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有如下hive记录表records，记录车辆的过车信息： create table records( id string, //记录编号 indate string, //过车记录时间 plate_no string, //车辆号牌 device_id int, //经过的设备编号 ) partitioned by(month string,day string) row format delimited fields terminated by ‘t’ stored as ORC;
1请使用HQL得到最近一个月内晚上(晚22点-早6点)出现记录最多的车辆号牌Top10及次数 select plate_no, rank() over(sort by total desc) rk, total from ( select plate_no, count(1) total from ( select indate, plate_no from records where From_unixtime(unix_timestamp(indate,”yyyy-MM-dd HH:mm:ss”),”yyyy-MM-dd”)>=date_sub(‘2020-04-26’,30) and From_unixtime(unix_timestamp(indate,”yyyy-MM-dd HH:mm:ss”),”yyyy-MM-dd”) <=’2020-04-26′ and From_unixtime(unix_timestamp(indate,”yyyy-MM-dd HH:mm:ss”),”HH”) <=22 and From_unixtime(unix_timestamp(indate,”yyyy-MM-dd HH:mm:ss”),”HH”) >=6 ) t1 group by plate_no ) t2 where rk = 10
2请用spark RDD将上述表中indate、plate_no、device_id三个字段记录重复的数据只保留一条 object interview02 { def main(args: Array[String]): Unit = { val session: SparkSession = SparkSession.builder().master(“local[*]”).appName(“interview”).getOrCreate() val rdd: RDD[Row] = session.read.csv(“records”).rdd
val distinct = rdd.map(row => ((“indate”,”plate_no”,”device_id”),(“id”))) .groupByKey() distinct.distinct().foreach(println) session.stop() }
}
3有string A和string B，分别由4096个随机的0或1组成，样例为String A = “01010111001….110010”,String B=”10101110….100101″， 现有公式double C=(A和B逐位与的和)/(A中1的个数*B中1的个数)，请用自己熟悉的语言实现出满足此公式的方法。 public class Method { public static void main(String[] args) {
代码语言：javascript复制 String A = "0101010101101010";
String B = "1010101010101011";
char[] chars1 = A.toCharArray();
char[] chars2 = B.toCharArray();
int sum;
Double sum2 = 0.0;
int c = 0;
int d = 0;
for (int i = 0; i < chars1.length; i ) {
char a = chars1[i];
// for (int j= 0;j< chars2.length;j ){
char b = chars2[i];
sum = Integer.parseInt(String.valueOf(a)) & Integer.parseInt(String.valueOf(b));
/* System.out.println(sum); */
sum2 = sum;
}
for (int j = 0; j < chars1.length; j ) {
char a = chars1[j];
char b = chars2[j];
if (Integer.parseInt(String.valueOf(a)) == 1) {
c = 1;
}
if (Integer.parseInt(String.valueOf(b)) == 1) {
d = 1;
}
}
Double C = sum2/(c*d);
System.out.println(sum2);
System.out.println(c);
System.out.println(d);
System.out.println(C);} }
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