String Hive function for split key-value pair into two columns - scala

Question

How to split this data T_32_P_1_A_420_H_60_R_0.30841494477846165_S_0 into two columns using hive function
For example
T 32
P 1
A 420
H 60
R 0.30841494477846165
S 0

Answer 1

You can do this with a regex implementation:
def main(args: Array[String]) {
val s = "T_32_P_1_A_420_H_60_R_0.30841494477846165_S_0"
val pattern = "[A-Z]\\_\\d+\\.?\\d*"
var buff = new String()
val r = Pattern.compile(pattern)
val m = r.matcher(s)
while (m.find()) {
buff = buff + (m.group(0))
buff = buff + "\n"
}
buff = buff.toString.replaceAll("\\_", " ")
println("output:\n" + buff)
}
Output:
output:
T 32
P 1
A 420
H 60
R 0.30841494477846165
S 0

Answer 2

If you need to collect the data for further processing, and you're guaranteed it's always paired correctly, you could do something like this.
scala> val str = "T_32_P_1_A_420_H_60_R_0.30841494477846165_S_0"
str: String = T_32_P_1_A_420_H_60_R_0.30841494477846165_S_0
scala> val data = str.split("_").sliding(2,2)
data: Iterator[Array[String]] = non-empty iterator
scala> data.toList // just to see it
res29: List[Array[String]] = List(Array(T, 32), Array(P, 1), Array(A, 420), Array(H, 60), Array(R, 0.30841494477846165), Array(S, 0))

Answer 3

You can split your string, get an array, zipWithIndex and filter based on index to get two arrays col1 and col2 and then use it for printing:
val str = "T_32_P_1_A_420_H_60_R_0.30841494477846165_S_0"
val tmp = str.split('_').zipWithIndex
val col1 = tmp.filter( p => p._2 % 2 == 0 ).map( p => p._1)
val col2 = tmp.filter( p => p._2 % 2 != 0 ).map( p => p._1)
//col1: Array[String] = Array(T, P, A, H, R, S)
//col2: Array[String] = Array(32, 1, 420, 60, ...