How to create a two dimensional list using for loop in Scala?
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0
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If I write the following code in scala I get one dimensional list, as such:
scala> for (a <- (1 to 2).toList; b <- (1 to 3).toList) yield (a, b)
res1 = List((1,1), (1,2), (1,3), (2,1), (2,2), (2,3))
But I'm expecting :-
List(List((1,1), (1,2), (1,3)), List((2,1), (2,2), (2,3)))
Is it possible to do this using a for loop in scala or is some kind of other construct needed?
scala
asked Nov 10 at 16:04
Joe Johnes
33
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up vote
0
down vote
favorite
If I write the following code in scala I get one dimensional list, as such:
scala> for (a <- (1 to 2).toList; b <- (1 to 3).toList) yield (a, b)
res1 = List((1,1), (1,2), (1,3), (2,1), (2,2), (2,3))
But I'm expecting :-
List(List((1,1), (1,2), (1,3)), List((2,1), (2,2), (2,3)))
Is it possible to do this using a for loop in scala or is some kind of other construct needed?
scala
asked Nov 10 at 16:04
Joe Johnes
33
New contributor
Joe Johnes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I write the following code in scala I get one dimensional list, as such:
scala> for (a <- (1 to 2).toList; b <- (1 to 3).toList) yield (a, b)
res1 = List((1,1), (1,2), (1,3), (2,1), (2,2), (2,3))
But I'm expecting :-
List(List((1,1), (1,2), (1,3)), List((2,1), (2,2), (2,3)))
Is it possible to do this using a for loop in scala or is some kind of other construct needed?
scala
asked Nov 10 at 16:04
Joe Johnes
33
New contributor
Joe Johnes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If I write the following code in scala I get one dimensional list, as such:
scala> for (a <- (1 to 2).toList; b <- (1 to 3).toList) yield (a, b)
res1 = List((1,1), (1,2), (1,3), (2,1), (2,2), (2,3))
But I'm expecting :-
List(List((1,1), (1,2), (1,3)), List((2,1), (2,2), (2,3)))
Is it possible to do this using a for loop in scala or is some kind of other construct needed?
scala
scala
asked Nov 10 at 16:04
Joe Johnes
33
New contributor
Joe Johnes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 10 at 16:04
Joe Johnes
33
New contributor
Joe Johnes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 10 at 17:31
asked Nov 10 at 16:04
Joe Johnes
33
New contributor
Joe Johnes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 10 at 16:04
Joe Johnes
33
asked Nov 10 at 16:04
Joe Johnes
33
33
New contributor
Joe Johnes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Joe Johnes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
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3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
You could do it in 2 for comprehensions:
for (n <- (1 to 4).toList) yield (for (m <- ('a' to 'c').toList) yield (n, m))
answered Nov 10 at 16:25
Harald
3,71221727
1
While this is the correct answer, I would like to point out that since in scalafor/yieldis not really "iterating", as the name may suggest to newcomers (specially for people coming from an OOP/Imperative background), and is just syntactic sugar for calls tomap,flatMap&filter, @Tom's answer is also correct.
– Luis Miguel Mejía Suárez
Nov 10 at 17:48
add a comment |
up vote
1
down vote
You could also use map
(1 to 4).toList.map(n => ('a' to 'c').toList.map(m => (n, m)))
answered Nov 10 at 17:31
Tom
1,6042924
add a comment |
up vote
0
down vote
{for (i <- 1 to 2; j <- 1 to 3) yield (i,j)}.grouped(3).toList
You get a List of Vector, but that's fine, Vector is usually preferred in most circumstances. Fast random access, append, prepend, and updates. You can forgo converting toList until the end. If you are OK with Vector Tom's answer is really good and you can just refactor down to:
(1 to 4).map(n => ('a' to 'c').map(m => (n, m)))
answered Nov 10 at 18:07
Daniel Hinojosa
78237
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You could do it in 2 for comprehensions:
for (n <- (1 to 4).toList) yield (for (m <- ('a' to 'c').toList) yield (n, m))
answered Nov 10 at 16:25
Harald
3,71221727
1
While this is the correct answer, I would like to point out that since in scalafor/yieldis not really "iterating", as the name may suggest to newcomers (specially for people coming from an OOP/Imperative background), and is just syntactic sugar for calls tomap,flatMap&filter, @Tom's answer is also correct.
– Luis Miguel Mejía Suárez
Nov 10 at 17:48
add a comment |
up vote
1
down vote
accepted
You could do it in 2 for comprehensions:
for (n <- (1 to 4).toList) yield (for (m <- ('a' to 'c').toList) yield (n, m))
answered Nov 10 at 16:25
Harald
3,71221727
1
While this is the correct answer, I would like to point out that since in scalafor/yieldis not really "iterating", as the name may suggest to newcomers (specially for people coming from an OOP/Imperative background), and is just syntactic sugar for calls tomap,flatMap&filter, @Tom's answer is also correct.
– Luis Miguel Mejía Suárez
Nov 10 at 17:48
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You could do it in 2 for comprehensions:
for (n <- (1 to 4).toList) yield (for (m <- ('a' to 'c').toList) yield (n, m))
answered Nov 10 at 16:25
Harald
3,71221727
You could do it in 2 for comprehensions:
for (n <- (1 to 4).toList) yield (for (m <- ('a' to 'c').toList) yield (n, m))
answered Nov 10 at 16:25
Harald
3,71221727
answered Nov 10 at 16:25
Harald
3,71221727
answered Nov 10 at 16:25
Harald
3,71221727
answered Nov 10 at 16:25
Harald
3,71221727
3,71221727
1
While this is the correct answer, I would like to point out that since in scalafor/yieldis not really "iterating", as the name may suggest to newcomers (specially for people coming from an OOP/Imperative background), and is just syntactic sugar for calls tomap,flatMap&filter, @Tom's answer is also correct.
– Luis Miguel Mejía Suárez
Nov 10 at 17:48
add a comment |
1
While this is the correct answer, I would like to point out that since in scalafor/yieldis not really "iterating", as the name may suggest to newcomers (specially for people coming from an OOP/Imperative background), and is just syntactic sugar for calls tomap,flatMap&filter, @Tom's answer is also correct.
– Luis Miguel Mejía Suárez
Nov 10 at 17:48
1
1
While this is the correct answer, I would like to point out that since in scala
for/yield is not really "iterating", as the name may suggest to newcomers (specially for people coming from an OOP/Imperative background), and is just syntactic sugar for calls to map, flatMap & filter, @Tom's answer is also correct.– Luis Miguel Mejía Suárez
Nov 10 at 17:48
While this is the correct answer, I would like to point out that since in scala
for/yield is not really "iterating", as the name may suggest to newcomers (specially for people coming from an OOP/Imperative background), and is just syntactic sugar for calls to map, flatMap & filter, @Tom's answer is also correct.– Luis Miguel Mejía Suárez
Nov 10 at 17:48
add a comment |
up vote
1
down vote
You could also use map
(1 to 4).toList.map(n => ('a' to 'c').toList.map(m => (n, m)))
answered Nov 10 at 17:31
Tom
1,6042924
add a comment |
up vote
1
down vote
You could also use map
(1 to 4).toList.map(n => ('a' to 'c').toList.map(m => (n, m)))
answered Nov 10 at 17:31
Tom
1,6042924
add a comment |
up vote
1
down vote
up vote
1
down vote
You could also use map
(1 to 4).toList.map(n => ('a' to 'c').toList.map(m => (n, m)))
answered Nov 10 at 17:31
Tom
1,6042924
You could also use map
(1 to 4).toList.map(n => ('a' to 'c').toList.map(m => (n, m)))
answered Nov 10 at 17:31
Tom
1,6042924
answered Nov 10 at 17:31
Tom
1,6042924
answered Nov 10 at 17:31
Tom
1,6042924
answered Nov 10 at 17:31
Tom
1,6042924
1,6042924
add a comment |
add a comment |
up vote
0
down vote
{for (i <- 1 to 2; j <- 1 to 3) yield (i,j)}.grouped(3).toList
You get a List of Vector, but that's fine, Vector is usually preferred in most circumstances. Fast random access, append, prepend, and updates. You can forgo converting toList until the end. If you are OK with Vector Tom's answer is really good and you can just refactor down to:
(1 to 4).map(n => ('a' to 'c').map(m => (n, m)))
answered Nov 10 at 18:07
Daniel Hinojosa
78237
add a comment |
up vote
0
down vote
{for (i <- 1 to 2; j <- 1 to 3) yield (i,j)}.grouped(3).toList
You get a List of Vector, but that's fine, Vector is usually preferred in most circumstances. Fast random access, append, prepend, and updates. You can forgo converting toList until the end. If you are OK with Vector Tom's answer is really good and you can just refactor down to:
(1 to 4).map(n => ('a' to 'c').map(m => (n, m)))
answered Nov 10 at 18:07
Daniel Hinojosa
78237
add a comment |
up vote
0
down vote
up vote
0
down vote
{for (i <- 1 to 2; j <- 1 to 3) yield (i,j)}.grouped(3).toList
You get a List of Vector, but that's fine, Vector is usually preferred in most circumstances. Fast random access, append, prepend, and updates. You can forgo converting toList until the end. If you are OK with Vector Tom's answer is really good and you can just refactor down to:
(1 to 4).map(n => ('a' to 'c').map(m => (n, m)))
answered Nov 10 at 18:07
Daniel Hinojosa
78237
{for (i <- 1 to 2; j <- 1 to 3) yield (i,j)}.grouped(3).toList
You get a List of Vector, but that's fine, Vector is usually preferred in most circumstances. Fast random access, append, prepend, and updates. You can forgo converting toList until the end. If you are OK with Vector Tom's answer is really good and you can just refactor down to:
(1 to 4).map(n => ('a' to 'c').map(m => (n, m)))
answered Nov 10 at 18:07
Daniel Hinojosa
78237
answered Nov 10 at 18:07
Daniel Hinojosa
78237
answered Nov 10 at 18:07
Daniel Hinojosa
78237
answered Nov 10 at 18:07
Daniel Hinojosa
78237
78237
add a comment |
add a comment |
Joe Johnes is a new contributor. Be nice, and check out our Code of Conduct.
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