可以建议使用np.unique和np.bincount获得最主要颜色的两种方法。另外，在链接的页面中，它是bincount作为一种更快的替代方法讨论的，因此这可能是可行的方法。
def unique_count_app(a):
colors, count = np.unique(a.reshape(-1,a.shape[-1]), axis=0, return_counts=True)
return colors[count.argmax()]
def bincount_app(a):
a2D = a.reshape(-1,a.shape[-1])
col_range = (256, 256, 256) # generically : a2D.max(0)+1
a1D = np.ravel_multi_index(a2D.T, col_range)
return np.unravel_index(np.bincount(a1D).argmax(), col_range)
1000 x 1000在宽范围内对彩色图像进行验证和计时，以[0,9)确保可重现的结果-
In [28]: np.random.seed(0)
...: a = np.random.randint(0,9,(1000,1000,3))
...:
...: print unique_count_app(a)
...: print bincount_app(a)
[4 7 2]
(4, 7, 2)
In [29]: %timeit unique_count_app(a)
1 loop, best of 3: 820 ms per loop
In [30]: %timeit bincount_app(a)
100 loops, best of 3: 11.7 ms per loop
在利用进一步推动multi- core与numexpr模块大数据-
import numexpr as ne
def bincount_numexpr_app(a):
a2D = a.reshape(-1,a.shape[-1])
col_range = (256, 256, 256) # generically : a2D.max(0)+1
eval_params = {'a0':a2D[:,0],'a1':a2D[:,1],'a2':a2D[:,2],
's0':col_range[0],'s1':col_range[1]}
a1D = ne.evaluate('a0*s0*s1+a1*s0+a2',eval_params)
return np.unravel_index(np.bincount(a1D).argmax(), col_range)
时间-
In [90]: np.random.seed(0)
...: a = np.random.randint(0,9,(1000,1000,3))
In [91]: %timeit unique_count_app(a)
...: %timeit bincount_app(a)
...: %timeit bincount_numexpr_app(a)
1 loop, best of 3: 843 ms per loop
100 loops, best of 3: 12 ms per loop
100 loops, best of 3: 8.94 ms per loop
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