Let say we have :
Map hm = new HashMap();
How to avoid putting duplicate values(Emplyees) in this HashMap?
I assume you are coding in Java, so:
if(!myMap.containsKey(myKey)){
myMap.put(myKey, myValue);
}
The good thing with HashMap is that the containsKey method takes constant time (or constant amortized time) regardless of the number of elements in your map so you can call it without bothering of the time it may take!
If you use an other language, the logic remains the same.
I think duplicate values in Map can be removed using this generic method if your userdefined object is overridden with equals and hashcode from object class
public static <K, V > Map<K,V> genericMethodtoDeleteMapduplicate(Map<K, V> pMap) {
Map<K,V> mapWithoutDup=new HashMap<>();
Set<V> totalvaluesPresent=new HashSet<>();
for (Map.Entry<K, V> a : pMap.entrySet()) {
if(totalvaluesPresent.add(a.getValue())){
mapWithoutDup.put(a.getKey(), a.getValue());
}
}
return mapWithoutDup;
}
Not sure what language you are using but in java for Hashmap their are:
boolean containsKey(Object key)
- Returns true if this map contains a mapping for the
specified key.
and
boolean containsValue(Object value)
- Returns true if this map maps one or more keys to the
specified value.
Just call which ever one makes more sense for you to check if the entry is already in the map, if it is then you know its a duplicate, otherwise put it in. I'm certain whatever language you are using will have something similar!!
Related
I am new to Groovy so I am a bit confused by the run time binding, typed and not typed attributes of the language. Personally I prefer types to be declared.
However, I have a question.
I have a small method that takes some variable from maps, input, whatever, that I know are numbers. Let's say that I don't know what the initial type was (it's somewhere deep in the code or comes from an external source), other that it was a number. Now I have a method that takes two of these arguments and I have to do a modulo operation on them. Because they might be decimal or not, I wrote a small method using the remainder of BigDecimal so to enforce the type I used the type BigDecimal on the method signature.
def callerMethod(Map map) {
...
map.each{
calledMethod(it.val1, it.val2)
...
}
...
}
def calledMethod(BigDecimal val1, BigDecimal val2) {
...
vl1.remainder(val2)
...
}
Is this correct? If the incoming argument is Integer (most of the time the primitives are boxed if I understand it correctly), will it be implicitly cast or turned into a BigDecimal?
How does this work in Groovy.
I still think that since I have the option to use types, I want to use them rather than declaring everything def. It also makes it easier to read code or see what something is if you reading already existing code
The problem in this methods are not the type of variables, is the each of your map
In a groovy Map, the each have two signatures.
One receive a Map.Entry of parameter and other receive key and value
Ex.:
Map map = [key1:'value1',key2:'value2']
map.each{ Map.Entry entryMap ->
println "The value of key: ${entryMap.key} is ${entryMap.value}"
}
The result of this each will be:
The value of key: key1 is value1
The value of key: key2 is value2
Or could be like this
Map map = [key1:'value1',key2:'value2']
map.each{ def key, def value ->
println "The value of key: ${key} is ${value}"
}
And the result of this second will be the same of the first.
If you want to pass two specific arguments to you calledMethod, pass both outside of the each like this:
def callerMethod(Map map) {
calledMethod(map.val1, map.val2)
}
I don't understand perfectly what you want.. I hope that's help you to do you code.
I want to understand how hashmap works if an object has enum, because everytime an enum provides a random number. I provide below the code.
public class Person {
private String name;
private EmpType eType;
//equals() & hashcode() implementation
}
public enum EmpType {
PERMANENT,TEMPORARY
}
If an object has enum, everytime it provides a different hashcode. If the map is serialized, can we ensure that we can retrieve the same value for a particular object ?
The hash code is varies only from one JVM execution to another. Within one program it remains constant, so hash maps will work fine. Serialization of hash maps doesn't involve storing hash code at all. It just serializes its elements (keys and values) and recalculates hash code when those are deserialized.
While studying the Groovy (2.4.4) syntax in the official documentation, I came across the special behavior concerning maps with GStrings as identifiers. As described in the documentation, GStrings are a bad idea as (hash)map identifiers, because the hashcodes of a non-evaluated GString-object differs from a regular String-object with the same representation as the evaluated GString.
Example:
def key = "id"
def m = ["${key}": "value for ${key}"]
println "id".hashCode() // prints "3355"
println "${key}".hashCode() // prints "3392", different hashcode
assert m["id"] == null // evaluates true
However, my intuitive expectation was that using the actual GString identifier to address a key in the map will in fact deliver the value - but it does not.
def key = "id"
def m = ["${key}": "value for ${key}"]
assert m["${key}"] == null // evaluates also true, not expected
That made me curious. So I had several suggestions concerning this issue and did some experiments.
(pls keep in my mind that I am new to Groovy and I was just brainstorming on the fly - continue to Suggestion #4 if you do not want to read how I tried to examine the cause of the issue)
Suggestion #1. hashcode for GString objects works/is implemented somewhat non-deterministic for whatever reason and delivers different results depending on the context or the actual object.
That turned out to be nonsense quite fast:
println "${key}".hashCode() // prints "3392"
// do sth else
println "${key}".hashCode() // still "3392"
Suggestion #2. The actual key in the map or the map item does not have the expected representation or hashcode.
I took a closer look at the item in the map, the key, and its hashcode.
println m // prints "[id:value for id]", as expected
m.each {
it -> println key.hashCode()
} // prints "3355" - hashcode of the String "id"
So the hashcode of the key inside the map is different from the GString hashcode. HA! or not. Though it is nice to know, it is actually not relevant because I still do know the actual hashcodes in the map index. I just rehashed a key that has been transformed to a string after being put into the index. So what else?
Suggestion #3. The equals-method of a GString has an unknown or non- implemented behavior.
No matter whether two hashcodes are equal, they may not represent the same object in a map. That depends on the implementation of the equals method for the class of the key-object. If the equals-method is, for instance, not implemented, two objects are not equal even if the hashcode is identical and therefore the desired map key cannot be adressed properly. So I tried:
def a = "${key}"
def b = "${key}"
assert a.equals(b) // returns true (unfortunate but expected)
So two representations of the same GString are equal by default.
I skip some others ideas I tried and continue with the last thing I tried just before I was going to write this post.
Suggestion #4. The syntax of access matters.
That was a real killer of understanding. I knew before: There are syntactically different ways two access map values. Each way has its restrictions, but I thought the results stay the same. Well, this came up:
def key = "id"
def m = ["${key}": "value for ${key}"]
assert m["id"] == null // as before
assert m["${key}"] == null // as before
assert m.get("${key}") == null // assertion fails, value returned
So if I use the get-method of a map, I get the actual value in the way I expected it to in the first place.
What is the explanation for this map access behavior concerning GStrings? (or what kind of rookie mistake is hidden here?)
Thanks for your patience.
EDIT: I am afraid that my actual question is not clearly stated, so here is the case in short and concise:
When I have a map with a GString as a key like this
def m = ["${key}": "value for ${key}"]
why does this return the value
println m.get("${key}")
but that does not
println m["${key}"]
?
You can look at this matter with a very different approach. A map is supposed to have immutable keys (at least for hashcode and equals), because the map implementation depends on this. GString is mutable, thus not really suited for map keys in general. There is also the problem of calling String#equals(GString). GString is a Groovy class, so we can influence the equals method to equal to a String just fine. But String is very different. That means calling equals on a String with a GString will always be false in the Java world, even if hashcode() would behave the same for String and GString. And now imagine a map with String keys and you ask the map for a value with a GString. It would always return null. On the other hand a map with GString keys queried with a String could return the "proper" value. This means there will always be a disconnection.
And because of this problem GString#hashCode() is not equal to String#hashCode() on purpose.
It is in no way non-deterministic, but a GString hashcode can change, if the participating objects change their toString representation:
def map = [:]
def gstring = "$map"
def hashCodeOld = gstring.hashCode()
assert hashCodeOld == gstring.hashCode()
map.foo = "bar"
assert hashCodeOld != gstring.hashCode()
Here the toString representation of map will change for Groovy and GString, thus the GString will produce a different hashcode
I have a question about using the new Objects.compare(o1, o2, Comparator) method - from my own testing of it, if both o1 and o2 are null then it returns 0, however, if one of them is null then it still throws a null pointer exception. I have found a lot of material on Objects.equals and some of the other Objects utility methods but not much at all on Objects.compare and when we are expected to use it / replace old code with it.
So here I could do this:
String s1 = "hi";
String s2 = "hi";
int x = Objects.compare(s1, s2, Comparator.naturalOrder());
System.out.println("x = " + x);
That works fine, returns 0, now this:
String s1 = null;
String s2 = null;
Also works fine and returns 0. However, this:
String s1 = "hi";
Strng s2 = null;
Throws a NullPointerException. I'm guessing the benefit of Objects.compare(o1,o2,Comparator) vs o1.compareTo(o2) is that it at least handles circumstances where both objects are null and when one of them is null it allows you to design a Comparator to handle it. I'm supposing, e.g.
int x = Objects.compare(s1, s2, Comparator.nullsFirst(Comparator.naturalOrder()));
Whereas with x.compareTo(y) there's no way to handle null unless you do so beforehand? So do the Java library developers now intend us to replace all calls to compareTo with Objects.compare, when we're concerned about nulls? e.g. would we do this in our Comparable implementations?
Side query 1: With regards to using nullsFirst if you use it then pass in a Comparator, which is chained using comparing, thenComparing, etc, does it apply to all of the inner comparators? e.g.
Comparator.nullsFirst(Comparator.comparing(Song::getTitle)
.thenComparing(Song::getArtist)
.thenComparing(Song::getDuration)
)
Would that apply nullsFirst to everything inside or do you need to use nullsFirst individually on each of them? I think from testing that it only applies to the actual Song objects being null, not for the fields of title or artist being null, i.e. if they are null then a NullPointerException is still thrown. Anyway around that?
Side query 2: final question is that because I like the Comparator.comparing syntax, I'm proposing to start to write my compareTo implementions using it - I was struggling to think how to replace this traditional approach, e.g.
public int compareTo(Song other) {
int result = this.title.compareTo(other.title);
if (result == 0) {
result = this.artist.compareTo(other.artist);
if (result == 0) {
result = Integer.compare(this.duration, other.duration);
}
}
return result;
}
then I thought I could use Objects.compare(...) as follows:
public int compareTo(Song other) {
return Objects.compare(this, other, Comparator.nullsFirst(
Comparator.comparing(Song::getTitle)
.thenComparing(Song::getArtist)
.thenComparingInt(Song::getDuration)
));
}
I thought this version was more elegant - I am assuming it is working as I think it is, e.g. by passing this and other as the first 2 arguments then the comparator, it has the same effect as the traditional compareTo approach with if statements? Whilst I can see that the benefit of Objects.compare catching two nulls would never occur as if this was null then the compareTo method call would never be reached (either by handling the exception or it being thrown). But by using nullsFirst I suppose if the argument passed in, i.e. other, was null, then this would handle this safely?
Many thanks in advance for any help.
Objects.compare is not meant to provide a null safe comparison, since there is no default behavior that could be implemented. It just implements a shortcut of not invoking the Comparator’s method when both objects are identical. In other words, it does a==b? 0: c.compare(a, b), nothing more. So not breaking when both objects are null is just a side-effect. The encapsulated code might look trivial but the other methods in this class are of a similar category. Using small utility methods a lot might still result in a notable win.
By the way, it’s not a Java 8 method at all. It exists since Java 7.
Regarding your second question, Comparator.nullsFirst(…) decorates an existing Comparator and will enforce the rule for null values before delegating to the provided comparator as it is the purpose of this comparator to shield the existing one from ever seeing null values. It doesn’t matter whether the decorated comparator is a chained one or not. As long as it is what you called the “inner comparator”, as
you must not invoke thenComparing on the result of nullsFirst as that would imply calling the next comparator when both values are null.
Comparator.nullsFirst(Comparator.comparing(a).thenComparing(b)) // perfect
Comparator.nullsFirst(Comparator.comparing(a)).thenComparing(b) // ouch
Now to your third question, implementing a compareTo method using a nullsFirst comparator is violating the interface specification:
The implementor must ensure sgn(x.compareTo(y)) == -sgn(y.compareTo(x)) for all x and y. (This implies that x.compareTo(y) must throw an exception iff y.compareTo(x) throws an exception.)
This implies that passing null as argument should always result in a NullPointerException as swapping argument and receiver would throw as well, unconditionally.
Orders including a null policy should always be provided as separate Comparators.
Note that it would also be quite inefficient as you would create a new Comparator (multiple Comparators, to be precise) for every compareTo call. Now image sorting a rather large list of these objects…
What I normally do for your final question is to first create a static comparator reference within the class:
public static final Comparator<Song> COMP_DEFAULT
= nullsFirst(comparing(Song::getTitle, nullsFirst(naturalOrder()))
.thenComparing(Song::getArtist, nullsFirst(naturalOrder()))
.thenComparingInt(Song::getDuration));
And then refer to this comparator in compareTo
public int compareTo(Song other) {
return COMP_DEFAULT.compare(this, other);
}
This way you're not recreating your comparator for each compareTo call, null safety of Song is guaranteed as is the result of a.comparetTo(b) == b.compareTo(a).
We also ensure null safety of each property by using nullsFirst(naturalOrder()) for the passed in key comparator (second argument).
As the Comparator returned is immutable it can be made public which can be handy for bundling some alternate Comparators with the class that consumers may use.
I have the following code (note the code below doesnt update the property)
private void queryResultsFilePath_Click(object sender, EventArgs e)
{
Library.SProc.Browse browser = new Browse();
browser.selectFile(QueryResultFilePath);
}
and
public class Browse
{
public void selectFile(string propertyName)
{
...
propertyName = browserWindow.FileName;
}
}
Now i realise that i need to change the second method so that it returns a string and manually assign it to the property in the first example.
What im unsure of is that i thought that when i assigned a ref type as an actual parameter of a method, a copy of its value on the stack (ie its memory address in the heap) was copied to the new location on the stack for the methods formal parameter, so they are both pointing to the same memory address on the heap. So when i changed the value of the formal parameter, it would actually change the value stored on the heap and thus the actual parameters value.
Obviously im missing something since im having to return a string and manually assign it to the property. If someone could point out what ive misunderstood id appreciate it.
Thanks.
I believe the missing piece here is: strings are immutable.
Although you pass it by reference, as soon as anything attempts to mutate the string, a new string is created leaving the old one intact.
I believe it is the only reference type that has enforced immutability.
From MSDN:
Strings are immutable--the contents of a string object cannot be
changed after the object is created, although the syntax makes it
appear as if you can do this. For example, when you write this code,
the compiler actually creates a new string object to hold the new
sequence of characters, and that new object is assigned to b. The
string "h" is then eligible for garbage collection.
Further reading:
http://social.msdn.microsoft.com/Forums/en/netfxbcl/thread/e755cbcd-4b09-4a61-b31f-e46e48d1b2eb
If you wish the method to "change" the caller's string then you can simulate this using the ref keyword:
public void SelectFile(ref string propertyName)
{
propertyName = browserWindow.FileName;
}
In this example, the parameter propertyName will be assigned to in the method, because of ref being used, this also changes the string that the caller is pointing to. Note here that immutability is still enforced. propertyName used to point to string A, but after assignment now points to string B - the old string A is now unreferenced and will be garbage collected (but importantly still exists and wasn't changed - immutable). If the ref keyword wasn't used, the caller would still point at A and the method would point at B. However, because the ref keyword was used the callers variable now points to string B.
This is the same effect as the following example:
static void Main(string[] args)
{
MyClass classRef = new MyClass("A");
PointToANewClass(ref classRef);
// classRef now points to a brand new instance containing "B".
}
public static void PointToANewClass(ref MyClass classRef)
{
classRef = new MyClass("B");
}
If you try the above without the ref keyword, classRef would still point to an object containing "A" even though the class was passed by reference.
Don't get confused between string semantics and ref semantics. And also don't get confused between passing something by reference and assignment. Stuff is technically never passed by reference, the pointer to the object on the heap is passed by value - hence ref on a reference type has the behaviour specified above. Also hence not using ref will not allow a new assignment to be "shared" between caller and method, the method has received its own copy of the pointer to the object on the heap, dereferencing the pointer has the usual effect (looking at the same underlying object), but assigning to the pointer will not affect the callers copy of the pointer.
I'm really grateful to Adam Houldsworth, because I've finally understood how the .NET framework uses reference parameters and what happens with the string.
In .NET there are two kind of data types:
value type: primitive types like int, float, bool, and so on
reference type: all the other objects, including string
In the case of reference type, the object is stored in the heap, and a variable only holds a reference pointing to this object. You can access the object's properties through the reference and modify them. When you pass one of this variables as parameter, a copy of the reference pointing to the same object is passed on to the method body. So, when you access and modify properties, you are modifyin gthe same object stored on the heap. I.e, this class is a reference object:
public class ClassOne
{
public string Desc { get; set; }
}
When you do this
ClassOne one = new { Desc = "I'm a class one!" };
there's an object on the heap pointed to by the reference one. If you do this:
one.Desc = "Changed value!";
the object on the heap has been modified. If you pass this reference as a parameter:
public void ChangeOne(ClassOne one)
{
one.Desc = "Changed value!"
}
The original object on the heap is also changed, because one helds a copy of the original reference that points to the same object on the heap.
But if you do this:
public void ChangeOne(ClassOne one)
{
one = new ClassOne { Desc ="Changed value!" };
}
The original object is unchanged. That's because one was a copy of the reference that it's now pointing to a different object.
If you pass it explicitly by reference:
public void ChangeOne(ref ClassOne one)
{
one = new ClassOne { Desc ="Changed value!" };
}
one inside this method is not a copy of the outer refernce, but the reference itself, so, the original reference now points to this new object.
strings are inmutable. This means that you cannot change a string. if you try to do so, a new string is created. So, if you do this:
string s = "HELL";
s = s + "O";
The second line creates a new instance of string, with the value "HELLO" and "HELL" is abandoned on the heap (left to be garbage collected).
So it's not possible to change it if you pass it as a parameter like this:
public void AppendO(string one)
{
one = one + "O";
}
string original = "HELL";
AppendO(original);
the original string is left as is. The code inside the function creates a new object, and assign it to one, which is a copy of original reference. But original keeps pointing to "HELL".
In the case of value types, when they are passed as parameters to a function, they are passed by value, i.e. the function receives a copy of the original value. So, any modification done to the object inside the function body won't affect the original value outside the function.
The problem is that, although string is a reference type, it looks as if it behaves like a value type (this applies to comparisons, passing parameters, and so on).
However, as explained above, it's possible to make the compiler pass a reference type by reference using the ref keyword. This also also works for strings.
You can check this code, and you'll see that the string is modified (this would also apply to an int, float or any other value type):
public static class StringTest
{
public static void AppednO(ref string toModify)
{
toModify = toModify + "O";
}
}
// test:
string hell = "HELL";
StringTest.AppendO(ref hell);
if (hell == "HELLO")
{
// here, hell is "HELLO"
}
Note that, for avoiding errors, when you define a parameter as ref, you also have to pass the parameter with this modifier.
Anyway, for this case (and similar cases) I'd recommend you to use the more natural functional syntax:
var hell = StringTest.AppendO(hell);
(Of course, in this case, the function will have this signature and corresponding implementation:
public static string AppendO(string value)
{
return value + "O";
}
If you're going to make many changes to a string, you should use the StringBuilder class, which works with "mutable strings".
How a property, of type string, is passed
Strings are immutable and therefore you are passing copies of them to methods. This means that the copy changes but the original parameter stays the same.