Index of word in string 'covering' certain position
Not sure if this is the right place to ask but I couldn't find any related or similar questions.
Anyway: imagine you have a certain string like
val exampleString = "Hello StackOverflow this is my question, cool right?"
If given a position in this string, for example 23, return the word that 'occupies' this position in the string. If we look at the example string, we can see that the 23rd character is the letter 's' (the last character of 'this'), so we should return index = 5 (because 'this' is the 5th word). In my question spaces are counted as words. If, for example, we were given position 5, we land on the first space and thus we should return index = 1.
I'm implementing this in Scala (but this should be quite language-agnostic and I would love to see implementations in other languages).
Currently I have the following approach (assume exampleString is the given string and charPosition the given position):
exampleString.split("((?<= )|(?= ))").scanLeft(0)((a, b) => a + b.length()).drop(1).zipWithIndex.takeWhile(_._1 <= charPosition).last._2 + 1
This works, but it is way too complex to be honest. Is there a better (more efficient?) way to achieve this. I'm fairly new to functions like fold, scan, map, filter ... but I would love to learn more.
Thanks in advance.
scala
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
add a comment |
Not sure if this is the right place to ask but I couldn't find any related or similar questions.
Anyway: imagine you have a certain string like
val exampleString = "Hello StackOverflow this is my question, cool right?"
If given a position in this string, for example 23, return the word that 'occupies' this position in the string. If we look at the example string, we can see that the 23rd character is the letter 's' (the last character of 'this'), so we should return index = 5 (because 'this' is the 5th word). In my question spaces are counted as words. If, for example, we were given position 5, we land on the first space and thus we should return index = 1.
I'm implementing this in Scala (but this should be quite language-agnostic and I would love to see implementations in other languages).
Currently I have the following approach (assume exampleString is the given string and charPosition the given position):
exampleString.split("((?<= )|(?= ))").scanLeft(0)((a, b) => a + b.length()).drop(1).zipWithIndex.takeWhile(_._1 <= charPosition).last._2 + 1
This works, but it is way too complex to be honest. Is there a better (more efficient?) way to achieve this. I'm fairly new to functions like fold, scan, map, filter ... but I would love to learn more.
Thanks in advance.
scala
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
add a comment |
Not sure if this is the right place to ask but I couldn't find any related or similar questions.
Anyway: imagine you have a certain string like
val exampleString = "Hello StackOverflow this is my question, cool right?"
If given a position in this string, for example 23, return the word that 'occupies' this position in the string. If we look at the example string, we can see that the 23rd character is the letter 's' (the last character of 'this'), so we should return index = 5 (because 'this' is the 5th word). In my question spaces are counted as words. If, for example, we were given position 5, we land on the first space and thus we should return index = 1.
I'm implementing this in Scala (but this should be quite language-agnostic and I would love to see implementations in other languages).
Currently I have the following approach (assume exampleString is the given string and charPosition the given position):
exampleString.split("((?<= )|(?= ))").scanLeft(0)((a, b) => a + b.length()).drop(1).zipWithIndex.takeWhile(_._1 <= charPosition).last._2 + 1
This works, but it is way too complex to be honest. Is there a better (more efficient?) way to achieve this. I'm fairly new to functions like fold, scan, map, filter ... but I would love to learn more.
Thanks in advance.
scala
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
Not sure if this is the right place to ask but I couldn't find any related or similar questions.
Anyway: imagine you have a certain string like
val exampleString = "Hello StackOverflow this is my question, cool right?"
If given a position in this string, for example 23, return the word that 'occupies' this position in the string. If we look at the example string, we can see that the 23rd character is the letter 's' (the last character of 'this'), so we should return index = 5 (because 'this' is the 5th word). In my question spaces are counted as words. If, for example, we were given position 5, we land on the first space and thus we should return index = 1.
I'm implementing this in Scala (but this should be quite language-agnostic and I would love to see implementations in other languages).
Currently I have the following approach (assume exampleString is the given string and charPosition the given position):
exampleString.split("((?<= )|(?= ))").scanLeft(0)((a, b) => a + b.length()).drop(1).zipWithIndex.takeWhile(_._1 <= charPosition).last._2 + 1
This works, but it is way too complex to be honest. Is there a better (more efficient?) way to achieve this. I'm fairly new to functions like fold, scan, map, filter ... but I would love to learn more.
Thanks in advance.
scala
scala
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
asked Nov 27 '18 at 14:31
Robin HaveneersRobin Haveneers
426
426
add a comment |
add a comment |
1 Answer
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oldest
votes
def wordIndex(exampleString: String, index: Int): Int = {
exampleString.take(index + 1).foldLeft((0, exampleString.head.isWhitespace)) {
case ((n, isWhitespace), c) =>
if (isWhitespace == c.isWhitespace) (n, isWhitespace)
else (n + 1, !isWhitespace)
}._1
}
This will fold over the string, keeping track of whether the previous character was a whitespace or not, and if it detects a change, it will flip the boolean and add 1 to the count (n).
This will be able to handle groups of spaces (e.g. in hello world, world would be at position 2), and also spaces at the start of the string would count as index 0 and the first word would be index 1.
Note that this can't handle when the input is an empty string, I'll let you decide what you want to do in that case.
answered Nov 27 '18 at 15:03
TomTom
1,73921024
add a comment |
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1 Answer
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1 Answer
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oldest
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def wordIndex(exampleString: String, index: Int): Int = {
exampleString.take(index + 1).foldLeft((0, exampleString.head.isWhitespace)) {
case ((n, isWhitespace), c) =>
if (isWhitespace == c.isWhitespace) (n, isWhitespace)
else (n + 1, !isWhitespace)
}._1
}
This will fold over the string, keeping track of whether the previous character was a whitespace or not, and if it detects a change, it will flip the boolean and add 1 to the count (n).
This will be able to handle groups of spaces (e.g. in hello world, world would be at position 2), and also spaces at the start of the string would count as index 0 and the first word would be index 1.
Note that this can't handle when the input is an empty string, I'll let you decide what you want to do in that case.
answered Nov 27 '18 at 15:03
TomTom
1,73921024
add a comment |
def wordIndex(exampleString: String, index: Int): Int = {
exampleString.take(index + 1).foldLeft((0, exampleString.head.isWhitespace)) {
case ((n, isWhitespace), c) =>
if (isWhitespace == c.isWhitespace) (n, isWhitespace)
else (n + 1, !isWhitespace)
}._1
}
This will fold over the string, keeping track of whether the previous character was a whitespace or not, and if it detects a change, it will flip the boolean and add 1 to the count (n).
This will be able to handle groups of spaces (e.g. in hello world, world would be at position 2), and also spaces at the start of the string would count as index 0 and the first word would be index 1.
Note that this can't handle when the input is an empty string, I'll let you decide what you want to do in that case.
answered Nov 27 '18 at 15:03
TomTom
1,73921024
add a comment |
def wordIndex(exampleString: String, index: Int): Int = {
exampleString.take(index + 1).foldLeft((0, exampleString.head.isWhitespace)) {
case ((n, isWhitespace), c) =>
if (isWhitespace == c.isWhitespace) (n, isWhitespace)
else (n + 1, !isWhitespace)
}._1
}
This will fold over the string, keeping track of whether the previous character was a whitespace or not, and if it detects a change, it will flip the boolean and add 1 to the count (n).
This will be able to handle groups of spaces (e.g. in hello world, world would be at position 2), and also spaces at the start of the string would count as index 0 and the first word would be index 1.
Note that this can't handle when the input is an empty string, I'll let you decide what you want to do in that case.
answered Nov 27 '18 at 15:03
TomTom
1,73921024
def wordIndex(exampleString: String, index: Int): Int = {
exampleString.take(index + 1).foldLeft((0, exampleString.head.isWhitespace)) {
case ((n, isWhitespace), c) =>
if (isWhitespace == c.isWhitespace) (n, isWhitespace)
else (n + 1, !isWhitespace)
}._1
}
This will fold over the string, keeping track of whether the previous character was a whitespace or not, and if it detects a change, it will flip the boolean and add 1 to the count (n).
This will be able to handle groups of spaces (e.g. in hello world, world would be at position 2), and also spaces at the start of the string would count as index 0 and the first word would be index 1.
Note that this can't handle when the input is an empty string, I'll let you decide what you want to do in that case.
answered Nov 27 '18 at 15:03
TomTom
1,73921024
answered Nov 27 '18 at 15:03
TomTom
1,73921024
answered Nov 27 '18 at 15:03
TomTom
1,73921024
answered Nov 27 '18 at 15:03
TomTom
1,73921024
1,73921024
add a comment |
add a comment |
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