I have the following code:
object testLines extends App {
val items = Array("""a-b-c d-e-f""","""a-b-c th-i-t""")
val lines = items.map(_.replaceAll("-", "")split("\t"))
print(lines.map(_.mkString(",")).mkString("\n"))
}
By mistake i did not put a dot between replaceAll and split but it worked.
By contrary when putting a dot between replaceAll and split i got an error
identifier expected but ';' found.
Implicit conversions found: items =>
What is going on?
Why does it work without a dot but is not working with a dot.
Update:
It works also with dot. The error message is a bug in the scala ide. The first part of the question is still valid
Thanks,
David
You have just discovered that Operators are methods. x.split(y) can also be written x split y in cases where the method is operator-like and it looks nicer. However there is nothing stopping you putting either side in parentheses like x split (y), (x) split y, or even (x) split (y) which may be necessary (and is a good idea for readability even if not strictly necessary) if you are passing in a more complex expression than a simple variable or constant and need parentheses to override the precedence.
With the example code you've written, it's not a bad idea to do the whole thing in operator style for clarity, using parentheses only where the syntax requires and/or they make groupings more obvious. I'd probably have written it more like this:
object testLines extends App {
val items = Array("a-b-c d-e-f", "a-b-c th-i-t")
val lines = items map (_ replaceAll ("-", "") split "\t")
print(lines map (_ mkString ",") mkString "\n")
}
Related
Sorry, I ran into another question about using PetitParser. I've figured out my recursive issues, but now I have a problem with parentheses. If I need to be able to parse the following two expressions:
'(use = "official").empty()'
'(( 5 + 5 ) * 5) + 5'
I've tried doing something like the following:
final expression = (char('(') & any().starGreedy(char(')')).flatten() & char(')')).map((value) => ParenthesesParser(value));
But that doesnt' work on the first expression.
If I try this:
final expression = (char('(') & any().starLazy(char(')')).flatten() & char(')')).map((value) => ParenthesesParser(value));
It doesn't work on the second expression. Any suggestions on how to parse both?
I think neither of the parsers does what you want: The first parser, the greedy one with starGreedy, will consume up to the last closing parenthesis. The second parser, the lazy one with starLazy, will consume up to the first closing parenthesis.
To parse a balanced parenthesis you need recursion, so that each opening parenthesis is followed by a matching closing one:
final inner = undefined();
final parser = char('(') & inner.star().flatten() & char(')');
inner.set(parser | pattern('^)'));
In the snippet above, the inner parser is recursively trying to either parse another parenthesis pair, or otherwise it simply consumes any character that is not a closing parenthesis.
In sales column i have values with pound sign £1200. It is not readable by Data frame in scala, please help me for the same. i want column value in double, 1200. I am using below method but its not working.
def getRemovedDollarValue = udf(
(actualSales: String) => {
val actualSalesDouble = actualSales
.replace(",", "")
.replace("$", "")
.replace("\\u00A3","")
.replace("\\U00A3","")
.replaceAll("\\s", "_").trim().toDouble
java.lang.Double.parseDouble(actualSalesDouble.toString)
}
)
You need write: .replace("\u00A3","") instead of escaping .replace("\\u00A3","").
But I prefer just: .replace("£", "") - it is more readable.
I think the proposed solutions and comments all work but don't address the confusion behind why your code isn't working.
From the Pattern docs:
Thus the strings "\u2014" and "\\u2014", while not equal, compile into the same pattern, which matches the character with hexadecimal value 0x2014.
replace and replaceAll are both replacing all occurrences in a String, but only replaceAll is taking in a regular expression. You're passing in "\\u00A3" which will work as a pattern, but not a unicode literal due to the added backslash. As already suggested, either use replace with a unicode literal or the actual symbol, or change to replaceAll.
I have a List[String], for example:
val test=List("this is, an extremely long sentence. Check; But. I want this sentence.",
"Another. extremely. long. (for eg. description). But I want this sentence.",
..)
I want the result to be like:
List("I want this sentence", "But I want this sentence"..)
I tried few approaches but didn't work
test.map(x=>x.split(".").reverse.head)
test.map(x=>x.split(".").last)
Try using this
test.reverse.head.split("\\.").last
To handle any Exception
Try(List[String]().reverse.head.split("\\.").last).getOrElse("YOUR_DEFAULT_STRING")
You can map over you List, split each String and then take the last element. Try the below code.
val list = List("this is, an extremely long sentence. Check; But. I want this sentence.",
"Another. extremely. long. (for eg. description). But I want this sentence.")
list.map(_.split("\\.").last.trim)
It will give you
List(I want this sentence, But I want this sentence)
test.map (_.split("\\.").last)
Split takes a regular expression, and in such, the dot stands for every character, so you have to mask it.
Maybe you want to include question marks and bangs:
test.map (_.split("[!?.]").last)
and trim surrounding whitespace:
test.map (_.split("[!?.]").last.trim).
The reverse.head would have been a good idea, if there wasn't the last:
scala> test.map (_.split("[!?.]").reverse.head.trim)
res138: List[String] = List(I want this sentence, But I want this sentence)
You can do this a number of ways:
For each string in your original list: split by ., reverse the list, take the first value
test.map(_.split('.').reverse.headOption)
// List(Some( I want this sentence), Some( But I want this sentence))
.headOption results in Some("string") or None, and you can do something like a .getOrElse("no valid string found") on it. You can trim the unwanted whitespace if you want.
Regex match
test.map { sentence =>
val regex = ".*\\.\\s*([^.]*)\\.$".r
val regex(value) = sentence
value
}
This will fetch any string at the end of a longer string which is preceded by a full stop and a space and followed by a full stop. You can modify the regex to change the exact rules of the regex, and I recommend playing around with regex101.com if you fancy learning more regex. It's very good.
This solution is better for more complicated examples and requirements, but it's worth keeping in mind. If you are worried that the regex might not match, you can do something like checking if the regex matches before extracting it:
test.map { sentence =>
val regexString = ".*\\.\\s*([^.]*)\\.$"
val regex = regexString.r
if(sentence.matches(regexString)) {
val regex(value) = sentence
value
} else ""
}
Take the last after splitting the string by .
test.map(_.split('.').map(_.trim).lastOption)
This is driving me nuts... there must be a way to strip out all non-digit characters (or perform other simple filtering) in a String.
Example: I want to turn a phone number ("+72 (93) 2342-7772" or "+1 310-777-2341") into a simple numeric String (not an Int), such as "729323427772" or "13107772341".
I tried "[\\d]+".r.findAllIn(phoneNumber) which returns an Iteratee and then I would have to recombine them into a String somehow... seems horribly wasteful.
I also came up with: phoneNumber.filter("0123456789".contains(_)) but that becomes tedious for other situations. For instance, removing all punctuation... I'm really after something that works with a regular expression so it has wider application than just filtering out digits.
Anyone have a fancy Scala one-liner for this that is more direct?
You can use filter, treating the string as a character sequence and testing the character with isDigit:
"+72 (93) 2342-7772".filter(_.isDigit) // res0: String = 729323427772
You can use replaceAll and Regex.
"+72 (93) 2342-7772".replaceAll("[^0-9]", "") // res1: String = 729323427772
Another approach, define the collection of valid characters, in this case
val d = '0' to '9'
and so for val a = "+72 (93) 2342-7772", filter on collection inclusion for instance with either of these,
for (c <- a if d.contains(c)) yield c
a.filter(d.contains)
a.collect{ case c if d.contains(c) => c }
I want to implement a Scala-style string interpolation in Scala. Here is an example,
val str = "hello ${var1} world ${var2}"
At runtime I want to replace "${var1}" and "${var2}" with some runtime strings. However, when trying to use Regex.replaceAllIn(target: CharSequence, replacer: (Match) ⇒ String), I ran into the following problem:
import scala.util.matching.Regex
val placeholder = new Regex("""(\$\{\w+\})""")
placeholder.replaceAllIn(str, m => s"A${m.matched}B")
java.lang.IllegalArgumentException: No group with name {var1}
at java.util.regex.Matcher.appendReplacement(Matcher.java:800)
at scala.util.matching.Regex$Replacement$class.replace(Regex.scala:722)
at scala.util.matching.Regex$MatchIterator$$anon$1.replace(Regex.scala:700)
at scala.util.matching.Regex$$anonfun$replaceAllIn$1.apply(Regex.scala:410)
at scala.util.matching.Regex$$anonfun$replaceAllIn$1.apply(Regex.scala:410)
at scala.collection.Iterator$class.foreach(Iterator.scala:743)
at scala.collection.AbstractIterator.foreach(Iterator.scala:1174)
at scala.util.matching.Regex.replaceAllIn(Regex.scala:410)
... 32 elided
However, when I removed '$' from the regular expression, it worked:
val placeholder = new Regex("""(\{\w+\})""")
placeholder.replaceAllIn(str, m => s"A${m.matched}B")
res2: String = hello $A{var1}B world $A{var2}B
So my question is that whether this is a bug in Scala Regex. And if so, are there other elegant ways to achieve the same goal (other than brutal force replaceAllLiterally on all placeholders)?
$ is a treated specially in the replacement string. This is described in the documentation of replaceAllIn:
In the replacement String, a dollar sign ($) followed by a number will be interpreted as a reference to a group in the matched pattern, with numbers 1 through 9 corresponding to the first nine groups, and 0 standing for the whole match. Any other character is an error. The backslash (\) character will be interpreted as an escape character and can be used to escape the dollar sign. Use Regex.quoteReplacement to escape these characters.
(Actually, that doesn't mention named group references, so I guess it's only sort of documented.)
Anyway, the takeaway here is that you need to escape the $ characters in the replacement string if you don't want them to be treated as references.
new scala.util.matching.Regex("""(\$\{\w+\})""")
.replaceAllIn("hello ${var1} world ${var2}", m => s"A\\${m.matched}B")
// "hello A${var1}B world A${var2}B"
It's hard to tell what you're expecting the behavior to do. The issue is that s"${m.matched}" is turning into "${var1}" (and "${var2}"). The '$' is special character to say "place the group with name {var1} here instead".
For example:
scala> placeholder.replaceAllIn(str, m => "$1")
res0: String = hello ${var1} world ${var2}
It replaces the match with the first capturing group (which is m itself).
It's hard to tell exactly what you're doing, but you could escape any $ like so:
scala> placeholder.replaceAllIn(str, m => s"${m.matched.replace("$","\\$")}")
res1: String = hello ${var1} world ${var2}
If what you really want to do is evaluate var1/var2 for some variables in the local scope of the method; that's not possible. In fact, the s"Hello, $name" pattern is actually converted into new StringContext("Hello, ", "").s(name) at compile time.