that's good. thanks
On 2022/2/12 12:11, Raghavendra Ganesh wrote:
.withColumn("newColumn",expr(s"case when score>3 then 'good' else 'bad'
end"))
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You could use expr() function to achieve the same.
.withColumn("newColumn",expr(s"case when score>3 then 'good' else 'bad'
end"))
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Raghavendra
On Fri, Feb 11, 2022 at 5:59 PM frakass wrote:
> Hello
>
> I have a column whose value (Int type as score) is from 0 to 5.
> I want to query that,
Hello
I have a column whose value (Int type as score) is from 0 to 5.
I want to query that, when the score > 3, classified as "good". else
classified as "bad".
How do I implement that? A UDF like something as this?
scala> implicit class Foo(i:Int) {
| def classAs(f:Int=>String) = f(i)
