Reasonable optimized chart scaling

Question

I need to make a chart with an optimized y axis maximum value.

The current method I have of making charts simply uses the maximum value of all the graphs, then

Answer 1

In the past I've done this in a brute force-ish sort of way. Here's a chunk of C++ code that works well... but for a hardcoded lower and upper limits (0 and 5000):

int PickYUnits()
{
    int MinSize[8] = {20, 20, 20, 20, 20, 20, 20, 20};
    int ItemsPerUnit[8] = {5, 10, 20, 25, 50, 100, 250, 500};
    int ItemLimits[8] = {20, 50, 100, 250, 500, 1000, 2500, 5000};
    int MaxNumUnits = 8;
    double PixelsPerY;
    int PixelsPerAxis;
    int Units;

    //
    // Figure out the max from the dataset
    //  - Min is always 0 for a bar chart
    //
    m_MinY = 0;
    m_MaxY = -9999999;
    m_TotalY = 0;
    for (int j = 0; j < m_DataPoints.GetSize(); j++) {
        if (m_DataPoints[j].m_y > m_MaxY) {
            m_MaxY = m_DataPoints[j].m_y;
        }

        m_TotalY += m_DataPoints[j].m_y;
    }

    //
    // Give some space at the top
    //
    m_MaxY = m_MaxY + 1;


    //
    // Figure out the size of the range
    //
    double yRange = (m_MaxY - m_MinY);

    //
    // Pick the initial size
    //
    Units = MaxNumUnits;
    for (int k = 0; k < MaxNumUnits; k++)
    {
        if (yRange < ItemLimits[k])
        {
            Units = k;
            break;
        }
    }

    //
    // Adjust it upwards based on the space available
    //
    PixelsPerY = m_rcGraph.Height() / yRange;
    PixelsPerAxis = (int)(PixelsPerY * ItemsPerUnit[Units]);

    while (PixelsPerAxis < MinSize[Units]){
        Units += 1;
        PixelsPerAxis = (int)(PixelsPerY * ItemsPerUnit[Units]);
        if (Units == 5)
            break;
    }


    return ItemsPerUnit[Units];
}

However something in what you've said tweaked me. To pick nice axis numbers a definition of "nice number" would help:

• A "nice" number is one that has 3 or fewer non-zero digits (eg. 1230000)
• A "nice" number has the same or few non-zero digits than zero digits (eg 1230 is not nice, 1200 is nice)
• The nicest numbers are ones with multiples of 3 zeros (eg. "1,000", "1,000,000")
• The second nicest numbers are onces with multples of 3 zeros plus 2 zeros (eg. "1,500,000", "1,200")

Not sure if the above definition is "right" or actually helpful (but with the definition in hand it then becomes a simpler task to devise an algorithm).

Answer 2

You could round up to two significant figures. The following pseudocode should work:

// maxValue is the largest value in your chart
magnitude = floor(log10(maxValue))
base = 10^(magnitude - 1)
chartHeight = ceiling(maxValue / base) * base

For example, if maxValue is 1357, then magnitude is 3 and base is 100. Dividing by 100, rounding up, and multiplying by 100 has the result of rounding up to the next multiple of 100, i.e. rounding up to two significant figures. In this case, the result if 1400 (1357 ⇒ 13.57 ⇒ 14 ⇒ 1400).

Answer 3

A slight refinement and tested... (works for fractions of units and not just integers)

public void testNumbers() {
        double test = 0.20000;

        double multiple = 1;
        int scale = 0;
        String[] prefix = new String[]{"", "m", "u", "n"};
        while (Math.log10(test) < 0) {
            multiple = multiple * 1000;
            test = test * 1000;
            scale++;
        }

        double tick;
        double minimum = test / 10;
        double magnitude = 100000000;
        while (minimum <= magnitude){
            magnitude = magnitude / 10;
        }

        double residual = test / (magnitude * 10);
        if (residual > 5) {
            tick = 10 * magnitude;
        } else if (residual > 2) {
            tick = 5 * magnitude;
        } else if (residual > 1) {
            tick = 2 * magnitude;
        } else {
            tick = magnitude;
        }

        double curAmt = 0;

        int ticks = (int) Math.ceil(test / tick);

        for (int ix = 0; ix < ticks; ix++) {
            curAmt += tick;
            BigDecimal bigDecimal = new BigDecimal(curAmt);
            bigDecimal.setScale(2, BigDecimal.ROUND_HALF_UP);
            System.out.println(bigDecimal.stripTrailingZeros().toPlainString() + prefix[scale] + "s");
        }

        System.out.println("Value = " + test + prefix[scale] + "s");
        System.out.println("Tick = " + tick + prefix[scale] + "s");
        System.out.println("Ticks = " + ticks);
        System.out.println("Scale = " +  multiple + " : " + scale);


    }

Answer 4

If you want 1400 at the top, how about adjusting the last two parameters to 1400 instead of 1357:

Answer 5

You could use div and mod. For example.

Let's say you want your chart to round up by increments of 20 (just to make it more a more arbitrary number than your typical "10" value).

So I would assume that 1, 11, 18 would all round up to 20. But 21, 33, 38 would round to 40.

To come up with the right value do the following:

Where divisor = your rounding increment.

divisor = 20
multiple = maxValue / divisor;  // Do an integer divide here. 
if (maxValue modulus divisor > 0)
   multiple++;

graphMax = multiple * maxValue;

So now let's plugin real numbers:

divisor = 20;
multiple = 33 / 20; (integer divide)
so multiple = 1
if (33 modulus 20 > 0)  (it is.. it equals 13) 
   multiple++;

so multiple = 2;
graphMax = multiple (2) * maxValue (20);
graphMax = 40;

Answer 6

This is from a previous similar question:

Algorithm for "nice" grid line intervals on a graph

I've done this with kind of a brute force method. First, figure out the maximum number of tick marks you can fit into the space. Divide the total range of values by the number of ticks; this is the minimum spacing of the tick. Now calculate the floor of the logarithm base 10 to get the magnitude of the tick, and divide by this value. You should end up with something in the range of 1 to 10. Simply choose the round number greater than or equal to the value and multiply it by the logarithm calculated earlier. This is your final tick spacing.

Example in Python:

import math

def BestTick(largest, mostticks):
    minimum = largest / mostticks
    magnitude = 10 ** math.floor(math.log(minimum) / math.log(10))
    residual = minimum / magnitude
    if residual > 5:
        tick = 10 * magnitude
    elif residual > 2:
        tick = 5 * magnitude
    elif residual > 1:
        tick = 2 * magnitude
    else:
        tick = magnitude
    return tick