Hello everyone i am trying to remove an name that the user has put in from an String Array, i am new to programming and i have tried this but it doesn't work. Can someone help me or tell me what i am doing wrong?
String [] myName = {"Testname","Charel","melissa","Kelly"};
removeName(myName);
public void removeName(String[] names )
{
Scanner sc = new Scanner(System.in);
String name = "";
name = sc.nextLine();
for (int i = 0; i < names.length; i++) {
name = names[i-1];
}
}
How can i do this?
You probably need to use Lists for this. Your list will be a list of String, and use remove() method to do this.
An array's length is fixed and can't be changed this way.
Useful Link : Removing items from a list
First off, an array does not change size after it is initialized, the only way to change the size of an array is to replace it with a new array! So in order to not end up with a double entry or an empty field, you would need to make a new array that is one size shorter, and write the names you want to keep into that.
An array might be ill-suited for your purposes, so consider using a list or an ArrayList. A list can be resized, so removing an element will automatically shorten the list. I recommend you look into that.
Lastly, you currently aren't even comparing your input to your fields. Replace name = names[i-1]; with something along the lines of
if(name.equals(names[i]))
//TODO: Remove from list
See here for more details about String.equals()!
Also, keep in mind that the user input might not match any name at all, so prepare for that case as well!
To remove an element from an array in Java, you need to create a new array and copy over all the elements you want to keep. That is because Java arrays are fixed-size.
For example, to remove an element at a particular index, you could do it like this:
public static String[] remove(String[] array, int index) {
String[] result = new String[array.length - 1];
System.arraycopy(array, 0, result, 0, index);
System.arraycopy(array, index + 1, result, index, result.length - index);
return result;
}
You would then remove melissa from your array as follows:
String[] names = { "Testname", "Charel", "Melissa", "Kelly" };
names = remove(names, 2);
System.out.println(Arrays.toString(names));
Output
[Testname, Charel, Kelly]
Of course, it would be much easier to do it using a List:
List<String> names = new ArrayList<>(Arrays.asList("Testname", "Charel", "Melissa", "Kelly"));
names.remove(2);
System.out.println(names);
Or:
List<String> names = new ArrayList<>(Arrays.asList("Testname", "Charel", "Melissa", "Kelly"));
names.remove("Melissa");
System.out.println(names);
Output of both is the same as above.
There are some simple methods using java api provide by jdk, for example:
String [] myName = {"Testname","Charel","melissa","Kelly"};
List<String> container = new ArrayList(Arrays.asList(myName));
container.remove("Charel");
String[] result = new String[myName.length - 1];
container.toArray(result);
Alternatively you can also use this to convert array to list,
Collections.addAll(container, myName);
String [] myName = {"Testname","Charel","melissa","Kelly"};
removeName(myName);
public void removeName(String[] names )
{
Scanner sc = new Scanner(System.in);
String name = sc.nextLine();
for (int i = 0; i < names.length; i++) {
if(names[i]==name)
{
for(int j=i;j<names.length-1;j++)
{
names[j]=names[j+1];
}
}
}
}
I have a String[] with values like so:
public static final String[] VALUES = new String[] {"AB","BC","CD","AE"};
Given String s, is there a good way of testing whether VALUES contains s?
Arrays.asList(yourArray).contains(yourValue)
Warning: this doesn't work for arrays of primitives (see the comments).
Since java-8 you can now use Streams.
String[] values = {"AB","BC","CD","AE"};
boolean contains = Arrays.stream(values).anyMatch("s"::equals);
To check whether an array of int, double or long contains a value use IntStream, DoubleStream or LongStream respectively.
Example
int[] a = {1,2,3,4};
boolean contains = IntStream.of(a).anyMatch(x -> x == 4);
Concise update for Java SE 9
Reference arrays are bad. For this case we are after a set. Since Java SE 9 we have Set.of.
private static final Set<String> VALUES = Set.of(
"AB","BC","CD","AE"
);
"Given String s, is there a good way of testing whether VALUES contains s?"
VALUES.contains(s)
O(1).
The right type, immutable, O(1) and concise. Beautiful.*
Original answer details
Just to clear the code up to start with. We have (corrected):
public static final String[] VALUES = new String[] {"AB","BC","CD","AE"};
This is a mutable static which FindBugs will tell you is very naughty. Do not modify statics and do not allow other code to do so also. At an absolute minimum, the field should be private:
private static final String[] VALUES = new String[] {"AB","BC","CD","AE"};
(Note, you can actually drop the new String[]; bit.)
Reference arrays are still bad and we want a set:
private static final Set<String> VALUES = new HashSet<String>(Arrays.asList(
new String[] {"AB","BC","CD","AE"}
));
(Paranoid people, such as myself, may feel more at ease if this was wrapped in Collections.unmodifiableSet - it could then even be made public.)
(*To be a little more on brand, the collections API is predictably still missing immutable collection types and the syntax is still far too verbose, for my tastes.)
You can use ArrayUtils.contains from Apache Commons Lang
public static boolean contains(Object[] array, Object objectToFind)
Note that this method returns false if the passed array is null.
There are also methods available for primitive arrays of all kinds.
Example:
String[] fieldsToInclude = { "id", "name", "location" };
if ( ArrayUtils.contains( fieldsToInclude, "id" ) ) {
// Do some stuff.
}
Just simply implement it by hand:
public static <T> boolean contains(final T[] array, final T v) {
for (final T e : array)
if (e == v || v != null && v.equals(e))
return true;
return false;
}
Improvement:
The v != null condition is constant inside the method. It always evaluates to the same Boolean value during the method call. So if the input array is big, it is more efficient to evaluate this condition only once, and we can use a simplified/faster condition inside the for loop based on the result. The improved contains() method:
public static <T> boolean contains2(final T[] array, final T v) {
if (v == null) {
for (final T e : array)
if (e == null)
return true;
}
else {
for (final T e : array)
if (e == v || v.equals(e))
return true;
}
return false;
}
Four Different Ways to Check If an Array Contains a Value
Using List:
public static boolean useList(String[] arr, String targetValue) {
return Arrays.asList(arr).contains(targetValue);
}
Using Set:
public static boolean useSet(String[] arr, String targetValue) {
Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);
}
Using a simple loop:
public static boolean useLoop(String[] arr, String targetValue) {
for (String s: arr) {
if (s.equals(targetValue))
return true;
}
return false;
}
Using Arrays.binarySearch():
The code below is wrong, it is listed here for completeness. binarySearch() can ONLY be used on sorted arrays. You will find the result is weird below. This is the best option when array is sorted.
public static boolean binarySearch(String[] arr, String targetValue) {
return Arrays.binarySearch(arr, targetValue) >= 0;
}
Quick Example:
String testValue="test";
String newValueNotInList="newValue";
String[] valueArray = { "this", "is", "java" , "test" };
Arrays.asList(valueArray).contains(testValue); // returns true
Arrays.asList(valueArray).contains(newValueNotInList); // returns false
If the array is not sorted, you will have to iterate over everything and make a call to equals on each.
If the array is sorted, you can do a binary search, there's one in the Arrays class.
Generally speaking, if you are going to do a lot of membership checks, you may want to store everything in a Set, not in an array.
For what it's worth I ran a test comparing the 3 suggestions for speed. I generated random integers, converted them to a String and added them to an array. I then searched for the highest possible number/string, which would be a worst case scenario for the asList().contains().
When using a 10K array size the results were:
Sort & Search : 15
Binary Search : 0
asList.contains : 0
When using a 100K array the results were:
Sort & Search : 156
Binary Search : 0
asList.contains : 32
So if the array is created in sorted order the binary search is the fastest, otherwise the asList().contains would be the way to go. If you have many searches, then it may be worthwhile to sort the array so you can use the binary search. It all depends on your application.
I would think those are the results most people would expect. Here is the test code:
import java.util.*;
public class Test {
public static void main(String args[]) {
long start = 0;
int size = 100000;
String[] strings = new String[size];
Random random = new Random();
for (int i = 0; i < size; i++)
strings[i] = "" + random.nextInt(size);
start = System.currentTimeMillis();
Arrays.sort(strings);
System.out.println(Arrays.binarySearch(strings, "" + (size - 1)));
System.out.println("Sort & Search : "
+ (System.currentTimeMillis() - start));
start = System.currentTimeMillis();
System.out.println(Arrays.binarySearch(strings, "" + (size - 1)));
System.out.println("Search : "
+ (System.currentTimeMillis() - start));
start = System.currentTimeMillis();
System.out.println(Arrays.asList(strings).contains("" + (size - 1)));
System.out.println("Contains : "
+ (System.currentTimeMillis() - start));
}
}
Instead of using the quick array initialisation syntax too, you could just initialise it as a List straight away in a similar manner using the Arrays.asList method, e.g.:
public static final List<String> STRINGS = Arrays.asList("firstString", "secondString" ...., "lastString");
Then you can do (like above):
STRINGS.contains("the string you want to find");
With Java 8 you can create a stream and check if any entries in the stream matches "s":
String[] values = {"AB","BC","CD","AE"};
boolean sInArray = Arrays.stream(values).anyMatch("s"::equals);
Or as a generic method:
public static <T> boolean arrayContains(T[] array, T value) {
return Arrays.stream(array).anyMatch(value::equals);
}
You can use the Arrays class to perform a binary search for the value. If your array is not sorted, you will have to use the sort functions in the same class to sort the array, then search through it.
ObStupidAnswer (but I think there's a lesson in here somewhere):
enum Values {
AB, BC, CD, AE
}
try {
Values.valueOf(s);
return true;
} catch (IllegalArgumentException exc) {
return false;
}
Actually, if you use HashSet<String> as Tom Hawtin proposed you don't need to worry about sorting, and your speed is the same as with binary search on a presorted array, probably even faster.
It all depends on how your code is set up, obviously, but from where I stand, the order would be:
On an unsorted array:
HashSet
asList
sort & binary
On a sorted array:
HashSet
Binary
asList
So either way, HashSet for the win.
Developers often do:
Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);
The above code works, but there is no need to convert a list to set first. Converting a list to a set requires extra time. It can as simple as:
Arrays.asList(arr).contains(targetValue);
or
for (String s : arr) {
if (s.equals(targetValue))
return true;
}
return false;
The first one is more readable than the second one.
If you have the google collections library, Tom's answer can be simplified a lot by using ImmutableSet (http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/ImmutableSet.html)
This really removes a lot of clutter from the initialization proposed
private static final Set<String> VALUES = ImmutableSet.of("AB","BC","CD","AE");
In Java 8 use Streams.
List<String> myList =
Arrays.asList("a1", "a2", "b1", "c2", "c1");
myList.stream()
.filter(s -> s.startsWith("c"))
.map(String::toUpperCase)
.sorted()
.forEach(System.out::println);
One possible solution:
import java.util.Arrays;
import java.util.List;
public class ArrayContainsElement {
public static final List<String> VALUES = Arrays.asList("AB", "BC", "CD", "AE");
public static void main(String args[]) {
if (VALUES.contains("AB")) {
System.out.println("Contains");
} else {
System.out.println("Not contains");
}
}
}
Using a simple loop is the most efficient way of doing this.
boolean useLoop(String[] arr, String targetValue) {
for(String s: arr){
if(s.equals(targetValue))
return true;
}
return false;
}
Courtesy to Programcreek
the shortest solution
the array VALUES may contain duplicates
since Java 9
List.of(VALUES).contains(s);
Use the following (the contains() method is ArrayUtils.in() in this code):
ObjectUtils.java
public class ObjectUtils {
/**
* A null safe method to detect if two objects are equal.
* #param object1
* #param object2
* #return true if either both objects are null, or equal, else returns false.
*/
public static boolean equals(Object object1, Object object2) {
return object1 == null ? object2 == null : object1.equals(object2);
}
}
ArrayUtils.java
public class ArrayUtils {
/**
* Find the index of of an object is in given array,
* starting from given inclusive index.
* #param ts Array to be searched in.
* #param t Object to be searched.
* #param start The index from where the search must start.
* #return Index of the given object in the array if it is there, else -1.
*/
public static <T> int indexOf(final T[] ts, final T t, int start) {
for (int i = start; i < ts.length; ++i)
if (ObjectUtils.equals(ts[i], t))
return i;
return -1;
}
/**
* Find the index of of an object is in given array, starting from 0;
* #param ts Array to be searched in.
* #param t Object to be searched.
* #return indexOf(ts, t, 0)
*/
public static <T> int indexOf(final T[] ts, final T t) {
return indexOf(ts, t, 0);
}
/**
* Detect if the given object is in the given array.
* #param ts Array to be searched in.
* #param t Object to be searched.
* #return If indexOf(ts, t) is greater than -1.
*/
public static <T> boolean in(final T[] ts, final T t) {
return indexOf(ts, t) > -1;
}
}
As you can see in the code above, that there are other utility methods ObjectUtils.equals() and ArrayUtils.indexOf(), that were used at other places as well.
For arrays of limited length use the following (as given by camickr). This is slow for repeated checks, especially for longer arrays (linear search).
Arrays.asList(...).contains(...)
For fast performance if you repeatedly check against a larger set of elements
An array is the wrong structure. Use a TreeSet and add each element to it. It sorts elements and has a fast exist() method (binary search).
If the elements implement Comparable & you want the TreeSet sorted accordingly:
ElementClass.compareTo() method must be compatable with ElementClass.equals(): see Triads not showing up to fight? (Java Set missing an item)
TreeSet myElements = new TreeSet();
// Do this for each element (implementing *Comparable*)
myElements.add(nextElement);
// *Alternatively*, if an array is forceably provided from other code:
myElements.addAll(Arrays.asList(myArray));
Otherwise, use your own Comparator:
class MyComparator implements Comparator<ElementClass> {
int compareTo(ElementClass element1; ElementClass element2) {
// Your comparison of elements
// Should be consistent with object equality
}
boolean equals(Object otherComparator) {
// Your equality of comparators
}
}
// construct TreeSet with the comparator
TreeSet myElements = new TreeSet(new MyComparator());
// Do this for each element (implementing *Comparable*)
myElements.add(nextElement);
The payoff: check existence of some element:
// Fast binary search through sorted elements (performance ~ log(size)):
boolean containsElement = myElements.exists(someElement);
If you don't want it to be case sensitive
Arrays.stream(VALUES).anyMatch(s::equalsIgnoreCase);
Try this:
ArrayList<Integer> arrlist = new ArrayList<Integer>(8);
// use add() method to add elements in the list
arrlist.add(20);
arrlist.add(25);
arrlist.add(10);
arrlist.add(15);
boolean retval = arrlist.contains(10);
if (retval == true) {
System.out.println("10 is contained in the list");
}
else {
System.out.println("10 is not contained in the list");
}
Check this
String[] VALUES = new String[]{"AB", "BC", "CD", "AE"};
String s;
for (int i = 0; i < VALUES.length; i++) {
if (VALUES[i].equals(s)) {
// do your stuff
} else {
//do your stuff
}
}
Arrays.asList() -> then calling the contains() method will always work, but a search algorithm is much better since you don't need to create a lightweight list wrapper around the array, which is what Arrays.asList() does.
public boolean findString(String[] strings, String desired){
for (String str : strings){
if (desired.equals(str)) {
return true;
}
}
return false; //if we get here… there is no desired String, return false.
}
Use below -
String[] values = {"AB","BC","CD","AE"};
String s = "A";
boolean contains = Arrays.stream(values).anyMatch(v -> v.contains(s));
Use Array.BinarySearch(array,obj) for finding the given object in array or not.
Example:
if (Array.BinarySearch(str, i) > -1)` → true --exists
false --not exists
Try using Java 8 predicate test method
Here is a full example of it.
import java.util.Arrays;
import java.util.List;
import java.util.function.Predicate;
public class Test {
public static final List<String> VALUES =
Arrays.asList("AA", "AB", "BC", "CD", "AE");
public static void main(String args[]) {
Predicate<String> containsLetterA = VALUES -> VALUES.contains("AB");
for (String i : VALUES) {
System.out.println(containsLetterA.test(i));
}
}
}
http://mytechnologythought.blogspot.com/2019/10/java-8-predicate-test-method-example.html
https://github.com/VipulGulhane1/java8/blob/master/Test.java
Create a boolean initially set to false. Run a loop to check every value in the array and compare to the value you are checking against. If you ever get a match, set boolean to true and stop the looping. Then assert that the boolean is true.
As I'm dealing with low level Java using primitive types byte and byte[], the best so far I got is from bytes-java https://github.com/patrickfav/bytes-java seems a fine piece of work
You can check it by two methods
A) By converting the array into string and then check the required string by .contains method
String a = Arrays.toString(VALUES);
System.out.println(a.contains("AB"));
System.out.println(a.contains("BC"));
System.out.println(a.contains("CD"));
System.out.println(a.contains("AE"));
B) This is a more efficent method
Scanner s = new Scanner(System.in);
String u = s.next();
boolean d = true;
for (int i = 0; i < VAL.length; i++) {
if (VAL[i].equals(u) == d)
System.out.println(VAL[i] + " " + u + VAL[i].equals(u));
}
I'm trying to bubble sort string data that was input into an array in descending and ascending order.
The following is the code so far:
import java.util.*;
public class nextLineArray
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
String names[]=new String[12];
System.out.println("Enter the 12 names: ");
//Load Array
for(int i = 0; i < 12; i++)
{
names[i] = input.nextLine();
}
//Print initial list
System.out.println("List of names via input:"+ names);
//Print descending order list
String descSort;
descSort=bubbleSortDesc(names);
System.out.println("Names listed sorted in descending order (via BubbleSort): "+descSort);
}
public static String bubbleSortDesc(String[] names)
{
String temp;
int passNum, i, result;
for(passNum=1; passNum <= 11; passNum++)
{
for(i = 0; i<=(11-passNum); i++)
{
result=names[i].compareToIgnoreCase(names[i+1]);
if(result>0)
{
temp=names[i];
names[i]=names[i+1];
names[i+1]=temp;
}
}
}
return names;
}
}
When I try to return the sorted array to the main method it gives me the following error on the return line:
Incompatible Types
Our online instructor just started us out with using multiple methods and arrays at the same time and it is quite confusing...please excuse me if any of my mistakes appear to be obvious.
Edit: I have fixed the initial problem thanks to Alexandre Santos in the comments, I am now running into a problem when executing the program after inputting the data, instead of printing the strings in the array it prints out
[Ljava.lang.String;#6d782f7c
Take a look at the method
public static String bubbleSortDesc(String[] names)
The return of that method is supposed to be a String (only one), but you are returning the parameter "names", which is an array of strings. The "[]" after the String identifies it as an array.
I am not going to do your homework for you, so a hint: check if the return type of the method bubbleSortDesc should be one String or an array of Strings.
Good luck.
There are 2 points to fix. First you should return String array
public static String[] bubbleSortDesc(String[] names)
and therefore you should define it like this:
String descSort[];
public static String bubbleSortDesc(String[] names)
should be
public static String[] bubbleSortDesc(String[] names)
and also declare descSort as String array.
Also you are just printing the array objects. This will not print the list for you. You have iterate over the array.
Include this in you code:
for (String name:names)
{
System.out.println(name);
}
Do the same for descSort too....
You can fix your print command by changing it to the following:
System.out.println("Names listed sorted in descending order (via BubbleSort): "+ java.util.Arrays.deepToString(descSort));
If you want the nitty gritty, descSort is a String[]. In Java when you convert String[] into a String it gives you that crazy string representation. You have to instead converte each entry in the array to a String individually. Fortunately the deepToString method will do that for you.
Hi community I have a question, I happen to have an array of objects loaded on startup, through that generate array another array of integers that contains your code, it appears that array of integers'm removing their values, what I want is to compare the list of integer array currently have with the array of objects, and remove all code object that whole array mentioned is found.
My code java:
private List<ValidColumnKey> columnCustomer;
private int[] selectedCustomer;
public void init(){
this.setColumnCustomer(new ArrayList<ValidColumnKey>());
this.getColumnCustomer().add(new ValidColumnKey(1, "Codigo", "code"));
this.getColumnCustomer().add(new ValidColumnKey(2, "Nombre", "name"));
this.getColumnCustomer().add(new ValidColumnKey(3, "Nombre Comercial", "comercialName"));
this.getColumnCustomer().add(new ValidColumnKey(4, "Estado", "isActive"));
this.setSelectedCustomer(new int [this.getColumnCustomer().size()]);
int i = 0;
for(ValidColumnKey column : this.getColumnCustomer()){
this.getSelectedCustomer()[i] = column.getCodigo();
i++;
}
}
I mean I would have my array of integers with codes removed, like this:
selectedCustomer = [1, 2, 3];
What I wanted was to remove from the list of objects that do not have codes in the array of integers, but it is not my code:
List<ValidColumnKey> auxRemoColumnKeys = new ArrayList<ValidColumnKey>();
for(ValidColumnKey column : this.getColumnCustomer()){
for(Integer codigo : this.getSelectedCustomer()){
if (column.getCodigo() != codigo) {
auxRemoColumnKeys.add(column);
break;
}
}
}
this.getColumnCustomer().remove(auxRemoColumnKeys);
I could guide the solution.
this.getColumnCustomer().remove(auxRemoColumnKeys);
This statement assumes you have a valid equals method for your class ValidColumnKey, which I suspect is probably not the case.
What you want to do is iterate with a Iterator. Some sample code could be like
Set<Integer> toRemoveCodes = new HashSet<Integer>(Arrays.asList(1, 2, 3));
for (Iterator<ValidColumnKey> it = this.getColumnCustomer().iterator(); it.hasNext(); ) {
ValidColumnKey curColumnKey = it.next();
Integer code = curColumnKey.codigo();
if (toRemoveCodes.contains(code)) {
it.remove();
}
}
There are multiple reasons your current attempt is failing. The first is that this line:
if (column.getCodigo() != codigo) {
Is testing for object equivalence between Integers, not value equavalence between ints. If you want to compare Integers, you have to use the equals method:
if (!column.getCodigo().equals(codigo)) {
However, if getCodigo returns an int and getSelectedCustomer returns an int[] then this line should be changed instead:
for(int codigo : this.getSelectedCustomer()){
Because you didn't need to use Integer in the first place.
Secondly, this line attempts to remove auxRemoColumnKeys itself so you probably mean removeAll:
this.getColumnCustomer().remove(auxRemoColumnKeys);
Lastly, your logic is generally flawed. It basically says "for each element in getColumnCustomer, if getCodigo is not equal to all of getSelectedCustomer remove it". I don't think that's what you've intended.
This is a modified loop that uses the same "add to a list and remove the list items" procedure but the logic will work:
List<ValidColumnKey> auxRemoColumnKeys = new ArrayList<ValidColumnKey>();
int[] selected = this.getSelectedCustomer();
for (ValidColumnKey column : this.getColumnCustomer()) {
int i = 0;
for ( ; i < selected.length; i++) {
/* note: if getCodigo returns an Integer change this check to
* "if (column.getCodigo().equals(selected[i])) {"
*/
if (column.getCodigo() == selected[i]) {
break;
}
}
/* this says "if the search loop did not break early" */
if (i == selected.length) {
auxRemoColumnKeys.add(column);
}
}
this.getColumnCustomer().removeAll(auxRemoColumnKeys);