How do I trim the starting and ending character of a string in Scala
For inputs such as ",hello" or "hello,", I need the output as "hello".
Is there is any built-in method to do this in Scala?
Try
val str = " foo "
str.trim
and have a look at the documentation. If you need to get rid of the , character, too, you could try something like:
str.stripPrefix(",").stripSuffix(",").trim
Another way to clean up the front-end of the string would be
val ignoreable = ", \t\r\n"
str.dropWhile(c => ignorable.indexOf(c) >= 0)
which would also take care of strings like ",,, ,,hello"
And for good measure, here's a tiny function, which does it all in one sweep from left to right through the string:
def stripAll(s: String, bad: String): String = {
#scala.annotation.tailrec def start(n: Int): String =
if (n == s.length) ""
else if (bad.indexOf(s.charAt(n)) < 0) end(n, s.length)
else start(1 + n)
#scala.annotation.tailrec def end(a: Int, n: Int): String =
if (n <= a) s.substring(a, n)
else if (bad.indexOf(s.charAt(n - 1)) < 0) s.substring(a, n)
else end(a, n - 1)
start(0)
}
Use like
stripAll(stringToCleanUp, charactersToRemove)
e.g.,
stripAll(" , , , hello , ,,,, ", " ,") => "hello"
To trim the start and ending character in a string, use a mix of drop and dropRight:
scala> " hello,".drop(1).dropRight(1)
res4: String = hello
The drop call removes the first character, dropRight removes the last. Note that this isn't "smart" like trim is. If you don't have any extra character at the start of "hello,", you will trim it to "ello". If you need something more complicated, regex replacement is probably the answer.
If you want to trim only commas and might have more than one on either end, you could do this:
str.dropWhile(_ == ',').reverse.dropWhile(_ == ',').reverse
The use of reverse here is because there is no dropRightWhile.
If you're looking at a single possible comma, stripPrefix and stripSuffix are the way to go, as indicated by Dirk.
Given you only want to trim off invalid characters from the prefix and the suffix of a given string (not scan through the entire string), here's a tiny trimPrefixSuffixChars function to quickly perform the desired effect:
def trimPrefixSuffixChars(
string: String
, invalidCharsFunction: (Char) => Boolean = (c) => c == ' '
): String =
if (string.nonEmpty)
string
.dropWhile(char => invalidCharsFunction(char)) //trim prefix
.reverse
.dropWhile(char => invalidCharsFunction(char)) //trim suffix
.reverse
else
string
This function provides a default for the invalidCharsFunction defining only the space (" ") character as invalid. Here's what the conversion would look like for the following input strings:
trimPrefixSuffixChars(" Tx ") //returns "Tx"
trimPrefixSuffixChars(" . Tx . ") //returns ". Tx ."
trimPrefixSuffixChars(" T x ") //returns "T x"
trimPrefixSuffixChars(" . T x . ") //returns ". T x ."
If you have you would prefer to specify your own invalidCharsFunction function, then pass it in the call like so:
trimPrefixSuffixChars(",Tx. ", (c) => !c.isLetterOrDigit) //returns "Tx"
trimPrefixSuffixChars(" ! Tx # ", (c) => !c.isLetterOrDigit) //returns "Tx"
trimPrefixSuffixChars(",T x. ", (c) => !c.isLetterOrDigit) //returns "T x"
trimPrefixSuffixChars(" ! T x # ", (c) => !c.isLetterOrDigit) //returns "T x"
This attempts to simplify a number of the example solutions provided in other answers.
Someone requested a regex-version, which would be something like this:
val result = " , ,, hello, ,,".replaceAll("""[,\s]+(|.*[^,\s])[,\s]+""", "'$1'")
Result is: result: String = hello
The drawback with regexes (not just in this case, but always), is that it is quite hard to read for someone who is not already intimately familiar with the syntax. The code is nice and concise, though.
Another tailrec function:
def trim(s: String, char: Char): String = {
if (s.stripSuffix(char.toString).stripPrefix(char.toString) == s)
{
s
} else
{
trim(s.stripSuffix(char.toString).stripPrefix(char.toString), char)
}
}
scala> trim(",hello",',')
res12: String = hello
scala> trim(",hello,,,,",',')
res13: String = hello
Related
I'm trying to iterate on a String value to change each occurence of it.
For example i want that "1" become "one", "2" become "two", etc.
I've done this :
override def toString = {
val mapXX = init.map(_.clone);
var returnVALUE = mapXX.map(_.mkString).mkString("\n")
for(c <- returnVALUE){
c match {
case 1 => "one";
case 2 => "two";
...
case _ => "";
}
}
returnVALUE
}
}
It didn't change anything of my list, i have the same display of my list. Nothing has changed.
Did someone knows how can we iterate on each character of a String value in order to replace each character by something else ?
Thanks
It's not completely clear what you're doing. Try
returnVALUE.map {
case '1' => "one"
case '2' => "two"
case '3' => "three"
// ...
case _ => " "
}.mkString
and this should be the last line of toString.
String#map accepts a function from Char to something (e.g. to String).
If returnVALUE is "1 2 3" then this produces "one two three".
When the last line is returnVALUE this means you return the original value of returnVALUE, not the modified value.
A for comprehension without the yield clause doesn't create any results. It can only be used for side effects, which good Scala programmers try to avoid.
Maybe something like this.
val numberNames = Map(0 -> "zero", 1 -> "one", 2 -> "two").withDefaultValue("too big")
val result = List(2,0,1,4).map(numberNames)
//result: List[String] = List(two, zero, one, too big)
I want to write an algorithm that removes every word started by an uppercase character in a string.
For example:
Original string: "Today is Friday the 29Th."
Desired result: "is the 29Th."
I wrote this algorithm, but it is not complete:
def removeUpperCaseChars(str: String) = {
for (i <- 0 to str.length - 1) {
if (str.charAt(i).isUpper) {
var j = i
var cont = i
while (str.charAt(j) != " ") {
cont += 1
}
val subStr = str.substring(0, i) + str.substring(cont, str.length - 1)
println(subStr)
}
}
}
It (supposedly) removes every word with uppercase characters instead of removing only the words that start with uppercase characters. And worse than that, Scala doesn't give any result.
Can anyone help me with this problem?
With some assumptions, like words are always split with a space you can implement it like this:
scala> "Today is Friday the 29Th.".split("\\s+").filterNot(_.head.isUpper).mkString(" ")
res2: String = is the 29Th.
We don't really want to write algorithms in the way you did in scala. This is reather a way you would do this in C.
How about string.replaceAll("""\b[A-Z]\w+""", "")?
I have a string x: x = "{abc}{def}{ghi}"
And I need to print the string between second { and second }, in this case def. How can I do this without knowing the length of the string? For example, the string x could also be {abcde}{fghij}{klmno}"
This is where pattern matching is useful:
local x = "{abc}{def}{ghi}"
local result = x:match(".-{.-}.-{(.-)}")
print(result)
.- matches zero or more characters, non-greedy. The whole pattern .-{.-}.-{(.-)} captures what's between the second { and the second }.
Try also x:match(".-}{(.-)}"), which is simpler.
I would go about it in a different manner:
local i, x, result = 1, "{abc}{def}{ghi}"
for w in x:gmatch '{(.-)}' do
if i == 2 then
result = w
break
else
i = i + 1
end
end
print( result )
I've spend whole day on the internet and I wasn't able to find any built-in predicate in Win-Prolog that can single out every word in a String.
Example:
| ?- read(X).
|: 'this is a string'.
X = 'this is a string'.
is there any predicate I can use that will help me, single out every word of in the string?
like
A = this
B = is
C = a
D = string
or a list
A = [This, is, a, string]
is it possible?
In Win-Prolog you might use:
split_string(String, List):-
string_chars(String, LChar),
split_string(List, LWord-LWord, LChar, []).
split_string([Word|TailWords], LWord-[])-->
" ",
{string_chars(Word, LWord)},
split_string(TailWords, NLWord-NLWord).
split_string(Words, Head-[Char|LWord]) -->
[Char],
{[Char] \= " "},
split_string(Words, Head-LWord).
split_string([Word], LWord-[])-->
[],
{ string_chars(Word, LWord)}.
In SWI-Prolog you can use atomic_list_concat(List, ' ', 'this is a string').
I have a string stored in sqlite database and I've assigned it to a var, e.g. string
string = "First line and string. This should be another string in a new line"
I want to split this string into two separated strings, the dot (.) must be replace with (\n) new line char
At the moment I'm stuck and any help would be great!!
for row in db:nrows("SELECT * FROM contents WHERE section='accounts'") do
tabledata[int] = string.gsub(row.contentName, "%.", "\n")
int = int+1
end
I tried the other questions posted here in stachoverflow but with zero luck
What about this solution:`
s = "First line and string. This should be another string in a new line"
a,b=s:match"([^.]*).(.*)"
print(a)
print(b)
Are you looking to actually split the string into two different string objects? If so maybe this can help. It's a function I wrote to add some additional functionality to the standard string library. You can use it as-is or rename it to what ever you like.
--[[
string.split (s, p)
====================================================================
Splits the string [s] into substrings wherever pattern [p] occurs.
Returns: a table of substrings or, if no match is made [nil].
--]]
string.split = function(s, p)
local temp = {}
local index = 0
local last_index = string.len(s)
while true do
local i, e = string.find(s, p, index)
if i and e then
local next_index = e + 1
local word_bound = i - 1
table.insert(temp, string.sub(s, index, word_bound))
index = next_index
else
if index > 0 and index <= last_index then
table.insert(temp, string.sub(s, index, last_index))
elseif index == 0 then
temp = nil
end
break
end
end
return temp
end
Using it is very simple, it returns a tables of strings.
Lua 5.1.4 Copyright (C) 1994-2008 Lua.org, PUC-Rio
> s = "First line and string. This should be another string in a new line"
> t = string.split(s, "%.")
> print(table.concat(t, "\n"))
First line and string
This should be another string in a new line
> print(table.maxn(t))
2