I have some questions:
(1)
for i, j in pixels:
…
for i, j in pixels:
…
第二个循环会被并行吗？如果第二个是在tf.func里面呢？
(2)
for i, j in pixels:
pixels[i, j] = xxx
for i, j in pixels:
pixels[(i + offset)%size, (j + offset)%size] = xxx
在效率上会有明显区别吗？
(3)
pixels = ti.var(dt=ti.f32, shape=(size, size))
怎样高效的得到pixels里面最大的值
DFT code
# Follow https://homepages.inf.ed.ac.uk/rbf/HIPR2/fourier.htm
import taichi as ti
from PIL import Image
import numpy as np
from math import pi
ti.init(arch=ti.gpu)
size = 512
def read_image(path):
image = Image.open(path).convert('LA')
# resize to square image
image = image.resize([size, size])
data = np.asarray(image)
data = np.cast[np.float32](data)
# Convert to grey value within [0, 1]
return data[:, :, 0] / data[:, :, 1]
def show_image(data, scale=True):
if scale:
data = data / np.max(data)
data = data * 255.0
c = np.ones_like(data) * 255.0
data = np.stack([data, c], -1)
data = np.cast[np.uint8](data)
img = Image.fromarray(data, 'LA')
img.show()
pixels = ti.var(dt=ti.f32, shape=(size, size))
results = ti.var(dt=ti.f32, shape=(size, size))
path = input('image path: ')
pixels.from_numpy(read_image(path))
@ti.kernel
def fourier():
# parallel
for k, l in results:
v = ti.Vector([0.0, 0.0])
for i in range(size):
for j in range(size):
center = size // 2
kk = (k + center) % size
ll = (l + center) % size
angle = -2.0 * pi * (kk * i + ll * j) / float(size)
p = ti.Vector([ti.cos(angle), ti.sin(angle)])
v += pixels[i, j] * p
center = size // 2
results[k, l] = ti.log(1.0 + v.norm())
fourier()
data = results.to_numpy()
show_image(data)
Lenna的结果
