Map<String, List<SchemaRowFilterCondition>> groups = new HashMap<>(); 帮我把groups喝 Function<SchemaRowFilterCondition, String> groupByFunction = SchemaRowFilterCondition::getField; for (SchemaRowRuleEo schemaRowRuleEo : schemaRowRuleList) { String filterConditionList = schemaRowRuleEo.getFilterConditionList(); //拿到要素 List<SchemaRowFilterCondition> schemaRowFilterConditions = JSONObject.parseArray(filterConditionList, SchemaRowFilterCondition.class); if(!CollectionUtils.isEmpty(groups)){ Map<String, List<SchemaRowFilterCondition>> collect = schemaRowFilterConditions.stream().collect(Collectors.groupingBy(groupByFunction)); groups = Stream.concat(groups.entrySet().stream(), collect.entrySet().stream()) .collect(Collectors.toMap(SchemaRowFilterCondition::getField, List<SchemaRowFilterCondition::getKey>)); } //根据要素进行分组 groups = schemaRowFilterConditions.stream().collect(Collectors.groupingBy(groupByFunction)); }帮我把collect和group两个map合并

你可以使用以下代码将`collect`和`groups`两个Map合并：

groups = Stream.concat(groups.entrySet().stream(), collect.entrySet().stream())
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
                        (list1, list2) -> {
                            List<SchemaRowFilterCondition> combinedList = new ArrayList<>(list1);
                            combinedList.addAll(list2);
                            return combinedList;
                        }));

在上面的代码中，我们使用`Stream.concat()`方法将`groups`和`collect`两个Map的entrySet流合并，然后使用`Collectors.toMap()`方法将结果收集到`groups`这个新的Map中。在收集器中，我们使用entry的key作为新Map的key，value作为新Map的value，并使用一个合并函数来解决key重复时如何合并value的问题。在合并函数中，我们将两个List合并为一个新的List。