Java 8流：条件收集器 [英] Java 8 streams: conditional Collector

问题描述

我想使用Java 8流将List of String值转换为单个String。像A，B这样的值列表应返回类似值：'A'，'B'添加的字符串。这工作正常，但我想根据值的数量更改前缀和后缀。例如，如果我有一个只有A的列表，我希望生成的字符串为值'A'已添加。

I want to use Java 8 streams to convert a List of String values to a single String. A List of values like "A", "B" should return a String like "Values: 'A', 'B' added". This works fine, however I want to change the Pre- and Postfix depending on the amount of values. For example, if I have a List of only "A" I want the resulting String to be "Value 'A' added".

import java.util.stream.Collectors;
import java.util.ArrayList;
import java.util.List;

public class HelloWorld
{
  public static void main(String[] args)
  {
    List<String> values = new ArrayList<>();
    values.add("A");
    values.add("B");
    values.add("C");
    List<String> value = new ArrayList<>();
    value.add("A");
    System.out.println(log(values));
    System.out.println(log(value));
  }
  public static String log(List<String> values){
    return values.stream()
                 //.filter(...)
                 .map(x -> "'" + x + "'")
                 .collect(Collectors.joining(",","values:"," added"));

  }
}

有没有办法改变Collctor，取决于结果列表的大小？然后我可以做类似的事情

Is there a way to change the Collctor, depending on the size of the resulting List? Then I could do something like

.collect(Collectors.joining(",", size = 1 ? "Value " : "Values: "," added"));

我更喜欢没有中间List结果的单流操作。我之前也不知道结果，因为我过滤了流。

I would prefer a single stream operation without an intermediate List result. I also do not know the result beforehand, because I filter the stream.

编辑：我最终使用了Eugene的建议。我想要做的是找到两个列表之间的差异，并返回人类可读形式的差异。很好用！

I ended up using Eugene's suggestion. What I wanted to do is find the differences between two Lists and return the differences in human readable form. Works nicely!

import java.util.stream.Collectors;
import java.util.ArrayList;
import java.util.List;


public class HelloWorld
{

  public static void main(String[] args)
  {
    List<String> oldValues = new ArrayList<>();
    oldValues.add("A");
    oldValues.add("B");
    oldValues.add("C");
    List<String> newValues = new ArrayList<>();
    newValues.add("A");
    newValues.add("C");
    newValues.add("D");
    newValues.add("E");
    System.out.println(HelloWorld.<String>log(oldValues, newValues, " deleted"));
    System.out.println(HelloWorld.<String>log(newValues, oldValues, " added"));
  }

    public static <T> String log(List<T> first, List<T> second, String postfix) {
        return  (String) first
                .stream()
                .filter(x -> !second.contains(x))
                .map(x -> "'" + x.toString() + "'").
                collect(Collectors.collectingAndThen(Collectors.toList(),
                    list -> {
                        if (list.size() == 1) {
                            return "Value " + list.get(0) + postfix;
                        }
                        if (list.size() > 1) {
                            List<String> strings = new ArrayList<>(list.size());
                            for (Object object : list) {
                                strings.add(object.toString());
                            }
                            return "Values: " + String.join(",", strings) +  postfix;
                        }
                        return "";
                    }));
    }
}

输出：

Value 'B' deleted
Values: 'D','E' added

推荐答案

你可以通过以下方式完成：

You could do it via:

 return values.stream()
            .map(x -> "'" + x + "'")
            .collect(Collectors.collectingAndThen(Collectors.toList(),
                    list -> {
                        if (list.size() == 1) {
                            return "value" + list.get(0);
                        }
                        if (list.size() > 1) {
                            return String.join(",", list);
                        }
                        return "Nothing found";
                    }));

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