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Java: Removing an empty Element from String Array [duplicate]

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Resize an Array while keeping current elements in Java?
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Closed 3 years ago.
I have an array String ar[] = {"HalloWelt", " "};, ar.length is 2.
It does register two values within "HalloWelt" on index 0, and a blank/empty string on index 1;
I wonder how can I remove empty space on the index 1 - > but also keep it as a String Array since it is necessary for next task. Or how to do bunch of conversions but end up with String Array in the end.
My attempt
public String[] arraysWhiteSpaceEliminator(String[] arr) {
int k=0; //Identify how big the array should be i.e. till it reaches an empty index.
for(int i=0; i<bsp.length;i++) {
arr[i].trim();
System.out.println(arr[i].isEmpty());
if(arr[i].isEmpty()) {
}
else {
k = k+1; //if the index isn't empty == +1;
}
}
String[] clearnArray = new String[k];
for(int s = 0; s<k; s++) {
clearnArray [s] = arr[s]; //define New Array till we reach the empty index.
//System.out.println(clearnArray [s]+" " +s);
}
return clearnArray ;
};
The logic is very simple:
Identify how big the clearnArray should be.
Iterate through original Array with .trim() to remove white Space and check wether isEmpty().
Add to the k if the index isnt Empty.
Create clearnArray with the k as size.
Loop through originial Array till k -> add all the items to cleanArray till k.
Issue: .trim() and .isEmpty() don't record that the index is empty. ?!

A solution with streams:
String[] clean = Arrays.stream(ar)
.map(String::trim)
.filter(Predicate.isEqual("").negate())
.toArray(String[]::new);
Note that this assumes none of the array elements are null. If this is a possibility, simply add the following stage before the map:
.filter(Objects::nonNull)

The problem with your code is that after counting to find k, you just write the first k elements from the original array. To solve the problem by your technique, you need to check each element of the input array again to see if it's empty (after trimming), and only write to the new array the elements which pass the test.
The code can be simplified using the "enhanced" for loop, since you don't need indices for the original array. (The variable i keeps track of the current index in the new array.) Note also that strings are immutable, so calling .trim() does nothing if you don't use the result anywhere. Your code also refers to bsp which is not defined, so I changed that.
int k = 0;
for(String s : arr) {
s = s.trim();
if(!s.isEmpty()) {
k++;
}
}
String[] cleanArray = new String[k];
int i = 0;
for(String s : arr) {
s = s.trim();
if(!s.isEmpty()) {
cleanArray[i] = s;
i++;
}
}
return cleanArray;

Calculate the number of non-null elements and create an array of that size, like
String[] strs = ...;
int count = 0;
for (int i = 0; i < strs.length; i++) {
if (strs[i] != null) count++;
}
String newStrArray[] = new String[count];
int idx = 0;
for (int i = 0; i < strs.length; i++) {
if (strs[i] != null) newStrArray[idx++] = strs[i];
}
return newStrArray;
You could also probably make this prettier using streams. However I haven't used streaming functionality in Java, so I can't help there.
Two things to note:
Unless you are serializing or the nulls are causing other problems, trimming the array just to get rid of the nulls probably won't have an impact on memory, as the size of an array entry (4 bytes) is very likely inconsequential to the memory block size allocated for the Array object
Converting first to an List and then back to an array is lazy and possibly inefficient. ArrayList, for example, will likely include extra space in the array it creates internally so that you can add more elements to the ArrayList and not have to create a whole new internal array.

in your method, create a new
List cleanedList = new ArrayList();
add iterate through ar[]ar[] = ["HalloWelt",""], and add only non-empty values to cleaned List....then return the array.
return cleanedList.toArray()
like below:
List<String> cleanedList = new ArrayList<>();
for(String s : arr) {
s = s.trim();
if(!s.isEmpty()) {
cleanedList.add(s);
}
}
return cleanArray.toArray();

Split a field through array in java

I would like to know how to split a field through array using Java. For example we have GLaccount like AAAA-BBBB-CCCC and we would like to split each component and store it in an variable however the GLaccount may have AAAA-BBBB (no third component) so in this case variable segment3 throws NULL POINTER exception so I am not sure on how to fix this since I am new to Java.
String GL = getOwner().getGL("GLACCT");
String segment1 = GL.split("-")[0];
String segment2 = GL.split("-")[1];
String segment3 = GL.split("-")[2];

Using split("-" ) will give you an array of strings.
before using array value, you can check the size of array that if it contains enough elements to use..
String GL = getOwner().getGL("GLACCT");
String[] array=GL.split("-");
String segment1 = array[0];
String segment2 = array[1];
//check if array have 3rd element
if(array.length >2)
String segment3 = array[2];
else
System.out.println("No third element") ;

Use split method (once) and check returned array length :
String[] values3 = "AAAA-BBBB-CCCC".split("-");
// values.length == 3
String[] values2 = "AAAA-BBBB".split("-");
// values2.length == 2

import java.util.Arrays;
List<String> list = Arrays.asList(GL.split("-"));
With this code you do not need to think if you have 2,3 or 10 strings, and to add new if for every new one.

How to get 2D array possible combinations

I have the following 2D array:
String[M][]
String[0]
"1","2","3"
String[1]
"A", "B"
.
.
.
String[M-1]
"!"
All the possible combinations should be in store in a resulting array
String[] combinations. So for example:
combinations[0] == {"1A....!")
combinations[1] == {"2A....!")
combinations[2] == {"3A....!")
combinations[3] == {"1B....!")
Notice that that the arrays are of variable length. Order of the elements in the output String doesn't matter. I also don't care if there are duplicates.
If the arrays were the same length, nested loops would do the trick, but they are not, and I really don't know how to approach the problem.

You can iterate through the combinations one at a time like clockwork by using an array to record the size of each inner array, and a counter array which keeps track of which member to use from each inner array. Something like this method:
/**
* Produce a List<String> which contains every combination which can be
* made by taking one String from each inner String array within the
* provided two-dimensional String array.
* #param twoDimStringArray a two-dimensional String array which contains
* String arrays of variable length.
* #return a List which contains every String which can be formed by taking
* one String from each String array within the specified two-dimensional
* array.
*/
public static List<String> combinations(String[][] twoDimStringArray) {
// keep track of the size of each inner String array
int sizeArray[] = new int[twoDimStringArray.length];
// keep track of the index of each inner String array which will be used
// to make the next combination
int counterArray[] = new int[twoDimStringArray.length];
// Discover the size of each inner array and populate sizeArray.
// Also calculate the total number of combinations possible using the
// inner String array sizes.
int totalCombinationCount = 1;
for(int i = 0; i < twoDimStringArray.length; ++i) {
sizeArray[i] = twoDimStringArray[i].length;
totalCombinationCount *= twoDimStringArray[i].length;
}
// Store the combinations in a List of String objects
List<String> combinationList = new ArrayList<String>(totalCombinationCount);
StringBuilder sb; // more efficient than String for concatenation
for (int countdown = totalCombinationCount; countdown > 0; --countdown) {
// Run through the inner arrays, grabbing the member from the index
// specified by the counterArray for each inner array, and build a
// combination string.
sb = new StringBuilder();
for(int i = 0; i < twoDimStringArray.length; ++i) {
sb.append(twoDimStringArray[i][counterArray[i]]);
}
combinationList.add(sb.toString()); // add new combination to list
// Now we need to increment the counterArray so that the next
// combination is taken on the next iteration of this loop.
for(int incIndex = twoDimStringArray.length - 1; incIndex >= 0; --incIndex) {
if(counterArray[incIndex] + 1 < sizeArray[incIndex]) {
++counterArray[incIndex];
// None of the indices of higher significance need to be
// incremented, so jump out of this for loop at this point.
break;
}
// The index at this position is at its max value, so zero it
// and continue this loop to increment the index which is more
// significant than this one.
counterArray[incIndex] = 0;
}
}
return combinationList;
}
How the method works
If you imagine the counter array being like a digital clock reading then the first String combination sees the counter array at all zeroes, so that the first String is made by taken the zero element (first member) of each inner array.
To get the next combination the counter array is incremented by one. So the least-significant counter index is increased by one. If this causes its value to become equal to the length of the inner array it represents then the index is zeroed, and the next index of greater significance is increased. A separate size array stores the length of each inner array, so that the counter array loop knows when an index has reached its maximum.
For example, if the size array was:
[3][3][2][1]
and the counter array was at:
[0][2][1][0]
then the increment would make the least significant (right-most) index equal to 1, which is its maximum value. So that index gets zeroed and the next index of greater significance (the second-from-right) gets increased to 2. But that is also the maximum of that index, so it gets zeroed and we move to the next index of greater significance. That gets increased to three, which is its maximum value so it gets zeroed and we move to the most significant (left-most) index. That gets increased to 1, which is less than its maximum so the incremented counter array becomes:
[1][0][0][0]
Which means the next String combination is made by taking the second member of the first inner array, and the first member of the next three inner arrays.
Dire warnings and notes
I wrote this just now in about forty minutes, and it's half-one in the morning, which means that even though it seems to do exactly what is needed, there are very likely bugs or bits of code which could be optimised. So be sure to unit test it thoroughly if its performance is critical.
Note that it returns a List rather than a String array because I think that Java Collections are vastly preferable to using arrays in most cases. Also, if you need a result set with no duplicates, you can simply change the List to a Set which will automatically drop duplicates and leave you with a unique set.
If you really need the result as a String array, don't forget you can use the List<String>.toArray(String[]) method to simply convert the returned List to what you need.

This problem has a very nice recursive structure to it (which also means it could explode in memory, the correct way should be using iterators such as the other answer, but this solution looks nicer imo and we can prove correctness inductively because of the recursive nature). A combination consists of an element from the first list attached to all possible combinations formed from the remaining (n-1) lists. The recursive work is done in AllCombinationsHelper, but you invoke AllCombinations. Note to test for empty lists and more extensively.
public static List<String> AllCombinations(List<List<Character>> aList) {
if(aList.size() == 0) { return new ArrayList<String>(); }
List<Character> myFirstSubList = aList.remove(0);
List<String> myStrings = new ArrayList<String>();
for(Character c : myFirstSubList) {
myStrings.add(c.toString());
}
return AllCombinationsHelper(aList, myStrings);
}
public static List<String> AllCombinationsHelper(List<List<Character>> aList,
List<String> aCollection) {
if(aList.size() == 0) { return aCollection; }
List<Character> myFirstList = aList.remove(0);
List<String> myReturnSet = new ArrayList<String>();
for(String s : aCollection) {
for(Character c : myFirstList) {
myReturnSet.add(c + s);
}
}
return AllCombinationsHelper(aList, myReturnSet);
}

Should be straight forward to do with recursion.
Let me rephrase a bit, so the terminology is less confusing.
We will call String[] as Token List, which is a list of Tokens
Now you have a List of Token List, you want to get one Token from each Token List available, and find out all combination.
What you need to do is, given a list of TokenList
If the List is having only one TokenList, the content of the Token List itself is all combinations
Else, make a sub-list by excluding the first Token List, and find out all combinations of that sub list. When you have the combinations, the answer is simply loop through your first token list, and generate all combinations using each token in the token list, and the result combinations.
I am only giving a psuedo code:
List<String> allCombinations(List<TokenList> listOfTokenList) {
if (length of strings == 1) {
return strings[0];
}
List<String> subListCombinations
= allCombination(listOfTokenList.subList(1)); // sublist from index 1 to the end
List<String> result;
for each (token in listOfTokenList[0]) {
for each (s in subListCombination) {
result.add(token + s);
}
}
return result;
}

I have been struggling with this problem for some time. But I finally solved it. My main obstacle was the SCOPE I used for declaring each variable. If you do not declare your variables in the correct scope, then the variable will retain changes made in the previous iteration.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class RecursiveAlgorithmTest {
private static int recursiveCallsCounter = 0;
public static ArrayList<ArrayList<String>> testCases = new ArrayList<ArrayList<String>>();
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
//set values for ArrayOfArrays
ArrayList<String> VariableA = new ArrayList<String>(Arrays.asList("red", "green"));
ArrayList<String> VariableB = new ArrayList<String>(Arrays.asList("A", "B", "C"));
ArrayList<String> VariableC = new ArrayList<String>(Arrays.asList("1", "2", "3", "4"));
ArrayList<ArrayList<String>> AofA = new ArrayList<ArrayList<String>>();
AofA.add(VariableA); AofA.add(VariableB); AofA.add(VariableC);
System.out.println("Array of Arrays: ToString(): " +AofA.toString());
ArrayList<String> optionsList = new ArrayList<String>();
//recursive call
recurse(optionsList, AofA, 0);
for (int i = 0 ; i < testCases.size() ; i++) {
System.out.println("Test Case " + (i+1) + ": " + testCases.get(i));
}
}//end main(String args[])
private static void recurse(ArrayList<String> newOptionsList,
ArrayList<ArrayList<String>> newAofA, int placeHolder){
recursiveCallsCounter++;
System.out.println("\n\tStart of Recursive Call: " + recursiveCallsCounter);
System.out.println("\tOptionsList: " + newOptionsList.toString());
System.out.println("\tAofA: " + newAofA.toString());
System.out.println("\tPlaceHolder: "+ placeHolder);
//check to see if we are at the end of all TestAspects
if(placeHolder < newAofA.size()){
//remove the first item in the ArrayOfArrays
ArrayList<String> currentAspectsOptions = newAofA.get(placeHolder);
//iterate through the popped off options
for (int i=0 ; i<currentAspectsOptions.size();i++){
ArrayList<String> newOptions = new ArrayList<String>();
//add all the passed in options to the new object to pass on
for (int j=0 ; j < newOptionsList.size();j++) {
newOptions.add(newOptionsList.get(j));
}
newOptions.add(currentAspectsOptions.get(i));
int newPlaceHolder = placeHolder + 1;
recurse(newOptions,newAofA, newPlaceHolder);
}
} else { // no more arrays to pop off
ArrayList<String> newTestCase = new ArrayList<String>();
for (int i=0; i < newOptionsList.size();i++){
newTestCase.add(newOptionsList.get(i));
}
System.out.println("\t### Adding: "+newTestCase.toString());
testCases.add(newTestCase);
}
}//end recursive helper
}// end of test class

In Python one uses itertools.product and argument unpacking (apply)
>>> import itertools
>>> S=[['1','2','3'],['A','B'],['!']]
>>> ["".join(x) for x in itertools.product(*S)]
['1A!', '1B!', '2A!', '2B!', '3A!', '3B!']

How can I store a string Matrix inside an array?

I have to store a String matrix(3x20) inside an array whose length may vary.
I am trying the following code but I am getting an incompatible types error.
How could I fix this error?
My code is:
int x=0;
String[] arrayF=new String[10];
arrayF[x]= new String[3][20];

You can't assign array this way. You should eventually assign each element of the first 2-array to the 1-d array.
Something like:
String[][] array2D =new String[M][N];
String[] array1D = new String[M * N];
for (int i = 0 ; i < M ; i++)
{
for (int j = 0 ; i < N ; i++)
{
array1D[(j * N) + i] = array2D[i][j];
}
}

arrayF is an array of strings, so each element in arrayF must be a string (by definition of the array).
What you are trying to do is is put an array (new String[3][20]), instead of a string, in each element of arrayF, which obviously contradicts it's definition (hence the incompatible types error).
One solution for what you want might be using a 3-d array of strings:
String[][][] arr = new String[10][3][20];

arrayF is one dimensional array with String type.
You cannot add two dimensional array to arrayF. For dynamic array size, you should use ArrayList.
List<String[][]> main = new ArrayList<String[][]>();
String[][] child1 = new String[3][20];
String[][] child2 = new String[3][20];
main.add(child1);
main.add(child2);
Refer to
Variable length (Dynamic) Arrays in Java

use something like this:
String [][] strArr = new String[3][20];
ArrayList<String[][]> tm = new ArrayList<String[][]>();
tm.add(strArr);

how do I ignore/delete values of an array in java

I have a list of words , there are 4 words, it cant contain more that 4 its just an example. I want to use just 2 of the words the rest of them should be ignored or deleted e.g :
String planets = "Moon,Sun,Jupiter,Mars";
String[] planetsArray = planets.split(",");
int numberOfPlanets = planetsArray.length;
the result i get is 4. How do i delete the rest of the words if my list contains more that 2 words ?

As suggested in your previous question, you can use
String[] fewPlanets = new String[]{planets[0], planets[1]};
Just make sure the planets array has 2 elements or more to avoid an ArrayIndexOutOfBoundsException. You can use length to check it: if (planets.length >= 2)
For a more sophisticated solution, you could also do this using System.arrayCopy() if you're using Java 1.5 or earlier,
int numberOfElements = 2;
String[] fewPlanets = new String[2];
System.arraycopy(planets, 0, fewPlanets, 0, numberOfElements);
or Arrays.copyOf() if you're using Java 1.6 or later:
int numberOfElements = 2;
String[] fewPlanets = Arrays.copyOf(planets, numberOfElements);

String planets = "Moon,Sun,Jupiter,Mars";
String[] planetsArray = planets.split(",");
if(planetsArray .length > 2){
String []newArr = new String[2];
newArr[0]=planetsArray [0];
newArr[1]=planetsArray [2];
planetsArray = newArr ;
}

Use Arrays.asList to get a List of Strings from String[] planetsArray.
Then use the methods of the List interface -contains,remove,add, ...- to simply do whatever you want on that List.

If you need to select the first 2 planets just copy the array:
String[] newPlanetsArray = Arrays.CopyOf(planetsArray, 2);
If you need to select 2 specific planets you can apply the following algorithm:
First, create a new array with 2 elements. Then, iterate through the elements in the original array and if the current element is a match add it to the new array (keep track of the current position in the new array to add the next element).
String[] newPlanetsArray = new String[2];
for(int i = 0, int j = 0; i < planetsArray.length; i++) {
if (planetsArray[i].equals("Jupiter") || planetsArray[i].equals("Mars")) {
newPlanetsArray[j++] = planetsArray[i];
if (j > 1)
break;
}
}

You could use an idea from How to find nth occurrence of character in a string? and avoid reading the remaining values from your comma separated string input. Simply locate the second comma and substring upto there
(Of course if your code snippet is just an example and you do not have a comma separated input, then please ignore this suggestion :)