如何在 spark scala 中迭代 array[string]？

Question

enter image description here here is my sample input:

val list=List("a;bc:de;f","uvw:xy;z","123:456")

I am applying following operation

val upper=list.map(x=>x.split(":")).map(x=>x.split(";"))

but it is throwing error- error: value split is not a member of Array[String]

can anyone help how to use both split so that i can get answer!

Thank you in advance.

Answer 1

Using list.map(x=>x.split(":")) will give you a list of Array.

upper: List[Array[String]] = List(Array(a;bc, de;f), Array(uvw, xy;z), Array(123, 456))

Mapping afterwards, you can see that the item will be an array where you are trying to run split on.

You might use flatMap instead which will first give you List(a;bc, de;f, uvw, xy;z, 123, 456) and then you can use map on those items splitting on ;

val upper = list.flatMap(_.split(":")).map(_.split(";"))

Output

upper: List[Array[String]] = List(Array(a, bc), Array(de, f), Array(uvw), Array(xy, z), Array(123), Array(456))

Answer 2

You can use split with multiple delimiters in one map iteration :

val upper = list.map(x => x.split("[:;]"))

//upper: List[Array[String]] = List(Array(a, bc, de, f), Array(uvw, xy, z), Array(123, 456))

Answer 3

Here is the code i have tried and it worked:

val upper=list.map(x=>x.split(":")).map(x=>x.map(x=>x.split(";")))

which gives the output:

upper: List[Array[Array[String]]] = List(Array(Array(a, bc), Array(de, f)), Array(Array(uvw), Array(xy, z)), Array(Array(123), Array(456)))