我试图重新实现我前一段时间成功完成的事情,但我还没有把它做得恰到好处。
分形高度图生成算法I使用的本质上是递归菱形平方算法。它看起来很好地完成了,但是地图生成的并不是很正确的......它似乎没有成功访问网格中的每个点来确定颜色,并且残余结构'在地图中似乎与网格递归的方式有关。我不确定问题在哪里或如何产生我所看到的问题。
到目前为止我的代码是,
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from math import sqrt
from collections import namedtuple
import random
Coord=namedtuple('Coord','x y')
class Grid(object):
'''grid handedness, 0,0=topleft max,max=bottomr right'''
def __init__(self,x,y):
self.size_x=x
self.size_y=y
self.data=[ [0 for _ in xrange(x)] for _ in xrange(y) ]
def _render_to_text(self):
print '\n\n'
for row in self.data:
print [ int(n) for n in row ]
def _render_to_colormap(self):
plt.imshow(self.data, interpolation='nearest',cmap=cm.gist_rainbow)
plt.show()
def render(self):
self._render_to_colormap()
#self._render_to_text()
def make(self,coordinate,value):
self.data[coordinate.x][coordinate.y]=value
def make_new(self,coordinate,value):
if self.data[coordinate.x][coordinate.y]==0:
self.make(coordinate,value)
def get(self,coordinate):
return self.data[coordinate.x][coordinate.y]
class FractalHeightmap(object):
'''populates a 'grid' with a fractal heightmap'''
def __init__(self,grid,rng_seed,roughness,
corner_seeds=[(0,100),(0,100),(0,100),(0,100)],
max_depth=3):
self.grid=grid
self.max_depth=max_depth
self._set_initial_corners(corner_seeds)
self.roughness=roughness
self.generate_heightmap([Coord(0,0),
Coord(self.grid.size_x-1,0),
Coord(0,self.grid.size_y-1),
Coord(self.grid.size_x-1,self.grid.size_y-1)],1
)
def _set_initial_corners(self,corner_seeds):
tl,tr,bl,br=corner_seeds
seeds=[[tl,tr],[bl,br]]
for x_idx,x in enumerate([0,self.grid.size_x-1]):
for y_idx,y in enumerate([0,self.grid.size_y-1]):
try:
minval,maxval=seeds[x_idx][y_idx]
val=minval+(random.random()*(maxval-minval))
except ValueError:
val=seeds[x_idx][y_idx]
self.grid.make_new(Coord(x,y),val)
def generate_heightmap(self,corners,depth):
'''corners = (Coord(),Coord(),Coord(),Coord() / tl/tr/bl/br'''
if depth>self.max_depth: return
#
tl,tr,bl,br=corners
center=Coord((tr.x-tl.x)/2,(br.y-tr.y)/2)
#define edge center coordinates
top_c=Coord(tl.x+((tr.x-tl.x)/2),tl.y)
left_c=Coord(tl.x,tl.y+((bl.y-tl.y)/2))
right_c=Coord(tr.x,tr.y+((br.y-tr.y)/2))
bot_c=Coord(bl.x+((br.x-bl.x)/2),bl.y)
#calc the center and edge_center heights
avg=sum([self.grid.get(tl),
self.grid.get(tr),
self.grid.get(bl),
self.grid.get(br)]
)/4.0 ###NOTE, we can choose to use the current corners, the new edge-centers, or all
#currenty we use the current corners
#then do the edge centers
offset=((random.random())-.5)*self.roughness
self.grid.make_new(center,avg+offset)
#top_c
avg=sum([self.grid.get(tl),
self.grid.get(tr)]
)/2.0
offset=((random.random())-.5)*self.roughness
self.grid.make_new(top_c,avg+offset)
#left_c
avg=sum([self.grid.get(tl),
self.grid.get(bl)]
)/2.0
offset=((random.random())-.5)*self.roughness
self.grid.make_new(left_c,avg+offset)
#right_c
avg=sum([self.grid.get(tr),
self.grid.get(br)]
)/2.0
offset=((random.random())-.5)*self.roughness
self.grid.make_new(right_c,avg+offset)
#bot_c
avg=sum([self.grid.get(bl),
self.grid.get(br)]
)/2.0
offset=((random.random())-.5)*self.roughness
self.grid.make_new(bot_c,avg+offset)
self.generate_heightmap((tl,top_c,left_c,center),depth+1)
self.generate_heightmap((top_c,tr,center,right_c),depth+1)
self.generate_heightmap((left_c,center,bl,bot_c),depth+1)
self.generate_heightmap((center,right_c,bot_c,br),depth+1)
if __name__ == '__main__':
g_size=32 #//must be power of 2
g=Grid(g_size+1,g_size+1)
f=FractalHeightmap(g,1,10,max_depth=sqrt(g_size))
g.render()
如果你按原样运行它,你应该看到色彩图,看看为什么它是 - 非常正确,将深度改为2的不同功率以不同的方式显示它 - 值256及以上需要一段时间来生成
非常感谢任何帮助。
答案 0 :(得分:3)
抱歉没有话题但是我想分享另一个很好的算法来生成地形,我开始使用后我意识到我不喜欢钻石和方形。 Here一个描述,这里是一个实现:
#/usr/bin/python
#coding=UTF-8
import random,math
class HillGrid:
def __init__(self,KRADIUS =(1.0/5.0),ITER=200,SIZE=40):
self.KRADIUS = KRADIUS
self.ITER = ITER
self.SIZE = SIZE
self.grid = [[0 for x in range(self.SIZE)] for y in range(self.SIZE)]
self.MAX = self.SIZE * self.KRADIUS
for i in range(self.ITER):
self.step()
def dump(self):
for ele in self.grid:
s = ''
for alo in ele:
s += '%s ' % str(alo)
print s
def __getitem__(self,n):
return self.grid[n]
def step(self):
point = [random.randint(0,self.SIZE-1),random.randint(0,self.SIZE-1)]
radius = random.uniform(0,self.MAX)
x2 = point[0]
y2 = point[1]
for x in range(self.SIZE):
for y in range(self.SIZE):
z = (radius**2) - ( math.pow(x2-x,2) + math.pow(y2-y,2) )
if z >= 0:
self.grid[x][y] += int(z)
if __name__ == '__main__':
h = HillGrid(ITER=50,SIZE = 20)
h.dump()
答案 1 :(得分:0)
我回答了自己的问题,经过多次逐步踩踏网格。
中心点的计算不正确,
中心坐标=((tr.x-tl.x)/ 2,(br.y-tr.y)/ 2)
应该是
中心坐标=(tl.x +((tr.x-tl.x)/ 2),tr.y +((br.y-tr.y)/ 2))
第一个版本是忘记将子网格原点的x / y位置添加到计算中点坐标的结果中。