如何从 Java 中的数组生成直方图输出？
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how do I generate histogram output from an array in Java?
提问by MaryK
so I have a method that I call to my main which takes an array of 10 numbers and creates a histogram output in a nested for loop. I can't figure out how to get the correct number of asterisks next to the row title. The array is passed to the method from a previous method. Thanks!!
所以我有一个方法可以调用我的 main，它接受一个包含 10 个数字的数组，并在嵌套的 for 循环中创建一个直方图输出。我不知道如何在行标题旁边获得正确数量的星号。数组从前一个方法传递给方法。谢谢！！
public static void outputHistogram(int [] list)
{
int k =0;
for(int i=0;i<=9;++i)
{
System.out.print((i*10+1) +"-"+(i*10+10)+":"+"\t");
for(int j=1; j<=i;++j)
System.out.print("*");
System.out.println();
}
}
回答by SudoRahul
If you plan to use your list, then you can probably have a forloop which looks like this:-
如果你打算使用你的list，那么你可能有一个for看起来像这样的循环：-
for(int i=0;i<list.length;i++)
for(int i=0;i<list.length;i++)
and the second forloop should be like this:-
第二个for循环应该是这样的：-
for(int j=0; j<list[i];j++)
for(int j=0; j<list[i];j++)
回答by Chris Cooper
Your test in the second for loop needs to be j < list[i]
您在第二个 for 循环中的测试需要是 j < list[i]
回答by ghdalum
You don't even use your list anywhere in the outputHistorgram method. I don't know exactly was this should produce, but maybe doing the for-loop like this might help:
您甚至不在 outputHistorgram 方法中的任何地方使用您的列表。我不知道这到底是不是应该产生，但也许像这样执行 for 循环可能会有所帮助：
for(int i : list)
If this is not correct then give example input and output.
如果这不正确，则给出示例输入和输出。
回答by vishal bhuva
I think you want to do this.
我想你想这样做。
public static void outputHistogram(int [] list)
{
{
int k =0;
for(int i=0;i<=9;++i)
{
System.out.print((i*10+1) +"-"+(i*10+10)+":"+"\t");
for(int j=1; j<=(i*10+10);++j)
System.out.print("*");
System.out.println();
}
}
}
回答by Riyar Ajay
import java.util.Scanner;
class Histogram
{
private int count[]=new int[10]; // count array will keep elements of element
// in particular range;
public void showHistogram(int elements[]) // for example 27 15 34 22 11 11 19
{ // in above input there is count[0]=0;
for(int i=0;i<elements.length;i++) // count[1]=4 and count[2]=2 and count[3]=1;
{
if(elements[i]>=0 && elements[i]<50)
{
if(elements[i]<10)
{
count[0]++;
}
else if(elements[i]>=10 && elements[i]<20)
{
count[1]++;
}
else if(elements[i]>=20 && elements[i]<30)
{
count[2]++;
}
else if(elements[i]>=30 && elements[i]<40)
{
count[3]++;
}
else
{
count[4]++;
}
}
else if(elements[i]>=50 &&elements[i]<=100)
{
if(elements[i]<60)
{
count[5]++;
}
else if(elements[i]>=60 && elements[i]<70)
{
count[6]++;
}
else if(elements[i]>=70 && elements[i]<80)
{
count[7]++;
}
else if(elements[i]>=80 && elements[i]<90)
{
count[8]++;
}
else
{
count[9]++;
}
}
}
showHistogram1();
}
private void showHistogram1()
{
System.out.println("Histogram of the elements:");
for(int i=0;i<count.length;i++) // this loop will print line
{
for(int j=0;j<count[i];j++) // this will print elements element(*)
{ // at each line.
System.out.print("* ");
}
if(count[i]!=0) // if line does'nt contain zero
System.out.println(""); // then if will change the row;
}
}
}
/*
in above code if count[i]=zero means if there is elements
element in particular range say [0-9] then it will
elementst jump on next line;
*/
class HistogramGenerator //Question3
{
public static void main(String args[])
{
Histogram hg=new Histogram();
System.out.println("Enter the elements of Elements want in a Histogram:");
Scanner sc=new Scanner(System.in);
int noOfElements=sc.nextInt();
int histogramElements[]=new int[noOfElements];
System.out.println("Enter the Elements for Histogram:");
for(int i=0;i<noOfElements;i++)
{
histogramElements[i]=sc.nextInt();
}
hg.showHistogram(histogramElements);
}
}
