我正在尝试使用Perlin噪声发生器来制作地图的瓷砖,但我注意到我的噪音太尖锐了,我的意思是,它有太多的高度,没有平坦的地方,而且它们看起来不像山,岛屿,湖泊或任何东西;它们似乎太随意而且有很多峰值.
在问题的最后,需要进行修改才能修复它.
该问题的重要代码是：
1D：
def Noise(self,x): # I wrote this noise function but it seems too random
random.seed(x)
number = random.random()
if number < 0.5:
final = 0 - number * 2
elif number > 0.5:
final = number * 2
return final
def Noise(self,x): # I found this noise function on the internet
x = (x<<13) ^ x
return ( 1.0 - ( (x * (x * x * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0)
2D：
def Noise(self,x,y): # I wrote this noise function but it seems too random
n = x + y
random.seed(n)
number = random.random()
if number < 0.5:
final = 0 - number * 2
elif number > 0.5:
final = number * 2
return final
def Noise(self,y): # I found this noise function on the internet
n = x + y * 57
n = (n<<13) ^ n
return ( 1.0 - ( (x * (x * x * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0)
我在代码中留下了1D和2D Perlin噪音,因为可能有人对此感兴趣：
(我花了很长时间才找到一些代码,所以我觉得有人会很高兴在这里找到一个例子).
你不需要Matplotlib或NumPy来制造噪音;我只是使用它们制作图表并更好地查看结果.
import random
import matplotlib.pyplot as plt # To make graphs
from mpl_toolkits.mplot3d import Axes3D # To make 3D graphs
import numpy as np # To make graphs
class D(): # Base of classes D1 and D2
def Cubic_Interpolate(self,v0,v1,v2,v3,x):
P = (v3 - v2) - (v0 - v1)
Q = (v0 - v1) - P
R = v2 - v0
S = v1
return P * x**3 + Q * x**2 + R * x + S
class D1(D):
def __init__(self,lenght,octaves):
self.result = self.Perlin(lenght,octaves)
def Noise(self,x): # I wrote this noise function but it seems too random
random.seed(x)
number = random.random()
if number < 0.5:
final = 0 - number * 2
elif number > 0.5:
final = number * 2
return final
def Noise(self,x): # I found this noise function on the internet
x = (x<<13) ^ x
return ( 1.0 - ( (x * (x * x * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0)
def Perlin(self,octaves):
result = []
for x in range(lenght):
value = 0
for y in range(octaves):
frequency = 2 ** y
amplitude = 0.25 ** y
value += self.Interpolate_Noise(x * frequency) * amplitude
result.append(value)
print(f"{x} / {lenght} ({x/lenght*100:.2f}%): {round(x/lenght*10) * '#'} {(10-round(x/lenght*10)) * ' '}. Remaining {lenght-x}.") # I don't use `os.system('cls')` because it slow down the code.
return result
def Smooth_Noise(self,x):
return self.Noise(x) / 2 + self.Noise(x-1) / 4 + self.Noise(x+1) / 4
def Interpolate_Noise(self,x):
round_x = round(x)
frac_x = x - round_x
v0 = self.Smooth_Noise(round_x - 1)
v1 = self.Smooth_Noise(round_x)
v2 = self.Smooth_Noise(round_x + 1)
v3 = self.Smooth_Noise(round_x + 2)
return self.Cubic_Interpolate(v0,frac_x)
def graph(self,*args):
plt.plot(np.array(self.result),'-',label = "Line")
for x in args:
plt.axhline(y=x,color='r',linestyle='-')
plt.xlabel('X')
plt.ylabel('Y')
plt.title("Simple Plot")
plt.legend()
plt.show()
class D2(D):
def __init__(self,octaves = 1):
self.lenght_axes = round(lenght ** 0.5)
self.lenght = self.lenght_axes ** 2
self.result = self.Perlin(self.lenght,y): # I wrote this noise function but it seems too random
n = x + y
random.seed(n)
number = random.random()
if number < 0.5:
final = 0 - number * 2
elif number > 0.5:
final = number * 2
return final
def Noise(self,y): # I found this noise function on the internet
n = x + y * 57
n = (n<<13) ^ n
return ( 1.0 - ( (x * (x * x * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0)
def Smooth_Noise(self,y):
corners = (self.Noise(x - 1,y - 1) + self.Noise(x + 1,y - 1) + self.Noise(x - 1,y + 1) + self.Noise(x + 1,y + 1) ) / 16
sides = (self.Noise(x - 1,y) + self.Noise(x + 1,y) + self.Noise(x,y - 1) + self.Noise(x,y + 1) ) / 8
center = self.Noise(x,y) / 4
return corners + sides + center
def Interpolate_Noise(self,y):
round_x = round(x)
frac_x = x - round_x
round_y = round(y)
frac_y = y - round_y
v11 = self.Smooth_Noise(round_x - 1,round_y - 1)
v12 = self.Smooth_Noise(round_x,round_y - 1)
v13 = self.Smooth_Noise(round_x + 1,round_y - 1)
v14 = self.Smooth_Noise(round_x + 2,round_y - 1)
i1 = self.Cubic_Interpolate(v11,v12,v13,v14,frac_x)
v21 = self.Smooth_Noise(round_x - 1,round_y)
v22 = self.Smooth_Noise(round_x,round_y)
v23 = self.Smooth_Noise(round_x + 1,round_y)
v24 = self.Smooth_Noise(round_x + 2,round_y)
i2 = self.Cubic_Interpolate(v21,v22,v23,v24,frac_x)
v31 = self.Smooth_Noise(round_x - 1,round_y + 1)
v32 = self.Smooth_Noise(round_x,round_y + 1)
v33 = self.Smooth_Noise(round_x + 1,round_y + 1)
v34 = self.Smooth_Noise(round_x + 2,round_y + 1)
i3 = self.Cubic_Interpolate(v31,v32,v33,v34,frac_x)
v41 = self.Smooth_Noise(round_x - 1,round_y + 2)
v42 = self.Smooth_Noise(round_x,round_y + 2)
v43 = self.Smooth_Noise(round_x + 1,round_y + 2)
v44 = self.Smooth_Noise(round_x + 2,round_y + 2)
i4 = self.Cubic_Interpolate(v41,v42,v43,v44,frac_x)
return self.Cubic_Interpolate(i1,i2,i3,i4,frac_y)
def Perlin(self,octaves):
result = []
for x in range(lenght):
value = 0
for y in range(octaves):
frequency = 2 ** y
amplitude = 0.25 ** y
value += self.Interpolate_Noise(x * frequency,x * frequency) * amplitude
result.append(value)
print(f"{x} / {lenght} ({x/lenght*100:.2f}%): {round(x/lenght*10) * '#'} {(10-round(x/lenght*10)) * ' '}. Remaining {lenght-x}.") # I don't use `os.system('cls')` because it slow down the code.
return result
def graph(self,color = 'viridis'):
# Other colors: https://matplotlib.org/examples/color/colormaps_reference.html
fig = plt.figure()
Z = np.array(self.result).reshape(self.lenght_axes,self.lenght_axes)
ax = fig.add_subplot(1,2,1,projection='3d')
X = np.arange(self.lenght_axes)
Y = np.arange(self.lenght_axes)
X,Y = np.meshgrid(X,Y)
d3 = ax.plot_surface(X,Y,Z,rstride=1,cstride=1,cmap=color,linewidth=0,antialiased=False)
fig.colorbar(d3)
ax = fig.add_subplot(1,2)
d2 = ax.imshow(Z,interpolation='none')
fig.colorbar(d2)
plt.show()
问题是输出似乎不适合地图.
使用以下命令查看此输出：
test = D2(1000,3) test.graph()
我正在寻找更顺畅的东西.
也许在二维噪音中很难注意到我在说什么,但在一维中它更容易：
test = D1(1000,3) test.graph()
来自互联网的噪音功能稍微小一点,频率较低,但它仍然有太多.我正在寻找更顺畅的东西.
这样的事情可能是：
或这个：
P.S：我是根据this pseudocode制作的.
编辑：
pikalek：
即使值较低,它也有峰值,没有曲线或平滑/平坦的线条.
geza：解决方案
感谢geza’s suggestions,我找到了解决问题的方法：
def Perlin(self,lenght_axes,octaves,zoom = 0.01,amplitude_base = 0.5):
result = []
for y in range(lenght_axes):
line = []
for x in range(lenght_axes):
value = 0
for o in range(octaves):
frequency = 2 ** o
amplitude = amplitude_base ** o
value += self.Interpolate_Noise(x * frequency * zoom,y * frequency * zoom) * amplitude
line.append(value)
result.append(line)
print(f"{y} / {lenght_axes} ({y/lenght_axes*100:.2f}%): {round(y/lenght_axes*20) * '#'} {(20-round(y/lenght_axes*20)) * ' '}. Remaining {lenght_axes-y}.")
return result
其他修改包括：
> Z = np.array(self.result)而不是图函数中的Z = np.array(self.result).reshape(self.lenght_axes,self.lenght_axes).
>在round_x和round_y变量的Interpolate_Noise函数中使用math.floor()(记住导入数学)而不是round().
>修改Noise(第二个)返回行返回(1.0 – ((n *(n * n * 15731 789221)1376312589)& 0x7fffffff)/ 1073741824.0).
D2(10000,10)
现在唯一奇怪的是山(黄色)总是在同一个地方,但我认为这是改变噪音功能中的数字的问题.
解决方法
我在你的代码中发现了这些错误：
>您需要将Interpolate_Noise参数乘以“缩放”到地图中(例如,将x乘以0.01).如果你在1D情况下这样做,你会发现生成的函数已经好多了
>将八度计数从3增加到更大(3个八度音程不会产生太多细节)
>使用振幅0.5 ^八度音程,而不是0.25 ^八度音程(但你可以使用这个参数,所以0.25本身并不坏,但它没有给出太多细节)
>对于2D情况,你需要有2个外环(一个用于水平,一个用于垂直.当然,你仍然需要有八度环).因此,您需要使用水平和垂直位置正确“索引”噪声,而不仅仅是x和x.
>完全去除平滑.柏林噪音不需要它.
> 2D噪声函数有一个错误：它在返回表达式中使用x而不是n
>在三次插值中,使用round而不是math.floor.
这是我的答案,简单的(C)实现Perlin-like(它不是正确的perlin)噪音：https://stackoverflow.com/a/45121786/8157187