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根据 Groovy 或 Z93F725A07423FE1C889F448B336

[英]Segregating a Map in two map based on value in Groovy or java

I have a map of type Map<Object,ArrayList> MapA.我有一个 Map<Object,ArrayList> MapA 类型的 map。 map can have values like map 可以具有如下值

MapA
{key=0003bb2d-f3cf-4b5c-9f34 , value = ["12345","Complete","5431"],
key=0003bb2d-f3cf-4b5c-9f71 , value = ["2311","InComplete","5321"],
key=0003bb2d-f3cf-4b5c-1d71 , value = ["4567","InComplete","4321"],
key=0003bb2d-f3cf-4b5c-1r51 , value = ["2345","Complete","5432"]}


output should be 

MapA
{key=0003bb2d-f3cf-4b5c-9f34 , value = ["12345","Complete","5431"],
key=0003bb2d-f3cf-4b5c-1r51 , value = ["2345","Complete","5432"]}

MapB
{key=0003bb2d-f3cf-4b5c-9f71 , value = ["2311","InComplete","5321"],
key=0003bb2d-f3cf-4b5c-1d71 , value = ["4567","InComplete","4321"]}

I want to create a new MapB from MapA having all key value pairs, where value has "InComplete" in it.我想从包含所有键值对的 MapA 创建一个新的 MapB,其中 value 中包含“InComplete”。 Either i can have two extra map having "Complete" and "InComplete" KeyValue pairs or if i can remove Incomplete from MapA.我可以有两个额外的 map 具有“完整”和“不完整”键值对,或者如果我可以从 MapA 中删除不完整。

So i want to Segregate MapA on the basis of Complete or InComplete into two seperate maps.所以我想根据 Complete 或 InComplete 将 MapA 分离成两个单独的地图。

I need to write it in groovy but if i can get some hint in java 8 as well, i can try to convert it.我需要将它写在 groovy 但如果我也能在 java 8 中得到一些提示,我可以尝试转换它。

Code snippet i am working on but this will create new Map but won't remove these values from MapA, how can i do that?我正在处理的代码片段,但这将创建新的 Map 但不会从 MapA 中删除这些值,我该怎么做?

MapA.entrySet().stream().filter(entry->entry.getValue().get(1).equals("InComplete")).collect(Collectors.toMap())

Appreciate your help感谢你的帮助

UPDATE更新

 def mapB = [:].withDefault {key -> return []}
    Iterator<Map.Entry<Object, ArrayList>> iterator = mapA.entrySet().iterator();
    while (iterator.hasNext()) {
        Map.Entry<Object, ArrayList> entry = iterator.next();
        if(entry.getValue().get(1).equals("InComplete")) {
            mapB.put(entry.getKey(), entry.getValue())
            iterator.remove()
        }
    }

There are thousands way of doing this.有成千上万种方法可以做到这一点。

Down below I put 3 options: 1 with filling 2 maps independently in 2 loops, 1 (slightly better performing) using the single loop, and 1 using groupBy :下面我放了 3 个选项:1 个在 2 个循环中独立填充 2 个地图,1 个(性能稍好)使用单循环,1 个使用groupBy

def map = [ '0003bb2d-f3cf-4b5c-9f34':["12345","Complete","5431"],
'0003bb2d-f3cf-4b5c-9f71':["2311","InComplete","5321"],
'0003bb2d-f3cf-4b5c-1d71':["4567","InComplete","4321"],
'0003bb2d-f3cf-4b5c-1r51':["2345","Complete","5432"]]

// 1. option
Map complete = map.findAll{ k, v -> 'Complete' == v[ 1 ] }
Map inComplete = map.findAll{ k, v -> 'InComplete' in v }
assert complete.toString() == '[0003bb2d-f3cf-4b5c-9f34:[12345, Complete, 5431], 0003bb2d-f3cf-4b5c-1r51:[2345, Complete, 5432]]'
assert inComplete.toString() == '[0003bb2d-f3cf-4b5c-9f71:[2311, InComplete, 5321], 0003bb2d-f3cf-4b5c-1d71:[4567, InComplete, 4321]]'

// 2. option
def twoMaps = map.inject( [:].withDefault{ [:] } ){ res, curr ->
  res[ curr.value[ 1 ] ] << curr
  res
}
assert twoMaps.toString() == '[Complete:[0003bb2d-f3cf-4b5c-9f34:[12345, Complete, 5431], 0003bb2d-f3cf-4b5c-1r51:[2345, Complete, 5432]], InComplete:[0003bb2d-f3cf-4b5c-9f71:[2311, InComplete, 5321], 0003bb2d-f3cf-4b5c-1d71:[4567, InComplete, 4321]]]'

// 3. option -> credits to @tim_yates!
def groupMaps = map.groupBy{ it.value[ 1 ] }
assert groupMaps.toString() == '[Complete:[0003bb2d-f3cf-4b5c-9f34:[12345, Complete, 5431], 0003bb2d-f3cf-4b5c-1r51:[2345, Complete, 5432]], InComplete:[0003bb2d-f3cf-4b5c-9f71:[2311, InComplete, 5321], 0003bb2d-f3cf-4b5c-1d71:[4567, InComplete, 4321]]]'

And for fun, this is the same but in Java 17为了好玩,这是相同的,但在 Java 17

var stringListMap = Map.of(
        "0003bb2d-f3cf-4b5c-9f34", List.of("12345", "Complete", "5431"),
        "0003bb2d-f3cf-4b5c-9f71", List.of("2311", "InComplete", "5321"),
        "0003bb2d-f3cf-4b5c-1d71", List.of("4567", "InComplete", "4321"),
        "0003bb2d-f3cf-4b5c-1r51", List.of("2345", "Complete", "5432")
);

var result = stringListMap.entrySet().stream().collect(
        Collectors.groupingBy(
                e -> e.getValue().get(1),
                Collectors.mapping(e -> e, Collectors.toList())
        )
);
result.forEach((k, v) -> System.out.println(k + " => " + v));

Which prints:哪个打印:

Complete => [0003bb2d-f3cf-4b5c-9f34=[12345, Complete, 5431], 0003bb2d-f3cf-4b5c-1r51=[2345, Complete, 5432]]
InComplete => [0003bb2d-f3cf-4b5c-9f71=[2311, InComplete, 5321], 0003bb2d-f3cf-4b5c-1d71=[4567, InComplete, 4321]]

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