Split a string if a substring is present - regex

Question

I have two strings like:
"Nikon Coolpix AW130 16MP Point and Shoot Digital Camera Black with 5x Optical Zoom"
"Nikon Coolpix AW130 16 MP Point & Shoot Camera Black"
I am trying to compare strings like these, as you can see both of them are same, when I tokenize based on space and compare each word the space between 16 and MP in 2nd string will cause a difference which is not actually there.
Is there anyway I can add a space in the 1st string where the 16MP is together so that I can tokenize based on space properly.
val productList=List("Nikon Coolpix AW130 16MP Point and Shoot Digital Camera Black with 5x Optical Zoom","Nikon Coolpix AW130 16 MP Point & Shoot Camera Black")
val tokens = ListBuffer[String]()
productList.split(" ").foreach(x => {
tokens += x
})
val res = tokens.toList

Answer 1

If you just want to remove the space between a number and a fixed MP string, you can use the following regex:
scala> "Nikon Coolpix AW130 16 MP Point & Shoot Camera Black".replaceAll("""(\d+) ?(MP)""", "$1$2")
res13: String = Nikon Coolpix AW130 16MP Point & Shoot Camera Black
The (\d+) part matches any number with at least 1 digit
The ? (note the space) matches 0 or one spaces
The (MP) part matches the string MP literally.
$1$2 prints the contents of the match of the first parentheses (\d+) appended to the match of the second parantheses (MP) - omitting the space, if there is one.
After that, the 16MP tokenS should be equal. You will still have the problem of and vs. &, though.

Answer 2

You can do it with RegEx.
Search both format and replace it to one specific.

Answer 3

Instead of split it is easier to do regex replaces; consecutively.
public static boolean equivalent(Sting a, String b) {
normalize(a).equalsIgnoreCase(normalize(b));
}
private static String normalize(String s) {
return s.replaceAll("(\\d+)", "$0 ") // At least one space after digits.
.replaceAll("\\bLimited\\b", "Ltd") // Example.
.replace("'", "") // Example.
.replace("&", " and ")
.replaceAll("\\s+", " ") // Multiple spaces to one.
.trim();
}
Or do a split on the normalized string (to get keywords).

Answer 4

You don't give enough details about the format of these strings but from this example I can infer something like that : (\w+) (\d+)\s*MP Point.*
You can then parse the strings and read the groups of the regex to compare the product.
Here is an example :
def main(args: Array[String]): Unit = {
val s0 = "Nikon Coolpix AW130 16MP Point and Shoot Digital Camera Black with 5x Optical Zoom"
val s1 = "Nikon Coolpix AW130 16 MP Point & Shoot Camera Black"
println(Product.parse(s0) == Product.parse(s1)) // prints true
}
case class Product(name: String, resolution: Int)
object Product {
private val regex = new Regex("(\\w+) (\\d+)\\s*MP Point.*", "productName", "resolution")
def parse(s: String) = regex.findFirstMatchIn(s) match {
case Some(m) => Product(m.group("productName"), m.group("resolution").toInt)
case None => throw new RuntimeException("Invalid format")
}
}