Develop Reference

Check if a set of points described a triangle

I tried to solve this question but couldn't find a simple solution without passing all rows and find which numbers are on the same line.
Is there a simple way to find triangles?
this is my solution for finding a triangle:
How can I change it to be more "pythonic"? (or even better method for solving it)
from sympy.solvers import solve
from sympy import Symbol
from collections import Counter
vals = [8,17,19] # the triangle
dicl = [] #list of dics
for v in vals:
dic = {}
dic['val'] = v
v1 = v
done = 0
stepsb = 0
while done == 0: #going backword untill reaching the big triabgle edges
x = Symbol('x')
k = solve((x**2 + x)/2 +1 - v1, x)
k = list(filter(lambda x:x>0, k))
if k[0]%1 == 0:
done = 1
else:
v1 -= 1
stepsb += 1
dic['line'] = k[0]
dic['stepsb'] = stepsb #dist from the left edge
dic['stepsf'] = (k[0]**2 + 3*k[0] + 2)/2 - v #dist from the right edge
dicl.append(dic)
print(dic)
lines = [l['line'] for l in dicl]
mc = Counter(lines).most_common(1)[0][0] #finding the numbers on the same line
minv = min([l['val'] for l in dicl if l['line'] == mc])
maxv = max([l['val'] for l in dicl if l['line'] == mc])
stb = [l['stepsb'] for l in dicl if l['val'] == minv][0]
stf = [l['stepsf'] for l in dicl if l['val'] == maxv][0]
for k in dicl:
if k['stepsb'] == stb and k['stepsf'] == stf:
print("good")
break

A first step could be to search for a formula that translates the one-dimensional point number t to an x,y coordinate.
So, search for an n such that n*(n+1)/2 < t:
from sympy import solve, Eq
from sympy.abc import n, t
f = Eq(n * (n + 1), 2 * t)
print(solve(f, n))
This shows as positive root: (sqrt(8*t + 1) - 1)/2.
To be strict smaller, a formula that copes with small approximation errors, could be:
floor((sqrt(8*t + 1) - 1)/2 - 0.0000001
The following idea is, given a list of indices:
convert them to xy coordinates
find their center (sum and divide by the length of the list)
find the distances of each xy to the center
check that all distances are equal
To convert to an xy position, note that the height of an equilateral triangle with base 1 is sqrt(3)/2, so the distances between the y-positions should be multiplied by that factor. The x-positions need to be centered which can be achieved by subtracting n/2.
import math
def find_xy(t):
# convert the numerical position into an xy coordinate in the plane
# first find largest n such that n*(n+1)/2 < t
n = math.floor((math.sqrt(8 * t + 1) - 1) / 2 - 0.0000001)
return (n + 1) * math.sqrt(3) / 2, t - n * (n + 1) // 2 - n/2
def sq_dist(p, q):
return (p[0] - q[0]) ** 2 + (p[1] - q[1]) ** 2
def center(points):
# find the center of a list of points
l = len(points)
x = sum(p[0] for p in points)
y = sum(p[1] for p in points)
return x / l, y / l
def is_regular(tri_points):
points = [find_xy(t) for t in tri_points]
cent = center(points)
dists = [sq_dist(cent, p) for p in points]
return max(dists) - min(dists) < 0.000001
Note that this code finds geometric figures for which all the points lie on a circle. This doesn't work for the parallelogram. The actual question also has some extra criteria: all edges should follow the grid lines, and all edges need to be equal in length.
Therefore, it is useful to have 3 coordinates for each point: the row, the column and the diagonal (the 3 directions of the grid).
The length in each direction, is just the maximum minus the minimum for that direction. These lengths are called d_r, d_c and d_d in the code below.
Checking for a valid triangle, the 3 lengths need to be equal. One way to check this, is to check that the minimum of the lengths is equal to the maximum.
For a valid parallelogram, two lengths need to be equal, and the third should be the double. Checking that the maximum length is twice the minimum length should cover this. But, because this can already be reached using 3 points, we should also check that for a given direction, there are exactly 2 points at the minimum and 2 at the maximum. Summing all points and comparing twice the sum of maximum and minimum should accomplish this.
For a valid hexagon, the 3 lengths should be equal. So, the same test as for the triangle: the minimum of the lengths equal to the maximum. And also the test on the sums is needed, as 4 points can already fulfil the length conditions.
import math
def find_row_col_diag(t):
# convert the numerical position into an row,col,diag coordinate in the plane
# first find largest n such that n*(n+1)/2 < t
n = math.floor((math.sqrt(8 * t + 1) - 1) / 2 - 0.0000001)
row, col = n + 1, t - n * (n + 1) // 2
return row, col, row - col
def check_valid_figure(tri_points):
points = [find_row_col_diag(t) for t in tri_points]
rs = [r for (r, c, d) in points]
cs = [c for (r, c, d) in points]
ds = [d for (r, c, d) in points]
sum_r = sum(rs)
min_r = min(rs)
max_r = max(rs)
d_r = max_r - min_r
sum_c = sum(cs)
min_c = min(cs)
max_c = max(cs)
d_c = max_c - min_c
sum_d = sum(ds)
min_d = min(ds)
max_d = max(ds)
d_d = max_d - min_d
if len(points) == 3:
is_ok = max(d_r, d_c, d_d) == min(d_r, d_c, d_d)
elif len(points) == 4:
is_ok = max(d_r, d_c, d_d) == 2 * min(d_r, d_c, d_d) \
and sum_r == 2 * (min_r + max_r) and sum_c == 2 * (min_c + max_c) and sum_d == 2 * (min_d + max_d)
elif len(points) == 6:
is_ok = max(d_r, d_c, d_d) == min(d_r, d_c, d_d) \
and len(set(rs)) == 3 and len(set(cs)) == 3 and len(set(ds)) == 3
else:
is_ok = False
print(" ".join([str(t) for t in tri_points]), end=" ")
if is_ok:
print("are the vertices of a",
"triangle" if len(points) == 3 else "parallelogram" if len(points) == 4 else "hexagon")
else:
print("are not the vertices of an acceptable figure")
tri_point_lists = [[1, 2, 3],
[11, 13, 22, 24],
[11, 13, 29, 31],
[11, 13, 23, 25],
[26, 11, 13, 24],
[22, 23, 30],
[4, 5, 9, 13, 12, 7]]
for lst in tri_point_lists:
check_valid_figure(lst)
The last code can be further compressed using list comprehensions:
def check_valid_figure_bis(tri_points):
points = [find_row_col_diag(t) for t in tri_points]
rs, cs, ds = [[p[i] for p in points] for i in range(3)]
sums = [sum(xs) for xs in (rs, cs, ds)]
mins = [min(xs) for xs in (rs, cs, ds)]
maxs = [max(xs) for xs in (rs, cs, ds)]
lens = [ma - mi for mi, ma in zip(mins, maxs)]
if len(points) == 3:
is_ok = max(lens) == min(lens)
elif len(points) == 4:
is_ok = max(lens) == 2 * min(lens) and all([su == 2 * (mi + ma) for su, mi, ma in zip(sums, mins, maxs)])
elif len(points) == 6:
is_ok = max(lens) == min(lens) and all([len(set(xs)) == 3 for xs in (rs, cs, ds)])
else:
is_ok = False
return is_ok

Splitting the state vector when solving a series of odes using python odeint scipy

I'm converting my code from Matlab to Python and stuck on how to split the state vector such that the result returns the two solution. I have a vector and a single value for the two initial conditions and I expect as the final result a matrix and a vector.
I tried joining the initial conditions (y0 = [c_pt_0, x_0]) in the same manner as the solution (soln = [dfdt,dcdt]) (which is shown below in the code). I also tried a similar approach that is used in matlab, which is concatenating the the initial conditions to one single array and unpacking the results but I think the problem is in the dimensions.
#Basic imports
import numpy as np
import pylab
import matplotlib. pyplot as plt
import scipy
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# define parameters
pi = 3.14159265
V_m = 9.09
m_V__M_Pt = 1e6/195.084
rho = 21.45
R0 = 10**(-8.19)
k_d = 10**(-13)
k_r = 10**(-5)
S = 0.314 #distribution parameter
M = 0.944 #distribution parameter
## geometry
# Finite Volume Method with equidistant elements
r_max = 30.1e-9 #maximum value
n = 301 #number of elements of FVM
dr = r_max/n #length of elements, equidistant
ini_r = np.linspace(5e-10,r_max,n+1) #boundaries of elements
mid_r = ini_r[0:n]+dr/2 #center of elements
## initial conditions
#initial distribution
x0 = 1/(S*np.sqrt(2*pi)*mid_r*1e9)*np.exp((-(np.log(mid_r*1e9)-M)**2)/(2*S**2))
c_pt_0 = 0
y0 = [x0, c_pt_0]
MN_0 = scipy.trapz(np.power(mid_r, 3)*x0,
x=mid_r) # initial mass
M_0 = 4/3*pi*rho*MN_0
def f(y, t):
r = y[0]
c_pt = y[1]
#materials balance
drdt = V_m * k_r * c_pt * np.exp(-R0/ mid_r) - V_m * k_d * np.exp(R0/ mid_r)
dmdt = 4*pi*rho*mid_r**2*drdt
dMdt = np.trapz(r*dmdt, x=mid_r)
dcdt = m_V__M_Pt*(-dMdt)/M_0
dfdt = -(np.gradient(r*drdt, dr))
soln = [dfdt, dcdt]
return soln
#------------------------------------------------------
#define timespace
time = np.linspace(0, 30, 500)
#solve ode system
sln_1 = odeint (f , y0 , time,
rtol = 1e-3, atol = 1e-5)
pylab.plot(mid_r, sln_1[1,:], color = 'r', marker = 'o')
pylab.plot(mid_r, sln_1[-1,:], color = 'b', marker = 'o')
plt.show()
Traceback:
ValueError: setting an array element with a sequence.
Any help is much appreciated.
EDIT: ADDED MATLAB CODE
Here is the MATLAB code that works that I want to convert to python where the state vector is split. I have three files (one main, the f function, and the parameters). Please excuse any face palm coding errors but I do appreciate any suggestions even for this.
modified_model.m:
function modified_model
% import parameters
p = cycling_parameters;
% initial conditions
c_pt_0 = 0;
y0 = [p.x0; c_pt_0];
% call integrator
options_ODE=odeset('Stats','on', 'RelTol',1e-3,'AbsTol',1e-5);
[~, y] = ode15s(#(t,y) f(t, y, p), p.time, y0, options_ODE);
%% Post processing
% split state vector
r = y(:,1:p.n);
c_Pt = y(:,p.n+1);
%% Plot results
figure
hold on;
plot(p.r_m, r(1,:));
plot(p.r_m, r(end,:));
xlabel({'size'},'FontSize',15)
ylabel({'counts'},'FontSize',15)
f.m
function soln = f(~, y, p)
%split state vector
r = y(1:p.n);
c_pt = y(p.n+1);
% materials balance
drdt = p.Vm_Pt.*p.k_rdp.*c_pt.*exp(-p.R0./p.r_m) - p.Vm_Pt.*p.k_dis.*exp(p.R0./p.r_m);
dmdt = 4*pi*p.rho*p.r_m.^2.*drdt;
dMdt = trapz(p.r_m, r.*dmdt);
dcdt = p.I_V*p.m_V__M_Pt*(-dMdt)/p.M_0;
dfdt = - gradient(r.*drdt,p.dr);
soln = [dfdt; dcdt];
and the parameters file: cycling_parameters.m
function p=cycling_parameters
p.M = 195.084;
p.rho = 21.45;
p.time = linspace(0, 30, 500);
p.m_V__M_Pt = 1e6/p.M;
p.Vm_Pt = 9.09;
p.R0_log = -8.1963;
p.k_dis_log = -13;
p.k_rdp_log = -11;
p.R0 = 10^(p.R0_log);
p.k_dis = 10^(p.k_dis_log);
p.k_rdp = 10^(p.k_rdp_log);
%%% geometry %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Finite Volume Method with equidistant elements
p.r_max = 10.1e-9; % [m] maximum radius of PRD
p.n = 301; % number of elements of FVM
p.dr = p.r_max/p.n; % [m] length of elements, equidistant
p.r = linspace(5e-10,p.r_max,p.n+1)'; % [m] boundaries of elements
p.r_m = p.r(1:p.n)+p.dr/2; % [m] center of elements
%log normal initial distribution
S = 0.314;
M = 0.944;
p.x0 = 1./(S.*sqrt(2.*pi).*p.r_m*1e9).*exp((-(log(p.r_m*1e9)-M).^2)./(2.*S.^2));
p.r_squared = p.r_m.^2; % [m^2] squares of the radius (center of elements)
p.r_cubed = p.r_m.^3; % [m^3] cubes of the radius (center of elements)
p.MN_0 = trapz(p.r_m, p.r_cubed.*p.x0); % Eq. 2.11 denominator
p.M_0 = 4/3*pi*p.rho*p.MN_0;
p.I_V = 1; %ionomer volume fraction in the catalyst layer

After looking at both codes, the issue is that the odeint solver only takes 1D array inputs and your y0 is [int, array(300,)] and odeint can't work with that. However, you can merge the y0 into a 1D array and then split it up in the function you are integrating over to do the calculation then recombine as the output. Here's a working code of that:
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
class P:
def __init__(self, S, M):
self.M = 195.084
self.rho = 21.45
self.m_V__M_Pt = (1*10**6)/self.M
self.Vm_Pt = 9.09
self.R0_log = -8.1963
self.k_dis_log = -13
self.k_rdp_log = -11
self.R0 = 10**(self.R0_log)
self.k_dis = 10**(self.k_dis_log)
self.k_rdp = 10**(self.k_rdp_log)
self.r_max = 10.1*10**(-9)
self.n = 301
self.dr = self.r_max / self.n
self.r = np.linspace(5*10**(-10), self.r_max, self.n)
self.r_m = self.r[0:self.n+1]+self.dr/2
self.x0 = self.compute_x0(S, M)
self.r_squared = np.power(self.r_m, 2)
self.r_cubed = np.power(self.r_m, 3)
self.MN_0 = np.trapz(self.r_m, np.multiply(self.r_cubed, self.x0))
self.M_0 = (4 / 3)* np.pi * self.rho * self.MN_0
self.I_V = 1
def compute_x0(self, S, M):
p1 = np.multiply(2, np.power(S, 2))
p2 = np.multiply(S, np.sqrt(np.multiply(2, np.pi)))
p3 = np.log(self.r_m*1*10**(9)) - M
p4 = np.multiply(p2, self.r_m*10**(9))
p5 = np.power(-p3, 2)
p6 = np.multiply(p4, np.exp(np.divide(p5,p1)))
p7 = np.divide(1, p6)
return p7
def cycling_parameters():
S = 0.314
M = 0.944
p = P(S, M)
return p
def f(y, t):
p = cycling_parameters()
c_pt = y[0]
r = np.delete(y, 0)
p1 = np.multiply(p.Vm_Pt, p.k_rdp)
p2 = np.multiply(p1, c_pt)
p3 = np.multiply(p.Vm_Pt, p.k_dis)
drdt = np.multiply(p2, np.exp(np.divide(-p.R0, p.r_m))) - np.multiply(p3, np.exp(np.divide(p.R0, p.r_m)))
dmdt = np.multiply(4*np.pi*p.rho*np.power(p.r_m, 2), drdt)
p4 = np.multiply(r, dmdt)
dMdt = np.trapz(p.r_m, p4)
dcdt = p.I_V*p.m_V__M_Pt*(-dMdt)/p.M_0
p5 = np.multiply(r, drdt)
dfdt = - np.gradient(p5,p.dr)
ans = np.insert(dfdt, 0, dcdt)
return ans
def modified_model():
p = cycling_parameters()
c_pt_0 = 0
y0 = np.insert(p.x0, 0, c_pt_0)
t = np.linspace(0, 30, 500)
ans = odeint(f, y0, t, rtol = 1e-3, atol = 1e-5)
r = ans[:, 1:p.n+1]
c_Pt = ans[:, 0]
print(r)
print(c_Pt)
plt.plot(p.r_m, r[0, :], color='r', linewidth=0.5)
plt.plot(p.r_m, r[r.shape[0]-1, :], color='b', linewidth=0.5)
plt.show()
if __name__ == '__main__':
modified_model()
Python plot (what this script outputs):
Original Matlab Plot:

How to compute the Mass distribution in Python?

Mass distribution is defined as follows.
f is the probability density function of a continuous variable.
Given a set of data values, which are saved in a list, how to approximate this function? Since the integrals in the numerator and the denominator are identical to the expected value of a distribution, can we use the sample mean based approach as follows?
def get_mass_distribution(values):
x = np.linspace(0, max(values), max(values))
mean = sum(values)/len(values)
mass = []
values.sort()
for i in range(len(values)):
mass.append(sum(values[0:i+1])/(mean*(i+1)))
return x, mass

You should use trapezoidal rule to approximate this integral.
def get_mass_distribution(data):
a = np.array(data)
ag = st.gaussian_kde(a)
denom_integral = trapezoidal(ag, 0, max(data), max(data)*10)
Fm = [0]
x = []
k = 0
while(k < max(data)):
x.append(k)
k = k+1
for i in x[1:]:
enum_integral = trapezoidal(ag, 0, i, i*10)
Fm.append(enum_integral/denom_integral)
return x, Fm
def trapezoidal(ag, a, b, n):
h = float(b - a) / n
s = 0.0
s += a*ag(a)[0]/2.0
for i in range(1, n):
s += (a + i*h)*ag(a + i*h)[0]
s += b*ag(b)[0]/2.0
return s * h

How can I modify my graph so it displays the proper information for the axes?

I have written a program to solve the Heat Equation (u_t = k * u_xx) numerically by method of Finite Differences.
For my problem, u is function of x and t, where 0 < x < L and t > 0. I have specified L = 1 (the length of the rod) and the terminal time T = 10 seconds for my problem, so I would like for the graph to be displayed on the domain (x,t) \in {(0,1) x (0, 10)}. However, my axes just don't make sense. It is plotting the x-axis from values of 0 - 40 and the t-axis is showing -0.25 - 0.00.
How can I edit my code so that when I plot u which depends on x, t the graph will display for values of x ranging from 0 - 1 and t ranging from 0 - 10 seconds??
Thanks in advance for any and all help. it is very greatly appreciated. Here is the code I am working with:
## This program is to implement a Finite Difference method approximation
## to solve the Heat Equation, u_t = k * u_xx,
## in 1D w/out sources & on a finite interval 0 < x < L. The PDE
## is subject to B.C: u(0,t) = u(L,t) = 0,
## and the I.C: u(x,0) = f(x).
import numpy as np
import matplotlib.pyplot as plt
# Parameters
L = 1 # length of the rod
T = 10 # terminal time
N = 40 # spatial values
M = 1600 # time values/hops; (M ~ N^2)
s = 0.25 # s := k * ( (dt) / (dx)^2 )
# uniform mesh
x_init = 0
x_end = L
dx = float(x_end - x_init) / N
x = np.arange(x_init, x_end, dx)
x[0] = x_init
# time discretization
t_init = 0
t_end = T
dt = float(t_end - t_init) / M
t = np.arange(t_init, t_end, dt)
t[0] = t_init
# time-vector
for m in xrange(0, M):
t[m] = m * dt
# spatial-vector
for j in xrange(0, N):
x[j] = j * dx
# definition of the solution u(x,t) to u_t = k * u_xx
u = np.zeros((N, M+1)) # array to store values of the solution
# Finite Difference Scheme:
u[:,0] = x * (x - 1) #initial condition
for m in xrange(0, M):
for j in xrange(1, N-1):
if j == 1:
u[j-1,m] = 0 # Boundary condition
elif j == N-1:
u[j+1,m] = 0 # Boundary Condition
else:
u[j,m+1] = u[j,m] + s * ( u[j+1,m] -
2 * u[j,m] + u[j-1,m] )
# for graph
print u, x, t
plt.plot(u)
plt.title('Finite Difference Approx. to Heat Equation')
plt.xlabel('x-axis')
plt.ylabel('time (seconds)')
plt.axis()
plt.show()
It appears that whatever displays for the x-axis reflects the number of step sizes in space that I take (N = 40) for my code. I thought np.arange(x_init, x_end, dx) would return evenly spaced values within the interval (x_init, x_end) with step size dx? So what am I doing wrong? Thanks again.

You have some issues with your code as your u turns out to be 40x1601 and not 40x1600. However, I think the plot you may be after (after correcting u) is
corrected_u = u[:,:-1:]
plt.pcolor(t, x, corrected_u)

Python 3.4.3 Diamond-Square Algorithm is producing odd results

I am currently stumped by an artefact in my code. It appears to produce very sharp points in a grid pattern that have a noticeable difference in value to their neighbours.
I am following the blog post at http://www.bluh.org/code-the-diamond-square-algorithm/ and converting from whichever language they are using (assuming either C# or Java), and have double-checked that what I am doing should match.
Is there any chance that someone could have a browse over this, and see what I'm doing wrong? I've stepped through it at smaller levels, and stopped it on specific iterations of the algorithm (by unrolling the top loop, and explicitly calling the algorithm a set number of times) and everything seems to work until we get to the very last set of points/pixels.
I use a class (called Matrix) to access the list, and wrap any out of bounds values.
The code for the algorithm is as follows:
class World :
def genWorld (self, numcells, cellsize, seed):
random.seed(seed)
self.dims = numcells*cellsize
self.seed = seed
self.cells = Matrix(self.dims, self.dims)
# set the cells at cellsize intervals
half = cellsize/2
for y in range(0, self.dims, cellsize):
for x in range(0, self.dims, cellsize):
self.cells[x,y] = random.random()
scale = 1.0
samplesize = cellsize
while samplesize > 1:
self._diamondSquare(samplesize, scale)
scale *= 0.8
samplesize = int(samplesize/2)
# I need to sort out the problem with the diamond-square algo that causes it to make the weird gridding pattern
def _sampleSquare(self, x, y, size, value):
half = size/2
a = self.cells[x-half, y-half]
b = self.cells[x+half, y-half]
c = self.cells[x-half, y+half]
d = self.cells[x+half, y+half]
res = min(((a+b+c+d+value)/5.0), 1.0)
self.cells[x, y] = res
def _sampleDiamond(self, x, y, size, value):
half = size/2
a = self.cells[x+half, y]
b = self.cells[x-half, y]
c = self.cells[x, y+half]
d = self.cells[x, y-half]
res = min(((a+b+c+d+value)/5.0), 1.0)
self.cells[x, y] = res
def _diamondSquare(self, stepsize, scale):
half = int(stepsize/2)
for y in range(half, self.dims+half, stepsize):
for x in range(half, self.dims+half, stepsize):
self._sampleSquare(x, y, stepsize, random.random()*scale)
for y in range(0, self.dims, stepsize):
for x in range(0, self.dims, stepsize):
self._sampleDiamond(x+half, y, stepsize, random.random()*scale)
self._sampleDiamond(x, y+half, stepsize, random.random()*scale)
and is called with:
w = World()
w.genWorld(16, 16, 1) # a 256x256 square world, since the numcells is multiplied by the cellsize to give us the length of ONE side of the resulting grid
then I save to file to check the result:
file = io.open("sample.raw",'wb')
arr = [int(i * 255) for i in w.cells.cells] # w.cells.cells should not have a value >= 1.0, so what's going on?
ind = 0
for a in arr:
if a > 255:
print ("arr["+str(ind)+"] ::= "+str(a))
ind += 1
file.write(bytearray(arr))
file.close()
which gives the result:
EDIT: Okay, so it appears that I managed to get it working. I swapped from using functions for working out the diamond and square steps to doing it all in the _diamondSquare() function, but this wasn't the only thing. I also found out that random.random() provides values in the range [0.0 ->1.0), when I was expecting values in the range [-1.0 -> 1.0). After I corrected this, everything started working properly, which was a relief.
Thanks for the advice everyone, here's the working code in case anyone else is struggling with something similar:
Random Function
# since random.random() gives a value in the range [0.0 -> 1.0), I need to change it to [-1.0 -> 1.0)
def rand():
mag = random.random()
sign = random.random()
if sign >=0.5:
return mag
return mag * -1.0
Matrix class
class Matrix:
def __init__(self, width, height):
self.cells = [0 for i in range(width*height)]
self.width = width
self.height = height
self.max_elems = width*height
def _getsingleindex(self, ind):
if ind < 0:
ind *= -1
while ind >= self.max_elems:
ind -= self.max_elems
return ind
def _getmultiindex(self, xind, yind):
if xind < 0:
xind *= -1
if yind < 0:
yind *= -1
while xind >= self.width:
xind -= self.width
while yind >= self.height:
yind -= self.height
return xind + (yind*self.height)
def __getitem__(self, inds):
# test that index is an integer, or two integers, and throw an indexException if not
if hasattr(inds, "__len__"):
if len(inds) > 1:
return self.cells[self._getmultiindex(int(inds[0]), int(inds[1]))]
return self.cells[self._getsingleindex(int(inds))]
def __setitem__(self, inds, object):
# test that index is an integer, or two integers, and throw an indexException if not
if hasattr(inds, "__len__"):
if len(inds) > 1:
self.cells[self._getmultiindex(int(inds[0]),int(inds[1]))] = object
return self.cells[self._getmultiindex(int(inds[0]),int(inds[1]))]
self.cells[self._getsingleindex(int(inds))] = object
return self.cells[self._getsingleindex(int(inds))]
def __len__(self):
return len(self.cells)
The Actual Diamond-Square Generation
# performs the actual 2D generation
class World:
def genWorld (self, numcells, cellsize, seed, scale = 1.0):
random.seed(seed)
self.dims = numcells*cellsize
self.seed = seed
self.cells = Matrix(self.dims, self.dims)
mountains = Matrix(self.dims, self.dims)
# set the cells at cellsize intervals
for y in range(0, self.dims, cellsize):
for x in range(0, self.dims, cellsize):
# this is the default, sets the heights randomly
self.cells[x,y] = random.random()
while cellsize > 1:
self._diamondSquare(cellsize, scale)
scale *= 0.5
cellsize = int(cellsize/2)
for i in range(len(mountains)):
self.cells[i] = self.cells[i]*0.4 + (mountains[i]*mountains[i])*0.6
def _diamondSquare(self, stepsize, scale):
half = int(stepsize/2)
# diamond part
for y in range(half, self.dims+half, stepsize):
for x in range(half, self.dims+half, stepsize):
self.cells[x, y] = ((self.cells[x-half, y-half] + self.cells[x+half, y-half] + self.cells[x-half, y+half] + self.cells[x+half, y+half])/4.0) + (rand()*scale)
# square part
for y in range(0, self.dims, stepsize):
for x in range(0, self.dims, stepsize):
self.cells[x+half,y] = ((self.cells[x+half+half, y] + self.cells[x+half-half, y] + self.cells[x+half, y+half] + self.cells[x+half, y-half])/4.0)+(rand()*scale)
self.cells[x,y+half] = ((self.cells[x+half, y+half] + self.cells[x-half, y+half] + self.cells[x, y+half+half] + self.cells[x, y+half-half])/4.0)+(rand()*scale)
Main Function (added for completeness)
# a simple main function that uses World to create a 2D array of diamond-square values, then writes it to a file
def main():
w = World()
w.genWorld(20, 16, 1)
mi = min(w.cells.cells)
ma = max(w.cells.cells) - mi
# save the resulting matrix to an image file
file = io.open("sample.raw",'wb')
maxed = [(i-mi)/ma for i in w.cells.cells]
arr = [int(i * 255) for i in maxed]
file.write(bytearray(arr))
file.close()