Iterating through array with condition to create new array

Question

How can I iterate through an array and assign its val? I tried the following two approaches but they did not work:

var a : Array[String] = Array("foo","bar")
var b : Array[String] = Array()
for (i <- 0 to (a.length-1)) {
    println(a(i))
    b :+ a(i)
}

and

var a : Array[String] = Array("foo","bar")
var b : Array[String] = Array()
for (element <- a){
    println(element)
    b :+ element
}

So the println works but they assignment doesn't.

I am really frustated since it seems so easy and I do note get it. :(

Try b = b ++ element

BioGenX
– BioGenX

2017-03-16 08:13:40 +00:00
Commented Mar 16, 2017 at 8:13 — BioGenX
– BioGenX, Commented Mar 16, 2017 at 8:13
That doesn't work unfortunately.

Ken Jiiii
– Ken Jiiii

2017-03-16 08:46:58 +00:00
Commented Mar 16, 2017 at 8:46 — Ken Jiiii
– Ken Jiiii, Commented Mar 16, 2017 at 8:46

jwvh · Accepted Answer · 2017-03-16 11:20:01Z

3

Let the Standard Library do the copying for you.

val a : Array[String] = Array("foo","bar")
val b : Array[String] = a.clone()

The reason your code doesn't work is because :+ isn't an assignment operator. This statement, for example, Array() :+ "str" doesn't modify the Array, it creates a new Array with the modified value. Unfortunately your code doesn't assign that new Array to a new or existing variable so the modification is lost.

You could do something like this ...

for (element <- a){
  b = b :+ element
}

... but there are better ways to get this done.

update

To filter string elements for multiple unrelated qualities you might try this.

val acceptableRE = "(f.*|ba.*)".r
val a: Array[String] = Array("foo", "bar", "buf", "bba")
val b: Array[String] = a.collect{case acceptableRE(s) => s}   // Array(foo, bar)

edited Mar 16, 2017 at 11:20

answered Mar 16, 2017 at 8:20

jwvh

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4 Comments

Ken Jiiii Over a year ago

Thanks for the answer. That makes sense. Actually I do not want to clone it but apply some cases afterwards to eliminate some values of the array. Thus I am using this for loop. Or what can you suggest to do that in a better way?

jwvh Over a year ago

So b is supposed to be a copy of a but with some values removed? If that's the case then go with Jyothi Babu Araja's suggestion and use filter (or filterNot, depending on the condition to be tested).

Ken Jiiii Over a year ago

This would mean that I have a seperate filter for each step? Like val b = a.filter(el => el.startsWith("f")).filter(el => el.startsWith("ba"))??

jwvh Over a year ago

Obviously not. No element will start with both "f" and "ba". See my update.

Jyothi Babu Araja · Accepted Answer · 2017-03-16 08:13:35Z

2

You can make use of .filter of Array to filter with condition.

Try this example with element length as condition

val a : Array[String] = Array("foo","bars")
val b = a.filter(el => el.length > 3)

Note: Better use val to avoid mutability.

answered Mar 16, 2017 at 8:13

Jyothi Babu Araja

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