如何在Java中实现列表折叠
我有一个列表,希望将其简化为一个值(函数式编程术语“fold”,Ruby术语如何在Java中实现列表折叠,java,collections,functional-programming,folding,Java,Collections,Functional Programming,Folding,我有一个列表,希望将其简化为一个值(函数式编程术语“fold”,Ruby术语inject),比如 由于我深受函数式编程思想(Scala)的影响,我正在寻找一种比函数式编程更简单/更短的编码方法 sb = new StringBuilder for ... { append ... } sb.toString 不幸的是,在Java中,您无法逃脱这个循环,但是有几个库。例如,您可以尝试几个库: 特别是在你的情况下,你可以重复使用我的 您需要的是一个字符串“join”函数,不幸的是,Jav
inject
),比如
由于我深受函数式编程思想(Scala)的影响,我正在寻找一种比函数式编程更简单/更短的编码方法
sb = new StringBuilder
for ... {
append ...
}
sb.toString
不幸的是,在Java中,您无法逃脱这个循环,但是有几个库。例如,您可以尝试几个库:
- 特别是在你的情况下,你可以重复使用我的
Edit:似乎有许多有用的字符串函数(包括join)。不幸的是,Java不是函数式编程语言,没有一种很好的方法来实现您想要的功能 我相信ApacheCommons库有一个可以实现您想要的功能 它必须足够好,才能在方法中隐藏循环
public static String combine(List<String> list, String separator){
StringBuilder ret = new StringBuilder();
for(int i = 0; i < list.size(); i++){
ret.append(list.get(i));
if(i != list.size() - 1)
ret.append(separator);
}
return ret.toString();
}
公共静态字符串组合(列表、字符串分隔符){
StringBuilder ret=新的StringBuilder();
对于(int i=0;i
我想你可以递归地做:
public static String combine(List<String> list, String separator){
return recursiveCombine("", list, 0, separator);
}
public static String recursiveCombine(String firstPart, List<String> list, int posInList, String separator){
if (posInList == list.size() - 1) return firstPart + list.get(posInList);
return recursiveCombine(firstPart + list.get(posInList) + separator, list, posInList + 1, seperator);
}
公共静态字符串组合(列表、字符串分隔符){
返回递归组合(“,列表,0,分隔符);
}
公共静态字符串recursiveCombine(字符串第一部分、列表、int-posInList、字符串分隔符){
if(posInList==list.size()-1)返回firstPart+list.get(posInList);
返回递归组合(firstPart+list.get(posInList)+分隔符,list,posInList+1,分隔符);
}
给定
public static <T,Y> Y fold(Collection<? extends T> list, Injector<T,Y> filter){
for (T item : list){
filter.accept(item);
}
return filter.getResult();
}
public interface Injector<T,Y>{
public void accept(T item);
public Y getResult();
}
公共静态Y折叠(集合回答您的原始问题:
public static <A, B> A fold(F<A, F<B, A>> f, A z, Iterable<B> xs)
{ A p = z;
for (B x : xs)
p = f.f(p).f(x);
return p; }
例3:
import static fj.data.Stream.fromString;
import static fj.data.Stream.asString;
...
String abc = asString(fromString("abc").intersperse(','));
如果您希望在不切换语言的情况下将一些功能方面应用于普通的旧Java,并且这些库可以帮助您添加语法糖
在的帮助下,您可以使用:
Joiner.on(“,”)
是一个不可变的对象,因此您可以自由地共享它(例如作为常量)
您还可以配置空处理,如Joiner.on(“,”).useForNull(“nil”);
或Joiner.on(“,”).skipNulls()
为了避免在生成大字符串时分配大字符串,可以使用它通过appendeable
接口或StringBuilder
类附加到现有流、StringBuilder等:
Joiner.on(",").appendTo(someOutputStream, "a", "b", "c");
在编写映射时,需要两个不同的分隔符来分隔条目和键+值:
Joiner.on(", ").withKeyValueSeparator(":")
.join(ImmutableMap.of(
"today", "monday"
, "tomorrow", "tuesday"))
首先,您需要一个Java函数库,它提供通用的函子和函数投影,比如fold。我在这里设计并实现了一个功能强大(凭借其优点)但简单的函数库:(我发现前面提到的其他库过于复杂)
然后,您的解决方案将如下所示:
Seq.of("","a",null,"b","",null,"c","").foldl(
new StringBuilder(), //seed accumulator
new Func2<StringBuilder,String,StringBuilder>(){
public StringBuilder call(StringBuilder acc,String elmt) {
if(acc.length() == 0) return acc.append(elmt); //do not prepend "," to beginning
else if(elmt == null || elmt.equals("")) return acc; //skip empty elements
else return acc.append(",").append(elmt);
}
}
).toString(); //"a,b,c"
(“,”a“,”空“,”b“,”空“,”c“,”)的顺序。foldl(
新建StringBuilder(),//种子累加器
新功能2(){
公共StringBuilder调用(StringBuilder acc、String elmt){
如果(acc.length()==0),则返回acc.append(elmt);//不在“,”之前加上前缀
else if(elmt==null | | elmt.equals(“”))返回acc;//跳过空元素
否则返回acc.append(“,”).append(elmt);
}
}
).toString();/“a、b、c”
注意,通过应用fold,真正需要考虑的唯一部分是Func2.call的实现,3行代码定义了一个接受累加器和一个元素并返回累加器的操作符(我的实现考虑了空字符串和空值,如果去掉这种情况,那么只需要2行代码)
下面是Seq.foldl的实际实现,Seq实现了Iterable:
public R foldl(R seed,final Func2没有这样的函数,但是您可以创建如下内容,并在需要时调用它
import java.util.Arrays;
import java.util.List;
public class FoldTest {
public static void main( String [] args ) {
List<String> list = Arrays.asList("a","b","c");
String s = fold( list, ",");
System.out.println( s );
}
private static String fold( List<String> l, String with ) {
StringBuilder sb = new StringBuilder();
for( String s: l ) {
sb.append( s );
sb.append( with );
}
return sb.deleteCharAt(sb.length() -1 ).toString();
}
}
导入java.util.array;
导入java.util.List;
公共类民俗{
公共静态void main(字符串[]args){
List=Arrays.asList(“a”、“b”、“c”);
字符串s=折叠(列表“,”);
系统输出打印项次;
}
私有静态字符串折叠(列表l,带的字符串){
StringBuilder sb=新的StringBuilder();
for(字符串s:l){
某人追加;
给某人加上;
}
返回sb.deleteCharAt(sb.length()-1.toString();
}
}
将注入
(如Ruby和Smalltalk)、生成字符串
和追加字符串
。以下内容适用于您的示例:
String result1 = FastList.newListWith("a", "b", "c").makeString(",");
StringBuilder sb = new StringBuilder();
FastList.newListWith("a", "b", "c").appendString(sb, ",");
String result2 = sb.toString();
Assert.assertEquals("a,b,c", result1);
Assert.assertEquals(result1, result2);
注意:我是Eclipse集合的提交者。您正在寻找的是Java自8.0以来一直使用的stringjoin()
方法。请尝试以下方法之一
静态方法:
您可能希望静态导入收集器。加入以使代码更清晰
顺便说一下,此收集器可以应用于任何特定对象的集合:
Collection<Integer> numbers = Arrays.asList(1, 2, 3);
String result = numbers.stream()
.map(Object::toString)
.collect(Collectors.joining(","));
collectionnumber=Arrays.asList(1,2,3);
字符串结果=numbers.stream()
.map(对象::toString)
.collect(收集器。连接(“,”);
现在,您可以在Java8中使用String.join()
List strings=Arrays.asList(“a”、“b”、“c”);
连接的字符串=字符串。连接(“,”,字符串);
System.out.println(已连接);
在lambdas的支持下,我们可以使用以下代码:
static <T, R> R foldL(BiFunction<R, T, R> lambda, R zero, List<T> theList){
if(theList.size() == 0){
return zero;
}
R nextZero = lambda.apply(zero,theList.get(0));
return foldL(lambda, nextZero, theList.subList(1, theList.size()));
}
static R foldL(双功能lambda,R zero,列表){
如果(列表大小()==0){
返回零;
}
R nextZero=lambda.apply(零,theList.get(0));
返回foldL(lambda,nextZero,theList.subList(1,theList.size());
}
下面是通过保留节点信息来折叠列表的代码
Joiner.on(", ").withKeyValueSeparator(":")
.join(ImmutableMap.of(
"today", "monday"
, "tomorrow", "tuesday"))
Seq.of("","a",null,"b","",null,"c","").foldl(
new StringBuilder(), //seed accumulator
new Func2<StringBuilder,String,StringBuilder>(){
public StringBuilder call(StringBuilder acc,String elmt) {
if(acc.length() == 0) return acc.append(elmt); //do not prepend "," to beginning
else if(elmt == null || elmt.equals("")) return acc; //skip empty elements
else return acc.append(",").append(elmt);
}
}
).toString(); //"a,b,c"
public <R> R foldl(R seed, final Func2<? super R,? super E,? extends R> binop)
{
if(binop == null)
throw new NullPointerException("binop is null");
if(this == EMPTY)
return seed;
for(E item : this)
seed = binop.call(seed, item);
return seed;
}
import java.util.Arrays;
import java.util.List;
public class FoldTest {
public static void main( String [] args ) {
List<String> list = Arrays.asList("a","b","c");
String s = fold( list, ",");
System.out.println( s );
}
private static String fold( List<String> l, String with ) {
StringBuilder sb = new StringBuilder();
for( String s: l ) {
sb.append( s );
sb.append( with );
}
return sb.deleteCharAt(sb.length() -1 ).toString();
}
}
String result1 = FastList.newListWith("a", "b", "c").makeString(",");
StringBuilder sb = new StringBuilder();
FastList.newListWith("a", "b", "c").appendString(sb, ",");
String result2 = sb.toString();
Assert.assertEquals("a,b,c", result1);
Assert.assertEquals(result1, result2);
Collection<String> source = Arrays.asList("a", "b", "c");
String result = String.join(",", source);
Collection<String> source = Arrays.asList("a", "b", "c");
String result = source.stream().collect(Collectors.joining(","));
Collection<Integer> numbers = Arrays.asList(1, 2, 3);
String result = numbers.stream()
.map(Object::toString)
.collect(Collectors.joining(","));
static <T, R> R foldL(BiFunction<R, T, R> lambda, R zero, List<T> theList){
if(theList.size() == 0){
return zero;
}
R nextZero = lambda.apply(zero,theList.get(0));
return foldL(lambda, nextZero, theList.subList(1, theList.size()));
}
public class FoldList {
public static void main(String[] args) {
Node a = new Node(1);
Node b = new Node(2);
Node c = new Node(3);
Node d = new Node(4);
Node e = new Node(5);
Node f = new Node(6);
Node g = new Node(7);
Node h = new Node(8);
Node i = new Node(9);
a.next = b;
b.next = c;
c.next = d;
d.next = e;
e.next = f;
f.next = g;
g.next = h;
h.next = i;
foldLinkedList(a);
}
private static void foldLinkedList(Node a) {
Node middle = getMiddleNodeOfTheList(a);
reverseListOnWards(middle);
foldTheList(a, middle);
}
private static Node foldTheList(Node a, Node middle) {
Node leftBackTracePtr = a;
Node leftForwardptr = null;
Node rightBackTrack = middle;
Node rightForwardptr = null;
Node leftCurrent = a;
Node rightCurrent = middle.next;
while (middle.next != null) {
leftForwardptr = leftCurrent.next;
rightForwardptr = rightCurrent.next;
leftBackTracePtr.next = rightCurrent;
rightCurrent.next = leftForwardptr;
rightBackTrack.next = rightForwardptr;
leftCurrent = leftForwardptr;
leftBackTracePtr = leftCurrent;
rightCurrent = middle.next;
}
leftForwardptr = leftForwardptr.next;
leftBackTracePtr.next = middle;
middle.next = leftForwardptr;
return a;
}
private static void reverseListOnWards(Node node) {
Node startNode = node.next;
Node current = node.next;
node.next = null;
Node previous = null;
Node next = node;
while (current != null) {
next = current.next;
current.next = previous;
previous = current;
current = next;
}
node.next = previous;
}
static Node getMiddleNodeOfTheList(Node a) {
Node slowptr = a;
Node fastPtr = a;
while (fastPtr != null) {
slowptr = slowptr.next;
fastPtr = fastPtr.next;
if (fastPtr != null) {
fastPtr = fastPtr.next;
}
}
return slowptr;
}
static class Node {
public Node next;
public int value;
public Node(int value) {
this.value = value;
}
}
}
// Given
List<String> arr = Arrays.asList("a", "b", "c");
String first = arr.get(0);
arr = arr.subList(1, arr.size());
String folded = arr.stream()
.reduce(first, (a, b) -> a + "," + b);
System.out.println(folded); //a,b,c