我正在尝试将此噪声模块调整到我正在处理的项目中，但我没有得到我期望的结果： https ://pypi.python.org/pypi/noise/
每行往往具有完全相同的值，而不是变化的高度图。如果我创建 250x250 的东西，它将生成相同的值 250 次，然后生成一个新值 250 次，直到它结束。
这是我目前正在使用的功能。我很好地理解了这个功能，但我只是不确定如何获得更多“有趣”的结果。谢谢您的帮助。
class SimplexNoise(BaseNoise):
def noise2(self, x, y):
"""2D Perlin simplex noise.
Return a floating point value from -1 to 1 for the given x, y coordinate.
The same value is always returned for a given x, y pair unless the
permutation table changes (see randomize above).
"""
# Skew input space to determine which simplex (triangle) we are in
s = (x + y) * _F2
i = floor(x + s)
j = floor(y + s)
t = (i + j) * _G2
x0 = x - (i - t) # "Unskewed" distances from cell origin
y0 = y - (j - t)
if x0 > y0:
i1 = 1; j1 = 0 # Lower triangle, XY order: (0,0)->(1,0)->(1,1)
else:
i1 = 0; j1 = 1 # Upper triangle, YX order: (0,0)->(0,1)->(1,1)
x1 = x0 - i1 + _G2 # Offsets for middle corner in (x,y) unskewed coords
y1 = y0 - j1 + _G2
x2 = x0 + _G2 * 2.0 - 1.0 # Offsets for last corner in (x,y) unskewed coords
y2 = y0 + _G2 * 2.0 - 1.0
# Determine hashed gradient indices of the three simplex corners
perm = BaseNoise.permutation
ii = int(i) % BaseNoise.period
jj = int(j) % BaseNoise.period
gi0 = perm[ii + perm[jj]] % 12
gi1 = perm[ii + i1 + perm[jj + j1]] % 12
gi2 = perm[ii + 1 + perm[jj + 1]] % 12
# Calculate the contribution from the three corners
tt = 0.5 - x0**2 - y0**2
if tt > 0:
g = _GRAD3[gi0]
noise = tt**4 * (g[0] * x0 + g[1] * y0)
else:
noise = 0.0
tt = 0.5 - x1**2 - y1**2
if tt > 0:
g = _GRAD3[gi1]
noise += tt**4 * (g[0] * x1 + g[1] * y1)
tt = 0.5 - x2**2 - y2**2
if tt > 0:
g = _GRAD3[gi2]
noise += tt**4 * (g[0] * x2 + g[1] * y2)
return noise * 70.0 # scale noise to [-1, 1]
win = pygcurse.PygcurseWindow(85, 70, 'Generate')
octaves = 2
ysize = 150
xsize = 150
freq = 32.0 * octaves
for y in range(ysize):
for x in range(xsize):
tile = SimplexNoise.noise2(x / freq, y / freq, octaves)
win.write(str(tile) + "\n")