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I am using Multimap<String, List<String>> in one of my API's. Now to get all the values means list of list I used the .values() method of the multimap. But this method is returning me Collection<List<String>>. Now to play on the index on this collection I want to convert it into List<List<String>> or ArrayList<List<String>>.
How to cast or convert without building new arrayList and explicit add list values from collections to that arraylist.
You can use the appropriate constructor:
List<List<String>> yourList = new ArrayList<>(yourCollection);
The order of the elements in the list is the order of the iterator of the collection.
If you don't want to build a new collection (and don't forget, that may not be costly since you'll create the new collection but not clone the actual members), why not just do
int i = 0;
for (List<String> list : collection) {
// whatever
i++;
}
Not hugely nice, I appreciate.
Or use google collection API
List<List<String>> = Lists.newArrayList(myMap.values());
2nd question, which is continue of first.
I have got two Lists of strings. There is an List of strings (asu) - M1, M2, M3 ... As well as an List of string (rzs) - M1, M2, M3 and all possible combinations thereof. The need for each element (asu) (for example M1) to find an element in (rzs) (M1, M1M2, ..), which contains (e.g. M1). Example: took M1 from (asu) and will start search for duplicate(contain) in (rzs). We found M1M2 in (rzs), it contains M1. After that we should delete both elements from lists. Great thanks to No Idea For Name helped for modification this code. But the program always fails because AbstractList.remove error. Please help to implementation logic and tuning code!
Imports..........
public class work{
List<string> asu = Arrays.asList("M1","M1","M1","M3","M4","M5","M1","M1","M1","M4","M5","M5");
List<string> rzs = Arrays.asList("M1","M2","M3","M4","M5",
"M1M2","M1M3","M1M4","M1M5","M2M3","M2M4","M2M5","M3M4","M3M5","M4M5"
,"M1M2M3","M1M2M4","M1M2M5","M1M3M4","M1M3M4","M1M4M5","M2M4","M2M5");
public static void main(String[] args) {
work bebebe = new work();
bebebe.mywork();
}
List<string> tmp1 = new ArrayList<string>();
List<string> tmp2 = new ArrayList<string>();
System.out.println(Arrays.deepToString(rzs));
System.out.println(Arrays.deepToString(asu));
for (string curr : asu){
for (string currRzs : rzs){
System.out.println("New iteration ");
if (currRzs.contains(curr)) {
System.out.println("Element ("+curr+") in ASU =
element ("+currRzs+") in RZS");
if(tmp1.contains(curr) == false)
tmp1.add(curr);
if(tmp2.contains(currRzs) == false)
tmp2.add(currRzs);
}
}
}
for (string curr : tmp1){
asu.remove(curr);
}
for (string currRzs : tmp2){
rzs.remove(currRzs);
}
You should try to make use of removeAll() or retainAll() methods of Collection.
For example:
List<String> aList = new ArrayList<String>();
aList.add("a");
aList.add("b");
aList.add("c");
aList.add("d");
aList.add("e");
List<String> bList = new ArrayList<String>();
bList.add("b");
bList.add("e");
bList.add("d");
aList.removeAll(bList);
will give you the "a" and "c" elements left in aList
While if you try to make use of retainAll() method:
aList.retainAll(bList);
will give you "b", "d" and "e" elements left in aList;
retainAll() is used to remove all the elements of the invoking collection which are not part of the given collection.
removeAll() is used to remove all the elements of a collection from another collection.
So, it all depends on your use-case.
EDIT
If in any case you want to remove some elements from these collections while iterating conditionally then you should first obtain the Iterator<Type> then call the remove() method over it.
Like:
while(iterator.hasNext()){
String str = iterator.next();
if(str.equals('test')){
iterator.remove();
}
}
Don't remove items from list using foreach loop. Use classic for and iterate over elements, and when removing item, decrease iterator.
To safely remove elements while iterating use Iterator.remove method:
The behavior of an iterator is unspecified if the underlying
collection is modified while the iteration is in progress in any way
other than by calling this method.
Iterator<String> i = tmp1.iterator();
while (i.hasNext()) {
i.next(); // must be called before remove
i.remove();
}
Also it is easier to remove all collection from another by simply calling:
asu.removeAll(tmp1);
instead of List you can use Set, which will remove automatically the duplicate elements...
You can use removeAll() method to remove collection of elements from the list instead of removing one by one.
use
asu.removeAll(tmp1);
instead of
for (string curr : tmp1)
{
asu.remove(curr);
}
and use
rzs.removeAll(tmp2);
instead of
for (string currRzs : tmp2)
{
rzs.remove(currRzs);
}
update
I trace out your problem.The problem lies in Arrays.asList() method.
According to Arrays#asList
asList() returns "a fixed-size list backed by the specified array". If you want to resize the array, you have to create a new one and copy the old data. Then the list won't be backed by the same array instance.
So create a duplicate ArrayList for the lists.Like this
List<string> asuDuplicat = new ArrayList<string>(asu);
List<string> rzsDuplicat = new ArrayList<string>(rzs);
use asuDuplicat,rzsDuplicat.
asuDuplicat.removeAll(tmp1);
rzsDuplicat.removeAll(tmp2);
I can not initialize a List as in the following code:
List<String> supplierNames = new List<String>();
supplierNames.add("sup1");
supplierNames.add("sup2");
supplierNames.add("sup3");
System.out.println(supplierNames.get(1));
I face the following error:
Cannot instantiate the type List<String>
How can I instantiate List<String>?
If you check the API for List you'll notice it says:
Interface List<E>
Being an interface means it cannot be instantiated (no new List() is possible).
If you check that link, you'll find some classes that implement List:
All Known Implementing Classes:
AbstractList, AbstractSequentialList, ArrayList, AttributeList, CopyOnWriteArrayList, LinkedList, RoleList, RoleUnresolvedList, Stack, Vector
Some of those can be instantiated (the ones that are not defined as abstract class). Use their links to know more about them, I.E: to know which fits better your needs.
The 3 most commonly used ones probably are:
List<String> supplierNames1 = new ArrayList<String>();
List<String> supplierNames2 = new LinkedList<String>();
List<String> supplierNames3 = new Vector<String>();
Bonus:
You can also instantiate it with values, in an easier way, using the Arrays class, as follows:
List<String> supplierNames = Arrays.asList("sup1", "sup2", "sup3");
System.out.println(supplierNames.get(1));
But note you are not allowed to add more elements to that list, as it's fixed-size.
Can't instantiate an interface but there are few implementations:
JDK2
List<String> list = Arrays.asList("one", "two", "three");
JDK7
//diamond operator
List<String> list = new ArrayList<>();
list.add("one");
list.add("two");
list.add("three");
JDK8
List<String> list = Stream.of("one", "two", "three").collect(Collectors.toList());
JDK9
// creates immutable lists, so you can't modify such list
List<String> immutableList = List.of("one", "two", "three");
// if we want mutable list we can copy content of immutable list
// to mutable one for instance via copy-constructor (which creates shallow copy)
List<String> mutableList = new ArrayList<>(List.of("one", "two", "three"));
Plus there are lots of other ways supplied by other libraries like Guava.
List<String> list = Lists.newArrayList("one", "two", "three");
List is an Interface, you cannot instantiate an Interface, because interface is a convention, what methods should have your classes. In order to instantiate, you need some realizations(implementations) of that interface. Try the below code with very popular implementations of List interface:
List<String> supplierNames = new ArrayList<String>();
or
List<String> supplierNames = new LinkedList<String>();
You will need to use ArrayList<String> or such.
List<String> is an interface.
Use this:
import java.util.ArrayList;
...
List<String> supplierNames = new ArrayList<String>();
List is an interface, and you can not initialize an interface. Instantiate an implementing class instead.
Like:
List<String> abc = new ArrayList<String>();
List<String> xyz = new LinkedList<String>();
In most cases you want simple ArrayList - an implementation of List
Before JDK version 7
List<String> list = new ArrayList<String>();
JDK 7 and later you can use the diamond operator
List<String> list = new ArrayList<>();
Further informations are written here Oracle documentation - Collections
List is just an interface, a definition of some generic list. You need to provide an implementation of this list interface. Two most common are:
ArrayList - a list implemented over an array
List<String> supplierNames = new ArrayList<String>();
LinkedList - a list implemented like an interconnected chain of elements
List<String> supplierNames = new LinkedList<String>();
Depending on what kind of List you want to use, something like
List<String> supplierNames = new ArrayList<String>();
should get you going.
List is the interface, ArrayList is one implementation of the List interface. More implementations that may better suit your needs can be found by reading the JavaDocs of the List interface.
If you just want to create an immutable List<T> with only one object in it, you can use this API:
List<String> oneObjectList = Collections.singletonList("theOnlyObject”);
More info: docs
List is an Interface . You cant use List to initialize it.
List<String> supplierNames = new ArrayList<String>();
These are the some of List impelemented classes,
ArrayList, LinkedList, Vector
You could use any of this as per your requirement. These each classes have its own features.
Just in case, any one still lingering around this question. Because, i see one or two new users again asking the same question and everyone telling then , No you can't do that, Dear Prudence, Apart from all the answers given here, I would like to provide additional Information -
Yes you can actually do, List list = new List();
But at the cost of writing implementations of all the methods of Interfaces.
The notion is not simply List list = new List(); but
List<Integer> list = new List<Integer>(){
#Override
public int size() {
// TODO Auto-generated method stub
return 0;
}
#Override
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
#Override
public boolean contains(Object o) {
// TODO Auto-generated method stub
return false;
}
..... and So on (Cant write all methods.)
This is an example of Anonymous class. Its correct when someone states , No you cant instantiate an interface, and that's right. But you can never say , You CANT write List list = new List(); but, evidently you can do that and that's a hard statement to make that You can't do.
Instead of :
List<String> supplierNames = new List<String>();
Write this if you are using latest JDK:
List<String> supplierNames = new ArrayList<>();
It's the correct way of initializing a List.
We created soyuz-to to simplify 1 problem: how to convert X to Y (e.g. String to Integer). Constructing of an object is also kind of conversion so it has a simple function to construct Map, List, Set:
import io.thedocs.soyuz.to;
List<String> names = to.list("John", "Fedor");
Please check it - it has a lot of other useful features
I have this code:
public static String SelectRandomFromTemplate(String template,int count) {
String[] split = template.split("|");
List<String> list=Arrays.asList(split);
Random r = new Random();
while( list.size() > count ) {
list.remove(r.nextInt(list.size()));
}
return StringUtils.join(list, ", ");
}
I get this:
06-03 15:05:29.614: ERROR/AndroidRuntime(7737): java.lang.UnsupportedOperationException
06-03 15:05:29.614: ERROR/AndroidRuntime(7737): at java.util.AbstractList.remove(AbstractList.java:645)
How would be this the correct way? Java.15
Quite a few problems with your code:
On Arrays.asList returning a fixed-size list
From the API:
Arrays.asList: Returns a fixed-size list backed by the specified array.
You can't add to it; you can't remove from it. You can't structurally modify the List.
Fix
Create a LinkedList, which supports faster remove.
List<String> list = new LinkedList<String>(Arrays.asList(split));
On split taking regex
From the API:
String.split(String regex): Splits this string around matches of the given regular expression.
| is a regex metacharacter; if you want to split on a literal |, you must escape it to \|, which as a Java string literal is "\\|".
Fix:
template.split("\\|")
On better algorithm
Instead of calling remove one at a time with random indices, it's better to generate enough random numbers in the range, and then traversing the List once with a listIterator(), calling remove() at appropriate indices. There are questions on stackoverflow on how to generate random but distinct numbers in a given range.
With this, your algorithm would be O(N).
This one has burned me many times. Arrays.asList creates an unmodifiable list.
From the Javadoc: Returns a fixed-size list backed by the specified array.
Create a new list with the same content:
newList.addAll(Arrays.asList(newArray));
This will create a little extra garbage, but you will be able to mutate it.
Probably because you're working with unmodifiable wrapper.
Change this line:
List<String> list = Arrays.asList(split);
to this line:
List<String> list = new LinkedList<>(Arrays.asList(split));
The list returned by Arrays.asList() might be immutable. Could you try
List<String> list = new ArrayList<>(Arrays.asList(split));
I think that replacing:
List<String> list = Arrays.asList(split);
with
List<String> list = new ArrayList<String>(Arrays.asList(split));
resolves the problem.
Just read the JavaDoc for the asList method:
Returns a {#code List} of the objects
in the specified array. The size of
the {#code List} cannot be modified,
i.e. adding and removing are
unsupported, but the elements can be
set. Setting an element modifies the
underlying array.
This is from Java 6 but it looks like it is the same for the android java.
EDIT
The type of the resulting list is Arrays.ArrayList, which is a private class inside Arrays.class. Practically speaking, it is nothing but a List-view on the array that you've passed with Arrays.asList. With a consequence: if you change the array, the list is changed too. And because an array is not resizeable, remove and add operation must be unsupported.
The issue is you're creating a List using Arrays.asList() method with fixed Length
meaning that
Since the returned List is a fixed-size List, we can’t add/remove elements.
See the below block of code that I am using
This iteration will give an Exception Since it is an iteration list Created by asList() so remove and add are not possible, it is a fixed array
List<String> words = Arrays.asList("pen", "pencil", "sky", "blue", "sky", "dog");
for (String word : words) {
if ("sky".equals(word)) {
words.remove(word);
}
}
This will work fine since we are taking a new ArrayList we can perform modifications while iterating
List<String> words1 = new ArrayList<String>(Arrays.asList("pen", "pencil", "sky", "blue", "sky", "dog"));
for (String word : words) {
if ("sky".equals(word)) {
words.remove(word);
}
}
Arrays.asList() returns a list that doesn't allow operations affecting its size (note that this is not the same as "unmodifiable").
You could do new ArrayList<String>(Arrays.asList(split)); to create a real copy, but seeing what you are trying to do, here is an additional suggestion (you have a O(n^2) algorithm right below that).
You want to remove list.size() - count (lets call this k) random elements from the list. Just pick as many random elements and swap them to the end k positions of the list, then delete that whole range (e.g. using subList() and clear() on that). That would turn it to a lean and mean O(n) algorithm (O(k) is more precise).
Update: As noted below, this algorithm only makes sense if the elements are unordered, e.g. if the List represents a Bag. If, on the other hand, the List has a meaningful order, this algorithm would not preserve it (polygenelubricants' algorithm instead would).
Update 2: So in retrospect, a better (linear, maintaining order, but with O(n) random numbers) algorithm would be something like this:
LinkedList<String> elements = ...; //to avoid the slow ArrayList.remove()
int k = elements.size() - count; //elements to select/delete
int remaining = elements.size(); //elements remaining to be iterated
for (Iterator i = elements.iterator(); k > 0 && i.hasNext(); remaining--) {
i.next();
if (random.nextInt(remaining) < k) {
//or (random.nextDouble() < (double)k/remaining)
i.remove();
k--;
}
}
This UnsupportedOperationException comes when you try to perform some operation on collection where its not allowed and in your case, When you call Arrays.asList it does not return a java.util.ArrayList. It returns a java.util.Arrays$ArrayList which is an immutable list. You cannot add to it and you cannot remove from it.
I've got another solution for that problem:
List<String> list = Arrays.asList(split);
List<String> newList = new ArrayList<>(list);
work on newList ;)
Replace
List<String> list=Arrays.asList(split);
to
List<String> list = New ArrayList<>();
list.addAll(Arrays.asList(split));
or
List<String> list = new ArrayList<>(Arrays.asList(split));
or
List<String> list = new ArrayList<String>(Arrays.asList(split));
or (Better for Remove elements)
List<String> list = new LinkedList<>(Arrays.asList(split));
Yes, on Arrays.asList, returning a fixed-size list.
Other than using a linked list, simply use addAll method list.
Example:
String idList = "123,222,333,444";
List<String> parentRecepeIdList = new ArrayList<String>();
parentRecepeIdList.addAll(Arrays.asList(idList.split(",")));
parentRecepeIdList.add("555");
You can't remove, nor can you add to a fixed-size-list of Arrays.
But you can create your sublist from that list.
list = list.subList(0, list.size() - (list.size() - count));
public static String SelectRandomFromTemplate(String template, int count) {
String[] split = template.split("\\|");
List<String> list = Arrays.asList(split);
Random r = new Random();
while( list.size() > count ) {
list = list.subList(0, list.size() - (list.size() - count));
}
return StringUtils.join(list, ", ");
}
*Other way is
ArrayList<String> al = new ArrayList<String>(Arrays.asList(template));
this will create ArrayList which is not fixed size like Arrays.asList
Arrays.asList() uses fixed size array internally.
You can't dynamically add or remove from thisArrays.asList()
Use this
Arraylist<String> narraylist=new ArrayList(Arrays.asList());
In narraylist you can easily add or remove items.
Arraylist narraylist=Arrays.asList(); // Returns immutable arraylist
To make it mutable solution would be:
Arraylist narraylist=new ArrayList(Arrays.asList());
Following is snippet of code from Arrays
public static <T> List<T> asList(T... a) {
return new ArrayList<>(a);
}
/**
* #serial include
*/
private static class ArrayList<E> extends AbstractList<E>
implements RandomAccess, java.io.Serializable
{
private static final long serialVersionUID = -2764017481108945198L;
private final E[] a;
so what happens is that when asList method is called then it returns list of its own private static class version which does not override add funcion from AbstractList to store element in array. So by default add method in abstract list throws exception.
So it is not regular array list.
Creating a new list and populating valid values in new list worked for me.
Code throwing error -
List<String> list = new ArrayList<>();
for (String s: list) {
if(s is null or blank) {
list.remove(s);
}
}
desiredObject.setValue(list);
After fix -
List<String> list = new ArrayList<>();
List<String> newList= new ArrayList<>();
for (String s: list) {
if(s is null or blank) {
continue;
}
newList.add(s);
}
desiredObject.setValue(newList);
What's the most efficient way to lower case every element of a List or Set?
My idea for a List:
final List<String> strings = new ArrayList<String>();
strings.add("HELLO");
strings.add("WORLD");
for(int i=0,l=strings.size();i<l;++i)
{
strings.add(strings.remove(0).toLowerCase());
}
Is there a better, faster way? How would this example look like for a Set? As there is currently no method for applying an operation to each element of a Set (or List) can it be done without creating an additional temporary Set?
Something like this would be nice:
Set<String> strings = new HashSet<String>();
strings.apply(
function (element)
{ this.replace(element, element.toLowerCase();) }
);
Thanks,
Yet another solution, but with Java 8 and above:
List<String> result = strings.stream()
.map(String::toLowerCase)
.collect(Collectors.toList());
This seems like a fairly clean solution for lists. It should allow for the particular List implementation being used to provide an implementation that is optimal for both the traversal of the list--in linear time--and the replacing of the string--in constant time.
public static void replace(List<String> strings)
{
ListIterator<String> iterator = strings.listIterator();
while (iterator.hasNext())
{
iterator.set(iterator.next().toLowerCase());
}
}
This is the best that I can come up with for sets. As others have said, the operation cannot be performed in-place in the set for a number of reasons. The lower-case string may need to be placed in a different location in the set than the string it is replacing. Moreover, the lower-case string may not be added to the set at all if it is identical to another lower-case string that has already been added (e.g., "HELLO" and "Hello" will both yield "hello", which will only be added to the set once).
public static void replace(Set<String> strings)
{
String[] stringsArray = strings.toArray(new String[0]);
for (int i=0; i<stringsArray.length; ++i)
{
stringsArray[i] = stringsArray[i].toLowerCase();
}
strings.clear();
strings.addAll(Arrays.asList(stringsArray));
}
You can do this with Google Collections:
Collection<String> lowerCaseStrings = Collections2.transform(strings,
new Function<String, String>() {
public String apply(String str) {
return str.toLowerCase();
}
}
);
If you are fine with changing the input list here is one more way to achieve it.
strings.replaceAll(String::toLowerCase)
Well, there is no real elegant solution due to two facts:
Strings in Java are immutable
Java gives you no real nice map(f, list) function as you have in functional languages.
Asymptotically speaking, you can't get a better run time than your current method. You will have to create a new string using toLowerCase() and you will need to iterate by yourself over the list and generate each new lower-case string, replacing it with the existing one.
Try CollectionUtils#transform in Commons Collections for an in-place solution, or Collections2#transform in Guava if you need a live view.
This is probably faster:
for(int i=0,l=strings.size();i<l;++i)
{
strings.set(i, strings.get(i).toLowerCase());
}
I don't believe it is possible to do the manipulation in place (without creating another Collection) if you change strings to be a Set. This is because you can only iterate over the Set using an iterator or a for each loop, and cannot insert new objects whilst doing so (it throws an exception)
Referring to the ListIterator method in the accepted (Matthew T. Staebler's) solution. How is using the ListIterator better than the method here?
public static Set<String> replace(List<String> strings) {
Set<String> set = new HashSet<>();
for (String s: strings)
set.add(s.toLowerCase());
return set;
}
I was looking for similar stuff, but was stuck because my ArrayList object was not declared as GENERIC and it was available as raw List type object from somewhere. I was just getting an ArrayList object "_products". So, what I did is mentioned below and it worked for me perfectly ::
List<String> dbProducts = _products;
for(int i = 0; i<dbProducts.size(); i++) {
dbProducts.add(dbProducts.get(i).toLowerCase());
}
That is, I first took my available _products and made a GENERIC list object (As I were getting only strings in same) then I applied the toLowerCase() method on list elements which was not working previously because of non-generic ArrayList object.
And the method toLowerCase() we are using here is of String class.
String java.lang.String.toLowerCase()
not of ArrayList or Object class.
Please correct if m wrong. Newbie in JAVA seeks guidance. :)
Using JAVA 8 parallel stream it becomes faster
List<String> output= new ArrayList<>();
List<String> input= new ArrayList<>();
input.add("A");
input.add("B");
input.add("C");
input.add("D");
input.stream().parallel().map((item) -> item.toLowerCase())
.collect(Collectors.toCollection(() -> output));