For a given list of strings and a key (a passphrase), is there a quick algorithm to get a permutation using the key, and the way back using the key again?
That means
list = {"abc", "def", "ghi", "jkl"})
key = "passphrase"
permute(Key, list) -> {"def","jkl", "abc", "ghi"}
(or similiar, just not the original), and
resolve(Key, {"def","jkl", "abc", "ghi"}) -> list
The list is not longer than 50 entries.
A java example would be preferred.
I have two lists of Strings and am removing duplicates like this:
List<String> list1 = Arrays.asList("1", "2", "3", "4");
List<String> list2 = Arrays.asList("1", "4", "5", "6");
List<String> duplicates = list1.stream().filter(s -> list2.contains(s)).collect(Collectors.toList());
list1.removeAll(duplicates);
list2.removeAll(duplicates);
So the result is:
list1 = 2, 3
list2 = 5, 6
Is there a better way to accomplish this? i.e. with fewer statements.
You can use removeAll which is defined in Collection interface.
boolean removeAll(Collection<?> c)
Removes all of this collection's elements that are also contained in
the specified collection (optional operation). After this call
returns, this collection will contain no elements in common with the
specified collection.
// init
List<String> sourceList1 = Arrays.asList("1", "2", "3", "4");
List<String> sourceList2 = Arrays.asList("1", "4", "5", "6");
// you need to create duplicate collection, because removeAll modify collection
List<String> resultList1 = new ArrayList(sourceList1);
List<String> resultList2 = new ArrayList(sourceList2);
//remove duplicates from collections
resultList1.removeAll(sourceList2); // second from first
resultList2.removeAll(sourceList1); // first from second
One of the possibilities worth considering is to create Set<String> and add these lists to it. Set allows adding only unique values to itself, it prevents adding duplicates.
The first way to use Set: Create a Set containing an intersection of both lists. Adding to new, getting rid of duplicates lists is taking place only if you checked that every object of the source is not present in previously created Set of duplicates.
Second way (only if your lists doesn't care about holding duplicates itself - for example in the first you have two times the same value existing): Create a Set for the first and for the second list, and add these lists to them and after that check for duplicates.
As I mentioned in comments I could misunderstood question and looked for "another", not for "more efficient" way of achieving what you're asking for, but maybe it could actually be helpful nonetheless.
I want a dictionary of values. The keys are all strings. Each key corresponds with some sort of list of strings. How do I make a list of strings for each key and update that accordingly? I'll explain:
I have a loop that is reading lines of a word list. The words are then converted into a string code and set as keys in the dictionary. Here is an example of the string code/word relationship.
123, [the]
456, [dog]
328, [bug]
...
However, my program keeps looping through the word list and eventually will run into a word with the same code as "the", but maybe a different word, lets say "cat". So I want the list to look like:
123, [the, cat]
456, [dog]
...
How do I get it to make an arraylist for every key that I can then add to on the fly when needed? My end goal is to be able to print out the list of words in that list for a called code (.get())
You can make a HashMap. In your case
HashMap<Integer, ArrayList<String>> works fine.
Like it has already been said, a MultiMap seems to be what you need. Guava that was already suggested and it's a good option. There is also and implementation from commons-collections you can use.
From commons-collections documentation:
MultiValuedMap<K, String> map = new MultiValuedHashMap<K, String>();
map.put(key, "A");
map.put(key, "B");
map.put(key, "C");
Collection<String> coll = map.get(key); // returns ["A", "B", "C"]
You can always implement your own MultiMap if you don't want to use an external library. Use a HashMap<String,List<String>> to store your values and wrap it with your own put, get and whatever other methods you see fit.
It sounds like you want a Multimap from the Guava library.
You can also go the route of using a Map<Integer, List<String>>, but then you will need to manually handle the case where the list is null (probably just allocate a new list in that case).
You can use a HashMap that links each id to a list of strings:
Map<String, List<String>> dictionary = new HashMap<String,List<String>>();
Now let's say you read two Strings: id and word . To add them to your dictionary, you can first verify if your id has already been read (using the containsKey() method)- in which case you just append the word to the list corresponding to that id - or, if this is not the case, you create a new list with this word:
//If the list already exists...
if(dictionary.containsKey(id)) {
List<String> appended = dictionary.get(id);
appended.add(word); //We add a new word to our current list
dictionary.remove(id); //We update the map by first removing the old list
dictionary.put(id, appended); //and then appending the new one
} else {
//Otherwise we create a new list for that id
List<String> newList = new ArrayList<String>();
newList.add(word);
dictionary.put(id, newList);
}
Then whenever you want to retrieve your list of strings for a certain id you can simply use dictionary.get(id);
You can find more information on HashMaps on the Java documentation
I assumed you didn't want repeats in your list so I used Set instead.
Map<String,Set<String>> mapToSet = new HashMap<>();
List<String []>keyvals = Arrays.asList(new String[][]{{"123","the"},{"123","cat"}});
for(String kv[] : keyvals) {
Set<String> s = mapToSet.get(kv[0]);
if(null == s) {
s = new HashSet<String>();
}
s.add(kv[1]);
mapToSet.put(kv[0], s);
}
I have multiple records received as string through a query. I want to store records having the same "long" key in a collection dynamically as I parse them in a loop. Like, insert key and value and if the key exists, it adds to the values and if not, a new key is created. What would be the most efficient way of doing this? I can do it using multiple arrays but I would prefer a cleaner way.
I cannot use a HashMap as I have to first store the records in an Array or ArrayList and then insert it which defeats the purpose as I have to group the lists by key first anyway. The no. of records will not more than 50 at a time.
E.g data:
for(i = 0; i < numRecords; i++ ) {
Data: 1 "A", 2 "B", 1 "C", 3 "D", 1 "E"
}
I want to have a collection where inside the loop I can just add: 1 "A" and so on..
I think Map<Long,List<String>> may help you.
Map<Long,List<String>> map = new HashMap<>();
...
if(map.get(key)!=null){
List<String> list = map.get(key);
list.add(value);
}else{
List<String> list = new ArrayList<>();
list.add(value);
map.put(key,list);
}
In Java, I have an ArrayList of Strings like:
[,Hi, ,How,are,you]
I want to remove the null and empty elements, how to change it so it is like this:
[Hi,How,are,you]
List<String> list = new ArrayList<String>(Arrays.asList("", "Hi", null, "How"));
System.out.println(list);
list.removeAll(Arrays.asList("", null));
System.out.println(list);
Output:
[, Hi, null, How]
[Hi, How]
Its a very late answer, but you can also use the Collections.singleton:
List<String> list = new ArrayList<String>(Arrays.asList("", "Hi", null, "How"));
// in one line
list.removeAll(Arrays.asList("", null))
// separately
list.removeAll(Collections.singleton(null));
list.removeAll(Collections.singleton(""));
Another way to do this now that we have Java 8 lambda expressions.
arrayList.removeIf(item -> item == null || "".equals(item));
If you are using Java 8 then try this using lambda expression and org.apache.commons.lang.StringUtils, that will also clear null and blank values from array input
public static String[] cleanArray(String[] array) {
return Arrays.stream(array).filter(x -> !StringUtils.isBlank(x)).toArray(String[]::new);
}
ref - https://stackoverflow.com/a/41935895/9696526
If you were asking how to remove the empty strings, you can do it like this (where l is an ArrayList<String>) - this removes all null references and strings of length 0:
Iterator<String> i = l.iterator();
while (i.hasNext())
{
String s = i.next();
if (s == null || s.isEmpty())
{
i.remove();
}
}
Don't confuse an ArrayList with arrays, an ArrayList is a dynamic data-structure that resizes according to it's contents. If you use the code above, you don't have to do anything to get the result as you've described it -if your ArrayList was ["","Hi","","How","are","you"], after removing as above, it's going to be exactly what you need - ["Hi","How","are","you"].
However, if you must have a 'sanitized' copy of the original list (while leaving the original as it is) and by 'store it back' you meant 'make a copy', then krmby's code in the other answer will serve you just fine.
Going to drop this lil nugget in here:
Stream.of("", "Hi", null, "How", "are", "you")
.filter(t -> !Strings.isNullOrEmpty(t))
.collect(ImmutableList.toImmutableList());
I wish with all of my heart that Java had a filterNot.
There are a few approaches that you could use:
Iterate over the list, calling Iterator.remove() for the list elements you want to remove. This is the simplest.
Repeatedly call List.remove(Object). This is simple too, but performs worst of all ... because you repeatedly scan the entire list. (However, this might be an option for a mutable list whose iterator didn't support remove ... for some reason.)
Create a new list, iterate over the old list, adding elements that you want to retain to a new list.
If you can't return the new list, as 3. above and then clear the old list and use addAll to add the elements of the new list back to it.
Which of these is fastest depends on the class of the original list, its size, and the number of elements that need to be removed. Here are some of the factors:
For an ArrayList, each individual remove operation is O(N), where N is the list size. It is expensive to remove multiple elements from a large ArrayList using the Iterator.remove() method (or the ArrayList.remove(element) method).
By contrast, the Iterator.remove method for a LinkedList is O(1).
For an ArrayList, creating and copying a list is O(N) and relatively cheap, especially if you can ensure that the destination list's capacity is large enough (but not too large).
By contrast, creating and copying to a LinkedList is also O(N), but considerably more expensive.
All of this adds up to a fairly complicated decision tree. If the lists are small (say 10 or less elements) you can probably get away with any of the approaches above. If the lists could be large, you need to weigh up all of the issues in the list of the expected list size and expected number of removals. (Otherwise you might end up with quadratic performance.)
This code compiles and runs smoothly.
It uses no iterator so more readable.
list is your collection.
result is filtered form (no null no empty).
public static void listRemove() {
List<String> list = Arrays.asList("", "Hi", "", "How", "are", "you");
List<String> result = new ArrayList<String>();
for (String str : list) {
if (str != null && !str.isEmpty()) {
result.add(str);
}
}
System.out.println(result);
}
If you get UnsupportedOperationException from using one of ther answer above and your List is created from Arrays.asList(), it is because you can't edit such List.
To fix, wrap the Arrays.asList() inside new LinkedList<String>():
List<String> list = new LinkedList<String>(Arrays.asList(split));
Source is from this answer.
Regarding the comment of Andrew Mairose - Although a fine solution, I would just like to add that this solution will not work on fixed size lists.
You could attempt doing like so:
Arrays.asList(new String[]{"a", "b", null, "c", " "})
.removeIf(item -> item == null || "".equals(item));
But you'll encounter an UnsupportedOperationException at java.util.AbstractList.remove(since asList returns a non-resizable List).
A different solution might be this:
List<String> collect =
Stream.of(new String[]{"a", "b", "c", null, ""})
.filter(item -> item != null && !"".equals(item))
.collect(Collectors.toList());
Which will produce a nice list of strings :-)
lukastymo's answer seems the best one.
But it may be worth mentioning this approach as well for it's extensibility:
List<String> list = new ArrayList<String>(Arrays.asList("", "Hi", null, "How"));
list = list.stream()
.filter(item -> item != null && !item.isEmpty())
.collect(Collectors.toList());
System.out.println(list);
What I mean by that (extensibility) is you could then add additional filters, such as:
.filter(item -> !item.startsWith("a"))
... although of course that's not specifically relevant to the question.
List<String> list = new ArrayList<String>(Arrays.asList("", "Hi", "", "How"));
Stream<String> stream = list .stream();
Predicate<String> empty = empt->(empt.equals(""));
Predicate<String> emptyRev = empty.negate();
list= stream.filter(emptyRev).collect(Collectors.toList());
OR
list = list .stream().filter(empty->(!empty.equals(""))).collect(Collectors.toList());
private List cleanInputs(String[] inputArray) {
List<String> result = new ArrayList<String>(inputArray.length);
for (String input : inputArray) {
if (input != null) {
String str = input.trim();
if (!str.isEmpty()) {
result.add(str);
}
}
}
return result;
}