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How can i print my array and my histogram on the same line?


How can i print my array and my histogram on the same line?

By : keshav coder
Date : November 22 2020, 02:59 PM
hop of those help? Here is a solution to your problem. The problem was that you were printing all of your Row data first and then all of your Integer data. You need to print them both at once. Not one after the other. Also just fyi, there is about a million different ways to make your initial code better. I would try and sit down and think of a better way to do this so you can learn (after you turn in your assignment so you are not rushed, we can all help. This is a good start though.)
First I erased the printing of the Row data, and added a constructor to your Counter class to take in your Row data.
code :
Row[] numbers = {
        new Row("1  -  10"), new Row("11 -  20"), new Row("21 -  30"),                 
        new Row("31 -  40"), new Row("41 -  50"), new Row("51 -  60"),
        new Row("61 -  70"), new Row("71 -  80"), new Row("81 -  90"),
        new Row("91 - 100")
};  
Counter section = new Counter(numbers);
section.StarCounter();
public class Counter
{
    Row[] numbers;
    public Counter(Row[] numbers) {
        this.numbers = numbers;
    }
    //Rest of your methods
}
System.out.print(numbers[0]);  //new line
for(int star = 0; star <= counted1  ; star++)
{
    System.out.print("*");
}
System.out.println();
System.out.print(numbers[1]);  //new line
for(int star = 0; star <= counted2 ; star++)
{
    System.out.print("*");
}
//Do the same for the rest
public String toString()
{
    return rows + "|";
}


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Plot as many circles as histogram values under the histogram line

Plot as many circles as histogram values under the histogram line


By : Aditya Das
Date : March 29 2020, 07:55 AM
Does that help What you have there is basically a nested selection -- for each bin, add the circles representing the values. The only difficulty in your case is that the property that determines the x position is only available in the parent and not the nesting. This is easily fixed by copying that value.
The relevant code is below.
code :
SansCircleGroup.selectAll("g.circle")
    .data(histogSans)
    .enter()
            .append("g")
            .attr("class", "circle")
            .each(function(d) { d.forEach(function(e) { e.x = d.x; }); })
            .selectAll("circle")
            .data(function (d) { return d; })
            .enter()
            .append("circle")
            .attr("cx", function (d, i) { return x(d.x); })
            .attr("cy", function (d, i) { return y(i+1); })
            .attr("r", 2)
            .style("fill", "white")
            .style("stroke", "#386cb0")
            .style("stroke-width", 1);
Reversed Digits Histogram: Given one integer N, for each digit from the end of N, print a line of stars which has number

Reversed Digits Histogram: Given one integer N, for each digit from the end of N, print a line of stars which has number


By : Orlando
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Given one integer N, for each digit from the end of N, print a line of stars which has number of stars equal to that digit C++ , Try this:
code :
void print(int num)
{
    while (num > 0)
    {
        int digit = num%10;
        for (int i=0; i<digit; i++)
            printf("*");
        printf("\n");
        num /= 10;
    }
}

int main()
{
    int num;
    scanf("%d",&num);
    print(num);
    return 0;
}
Fit the cumulative percentage line to the sorted histogram output with d3 for a pareto chart histogram

Fit the cumulative percentage line to the sorted histogram output with d3 for a pareto chart histogram


By : gaetano mercante
Date : March 29 2020, 07:55 AM
To fix the issue you can do The main issue with the line is that the cumulative distribution needs to be recalculated after the bar is sorted, or if you're gunning for a static pareto chart, the cumulative distribution needs to be calculated in the target sort order. For this purpose i've created a small function to do this calculation:
code :
function calcCDF(data){
  data.forEach(function(d,i){
      if(i === 0){
      d.cum = d.y/dataset.length
    }else{
      d.cum = (d.y/dataset.length) + data[i-1].cum
    }
  })
  return data
}
JavaScript - Print a Histogram from a given array

JavaScript - Print a Histogram from a given array


By : Nehal
Date : March 29 2020, 07:55 AM
wish help you to fix your issue // Given an array of integers [2, 1, 2, 101, 4, 95, 3, 250, 4, 1, 2, 2, 7, 98, 123, 99, ...]
code :
const dict = {}; // Empty dictionary
var min = Number.MAX_VALUE;
const maxRange = 5; // elements above maxRange will be clubbed in the same range.

//var arr = [2, 1, 2, 101, 4, 95, 3, 250, 4, 1, 2, 2, 7, 98, 123, 99];
const arr = [1, 2, 5, 3, 2, 2, 1, 5, 5, 6, 7, 1, 8, 10, 11, 12];

// iterate the array and set and update the counter in map
arr.forEach(function(num) {

  min = Math.min(min, num); // find min
  if (num > maxRange) {
    num = maxRange + 1;
  }
  dict[num] = dict[num] ? dict[num] + 1 : 1;
});

console.log("Num | Count");

// Print the occurrences per item in array starting from min to max
while (min <= maxRange + 1) {
  if (!dict[min]) { // print only those numbers which are defined in dictionary
    min++;
    continue;
  }
  var xArr = []
  var range = dict[min];
  for (i = 0; i < range; i++) {
    xArr.push('x');
  }

  var disp = (min <= maxRange) ? (min + "   | " + xArr.join("")) : (maxRange + "+  | " + xArr.join(""));
  console.log(disp);
  min = min + 1;
}
How to print histogram of ordered C++ string array?

How to print histogram of ordered C++ string array?


By : Maria Eloisa Salusti
Date : March 29 2020, 07:55 AM
I hope this helps you . The code below requires sorting and no limitation on number of words.
code :
const int wordCount = 6;
string stringText[wordCount] = {"alpha", "alpha", "bravo", "charlie", "charlie","charlie"};
int counter = 0;

while(counter<wordCount)
{
  cout<<stringText[counter];
  cout<<" : x";
  for(int i=counter+1;i<wordCount;++i)
  {
      if(stringText[i]==stringText[counter])
      {
          cout<<"x";
          counter++;
      }
  }
  cout<<endl;
  counter++;
}
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