需求
想要统计环比指标((本期数量-上期数量)/上期数量*100%
) 假设下面是统计9月份的数据 如下所示
品牌 | 数量 | 环比 |
---|---|---|
Bosh | 1561 | 311.87% |
Siemens | 2278 | -75.24% |
问题
查询的时候 需要同时查询8月的数据 统计出8月的数量 然后才能进行环比指标的计算
{ "count" : 379.0, "brand" : "Bosch", "month" : "2017-08" }
{ "count" : 1561.0, "brand" : "Bosch", "month" : "2017-09" }
{ "count" : 9202.0, "brand" : "Siemens", "month" : "2017-08" }
{ "count" : 2278.0, "brand" : "Siemens", "month" : "2017-09" }
怎么转换得到上图的结果呢? 即
{ "count" : 379.0, "brand" : "Bosch", "month" : "2017-08" }
{ "count" : 1561.0, "brand" : "Bosch", "month" : "2017-09" }
{ "count" : 9202.0, "brand" : "Siemens", "month" : "2017-08" }
{ "count" : 2278.0, "brand" : "Siemens", "month" : "2017-09" }
==>
{ "count" : 1561.0, "brand" : "Bosch", "month" : "2017-09","huanbi": 311.87 }
{ "count" : 2278.0, "brand" : "Siemens", "month" : "2017-09","huanbi":-75.24 }
我以为挺好实现的 没想到还挺折腾的 代码如下
Map<String,Object> record1 = new HashMap(ImmutableMap.of("count", 379, "brand", "Bosch", "month", "2017-08"));
Map<String,Object> record2 = new HashMap(ImmutableMap.of("count", 1561, "brand", "Bosch", "month", "2017-09"));
Map<String,Object> record3 = new HashMap(ImmutableMap.of("count", 9202, "brand", "Siemens", "month", "2017-08"));
Map<String,Object> record4 = new HashMap(ImmutableMap.of("count", 2278, "brand", "Siemens", "month", "2017-09"));
Map<String,Object> record5 = new HashMap(ImmutableMap.of("count", 2278, "brand", "foo", "month", "2017-09"));
List<Map<String, Object>> queryResult = Lists.newArrayList(record1, record4, record3, record2, record5);
// 先按品牌和日期排序
queryResult.sort((o1,o2)->{
int result = 0;
String[] keys = {"brand", "month"};
for (String key : keys) {
String val1 = o1.get(key).toString();
String val2 = o2.get(key).toString();
result = val1.compareTo(val2);
if(result != 0){
return result;
}
}
return result;
});
// 再按品牌分组
Map<String, List<Map<String, Object>>> brand2ListMap = queryResult.stream().collect(groupingBy(m -> m.get("brand").toString(), toList()));
/**
* 每组中第一条肯定是上一月的 找到上月的数目
* 第二条记录是本月的 找到本月的数据
* 计算环比 本期记录中添加环比
* 同时删除上一条记录
*/
for (String key : brand2ListMap.keySet()) {
List<Map<String, Object>> recordList = brand2ListMap.get(key);
if (recordList.size() > 1) {
Map<String, Object> prevRecord = recordList.get(0);
Map<String, Object> currentRecord = recordList.get(1);
Integer prevCount = (Integer) prevRecord.get("count");
Integer currentCount = (Integer) currentRecord.get("count");
BigDecimal huanbi = BigDecimal.valueOf((currentCount - prevCount) * 100).divide(BigDecimal.valueOf(prevCount), 2, ROUND_HALF_DOWN);
currentRecord.put("huanbi", huanbi);
recordList.remove(0);
}else{
// 不存在上期记录 环比默认为0
recordList.get(0).put("huanbi", 0);
}
}
// 生成一个新的List 只包含本期记录
List<Map<String, Object>> processedResult = new ArrayList(brand2ListMap.values().stream().flatMap(list->list.stream()).collect(toList()));
// 按照品牌排序
processedResult.sort(Comparator.comparing(o -> o.get("brand").toString()));
processedResult.forEach(System.out::println);
输出结果如下
{count=1561, month=2017-09, brand=Bosch, huanbi=311.87}
{count=2278, month=2017-09, brand=Siemens, huanbi=-75.24}
{count=2278, month=2017-09, brand=foo, huanbi=0}
应该不是我想的复杂了吧?应该没有更简单的方案了吧?
慕工程0101907
ITMISS
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