArrayArrayInt64,1,1:
[1, 2, 3, 4]
[1, 4]
[8, 9, 0, 0, 5, 6]
[7, 8, 9]
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
edited Nov 10 at 22:18
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
answered Nov 10 at 22:09
Przemyslaw Szufel
1,369110
1,369110
Thanks, but my problem is that this sequence is very long and I cannot do the first line of your code manually.
– Reabo
Nov 10 at 22:12
I found this and seems it works, thanks V=VectorVectorFloat64(100)
– Reabo
Nov 10 at 22:18
no problem - I edited my answer. You can create an emptyArraybyP = VectorVectorInt64()and then justpush!to it.
– Przemyslaw Szufel
Nov 10 at 22:18
if you know theArraysizeVectorVectorInt64(undef, N)and then usingP[n]=...is of course better (note theundef)
– Przemyslaw Szufel
Nov 10 at 22:21
1
Initialize empty:P = VectorInt. If you already know the entries:P = [[1,2,3,4],[1,4],[8,9,0,0,5,6]]. No need to specify the type in front of the literal.
– DNF
Nov 11 at 15:56
add a comment |
Thanks, but my problem is that this sequence is very long and I cannot do the first line of your code manually.
– Reabo
Nov 10 at 22:12
I found this and seems it works, thanks V=VectorVectorFloat64(100)
– Reabo
Nov 10 at 22:18
no problem - I edited my answer. You can create an emptyArraybyP = VectorVectorInt64()and then justpush!to it.
– Przemyslaw Szufel
Nov 10 at 22:18
if you know theArraysizeVectorVectorInt64(undef, N)and then usingP[n]=...is of course better (note theundef)
– Przemyslaw Szufel
Nov 10 at 22:21
1
Initialize empty:P = VectorInt. If you already know the entries:P = [[1,2,3,4],[1,4],[8,9,0,0,5,6]]. No need to specify the type in front of the literal.
– DNF
Nov 11 at 15:56
Thanks, but my problem is that this sequence is very long and I cannot do the first line of your code manually.
– Reabo
Nov 10 at 22:12
Thanks, but my problem is that this sequence is very long and I cannot do the first line of your code manually.
– Reabo
Nov 10 at 22:12
I found this and seems it works, thanks V=VectorVectorFloat64(100)
– Reabo
Nov 10 at 22:18
I found this and seems it works, thanks V=VectorVectorFloat64(100)
– Reabo
Nov 10 at 22:18
no problem - I edited my answer. You can create an empty
Array by P = VectorVectorInt64() and then just push! to it.– Przemyslaw Szufel
Nov 10 at 22:18
no problem - I edited my answer. You can create an empty
Array by P = VectorVectorInt64() and then just push! to it.– Przemyslaw Szufel
Nov 10 at 22:18
if you know the
Array size VectorVectorInt64(undef, N) and then using P[n]=... is of course better (note the undef)– Przemyslaw Szufel
Nov 10 at 22:21
if you know the
Array size VectorVectorInt64(undef, N) and then using P[n]=... is of course better (note the undef)– Przemyslaw Szufel
Nov 10 at 22:21
1
1
Initialize empty:
P = VectorInt. If you already know the entries: P = [[1,2,3,4],[1,4],[8,9,0,0,5,6]]. No need to specify the type in front of the literal.– DNF
Nov 11 at 15:56
Initialize empty:
P = VectorInt. If you already know the entries: P = [[1,2,3,4],[1,4],[8,9,0,0,5,6]]. No need to specify the type in front of the literal.– DNF
Nov 11 at 15:56
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