大家好，又见面了，我是你们的朋友全栈君。
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) {
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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