我有一个超过1000个名字的文件,它还包括性别和有多少人有这个名字。
例如
Sarah F 2000
我正在尝试打印从for循环创建的前10行,但出于某种原因,我尝试的只是打印最后一行10次。
import java.util.*;
import java.io.*;
import java.util.Collections;
public class NameYear
{
private String year;
ArrayList<OneName> oneName = new ArrayList<OneName>();
public NameYear(String year)
{
String line = "";
String Top = "";
Scanner sc = null;
try
{
sc = new Scanner(new File
("/home/mathcs/courses/cs225/koch/names/yob"+year+".txt"));
}
catch (Exception e)
{
System.out.println("Error Year should be between 1880 and 2013 not "+ year);
System.exit(1);
}
while(sc.hasNextLine())
{
// read a line from the input file via sc into line
line = sc.nextLine();
StringTokenizer stk = new StringTokenizer(line, ",");
String name = stk.nextToken();
char sex = stk.nextToken().charAt(0);
int count = Integer.parseInt(stk.nextToken());
OneName list = new OneName(name, sex, count);
oneName.add(list);
}
for (int i = 0 ; i < 10; i++)
{
System.out.println(descending());
}
public String descending()
{
String x = "";
Collections.sort(oneName, new OneNameCountCompare());
for(OneName b: oneName)
{
x = b.toString();
}
return x;
OneName文件
public class OneName
{
private String Name;
private char Sex;
private int Count;
public OneName(String name, char sex, int count)
{
Name = name;
Sex = sex;
Count = count;
}
public String getName()
{
return Name;
}
public char getSex()
{
return Sex;
}
public int getCount()
{
return Count;
}
public void setName(String name)
{
if (name.length() < 1)
{
throw new NullPointerException("Baby name is missing");
}
Name = name;
}
private char M;
private char F;
public void setSex(char sex)
{
if( sex != M)
{
if(sex != F)
{
throw new IllegalArgumentException("Sex has to be M or F");
}
}
Sex = sex;
}
public void setCount(int count)
{
if(count < 0)
{
throw new IllegalArgumentException("Count cant be negative");
}
Count = count;
}
public String toString()
{
return String.format("%s %c %d", Name, Sex, Count);
}
}
OneNameCount
import java.util.Comparator;
import java.util.Collections;
public class OneNameCountCompare implements Comparator<OneName>
{
public int compare(OneName b1, OneName b2)
{
if(b1.getCount() <b2.getCount())
{
return 1;
}
else
{
return -1;
}
}
}
主程序
import java.io.*;
import java.util.*;
public class TopNames
{
public static void main(String args[])
{
String line = ""; // string var to hold entire line
if (args.length < 1)
{
System.out.println("\nYou forgot to put a Year on the command line.");
System.exit(1);
};
String inFile = args[0]; // file name off command line
String year = inFile;
NameYear list = new NameYear(year);
}
}
答案 0 :(得分:3)
您的descending
函数返回一个字符串,并且始终使用相同的字符串(排序集合后的最后一个字符串)。你调用它的频率并不重要,如果数据没有变化,你总会得到同样的,最后一个字符串。
如果您想要排序后的前10个,descending
需要返回包含10个的List<String>
:
public List<String> descending()
{
List<String> x = new ArrayList<String>(10);
Collections.sort(oneName, new OneNameCountCompare());
for(OneName b: oneName)
{
x.add(b.toString());
if (x.size() == 10) // Or don't use enhanced for, use an index instead
{
break;
}
}
return x;
}
然后在打印时,将for (int i = 0 ; i < 10; i++)
循环替换为:
for (String s : descending())
{
System.out.println(s);
}
答案 1 :(得分:0)
您的错误在这里:
for (int i = 0 ; i < 10; i++) {
System.out.println(descending());
}
public String descending() {
String x = "";
Collections.sort(oneName, new OneNameCountCompare());
for(OneName b: oneName) {
x = b.toString();
}
return x;
}
首先,在for循环中,您没有使用作为计数指示符的 i 变量。这意味着descending()方法没有任何 i 的意识,他如何能够返回不同的东西?
尝试修改descending(),如下所示:
public String descending(int i) {
String x = "";
Collections.sort(oneName, new OneNameCountCompare());
OneName b = oneName.get(i);
x = b.toString();
return x;
}