我有一个具有不同年龄的1,000,000个用户列表,我想在Java中执行一个搜索,它将根据年龄范围输出组中的人数。例如:
Age Group Age Range
1 6 years old or younger
2 7 to 18 years old
3 19 to 26 years old
4 27 to 49 years old
5 50 to 64 years old
6 65 to 79 years old
7 80 years old or older
如果我输入一个特定的年龄组,我希望我的输出显示属于该年龄组的人数。这就是:
If I enter 1
输出应为:
**** users found (total number of users that falls within the
age range 6 years old or younger)
任何类型的数据结构都是完全可以的。
这就是我到目前为止所做的:
/**
A template used to read data lines into java.util.ArrayList data structure.
Input file: pjData.csv
Input file must be saved under the same directory/folder as the program.
Each line contains 5 fields, separated by commas. For example,
959695171, 64, AZ, M, 1
355480298, 101, TN, F, 1
**/
import java.io.*;
import java.util.*;
public class pj3Template2
{
public static void main(String args[])
{
String line;
String id, s, g;
Integer a, sa;
StringTokenizer st;
HealthDS2 records = new HealthDS2();
try {
FileReader f = new FileReader("pjData.csv");
BufferedReader in = new BufferedReader(f);
while ((line = in.readLine()) != null)
{
st = new StringTokenizer(line, ",");
id = st.nextToken(",").trim();
a = Integer.valueOf(st.nextToken(",").trim());
s = st.nextToken(",").trim().toUpperCase();
g = st.nextToken(",").trim().toUpperCase();
sa = Integer.valueOf(st.nextToken().trim());
records.add(new HealthRec2(id, a, s, g, sa));
} // loop until the end of file
in.close();
f.close();
}
catch (Exception e) { e.printStackTrace(); };
System.out.println(records.getSize() + " records processed.");
// Search by age
System.out.print("Enter 1-character age abbreviation to search: ");
String ui;
Scanner input = new Scanner(System.in);
ui = input.next().trim();
System.out.println("Searching all records in: " + ui);
ArrayList <HealthRec2> al = records.searchByAge(Integer.valueOf(ui.trim()));
System.out.println(al.size() + " records found.");
}
}
/**
Data class Sample records:
5501986, 31, WV, F, 1
1539057187, 5, UT, M, 2
**/
class HealthRec2
{
String ID;
Integer age;
String state;
String gender;
int status;
public HealthRec2() { }
public HealthRec2(String i, Integer a, String s, String g, int sa)
{ ID = i; age = a; state = s; gender = g; status = sa; }
// Reader methods
public String getID() { return ID; }
public Integer getAge() { return age; }
public String getState() { return state; }
public String getGender() { return gender; }
public int getStatus() { return status; }
// Writer methods
public void setAge(Integer a) { age = a; }
public void setState(String s) { state = s; }
public void setGender(String g) { gender = g; }
public void setStatus(int sa) { status = sa; }
public String toString()
{ return ID + " " + age + " " + state + " " + gender + " " + status; }
} // HealthRec
// Data structure used to implement the requirement
// This implementation uses java.util.ArrayList
class HealthDS2
{
ArrayList <HealthRec2> rec;
public HealthDS2()
{ rec = new ArrayList <HealthRec2>(); }
public HealthDS2(HealthRec2 r)
{
rec = new ArrayList <HealthRec2>();
rec.add(r);
}
public int getSize() { return rec.size(); }
public void add(HealthRec2 r) { rec.add(r); }
// Search by age
// No data validation is needed -- assuming the 1-character age is valid
// Returns an ArrayList of records
public ArrayList <HealthRec2> searchByAge(Integer a)
{
ArrayList <HealthRec2> temp = new ArrayList <HealthRec2>();
for (int k=0; k < rec.size(); ++k)
{
if (rec.get(k).getAge().equals(a))
temp.add(rec.get(k));
}
return temp;
} // searchByAge
} // HealthDS
我的目标是基于state
、status
、gender
和age
组进行搜索。我已经为其他人这样做了,但我只是在年龄组上遇到了一点问题,因为它是分组的,而不仅仅是在数据文件中搜索特定的年龄。我试着为每个组创建了七个数组列表,但在组之间切换仍然有一些问题。
发布于 2013-11-27 03:07:31
此代码执行以下操作:
对于非常大的数据集,你需要使用更好的数据结构,比如@kyticka提到的。
public static void main (String[] args) throws java.lang.Exception
{
int[] groupMin = new int[]{0, 10, 20};
int[] groupMax = new int[]{10, 20, 9999};
int[] ages = new int[]{ 1, 2, 3, 10, 12, 76, 56, 89 };
int targetGroup = 1;
int count = 0;
for( int age : ages ){
if( age >= groupMin[targetGroup] && age < groupMax[targetGroup] ){
count++;
}
}
System.out.println("Group " + targetGroup + " range is " +
groupMin[targetGroup] + " - " + groupMax[targetGroup]);
System.out.println("Count: " + count);
}
你可以在这里使用它:http://ideone.com/DAWGYX
发布于 2013-11-27 03:49:51
你可以使用一些方法来初始化你的1000000个用户,或者甚至下面的代码都会为用户生成随机的年龄:
import java.util.ArrayList;
import java.util.Random;
import java.util.Scanner;
public class UserListDemo {
int age;
class Users{
int age=0;
public Users(int a)
{
age=a;
}
public void setAge(int age)
{
this.age=age;
}
public int getAge()
{
return this.age;
}
}
public static void main(String a[])
{
UserListDemo uld=new UserListDemo();
ArrayList<Users> data=new ArrayList<Users>();
uld.initializeUsers(data);
System.out.println("Enter age group choice");
System.out.println("Enter 1 for age group 1-6");
System.out.println("Enter 2 for age group 7-18");
System.out.println("Enter 3 for age group 19-26");
System.out.println("Enter 4 for age group 27-49");
System.out.println("Enter 5 for age group 50-64");
System.out.println("Enter 6 for age group 65-79");
System.out.println("Enter 7 for age group 80-Older");
Scanner sc=new Scanner(System.in);
String choice=sc.nextLine();
int ch=Integer.valueOf(choice);
long result=0;
switch(ch)
{
case 1:
for(Users us:data)
{
if(us.age<=6)
result++;
}
case 2:
for(Users us:data)
{
if( us.age>=7 && us.age<=18 )
result++;
}
case 3:
for(Users us:data)
{
if( us.age>=19 && us.age<=26 )
result++;
}
case 4:
for(Users us:data)
{
if( us.age>=27 && us.age<=49 )
result++;
}
case 5:
for(Users us:data)
{
if( us.age>=50 && us.age<=64 )
result++;
}
case 6:
for(Users us:data)
{
if( us.age>=65 && us.age<=79 )
result++;
}
case 7:
for(Users us:data)
{
if( us.age>=80)
result++;
}
}
System.out.println("For the entered age group :"+ch+" ::"+result+" user has been found");
}
public void initializeUsers(ArrayList<Users> data)
{
Users us;
Random rand=new Random();
for(long l=0;l<1000000L;l++)
{
us=new Users(rand.nextInt(100));
data.add(us);
}
}
}
发布于 2018-10-09 16:46:07
有一百万条记录的有效答案是使用几个Map作为索引,甚至是一个实际的数据库。但是,由于练习中明确提到了ArrayLists,您可能还在学习基础知识,所以我将继续学习这些基础知识。
首先,您需要能够检索给定人员的组。您可以通过两种方式来完成此操作。
//在HealthRec2 int组内;//将组号存储为私有静态最终int[] ageGroups = //每个组的年龄限制新int[]{6,18,26,49,64,79};私有空updateGroup() { // <--从构造函数和setAge() int currentGroup =0调用;for (int limit : ageGroups) { currentGroup ++;//如果(age <= limit)中断,则前进到下一组;//一旦达到一个{ group = currentGroup;} private int getGroup() { return group;}
//在私有静态最终int[] ageGroups = //每个组的年龄限制新的int[]{6,18,26,49,64,79};公共的int getGroup() { int currentGroup = 0;for (int limit : ageGroups) { currentGroup ++;//如果(age <= limit)突破,则前进到下一组;//一旦达到1}int[] currentGroup;}就停止查看限制
无论使用哪种方法,您现在都可以使用非常类似的逻辑来查找给定年龄组中的人,因为您必须查找给定州或给定性别的记录。
选项A预先更昂贵,因为即使您不需要年龄组,您仍然必须计算它并将其存储在属性中,以防万一。如果您需要为同一记录多次调用getGroup
,则选项B的开销会更大-因为选项A的getGroup
要快得多。
https://stackoverflow.com/questions/20225257
复制相似问题