Sort by value (inverse order), take the first N digits
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public class RelationSpanSorter {
private RelationSpanSorter() {
}
// test
public static void main(String[] args) {
Map map = new HashMap();
(1, 4);
(5, 1);
(2, 3);
(4, 2);
(12, 12);
(21, 21);
(11, 11);
(41, 41);
(31, 31);
(15, 15);
(62, 62);
(14, 14);
(111, 111);
(523, 523);
(92, 92);
(40, 40);
Map sorted = sortByValue(map, 8);
(sorted);
}
@SuppressWarnings("unchecked")
public static Map sortByValue(Map map, int topN) {
List list = new LinkedList(());
(list, new Comparator() {
public int compare(Object o1, Object o2) {
return -((Comparable) (() (o1)).getValue()).compareTo((() (o2)).getValue());
}
});
Map result = new LinkedHashMap();
int i = 0;
for (Object it : list) {
entry = () it;
if (i >= topN) {
break;
}
((), ());
i++;
}
return result;
}
}
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The above code is adapted from: /srjklssj/article/details/6324880
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Map is a collection of key-value pairs, also known as dictionaries or associative arrays, etc., and is one of the most common data structures. How to sort a map by value in Java? It seems simple, but not easy!
For example, in Map, the key is String type, which represents a word, and the value is int type, which represents the number of times the word appears. Now we want to sort by the number of times the word appears:
Map map = new TreeMap();
("me", 1000);
("and", 4000);
("you", 3000);
("food", 10000);
("hungry", 5000);
("later", 6000);
The result of sorting by value should be:
key value
me 1000
you 3000
and 4000
hungry 5000
later 6000
food 10000
First of all, we cannot adopt the SortedMap structure, because SortedMap is a map sorted by keys, not a map sorted by value. What we want is a map sorted by value.
Couldn't you do this with a SortedMap?
No, because the map are being sorted by its keys.
Method 1:
The following Java code:
;
;
;
public classMain{
public static void main(String[]args) {
Setset= new TreeSet();(new Pair("me", "1000"));(new Pair("and", "4000"));(new Pair("you", "3000"));(new Pair("food", "10000"));(new Pair("hungry", "5000"));(new Pair("later", "6000"));(new Pair("myself", "1000"));
for (Iteratori=();();)
(());
}
}
classPairimplements Comparable {
private final Stringname;
private final intnumber;
public Pair(Stringname, intnumber) {
=name;
=number;
}
public Pair(Stringname, Stringnumber) throws NumberFormatException {
=name;
= (number);
}
public int compareTo(Objecto) {
if (oinstanceofPair) {
intcmp= (number, ((Pair)o).number);
if (cmp!= 0) {
returncmp;
}
(((Pair)o).name);
}
throw new ClassCastException("Cannot compare Pair with "
+().getName());
}
public String toString() {
returnname+ ' ' +number;
}
}
Similar C++ code:
typedef pair PAIR;
int cmp(const PAIR& x, const PAIR& y)
{
return > ;
}
map m;
vector vec;
for (map::iterator curr = (); curr != (); ++curr)
{
vec.push_back(make_pair(curr->first, curr->second));
}
sort((), (), cmp);
The essential significance of the above method is: encapsulate the key-value pairs () in the Map structure into a custom class (struct), or use the class directly. Custom classes know how to sort themselves, that is, sort by value, implement the Comparable interface or construct a Comparator object for themselves, and then use an ordered set (SortedSet,
TreeSet is an implementation of SortedSet), which realizes the purpose of sort by value in Map. That is to say, instead of using Map, it is regarded as an object, so that the problem becomes to implement an ordered set of the object or sorting the set of the object. You can use SortedSet, so that after insertion is completed, it will naturally be ordered, or use a List or array, and then sort it (()
or ())。
Encapsulate the information in its own class. Either implement
Comparable and write rules for the natural ordering or write a
Comparator based on your criteria. Store the information in a sorted
collection, or use the () method.
Method 2:
You can also use the following code to sort by value:
public static Map sortByValue(Mapmap)
{
Listlist= new LinkedList(());
(list, new Comparator() {
public int compare(Objecto1, Objecto2) {
return ((Comparable) (() (o1)).getValue())
.compareTo((() (o2)).getValue());
}
});
Mapresult= new LinkedHashMap();
for (Iteratorit=();();) {
entry= ()();((),());
}
returnresult;
}
public static Map sortByValue(Mapmap, final booleanreverse) {
Listlist= new LinkedList(());
(list, new Comparator() {
public int compare(Objecto1, Objecto2) {
if (reverse) {
return -((Comparable) (() (o1)).getValue())
.compareTo((() (o2)).getValue());
}
return ((Comparable) (() (o1)).getValue())
.compareTo((() (o2)).getValue());
}
});
Mapresult= new LinkedHashMap();
for (Iteratorit=();();) {
entry= ()();((),());
}
returnresult;
}
Mapmap= new HashMap();("a", 4);("b", 1);("c", 3);("d", 2);
Mapsorted= sortByValue(map);
(sorted);
// output : {b=1, d=2, c=3, a=4}
Or it can be like this:
Mapmap= new HashMap();("a", 4);("b", 1);("c", 3);("d", 2);
Set>treeSet= new TreeSet>(
new Comparator>() {
public int compare(Map.Entryo1,
Map.Entryo2) {
Integerd1=();
Integerd2=();
intr=(d1);
if (r!= 0)
returnr;
else
().compareTo(());
}
});(());
(treeSet);
// output : [a=4, c=3, d=2, b=1]
In addition, implementing sort map by value in Groovy is of course the essence is the same, but it is very simple:
Use the sort method of map in groovy (requires groovy 1.6).
def result = (){ a, b ->
()
}
like:
["a":3,"b":1,"c":4,"d":2].sort{ a,b -> - }
The result is: [b:1, d:2, a:3, c:4]
Similar in Python:
h = {"a":2,"b":1,"c":3}
i = () // i = [('a', 2), ('c', 3), ('b', 1)]
(lambda (k1,v1),(k2,v2): cmp(v2,v1) ) // i = [('c', 3), ('a', 2), ('b', 1)]