The CoffeeScript docs state that list comprehensions should be able to do the select/filter operations:
They should be able to handle most
places where you otherwise would use a
loop, each/forEach, map, or
select/filter.
You'd imagine you could do something in one line like result = item for item in list if item % 2 == 0
However the closest I can come is
list = [1,2,3,4]
result = []
for item in list
if item % 2 == 0 then result.push item
Whats the most concise way to filter a list in CoffeeScript?
result = (item for item in list when item % 2 == 0)
edit : added result =
Unless you are trying to squeeze every ounce of performance out, I would create a filter function like this:
filter = (list, func) -> x for x in list when func(x)
or if you want to add it to the prototype of every array:
Array::filter = (func) -> x for x in # when func(x)
and then write: (respectively)
result = filter mylist, (x) -> x % 2 == 0
or
result = mylist.filter (x) -> x % 2 == 0
For reference, this is the javascript that gets generated:
var filter, result;
filter = function(list, func) {
var x, _i, _len, _results;
_results = [];
for (_i = 0, _len = list.length; _i < _len; _i++) {
x = list[_i];
if (func(x)) {
_results.push(x);
}
}
return _results;
};
Array.prototype.filter = function(func) {
var x, _i, _len, _results;
_results = [];
for (_i = 0, _len = this.length; _i < _len; _i++) {
x = this[_i];
if (func(x)) {
_results.push(x);
}
}
return _results;
};
result = filter(mylist, function(x) {
return x % 2 === 0;
});
result = mylist.filter(function(x) {
return x % 2 === 0;
});
There is a similar question here
Related
I am new to groovy and trying to find the indexes of all sublists in a list.
I am trying to use something like Collections.indexOfSubList like in java but it gives exception saying it applies on Lists and not ArrayLists.
So I am trying to define my own function. I am finding all the indices of all the elements in the smaller list existing in the longer list and then subtracting the indices of the result array. If it comes to 1 then I am considering that index to a sublist.
I know that I have the logic a little twisted. Can somebody guide with a better and efficient way of doing this.
Below is my code:
List list1 = [1,2,3,4,5,6,1,2,3]
List list2 = [1,2]
index1 = list1.findIndexValues {
it == list2[0];
}
index2 = list1.findIndexValues {
it == list2[1];
}
println index1
println index2
result = []
for (int i = 0; i < index1.size(); i++) {
result.add(index2[i]-index1[i]);
}
println result
Edit: no longer uses Collections due to new issue re: Elastic Search.
The following code traverses along the source list, creating a sublist. It checks the sublist to see if it starts with the target list. See the asserts below (e.g. the indexes are 0-based):
def listStartsWithSubList = { source, target ->
def result = false
if (source.size() >= target.size()) {
result = true
target.eachWithIndex { item, index ->
result = result && (item == source[index])
}
}
result
}
def indexOfSubLists = { source, target ->
def results = []
source.eachWithIndex { item, index ->
def tmpList = source[index..source.size()-1]
if (listStartsWithSubList(tmpList, target)) {
results << index
}
}
results
}
assert [1] == indexOfSubLists([1,2,3], [2,3])
assert [2] == indexOfSubLists([1,2,3], [3])
assert [] == indexOfSubLists([1,2,3], [4])
assert [0,6] == indexOfSubLists([1,2,3,4,5,6,1,2,3], [1,2])
I am writing a Maximum Value Knapsack algorithm. It takes in a Knapsack object with Items that have a value and cost. I declare a 2D array for calculating the max value. For the base cases I have set the zeroth row values to 0 and zeroth column values to 0. I am running into trouble when I grab an item in the knapsack because when I want to grab the zeroth item, I am really grabbing the first item in the knapsack and am consequently getting the wrong values in the 2D array. Can someone check out my code and see what I am missing?
public static double MaximumKnapsack(Knapsack knapsack) {
int numItems = knapsack.getNumOfItems();
int budget = (int) knapsack.getBudget();
double[][] DP = new double[numItems+1][budget+1];
boolean taken = false;
for (int i = 0; i < numItems + 1; i++) {
for (int b = 0; b < budget + 1; b++) {
if (i == 0 || b == 0) {
DP[i][b] = 0;
}
else
{
Item item = knapsack.getItem(i);
if (item.getCost() > b) {
DP[i][b] = DP[i-1][b];
}
else
{
DP[i][b] = Math.max(DP[i-1][b-(int) item.getCost()] + item.getValue(),
DP[i-1][b]);
if (DP[i][b] == DP[i-1][b-(int) item.getCost()] + item.getValue() && item.getCost() != 0.0) {
taken = true;
}
}
}
}
taken = false;
}
return DP[numItems][budget];
}
I think the problem is in
Item item = knapsack.getItem(i);
beacuse your loop will start with i = 1. You should use:
Item item = knapsack.getItem(i-1);
I have been trying to implement parallel merge sort in Scala. But with 8 cores, using .sorted is still about twice as fast.
edit:
I rewrote most of the code to minimize object creation. Now it runs about as fast as the .sorted
Input file with 1.2M integers:
1.333580 seconds (my implementation)
1.439293 seconds (.sorted)
How should I parallelize this?
New implementation
object Mergesort extends App
{
//=====================================================================================================================
// UTILITY
implicit object comp extends Ordering[Any] {
def compare(a: Any, b: Any) = {
(a, b) match {
case (a: Int, b: Int) => a compare b
case (a: String, b: String) => a compare b
case _ => 0
}
}
}
//=====================================================================================================================
// MERGESORT
val THRESHOLD = 30
def inssort[A](a: Array[A], left: Int, right: Int): Array[A] = {
for (i <- (left+1) until right) {
var j = i
val item = a(j)
while (j > left && comp.lt(item,a(j-1))) {
a(j) = a(j-1)
j -= 1
}
a(j) = item
}
a
}
def mergesort_merge[A](a: Array[A], temp: Array[A], left: Int, right: Int, mid: Int) : Array[A] = {
var i = left
var j = right
while (i < mid) { temp(i) = a(i); i+=1; }
while (j > mid) { temp(i) = a(j-1); i+=1; j-=1; }
i = left
j = right-1
var k = left
while (k < right) {
if (comp.lt(temp(i), temp(j))) { a(k) = temp(i); i+=1; k+=1; }
else { a(k) = temp(j); j-=1; k+=1; }
}
a
}
def mergesort_split[A](a: Array[A], temp: Array[A], left: Int, right: Int): Array[A] = {
if (right-left == 1) a
if ((right-left) > THRESHOLD) {
val mid = (left+right)/2
mergesort_split(a, temp, left, mid)
mergesort_split(a, temp, mid, right)
mergesort_merge(a, temp, left, right, mid)
}
else
inssort(a, left, right)
}
def mergesort[A: ClassTag](a: Array[A]): Array[A] = {
val temp = new Array[A](a.size)
mergesort_split(a, temp, 0, a.size)
}
Previous implementation
Input file with 1.2M integers:
4.269937 seconds (my implementation)
1.831767 seconds (.sorted)
What sort of tricks there are to make it faster and cleaner?
object Mergesort extends App
{
//=====================================================================================================================
// UTILITY
val StartNano = System.nanoTime
def dbg(msg: String) = println("%05d DBG ".format(((System.nanoTime - StartNano)/1e6).toInt) + msg)
def time[T](work: =>T) = {
val start = System.nanoTime
val res = work
println("%f seconds".format((System.nanoTime - start)/1e9))
res
}
implicit object comp extends Ordering[Any] {
def compare(a: Any, b: Any) = {
(a, b) match {
case (a: Int, b: Int) => a compare b
case (a: String, b: String) => a compare b
case _ => 0
}
}
}
//=====================================================================================================================
// MERGESORT
def merge[A](left: List[A], right: List[A]): Stream[A] = (left, right) match {
case (x :: xs, y :: ys) if comp.lteq(x, y) => x #:: merge(xs, right)
case (x :: xs, y :: ys) => y #:: merge(left, ys)
case _ => if (left.isEmpty) right.toStream else left.toStream
}
def sort[A](input: List[A], length: Int): List[A] = {
if (length < 100) return input.sortWith(comp.lt)
input match {
case Nil | List(_) => input
case _ =>
val middle = length / 2
val (left, right) = input splitAt middle
merge(sort(left, middle), sort(right, middle + length%2)).toList
}
}
def msort[A](input: List[A]): List[A] = sort(input, input.length)
//=====================================================================================================================
// PARALLELIZATION
//val cores = Runtime.getRuntime.availableProcessors
//dbg("Detected %d cores.".format(cores))
//lazy implicit val ec = ExecutionContext.fromExecutorService(Executors.newFixedThreadPool(cores))
def futuremerge[A](fa: Future[List[A]], fb: Future[List[A]])(implicit order: Ordering[A], ec: ExecutionContext) =
{
for {
a <- fa
b <- fb
} yield merge(a, b).toList
}
def parallel_msort[A](input: List[A], length: Int)(implicit order: Ordering[A]): Future[List[A]] = {
val middle = length / 2
val (left, right) = input splitAt middle
if(length > 500) {
val fl = parallel_msort(left, middle)
val fr = parallel_msort(right, middle + length%2)
futuremerge(fl, fr)
}
else {
Future(msort(input))
}
}
//=====================================================================================================================
// MAIN
val results = time({
val src = Source.fromFile("in.txt").getLines
val header = src.next.split(" ").toVector
val lines = if (header(0) == "i") src.map(_.toInt).toList else src.toList
val f = parallel_msort(lines, lines.length)
Await.result(f, concurrent.duration.Duration.Inf)
})
println("Sorted as comparison...")
val sorted_src = Source.fromFile(input_folder+"in.txt").getLines
sorted_src.next
time(sorted_src.toList.sorted)
val writer = new PrintWriter("out.txt", "UTF-8")
try writer.print(results.mkString("\n"))
finally writer.close
}
My answer is probably going to be a bit long, but i hope that it will be useful for both you and me.
So, first question is: "how scala is doing sorting for a List?" Let's have a look at the code from scala repo!
def sorted[B >: A](implicit ord: Ordering[B]): Repr = {
val len = this.length
val b = newBuilder
if (len == 1) b ++= this
else if (len > 1) {
b.sizeHint(len)
val arr = new Array[AnyRef](len) // Previously used ArraySeq for more compact but slower code
var i = 0
for (x <- this) {
arr(i) = x.asInstanceOf[AnyRef]
i += 1
}
java.util.Arrays.sort(arr, ord.asInstanceOf[Ordering[Object]])
i = 0
while (i < arr.length) {
b += arr(i).asInstanceOf[A]
i += 1
}
}
b.result()
}
So what the hell is going on here? Long story short: with java. Everything else is just size justification and casting. Basically this is the line which defines it:
java.util.Arrays.sort(arr, ord.asInstanceOf[Ordering[Object]])
Let's go one level deeper into JDK sources:
public static <T> void sort(T[] a, Comparator<? super T> c) {
if (c == null) {
sort(a);
} else {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a, c);
else
TimSort.sort(a, 0, a.length, c, null, 0, 0);
}
}
legacyMergeSort is nothing but single threaded implementation of merge sort algorithm.
The next question is: "what is TimSort.sort and when do we use it?"
To my best knowledge default value for this property is false, which leads us to TimSort.sort algorithm. Description can be found here. Why is it better? Less comparisons that in merge sort according to comments in JDK sources.
Moreover you should be aware that it is all single threaded, so no parallelization here.
Third question, "your code":
You create too many objects. When it comes to performance, mutation (sadly) is your friend.
Premature optimization is the root of all evil -- Donald Knuth. Before making any optimizations (like parallelism), try to implement single threaded version and compare the results.
Use something like JMH to test performance of your code.
You should not probably use Stream class if you want to have the best performance as it does additional caching.
I intentionally did not give you answer like "super-fast merge sort in scala can be found here", but just some tips for you to apply to your code and coding practices.
Hope it will help you.
I need a fast method to find all indices of a search term that might occur in a string. I tried this 'brute force' String extension method:
// Note: makes use of ExSwift
extension String
{
var length: Int { return count(self) }
func indicesOf(searchTerm:String) -> [Int] {
var indices = [Int]()
for i in 0 ..< self.length {
let segment = self[i ... (i + searchTerm.length - 1)]
if (segment == searchTerm) {
indices.append(i)
}
}
return indices;
}
}
... But it's ridiculously slow, especially the shorter the search term is. What would be a better method to find all indices fast?
As Martin said you can implement some of the well known fastest algorithms in String Matching, The Knuth–Morris–Pratt string searching algorithm (or KMP algorithm) searches for occurrences of a "word" W within a main "text string" S.
The algorithm has complexity O(n), where n is the length of S and the O is big-O notation.
extension String {
// Build pi function of prefixes
private func build_pi(str: String) -> [Int] {
var n = count(str)
var pi = Array(count: n + 1, repeatedValue: 0)
var k = -1
pi[0] = -1
for (var i = 0; i < n; ++i) {
while (k >= 0 && str[k] != str[i]) {
k = pi[k]
}
pi[i + 1] = ++k
}
return pi
}
// Knuth-Morris Pratt algorithm
func searchPattern(pattern: String) -> [Int] {
var matches = [Int]()
var n = count(self)
var m = count(pattern)
var k = 0
var pi = build_pi(pattern)
for var i = 0; i < n; ++i {
while (k >= 0 && (k == m || pattern[k] != self[i])) {
k = pi[k]
}
if ++k == m {
matches.append(i - m + 1)
}
}
return matches
}
subscript (i: Int) -> Character {
return self[advance(self.startIndex, i)]
}
}
Then you can use it in the following way:
var string = "apurba mandal loves ayoshi loves"
var pattern = "loves"
println(string.searchPattern(pattern))
An the output should be :
[14, 27]
That belong to the start index of the pattern occurrences inside the the string. I hope this help you.
EDIT:
As Martin said in his comment you need to avoid the use of the advance function to index an String by an Int because it's O(position to index).
One possible solution is to convert the String to an array of Character and then access to the indexes is O(1).
Then the extension can be changed to this one :
extension String {
// Build pi function of prefixes
private func build_pi(str: [Character]) -> [Int] {
var n = count(str)
var pi = Array(count: n + 1, repeatedValue: 0)
var k = -1
pi[0] = -1
for (var i = 0; i < n; ++i) {
while (k >= 0 && str[k] != str[i]) {
k = pi[k]
}
pi[i + 1] = ++k
}
return pi
}
// Knuth-Morris Pratt algorithm
func searchPattern(pattern: String) -> [Int] {
// Convert to Character array to index in O(1)
var patt = Array(pattern)
var S = Array(self)
var matches = [Int]()
var n = count(self)
var m = count(pattern)
var k = 0
var pi = build_pi(patt)
for var i = 0; i < n; ++i {
while (k >= 0 && (k == m || patt[k] != S[i])) {
k = pi[k]
}
if ++k == m {
matches.append(i - m + 1)
}
}
return matches
}
}
Instead of checking for the search term at each position of the string
you could use rangeOfString() to find the next occurrence (hoping
that rangeOfString() uses more advanced algorithms):
extension String {
func indicesOf(searchTerm:String) -> [Int] {
var indices = [Int]()
var pos = self.startIndex
while let range = self.rangeOfString(searchTerm, range: pos ..< self.endIndex) {
indices.append(distance(self.startIndex, range.startIndex))
pos = range.startIndex.successor()
}
return indices
}
}
Generally, it depends on the size of the input string and the size
of the search string which algorithm is "the fastest". You'll find
an overview with links to various algorithms in
String searching algorithm.
Update for Swift 3:
extension String {
func indices(of searchTerm:String) -> [Int] {
var indices = [Int]()
var pos = self.startIndex
while let range = range(of: searchTerm, range: pos ..< self.endIndex) {
indices.append(distance(from: startIndex, to: range.lowerBound))
pos = index(after: range.lowerBound)
}
return indices
}
}
Using NSRegularExpression in Swift 4, you can do it like this. NSRegularExpression has been around forever and is probably a better choice than rolling your own algorithm for most cases.
let text = "The quieter you become, the more you can hear."
let searchTerm = "you"
let regex = try! NSRegularExpression(pattern: searchTerm, options: [])
let range: NSRange = NSRange(text.startIndex ..< text.endIndex, in: text)
let matches: [NSTextCheckingResult] = regex.matches(in: text, options: [], range: range)
let ranges: [NSRange] = matches.map { $0.range }
let indices: [Int] = ranges.map { $0.location }
let swiftRanges = ranges.map { Range($0, in: text) }
let swiftIndices: [String.Index] = swiftRanges.flatMap { $0?.lowerBound }
This is a question from The Algorithm Design Manual:
Suppose you are given three strings of characters: X, Y, and Z, where |X| = n,
|Y| = m, and |Z| = n+m. Z is said to be a shuffle of X and Y if and only if Z can be formed by interleaving the characters from X and Y in a way that maintains the left-to right ordering of the characters from each string.
Give an efficient dynamic programming algorithm that determines whether Z is a shuffle of X and Y.
Hint: the values of the dynamic programming matrix you construct should be Boolean, not numeric
This is what I tried:
Initially, I made a 1-D char array and pointers to the starting characters of X,Y,Z respectively. If Z-pointer with matches X-pointer store X in the char array else check the same with Y-pointer.If each entry in the char array is not different from its last entry, Z is not interleaved.
Can someone help me with this problem?
First, let's start with some definitions. I write X[i] for the ith element of X and X[i) for the substring of X starting at index i.
For example, if X = abcde, then X[2] = c and X[2) = cde.
Similar definitions hold for Y and Z.
To solve the problem by dynamic programming, you should keep a 2D boolean array A of size (n+1) x (m+1). In this array, A[i, j] = true if and only if X[i) and Y[j) can be interleaved to form Z[i+j).
For an arbitrary (i, j), somewhere in the middle of the 2D array, the recurrence relation is very simple:
A[i, j] := X[i] = Z[i+j] and A[i+1, j]
or Y[j] = Z[i+j] and A[i, j+1]
On the edges of the 2D array you have the case that either X or Y is already at its end, which means the suffix of the other should be equal to the suffix of Z:
A[m, j] := Y[j) = Z[m+j)
A[i, n] := X[i) = Z[i+n)
A[m, n] := true
If you first fill the border of the array (A[m, j] and A[i, n], for all i, j), you can then simply loop back towards A[0, 0] and set the entries appropriately. In the end A[0, 0] is your answer.
Following approach should give you an idea.
Define the condition d(s1,s2,s3) = (s1 + s2 == s3) { s3 is a shuffle of s1 and s2 }
We have to find d( X, Y, Z ).
if lengths of s1 and s2 are 1 each and length of s3 = 2,
d( s1,s2,s3 ) = { (s1[0] == s3[0] && s2[0] == s3[1]) || (s1[0] == s3[1] && s2[0] == s3[0])
Similarly d can be obtained for empty strings.
For strings of arbitrary length, following relation holds.
d( s1,s2,s3 ) = { ( d( s1-s1[last],s2,s3 - s3[last]) && s1[last] == s3[last] )
|| ( d( s1,s2 - s2[last],s3 - s3[last]) && s2[last] == s3[last] )
}
You can compute the d() entries starting from zero length strings and keep checking.
It is defined by following recurrence relation:-
S(i,j,k) = false
if(Z(i)==Y(k))
S(i,j,k) = S(i,j,k)||S(i+1,j,k+1)
if(Z(i)==X(j))
S(i,j,k) = S(i,j,k)||S(i+1,j+1,k)
Where S(i,j,k) corresponds to Z[i to end] formed by shuffle of X[j to end] and Y[K to end]
You should try to code this into DP on your own.
I think this is quite easy if you are solving this problem by using this approach with java
Java Based Solution
public class ValidShuffle {
public static void main(String[] args) {
String s1 = "XY";
String s2 = "12";
String results = "Y21XX";
validShuffle(s1, s2, results);
}
private static void validShuffle(String s1, String s2, String result) {
String s3 = s1 + s2;
StringBuffer s = new StringBuffer(s3);
boolean flag = false;
char[] ch = result.toCharArray();
if (s.length() != result.length()) {
flag = false;
} else {
for (int i = 0; i < ch.length; i++) {
String temp = Character.toString(ch[i]);
if (s3.contains(temp)) {
s = s.deleteCharAt(s.indexOf(temp));
s3 = new String(s);
flag = true;
} else {
flag = false;
break;
}
}
}
if (flag) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
If any problem in my code then comment me please. thank you
function checkShuffle(str1, str2, str3) {
var merge=str1+str2;
var charArr1= merge.split("").sort();
var charArr2= str3.split("").sort();
for(i=0;i<str3.length;i++){
if(charArr1[i] == charArr2[i]){
return true;
}
}
return false;
}
checkShuffle("abc", "def", "dfabce"); //output is true
JAVASCRIPT BASED SOLUTION
const first = "bac";
const second = "def"
const third = "dabecf";
function createDict(seq,str){
let strObj = {};
str = str.split("");
str.forEach((letter,index)=>{
strObj[letter] = {
wordSeq: seq,
index : index
} ;
})
return strObj;
}
function checkShuffleValidity(thirdWord,firstWord,secondWord){
let firstWordDict = createDict('first',firstWord);
let secondWordDict = createDict('second',secondWord);
let wordDict = {...firstWordDict,...secondWordDict};
let firstCount=0,secondCount = 0;
thirdWord = thirdWord.split("");
for(let i=0; i<thirdWord.length; i++){
let letter = thirdWord[i];
if(wordDict[letter].wordSeq == "first"){
if(wordDict[letter].index === firstCount){
firstCount++;
}else{
return false
}
}else{
if(wordDict[letter].index === secondCount){
secondCount++;
}else{
return false;
}
}
}
return true;
}
console.log(checkShuffleValidity(third,first,second));
Key points:
All strings shouldn't be null or empty.
The sum of the 2 strings length should be equal to the third string.
The third string should not contain the substrings of the 2 strings.
Else create arrays of characters , sort and compare.
Code:
public static boolean validShuffle(String first, String second, String third){
boolean status=false;
if((first==null || second==null || third==null) || (first.isEmpty()|| second.isEmpty() || third.isEmpty())){
status = false;
} else if((first.length()+second.length()) !=third.length()){
//check if the sum of 2 lengths equals to the third string length
status = false;
} else if(third.indexOf(first,0)!=-1 || third.indexOf(second,0)!=-1){
//check if the third string contains substrings
status = false;
} else {
char [] c1_2=(first+second).toCharArray();
char [] c3 =third.toCharArray();
Arrays.sort(c1_2);
Arrays.sort(c3);
status=Arrays.equals(c1_2, c3);
}
return status;
}