我们将从零初始数组开始。那么,那将是:np.zeros((heigh, width))
。接下来,我们将有几种方法来填充这种交替255s
的模式。
方法#1:使用开放范围与np.ogrid
阵列,以模拟这些迭代器和直接在量化方式利用翻译比较broadcasting
-
I,J = np.ogrid[:h,:w]
im[(I+J)%2==0] = 255
方法2:近距离看起来,你似乎将每行中的其他元素设置为255
,从第一行的第一个元素开始,第二行开始第二个元素,第三行开始返回第一个元素,依此类推。因此,我们可以用slicing
以及和应该是相当有效 -
im[::2,::2] = 255
im[1::2,1::2] = 255
运行测试
途径 -
def func(im):
h,w = im.shape
for i in range(h):
for j in range(w):
if (i + j) % 2 == 0:
im[i, j] = 255
return im
def app1(im):
h,w = im.shape
I,J = np.ogrid[:h,:w]
im[(I+J)%2==0] = 255
return im
def app2(im):
im[::2,::2] = 255
im[1::2,1::2] = 255
return im
验证 -
In [74]: im = np.random.randint(0,255,(1000,1000))
In [75]: im1 = im.copy()
...: im2 = im.copy()
...: im3 = im.copy()
...:
In [76]: func(im1)
...: app1(im2)
...: app2(im3)
...:
Out[76]:
array([[255, 133, 255, ..., 14, 255, 41],
[235, 255, 191, ..., 255, 40, 255],
[255, 151, 255, ..., 51, 255, 18],
...,
[ 50, 255, 177, ..., 255, 193, 255],
[255, 245, 255, ..., 114, 255, 27],
[223, 255, 148, ..., 255, 200, 255]])
In [77]: print np.allclose(im1,im2)
...: print np.allclose(im1,im3)
...:
True
True
计时 -
In [78]: %timeit func(im)
10 loops, best of 3: 106 ms per loop
In [79]: %timeit app1(im)
100 loops, best of 3: 14 ms per loop
In [80]: %timeit app2(im)
1000 loops, best of 3: 415 µs per loop
In [82]: 106/0.415 # Speedup with approach #2 over original one
Out[82]: 255.42168674698797