排序图< String，Integer>通过List< String>使用流

但是，以下两个代码示例在流执行之外修改了 nonOrderedData 的状态. 保留 nonOrderedData 中的键和相关值 only ，将不匹配的对放在最后: Map< String，Integer>nonOrderedData =新的HashMap<>(Map.of("b"，1，"e"，4，"z"，8，"a"，0，"q"，6"f"，5，"d"，7))；List< String>orderSequence = Arrays.asList("a"，"e"，"i"，"o"，"u"，"b"，"c"，"d")；Map< String，Integer>重新排序= orderSequence.溪流().filter(nonOrderedData :: containsKey).collect(Collectors.toMap(键->键，nonOrderedData :: remove，(v1，v2)->v1，LinkedHashMap :: new))；SortedMap< String，Integer>剩余=新的TreeMap<>(nonOrderedData);System.out.println(剩余:" +余数)；reordered.putAll(remainder);System.out.println(已重新排序); 输出: 保留:{f = 5，q = 6，z = 8}{a = 0，e = 4，b = 1，d = 7，f = 5，q = 6，z = 8} 类似于 orderSequence 和 nonOrderedData 之间的 RIGHT JOIN .保留 orderSequence 和 nonOrderedData 中的所有值，类似于 FULL JOIN 此处将为 orderSequence 中的未映射键提供默认值，并将 nonOrderedData 中的不匹配键添加到末尾. Map< String，Integer>reorderedFull = orderSequence.溪流().peek(key-> nonOrderedData.computeIfAbsent(key，(k)-> -1))//默认值.collect(Collectors.toMap(键->键，nonOrderedData :: remove，(v1，v2)->v1，LinkedHashMap :: new))；SortedMap< String，Integer>restderFull =新的TreeMap<>(nonOrderedData);System.out.println(剩余:" +"remainderFull")；reorderedFull.putAll(remainderFull);System.out.println(reorderedFull); 输出: 保留:{f = 5，q = 6，z = 8}{a = 0，e = 4，i = -1，o = -1，u = -1，b = 1，c = -1，d = 7，f = 5，q = 6，z = 8} Map<String, Integer> nonOrderedData = // {b=1, c=2, d=3, e=4, a=0}List<String> orderSequence = // a, b, c, d, eI need to apply the order sequence to get correctly ordered data, how could I achieve this using (prefered) stream?What I used is non stream way:Map<String, Integer> orderedData = new HashMap<>();for (Map.Entry<String, Integer> nod : nonOrderedData.entrySet()) { for (String os : orderSequence) { if (os == nod.getKey()) { // add nonOrderedData data } else { // add data by sequence } }}Would like to have more cleaner way of what I want.I've noticed that in my method I could just return new TreeMap<>(nonOrderedData) and it would work just fine, but I don't want to stick to just applying asc order - I would like to read actual sequence values and then change nonOrderedData. 解决方案 The HashMap cannot be used to store orderedData because it does not guarantee ordering of its keys, so LinkedHashMap should be used which maintains insertion order.Ordering by the data presented in orderSequence using streams may be implemented in two modes:Keep only the values available in nonOrderedData:Map<String, Integer> nonOrderedData = Map.of( "b", 1, "e", 4, "a", 0, "o", 5, "d", 7);List<String> orderSequence = Arrays.asList( "a", "e", "i", "o", "u", "b", "c", "d");Map<String, Integer> reordered = orderSequence .stream() .filter(nonOrderedData::containsKey) .collect(Collectors.toMap( key -> key, nonOrderedData::get, (v1, v2) -> v1, LinkedHashMap::new ));System.out.println(reordered);Output:{a=0, e=4, o=5, b=1, d=7}It is similar to INNER JOIN between orderSequence and nonOrderedData.Filling reorderedData with some default value if the key from orderSequence is missing in nonOrderedData:Map<String, Integer> reorderedWithDefault = orderSequence .stream() .collect(Collectors.toMap( key -> key, nonOrderedData.getOrDefault(key, -1), (v1, v2) -> v1, LinkedHashMap::new ));System.out.println(reorderedWithDefault);Output:{a=0, e=4, i=-1, o=5, u=-1, b=1, c=-1, d=7}It is similar to LEFT JOIN between orderSequence and nonOrderedData.UpdateIn the above-mentioned implementations the key-value pairs in nonOrderedData that do not match to the keys in orderSequence are skipped altogether. Such keys may be tracked (and added later to the reordered result) using Map::remove (Object key) which returns a value of the key being removed.However, the following two code examples modify the state of nonOrderedData outside the stream execution.Keep the keys and related values only from nonOrderedData, place the non-matched pairs to the end:Map<String, Integer> nonOrderedData = new HashMap<>(Map.of( "b", 1, "e", 4, "z", 8, "a", 0, "q", 6, "f", 5, "d", 7 ));List<String> orderSequence = Arrays.asList("a", "e", "i", "o", "u", "b", "c", "d");Map<String, Integer> reordered = orderSequence .stream() .filter(nonOrderedData::containsKey) .collect(Collectors.toMap( key -> key, nonOrderedData::remove, (v1, v2) -> v1, LinkedHashMap::new ));SortedMap<String, Integer> remainder = new TreeMap<>(nonOrderedData);System.out.println("remained: " + remainder);reordered.putAll(remainder);System.out.println(reordered);Output:remained: {f=5, q=6, z=8}{a=0, e=4, b=1, d=7, f=5, q=6, z=8}It is similar to RIGHT JOIN between orderSequence and nonOrderedData.Keep all values from both orderSequence and nonOrderedData similar to FULL JOINHere default values will be provided for the non-mapped keys in orderSequence and non-matched keys from nonOrderedData will be added to the end.Map<String, Integer> reorderedFull = orderSequence .stream() .peek(key -> nonOrderedData.computeIfAbsent(key, (k) -> -1)) // default value .collect(Collectors.toMap( key -> key, nonOrderedData::remove, (v1, v2) -> v1, LinkedHashMap::new ));SortedMap<String, Integer> remainderFull = new TreeMap<>(nonOrderedData);System.out.println("remained: " + remainderFull);reorderedFull.putAll(remainderFull);System.out.println(reorderedFull);Output:remained: {f=5, q=6, z=8}{a=0, e=4, i=-1, o=-1, u=-1, b=1, c=-1, d=7, f=5, q=6, z=8} 这篇关于排序图&lt; String，Integer&gt;通过List&lt; String&gt;使用流的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！