Background info (Optional reading):
I'm running simulations of reflections of sound waves in against boundaries. The medium conditions for the points in space are set using a matrix. Let's say the dimensions of the space is an N by N grid, and there are two speeds of sound I care about, c0 and c1.
Right now I'm using code like the following to generate barrier patterns
medium.sound_speed = c0*ones(N,N); % set the speed of sound to be c0 everywhere
medium.sound_speed(:, N/2:N) = c1; % set the right half of the grid to a different speed
medium.sound_speed(50:70, 50:70) = c1; % set a box to have a different speed
Or
% set all speeds to c0 except set the diagonal to c1
medium.sound_speed = c0*ones(N,N)-(c0*eye(N,N))+c1*eye(N,N);
However, I can't generate more complex boundaries with different curvatures.
Question
I want to programmatically create matrices with patterns reflecting functions. For instance, I want to enter f(x)=2 and for that to create a matrix that looked something like this, assuming N=6.
[ 0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
1 1 1 1 1 1
0 0 0 0 0 0
0 0 0 0 0 0 ]
Or f(x)=0.5*x+1
[ 0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 1 1
0 0 1 1 0 0
1 1 0 0 0 0
0 0 0 0 0 0]
I would also be able to generate curved patterns like f(x)=1/x, which seems to require some form of the Midpoint circle algorithm, used for drawing curvatures with pixels.
[ 1 0 0 0 0 0
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 1 0 0
0 0 0 0 1 1
0 0 0 0 0 0 ]
In reality, N is at least 128, so manually creating these matrices for shapes with some level of complexity is impractical, and I thought this was an interesting problem.
Does anyone know of some way to do this, or suggestions for alternative approaches?
Thank you in advance.
Edit:
I modified this implementation of Bresenham's algorithm to provide a matrix with the desired line given an origin and an ending point.
function M=bresenham_line(point)
if (abs(point(4)-point(2)) > abs(point(3)-point(1))) % If the line is steep
x0 = point(2);y0 = point(1); x1 = point(4);y1=point(3);% then it would be converted to
token =1; % non steep by changing coordinate
else
x0 = point(1);y0 = point(2); x1 = point(3);y1=point(4);
token = 0;
end
if(x0 >x1)
temp1 = x0; x0 = x1; x1 = temp1;
temp2 = y0; y0 = y1; y1 = temp2;
end
dx = abs(x1 - x0) ; % Distance to travel in x-direction
dy = abs(y1 - y0); % Distance to travel in y-direction
sx = sign(x1 - x0); % sx indicates direction of travel in X-dir
sy = sign(y1 - y0); % Ensures positive slope line
x = x0; y = y0; % Initialization of line
param = 2*dy - dx ; % Initialization of error parameter
for i = 0:dx-1 % FOR loop to travel along X
x_coord(i+1) = x; % Saving in matrix form for plot
y_coord(i+1) = y;
param = param + 2*dy; % parameter value is modified
if (param >0) % if parameter value is exceeded
y = y +1*sy; % then y coordinate is increased
param = param - 2*(dx ); % and parameter value is decreased
end
x = x + 1*sx; % X-coordinate is increased for next point
end
M = zeros(size(x_coord,2), size(y_coord,2));
for i=1:1:size(x_coord,2)
x = x_coord(i);
y = y_coord(i);
M(x,y) = 1;
end
M
Implemented like so:
c1 = 0;
M = bresenham_line([1 1 Nx/2+1 Ny+1]);
medium.sound_speed = c0*ones(Nx,Ny) - (c0*M) + c1*M;
No progress on curved function shapes yet.
A way to get some similar results:
f = #(x)0.5*x; %create the function (x should be written even if the function doesn't depend on x: #(x) 0*x + 2)
N = 6; %choose the size of the atrix
M = zeros(N,N); %create an empty matrix
x = (1:N);
y = round(f(x-1)); %discretization
x(y>N-1|y<0) = [];
y(y>N-1|y<0) = [];
M(sub2ind(size(M),y+1,x)) = 1;
M = flipud(M)
So you can choose your function, then the result in your matrix will look like a discretization of a normal plot.
This is a slightly 'dirty' way of getting something like this, although I you best bet might Bresenham's algorithm.
N = 128;
[X,Y] = meshgrid(1:N,1:N);
bound1 = Y<2*X;
bound2 = Y<2*X+1;
M = xor(bound1,bound2);
bound1 you can define any function y=f(x), and mark the area under it. with bound2 you select and area that is slightly higher (shifted up). Once you take and xor of the two area you get just the desired y=f(x) marked. I think that in order to get reasonable results the shift might be different for more complicated function.
For illustration I used imagesc (the flipud is just for make the (0,0) in the bottom left, instead of the top left):
imagesc(flipud(M));
Edit
Indeed for some function this might not be the best. For example for y=x^2, you have to increase the shift and still does not look great.
bound1 = Y<X.^2;
bound2 = Y<X.^2+15;
M = xor(bound1,bound2);
Recently I found this in some code I wrote a few years ago. It was used to rationalize a real value (within a tolerance) by determining a suitable denominator and then checking if the difference between the original real and the rational was small enough.
Edit to clarify : I actually don't want to convert all real values. For instance I could choose a max denominator of 14, and a real value that equals 7/15 would stay as-is. It's not as clear that as it's an outside variable in the algorithms I wrote here.
The algorithm to get the denominator was this (pseudocode):
denominator(x)
frac = fractional part of x
recip = 1/frac
if (frac < tol)
return 1
else
return recip * denominator(recip)
end
end
Seems to be based on continued fractions although it became clear on looking at it again that it was wrong. (It worked for me because it would eventually just spit out infinity, which I handled outside, but it would be often really slow.) The value for tol doesn't really do anything except in the case of termination or for numbers that end up close. I don't think it's relatable to the tolerance for the real - rational conversion.
I've replaced it with an iterative version that is not only faster but I'm pretty sure it won't fail theoretically (d = 1 to start with and fractional part returns a positive, so recip is always >= 1) :
denom_iter(x d)
return d if d > maxd
frac = fractional part of x
recip = 1/frac
if (frac = 0)
return d
else
return denom_iter(recip d*recip)
end
end
What I'm curious to know if there's a way to pick the maxd that will ensure that it converts all values that are possible for a given tolerance. I'm assuming 1/tol but don't want to miss something. I'm also wondering if there's an way in this approach to actually limit the denominator size - this allows some denominators larger than maxd.
This can be considered a 2D minimization problem on error:
ArgMin ( r - q / p ), where r is real, q and p are integers
I suggest the use of Gradient Descent algorithm . The gradient in this objective function is:
f'(q, p) = (-1/p, q/p^2)
The initial guess r_o can be q being the closest integer to r, and p being 1.
The stopping condition can be thresholding of the error.
The pseudo-code of GD can be found in wiki: http://en.wikipedia.org/wiki/Gradient_descent
If the initial guess is close enough, the objective function should be convex.
As Jacob suggested, this problem can be better solved by minimizing the following error function:
ArgMin ( p * r - q ), where r is real, q and p are integers
This is linear programming, which can be efficiently solved by any ILP (Integer Linear Programming) solvers. GD works on non-linear cases, but lack efficiency in linear problems.
Initial guesses and stopping condition can be similar to stated above. Better choice can be obtained for individual choice of solver.
I suggest you should still assume convexity near the local minimum, which can greatly reduce cost. You can also try Simplex method, which is great on linear programming problem.
I give credit to Jacob on this.
A problem similar to this is solved in the Approximations section beginning ca. page 28 of Bill Gosper's Continued Fraction Arithmetic document. (Ref: postscript file; also see text version, from line 1984.) The general idea is to compute continued-fraction approximations of the low-end and high-end range limiting numbers, until the two fractions differ, and then choose a value in the range of those two approximations. This is guaranteed to give a simplest fraction, using Gosper's terminology.
The python code below (program "simpleden") implements a similar process. (It probably is not as good as Gosper's suggested implementation, but is good enough that you can see what kind of results the method produces.) The amount of work done is similar to that for Euclid's algorithm, ie O(n) for numbers with n bits, so the program is reasonably fast. Some example test cases (ie the program's output) are shown after the code itself. Note, function simpleratio(vlo, vhi) as shown here returns -1 if vhi is smaller than vlo.
#!/usr/bin/env python
def simpleratio(vlo, vhi):
rlo, rhi, eps = vlo, vhi, 0.0000001
if vhi < vlo: return -1
num = denp = 1
nump = den = 0
while 1:
klo, khi = int(rlo), int(rhi)
if klo != khi or rlo-klo < eps or rhi-khi < eps:
tlo = denp + klo * den
thi = denp + khi * den
if tlo < thi:
return tlo + (rlo-klo > eps)*den
elif thi < tlo:
return thi + (rhi-khi > eps)*den
else:
return tlo
nump, num = num, nump + klo * num
denp, den = den, denp + klo * den
rlo, rhi = 1/(rlo-klo), 1/(rhi-khi)
def test(vlo, vhi):
den = simpleratio(vlo, vhi);
fden = float(den)
ilo, ihi = int(vlo*den), int(vhi*den)
rlo, rhi = ilo/fden, ihi/fden;
izok = 'ok' if rlo <= vlo <= rhi <= vhi else 'wrong'
print '{:4d}/{:4d} = {:0.8f} vlo:{:0.8f} {:4d}/{:4d} = {:0.8f} vhi:{:0.8f} {}'.format(ilo,den,rlo,vlo, ihi,den,rhi,vhi, izok)
test (0.685, 0.695)
test (0.685, 0.7)
test (0.685, 0.71)
test (0.685, 0.75)
test (0.685, 0.76)
test (0.75, 0.76)
test (2.173, 2.177)
test (2.373, 2.377)
test (3.484, 3.487)
test (4.0, 4.87)
test (4.0, 8.0)
test (5.5, 5.6)
test (5.5, 6.5)
test (7.5, 7.3)
test (7.5, 7.5)
test (8.534537, 8.534538)
test (9.343221, 9.343222)
Output from program:
> ./simpleden
8/ 13 = 0.61538462 vlo:0.68500000 9/ 13 = 0.69230769 vhi:0.69500000 ok
6/ 10 = 0.60000000 vlo:0.68500000 7/ 10 = 0.70000000 vhi:0.70000000 ok
6/ 10 = 0.60000000 vlo:0.68500000 7/ 10 = 0.70000000 vhi:0.71000000 ok
2/ 4 = 0.50000000 vlo:0.68500000 3/ 4 = 0.75000000 vhi:0.75000000 ok
2/ 4 = 0.50000000 vlo:0.68500000 3/ 4 = 0.75000000 vhi:0.76000000 ok
3/ 4 = 0.75000000 vlo:0.75000000 3/ 4 = 0.75000000 vhi:0.76000000 ok
36/ 17 = 2.11764706 vlo:2.17300000 37/ 17 = 2.17647059 vhi:2.17700000 ok
18/ 8 = 2.25000000 vlo:2.37300000 19/ 8 = 2.37500000 vhi:2.37700000 ok
114/ 33 = 3.45454545 vlo:3.48400000 115/ 33 = 3.48484848 vhi:3.48700000 ok
4/ 1 = 4.00000000 vlo:4.00000000 4/ 1 = 4.00000000 vhi:4.87000000 ok
4/ 1 = 4.00000000 vlo:4.00000000 8/ 1 = 8.00000000 vhi:8.00000000 ok
11/ 2 = 5.50000000 vlo:5.50000000 11/ 2 = 5.50000000 vhi:5.60000000 ok
5/ 1 = 5.00000000 vlo:5.50000000 6/ 1 = 6.00000000 vhi:6.50000000 ok
-7/ -1 = 7.00000000 vlo:7.50000000 -7/ -1 = 7.00000000 vhi:7.30000000 wrong
15/ 2 = 7.50000000 vlo:7.50000000 15/ 2 = 7.50000000 vhi:7.50000000 ok
8030/ 941 = 8.53347503 vlo:8.53453700 8031/ 941 = 8.53453773 vhi:8.53453800 ok
24880/2663 = 9.34284641 vlo:9.34322100 24881/2663 = 9.34322193 vhi:9.34322200 ok
If, rather than the simplest fraction in a range, you seek the best approximation given some upper limit on denominator size, consider code like the following, which replaces all the code from def test(vlo, vhi) forward.
def smallden(target, maxden):
global pas
pas = 0
tol = 1/float(maxden)**2
while 1:
den = simpleratio(target-tol, target+tol);
if den <= maxden: return den
tol *= 2
pas += 1
# Test driver for smallden(target, maxden) routine
import random
totalpass, trials, passes = 0, 20, [0 for i in range(20)]
print 'Maxden Num Den Num/Den Target Error Passes'
for i in range(trials):
target = random.random()
maxden = 10 + round(10000*random.random())
den = smallden(target, maxden)
num = int(round(target*den))
got = float(num)/den
print '{:4d} {:4d}/{:4d} = {:10.8f} = {:10.8f} + {:12.9f} {:2}'.format(
int(maxden), num, den, got, target, got - target, pas)
totalpass += pas
passes[pas-1] += 1
print 'Average pass count: {:0.3}\nPass histo: {}'.format(
float(totalpass)/trials, passes)
In production code, drop out all the references to pas (etc.), ie, drop out pass-counting code.
The routine smallden is given a target value and a maximum value for allowed denominators. Given maxden possible choices of denominators, it's reasonable to suppose that a tolerance on the order of 1/maxden² can be achieved. The pass-counts shown in the following typical output (where target and maxden were set via random numbers) illustrate that such a tolerance was reached immediately more than half the time, but in other cases tolerances 2 or 4 or 8 times as large were used, requiring extra calls to simpleratio. Note, the last two lines of output from a 10000-number test run are shown following the complete output of a 20-number test run.
Maxden Num Den Num/Den Target Error Passes
1198 32/ 509 = 0.06286837 = 0.06286798 + 0.000000392 1
2136 115/ 427 = 0.26932084 = 0.26932103 + -0.000000185 1
4257 839/2670 = 0.31423221 = 0.31423223 + -0.000000025 1
2680 449/ 509 = 0.88212181 = 0.88212132 + 0.000000486 3
2935 440/1853 = 0.23745278 = 0.23745287 + -0.000000095 1
6128 347/1285 = 0.27003891 = 0.27003899 + -0.000000077 3
8041 1780/4243 = 0.41951449 = 0.41951447 + 0.000000020 2
7637 3926/7127 = 0.55086292 = 0.55086293 + -0.000000010 1
3422 27/ 469 = 0.05756930 = 0.05756918 + 0.000000113 2
1616 168/1507 = 0.11147976 = 0.11147982 + -0.000000061 1
260 62/ 123 = 0.50406504 = 0.50406378 + 0.000001264 1
3775 52/3327 = 0.01562970 = 0.01562750 + 0.000002195 6
233 6/ 13 = 0.46153846 = 0.46172772 + -0.000189254 5
3650 3151/3514 = 0.89669892 = 0.89669890 + 0.000000020 1
9307 2943/7528 = 0.39094049 = 0.39094048 + 0.000000013 2
962 206/ 225 = 0.91555556 = 0.91555496 + 0.000000594 1
2080 564/1975 = 0.28556962 = 0.28556943 + 0.000000190 1
6505 1971/2347 = 0.83979548 = 0.83979551 + -0.000000022 1
1944 472/ 833 = 0.56662665 = 0.56662696 + -0.000000305 2
3244 291/1447 = 0.20110574 = 0.20110579 + -0.000000051 1
Average pass count: 1.85
Pass histo: [12, 4, 2, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
The last two lines of output from a 10000-number test run:
Average pass count: 1.77
Pass histo: [56659, 25227, 10020, 4146, 2072, 931, 497, 233, 125, 39, 33, 17, 1, 0, 0, 0, 0, 0, 0, 0]
I need a reasonably smart algorithm to come up with "nice" grid lines for a graph (chart).
For example, assume a bar chart with values of 10, 30, 72 and 60. You know:
Min value: 10
Max value: 72
Range: 62
The first question is: what do you start from? In this case, 0 would be the intuitive value but this won't hold up on other data sets so I'm guessing:
Grid min value should be either 0 or a "nice" value lower than the min value of the data in range. Alternatively, it can be specified.
Grid max value should be a "nice" value above the max value in the range. Alternatively, it can be specified (eg you might want 0 to 100 if you're showing percentages, irrespective of the actual values).
The number of grid lines (ticks) in the range should be either specified or a number within a given range (eg 3-8) such that the values are "nice" (ie round numbers) and you maximise use of the chart area. In our example, 80 would be a sensible max as that would use 90% of the chart height (72/80) whereas 100 would create more wasted space.
Anyone know of a good algorithm for this? Language is irrelevant as I'll implement it in what I need to.
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, 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
Edit: you are free to alter the selection of "nice" intervals. One commenter appears to be dissatisfied with the selections provided, because the actual number of ticks can be up to 2.5 times less than the maximum. Here's a slight modification that defines a table for the nice intervals. In the example, I've expanded the selections so that the number of ticks won't be less than 3/5 of the maximum.
import bisect
def BestTick2(largest, mostticks):
minimum = largest / mostticks
magnitude = 10 ** math.floor(math.log(minimum, 10))
residual = minimum / magnitude
# this table must begin with 1 and end with 10
table = [1, 1.5, 2, 3, 5, 7, 10]
tick = table[bisect.bisect_right(table, residual)] if residual < 10 else 10
return tick * magnitude
There are 2 pieces to the problem:
Determine the order of magnitude involved, and
Round to something convenient.
You can handle the first part by using logarithms:
range = max - min;
exponent = int(log(range)); // See comment below.
magnitude = pow(10, exponent);
So, for example, if your range is from 50 - 1200, the exponent is 3 and the magnitude is 1000.
Then deal with the second part by deciding how many subdivisions you want in your grid:
value_per_division = magnitude / subdivisions;
This is a rough calculation because the exponent has been truncated to an integer. You may want to tweak the exponent calculation to handle boundary conditions better, e.g. by rounding instead of taking the int() if you end up with too many subdivisions.
I use the following algorithm. It's similar to others posted here but it's the first example in C#.
public static class AxisUtil
{
public static float CalcStepSize(float range, float targetSteps)
{
// calculate an initial guess at step size
var tempStep = range/targetSteps;
// get the magnitude of the step size
var mag = (float)Math.Floor(Math.Log10(tempStep));
var magPow = (float)Math.Pow(10, mag);
// calculate most significant digit of the new step size
var magMsd = (int)(tempStep/magPow + 0.5);
// promote the MSD to either 1, 2, or 5
if (magMsd > 5)
magMsd = 10;
else if (magMsd > 2)
magMsd = 5;
else if (magMsd > 1)
magMsd = 2;
return magMsd*magPow;
}
}
CPAN provides an implementation here (see source link)
See also Tickmark algorithm for a graph axis
FYI, with your sample data:
Maple: Min=8, Max=74, Labels=10,20,..,60,70, Ticks=10,12,14,..70,72
MATLAB: Min=10, Max=80, Labels=10,20,,..,60,80
Here's another implementation in JavaScript:
var calcStepSize = function(range, targetSteps)
{
// calculate an initial guess at step size
var tempStep = range / targetSteps;
// get the magnitude of the step size
var mag = Math.floor(Math.log(tempStep) / Math.LN10);
var magPow = Math.pow(10, mag);
// calculate most significant digit of the new step size
var magMsd = Math.round(tempStep / magPow + 0.5);
// promote the MSD to either 1, 2, or 5
if (magMsd > 5.0)
magMsd = 10.0;
else if (magMsd > 2.0)
magMsd = 5.0;
else if (magMsd > 1.0)
magMsd = 2.0;
return magMsd * magPow;
};
I am the author of "Algorithm for Optimal Scaling on a Chart Axis". It used to be hosted on trollop.org, but I have recently moved domains/blogging engines.
Please see my answer to a related question.
Taken from Mark above, a slightly more complete Util class in c#. That also calculates a suitable first and last tick.
public class AxisAssists
{
public double Tick { get; private set; }
public AxisAssists(double aTick)
{
Tick = aTick;
}
public AxisAssists(double range, int mostticks)
{
var minimum = range / mostticks;
var magnitude = Math.Pow(10.0, (Math.Floor(Math.Log(minimum) / Math.Log(10))));
var residual = minimum / magnitude;
if (residual > 5)
{
Tick = 10 * magnitude;
}
else if (residual > 2)
{
Tick = 5 * magnitude;
}
else if (residual > 1)
{
Tick = 2 * magnitude;
}
else
{
Tick = magnitude;
}
}
public double GetClosestTickBelow(double v)
{
return Tick* Math.Floor(v / Tick);
}
public double GetClosestTickAbove(double v)
{
return Tick * Math.Ceiling(v / Tick);
}
}
With ability to create an instance, but if you just want calculate and throw it away:
double tickX = new AxisAssists(aMaxX - aMinX, 8).Tick;
I wrote an objective-c method to return a nice axis scale and nice ticks for given min- and max values of your data set:
- (NSArray*)niceAxis:(double)minValue :(double)maxValue
{
double min_ = 0, max_ = 0, min = minValue, max = maxValue, power = 0, factor = 0, tickWidth, minAxisValue = 0, maxAxisValue = 0;
NSArray *factorArray = [NSArray arrayWithObjects:#"0.0f",#"1.2f",#"2.5f",#"5.0f",#"10.0f",nil];
NSArray *scalarArray = [NSArray arrayWithObjects:#"0.2f",#"0.2f",#"0.5f",#"1.0f",#"2.0f",nil];
// calculate x-axis nice scale and ticks
// 1. min_
if (min == 0) {
min_ = 0;
}
else if (min > 0) {
min_ = MAX(0, min-(max-min)/100);
}
else {
min_ = min-(max-min)/100;
}
// 2. max_
if (max == 0) {
if (min == 0) {
max_ = 1;
}
else {
max_ = 0;
}
}
else if (max < 0) {
max_ = MIN(0, max+(max-min)/100);
}
else {
max_ = max+(max-min)/100;
}
// 3. power
power = log(max_ - min_) / log(10);
// 4. factor
factor = pow(10, power - floor(power));
// 5. nice ticks
for (NSInteger i = 0; factor > [[factorArray objectAtIndex:i]doubleValue] ; i++) {
tickWidth = [[scalarArray objectAtIndex:i]doubleValue] * pow(10, floor(power));
}
// 6. min-axisValues
minAxisValue = tickWidth * floor(min_/tickWidth);
// 7. min-axisValues
maxAxisValue = tickWidth * floor((max_/tickWidth)+1);
// 8. create NSArray to return
NSArray *niceAxisValues = [NSArray arrayWithObjects:[NSNumber numberWithDouble:minAxisValue], [NSNumber numberWithDouble:maxAxisValue],[NSNumber numberWithDouble:tickWidth], nil];
return niceAxisValues;
}
You can call the method like this:
NSArray *niceYAxisValues = [self niceAxis:-maxy :maxy];
and get you axis setup:
double minYAxisValue = [[niceYAxisValues objectAtIndex:0]doubleValue];
double maxYAxisValue = [[niceYAxisValues objectAtIndex:1]doubleValue];
double ticksYAxis = [[niceYAxisValues objectAtIndex:2]doubleValue];
Just in case you want to limit the number of axis ticks do this:
NSInteger maxNumberOfTicks = 9;
NSInteger numberOfTicks = valueXRange / ticksXAxis;
NSInteger newNumberOfTicks = floor(numberOfTicks / (1 + floor(numberOfTicks/(maxNumberOfTicks+0.5))));
double newTicksXAxis = ticksXAxis * (1 + floor(numberOfTicks/(maxNumberOfTicks+0.5)));
The first part of the code is based on the calculation I found here to calculate nice graph axis scale and ticks similar to excel graphs. It works excellent for all kind of data sets. Here is an example of an iPhone implementation:
Another idea is to have the range of the axis be the range of the values, but put the tick marks at the appropriate position.. i.e. for 7 to 22 do:
[- - - | - - - - | - - - - | - - ]
10 15 20
As for selecting the tick spacing, I would suggest any number of the form 10^x * i / n, where i < n, and 0 < n < 10. Generate this list, and sort them, and you can find the largest number smaller than value_per_division (as in adam_liss) using a binary search.
Using a lot of inspiration from answers already availible here, here's my implementation in C. Note that there's some extendibility built into the ndex array.
float findNiceDelta(float maxvalue, int count)
{
float step = maxvalue/count,
order = powf(10, floorf(log10(step))),
delta = (int)(step/order + 0.5);
static float ndex[] = {1, 1.5, 2, 2.5, 5, 10};
static int ndexLenght = sizeof(ndex)/sizeof(float);
for(int i = ndexLenght - 2; i > 0; --i)
if(delta > ndex[i]) return ndex[i + 1] * order;
return delta*order;
}
In R, use
tickSize <- function(range,minCount){
logMaxTick <- log10(range/minCount)
exponent <- floor(logMaxTick)
mantissa <- 10^(logMaxTick-exponent)
af <- c(1,2,5) # allowed factors
mantissa <- af[findInterval(mantissa,af)]
return(mantissa*10^exponent)
}
where range argument is max-min of domain.
Here is a javascript function I wrote to round grid intervals (max-min)/gridLinesNumber to beautiful values. It works with any numbers, see the gist with detailed commets to find out how it works and how to call it.
var ceilAbs = function(num, to, bias) {
if (to == undefined) to = [-2, -5, -10]
if (bias == undefined) bias = 0
var numAbs = Math.abs(num) - bias
var exp = Math.floor( Math.log10(numAbs) )
if (typeof to == 'number') {
return Math.sign(num) * to * Math.ceil(numAbs/to) + bias
}
var mults = to.filter(function(value) {return value > 0})
to = to.filter(function(value) {return value < 0}).map(Math.abs)
var m = Math.abs(numAbs) * Math.pow(10, -exp)
var mRounded = Infinity
for (var i=0; i<mults.length; i++) {
var candidate = mults[i] * Math.ceil(m / mults[i])
if (candidate < mRounded)
mRounded = candidate
}
for (var i=0; i<to.length; i++) {
if (to[i] >= m && to[i] < mRounded)
mRounded = to[i]
}
return Math.sign(num) * mRounded * Math.pow(10, exp) + bias
}
Calling ceilAbs(number, [0.5]) for different numbers will round numbers like that:
301573431.1193228 -> 350000000
14127.786597236991 -> 15000
-63105746.17236853 -> -65000000
-718854.2201183736 -> -750000
-700660.340487957 -> -750000
0.055717507097870114 -> 0.06
0.0008068701205775142 -> 0.00085
-8.66660070605576 -> -9
-400.09256079792976 -> -450
0.0011740548815578223 -> 0.0015
-5.3003294346854085e-8 -> -6e-8
-0.00005815960629843176 -> -0.00006
-742465964.5184875 -> -750000000
-81289225.90985894 -> -85000000
0.000901771713513881 -> 0.00095
-652726598.5496342 -> -700000000
-0.6498901364393532 -> -0.65
0.9978325804695487 -> 1
5409.4078950583935 -> 5500
26906671.095639467 -> 30000000
Check out the fiddle to experiment with the code. Code in the answer, the gist and the fiddle is slightly different I'm using the one given in the answer.
If you are trying to get the scales looking right on VB.NET charts, then I've used the example from Adam Liss, but make sure when you set the min and max scale values that you pass them in from a variable of type decimal (not of type single or double) otherwise the tick mark values end up being set to like 8 decimal places.
So as an example, I had 1 chart where I set the min Y Axis value to 0.0001 and the max Y Axis value to 0.002.
If I pass these values to the chart object as singles I get tick mark values of 0.00048000001697801, 0.000860000036482233 ....
Whereas if I pass these values to the chart object as decimals I get nice tick mark values of 0.00048, 0.00086 ......
In python:
steps = [numpy.round(x) for x in np.linspace(min, max, num=num_of_steps)]
Answer that can dynamically always plot 0, handle positive and negatives, and small and large numbers, gives the tick interval size and how many to plot; written in Go
forcePlotZero changes how the max values are rounded so it'll always make a nice multiple to then get back to zero. Example:
if forcePlotZero == false then 237 --> 240
if forcePlotZero == true then 237 --> 300
Intervals are calculated by getting the multiple of 10/100/1000 etc for max and then subtracting till the cumulative total of these subtractions is < min
Here's the output from the function, along with showing forcePlotZero
Force to plot zero
max and min inputs
rounded max and min
intervals
forcePlotZero=false
min: -104 max: 240
minned: -160 maxed: 240
intervalCount: 5 intervalSize: 100
forcePlotZero=true
min: -104 max: 240
minned: -200 maxed: 300
intervalCount: 6 intervalSize: 100
forcePlotZero=false
min: 40 max: 1240
minned: 0 maxed: 1300
intervalCount: 14 intervalSize: 100
forcePlotZero=false
min: 200 max: 240
minned: 190 maxed: 240
intervalCount: 6 intervalSize: 10
forcePlotZero=false
min: 0.7 max: 1.12
minned: 0.6 maxed: 1.2
intervalCount: 7 intervalSize: 0.1
forcePlotZero=false
min: -70.5 max: -12.5
minned: -80 maxed: -10
intervalCount: 8 intervalSize: 10
Here's the playground link https://play.golang.org/p/1IhiX_hRQvo
func getMaxMinIntervals(max float64, min float64, forcePlotZero bool) (maxRounded float64, minRounded float64, intervalCount float64, intervalSize float64) {
//STEP 1: start off determining the maxRounded value for the axis
precision := 0.0
precisionDampener := 0.0 //adjusts to prevent 235 going to 300, instead dampens the scaling to get 240
epsilon := 0.0000001
if math.Abs(max) >= 0 && math.Abs(max) < 2 {
precision = math.Floor(-math.Log10(epsilon + math.Abs(max) - math.Floor(math.Abs(max)))) //counting number of zeros between decimal point and rightward digits
precisionDampener = 1
precision = precision + precisionDampener
} else if math.Abs(max) >= 2 && math.Abs(max) < 100 {
precision = math.Ceil(math.Log10(math.Abs(max)+1)) * -1 //else count number of digits before decimal point
precisionDampener = 1
precision = precision + precisionDampener
} else {
precision = math.Ceil(math.Log10(math.Abs(max)+1)) * -1 //else count number of digits before decimal point
precisionDampener = 2
if forcePlotZero == true {
precisionDampener = 1
}
precision = precision + precisionDampener
}
useThisFactorForIntervalCalculation := 0.0 // this is needed because intervals are calculated from the max value with a zero origin, this uses range for min - max
if max < 0 {
maxRounded = (math.Floor(math.Abs(max)*(math.Pow10(int(precision)))) / math.Pow10(int(precision)) * -1)
useThisFactorForIntervalCalculation = (math.Floor(math.Abs(max)*(math.Pow10(int(precision)))) / math.Pow10(int(precision))) + ((math.Ceil(math.Abs(min)*(math.Pow10(int(precision)))) / math.Pow10(int(precision))) * -1)
} else {
maxRounded = math.Ceil(max*(math.Pow10(int(precision)))) / math.Pow10(int(precision))
useThisFactorForIntervalCalculation = maxRounded
}
minNumberOfIntervals := 2.0
maxNumberOfIntervals := 19.0
intervalSize = 0.001
intervalCount = minNumberOfIntervals
//STEP 2: get interval size (the step size on the axis)
for {
if math.Abs(useThisFactorForIntervalCalculation)/intervalSize < minNumberOfIntervals || math.Abs(useThisFactorForIntervalCalculation)/intervalSize > maxNumberOfIntervals {
intervalSize = intervalSize * 10
} else {
break
}
}
//STEP 3: check that intervals are not too large, safety for max and min values that are close together (240, 220 etc)
for {
if max-min < intervalSize {
intervalSize = intervalSize / 10
} else {
break
}
}
//STEP 4: now we can get minRounded by adding the interval size to 0 till we get to the point where another increment would make cumulative increments > min, opposite for negative in
minRounded = 0.0
if min >= 0 {
for {
if minRounded < min {
minRounded = minRounded + intervalSize
} else {
minRounded = minRounded - intervalSize
break
}
}
} else {
minRounded = maxRounded //keep going down, decreasing by the interval size till minRounded < min
for {
if minRounded > min {
minRounded = minRounded - intervalSize
} else {
break
}
}
}
//STEP 5: get number of intervals to draw
intervalCount = (maxRounded - minRounded) / intervalSize
intervalCount = math.Ceil(intervalCount) + 1 // include the origin as an interval
//STEP 6: Check that the intervalCount isn't too high
if intervalCount-1 >= (intervalSize * 2) && intervalCount > maxNumberOfIntervals {
intervalCount = math.Ceil(intervalCount / 2)
intervalSize *= 2
}
return}
This is in python and for base 10.
Doesn't cover all your questions but I think you can build on it
import numpy as np
def create_ticks(lo,hi):
s = 10**(np.floor(np.log10(hi - lo)))
start = s * np.floor(lo / s)
end = s * np.ceil(hi / s)
ticks = [start]
t = start
while (t < end):
ticks += [t]
t = t + s
return ticks