I have to write a piece of code for a class that counts the occurrences of characters within an input file and then sorts them by that, and I chose to do that by creating an ArrayList where each object[] has two elements, the character and the number of occurrences.
I was trying to increment the integer representing the number of occurrences and I just couldn't get that to work
My current attempt looks like this:
for(int i=0;i<=text.length();i++) {
if(freqlist.contains(text.charAt(i))) {
freqlist.indexOf(text.charAt(i))[1]=freqlist.get(freqlist.indexOf(text.charAt(i)))[1]+1;
}
}
text is just a string containing all of the input file
freqlist is declared earlier as
List<Object[]> freqlist=new ArrayList<Object[]>();
So, I was wondering how one could increment or modify an element of an array that is inside of an arraylist
In General there are 3 mistakes in your program which prevent it from working. It cannot work because the for loop has i<=text.length() and it should be i < text.length(), otherwise you will have exception. Second mistake is that you use freqlist.contains(...) where you assume both elements of object arrays are the same, or in other words the array is the equal, which is wrong assumption. Third mistake is using freqlist.indexOf(...) which relies on array equality again. I made the example working although this data structure List<Object[]> is inefficient for the task. It is best to use Map<Character,Integer>.
Here it is:
import java.util.ArrayList;
import java.util.List;
class Scratch {
public static void main(String[] args) {
String text = "abcdacd";
List<Object[]> freqlist= new ArrayList<>();
for(int i=0;i < text.length();i++) {
Object [] objects = find(freqlist, text.charAt(i));
if(objects != null) {
objects[1] = (Integer)objects[1] +1;
} else {
freqlist.add(new Object[]{text.charAt(i), 1});
}
}
for (Object[] objects : freqlist) {
System.out.println(String.format(" %s => %d", objects[0], objects[1]));
}
}
private static Object[] find(List<Object[]> freqlist, Character charAt) {
for (Object[] objects : freqlist) {
if (charAt.equals(objects[0])) {
return objects;
}
}
return null;
}
}
The way I would do this is first parse the file and convert it to an array of characters. This would then be sent to the charCounter() method which would count the number of times a letter occurs in the file.
/**
* Calculate the number of times a character is present in a character array
*
* #param myChars An array of characters from an input file, this should be parsed and formatted properly
* before sending to method
* #return A hashmap of all characters with their number of occurrences; if a
* letter is not in myChars it is not added to the HashMap
*/
public HashMap<Character, Integer> charCounter(char[] myChars) {
HashMap<Character, Integer> myCharCount = new HashMap<>();
if (myChars.length == 0) System.exit(1);
for (char c : myChars) {
if (myCharCount.containsKey(c)) {
//get the current number for the letter
int currentNum = myCharCount.get(c);
//Place the new number plus one to the HashMap
myCharCount.put(c, (currentNum + 1));
} else {
//Place the character in the HashMap with 1 occurrence
myCharCount.put(c, 1);
}
}
return myCharCount;
}
You could use some Stream magic, if you are using Java 8 for the grouping:
Map<String, Long> map = dummyString.chars() // Turn the String to an IntStream
.boxed() // Turn int to Integer to use Collectors.groupingBy
.collect(Collectors.groupingBy(
Character::toString, // Use the character as a key for the map
Collectors.counting())); // Count the occurrences
Now you could sort the result.
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));
}
How to delete the content of an array of objects. If there is other ways to delete a content of an array of objects , please do share.
import java.util.Arrays;
import java.util.Scanner;
import org.apache.commons.lang3.ArrayUtils;
public class Testing {
public static void deleteItem(ItemTracker[] listItems) {
System.out.println("Which item you want to delete? ");
for(int i=0; i < listItems.length; i++) {
if(input.equalsIgnoreCase("Quantity")) {
// Some Code
} else if(input.equalsIgnoreCase("Something"){
ArrayUtils.remove(listItems, i); // This is the part where it should delete .. but it doesnt delete.
}
break;
}
}
}
Change this
ArrayUtils.remove(listItems, i);
to
listItems = ArrayUtils.remove(listItems, i);
As you can see in the JavaDoc, the method does not change the argument listItems, rather it returns a new array with the remaining elements.
Edit
You also need to change your deletion method to
public static ItemTracker[] deleteItem(ItemTracker[] listItems) {
//..
}
So you could return the new array with the remaining elements.
Store the resulting array.
It won't change the original array object.
listItems = ArrayUtils.remove(listItems, i);
Edit: But for using this method you need the change to return type of your method
public static ItemTracker[] deleteItem(ItemTracker[] listItems){
System.out.println("Which item you want to delete? ");
for(int i=0; i < listItems.length; i++) {
if(input.equalsIgnoreCase("Quantity")) {
// Some Code
} else if(input.equalsIgnoreCase("Something"){
listItems = ArrayUtils.remove(listItems, i); // This is the part where it should delete .. but it doesnt delete.
}
break;
}
return listItems;
}
In your case usage of ArrayUtils is incorrect and redundant. You can delete element in next way:
// ...
listItems[i] = null;
// array will looks like [o1, o2, null, o3, o4, ...]
// ...
There is no other way without changing method's return type
Without additional libraries, with temporary list:
Element arrayToRemoveFrom[];
Element toRemove; // should be known already
ArrayList<Element> tmpList = new ArrayList<Element>(Arrays.asList(arrayToRemoveFrom));
tmpList.remove(toRemove);
// any other code processing and removing elements
arrayToRemoveFrom = tmpList.toArray(new Arrays[tmlList.size()]);
ArrayUtils.remove
This method returns a new array with the same elements of the input
array except the element on the specified position. The component type
of the returned array is always the same as that of the input array.
So,you should use it like this
listItems = ArrayUtils.remove(listItems, i);
NOTE
Here we have assign the returned array to current listItem.
As this method does not change the actual array but returns the changed array after removal same as #replace method works for String.
YES. I agree with zvdh I have missed the purpose of your method
because I was more concentrated on removal of element.Sorry for that!!
as this will not actually change the listItem and you need to return the new array which contains the change.
class Arrays
{
public static void main(String[] args)
{
double[] numbers = {6.0, 4.4, 1.9, 2.9, 3.4, 3.5};
java.util.Arrays.sort(numbers);
System.out.print("Ascending order= ");
for (int i = 0; i < numbers.length; i++)
System.out.print(numbers[i] + " ");
System.out.println();
System.out.print("Decending order= ");
for (int i = numbers.length -1; i >= 0; i--)
System.out.print(numbers[i] + " ");
}
}
This solution only displays in reverse order, but it can be changed to reorder the array using the same loop.