不必一一替换值，而是可以像这样重新映射整个数组：
import numpy as npa = np.array([1,2,2,1]).reshape(2,2)# palette must be given in sorted orderpalette = [1, 2]# key gives the new values you wish palette to be mapped to.key = np.array([0, 10])index = np.digitize(a.ravel(), palette, right=True)print(key[index].reshape(a.shape))
产量
[[ 0 10] [10 0]]
以上想法归功于@JoshAdel。它比我的原始答案快得多：
import numpy as npimport randompalette = np.arange(8)key = palette**2a = np.array([random.choice(palette) for i in range(514*504)]).reshape(514,504)def using_unique(): palette, index = np.unique(a, return_inverse=True) return key[index].reshape(a.shape)def using_digitize(): index = np.digitize(a.ravel(), palette, right=True) return key[index].reshape(a.shape)if __name__ == '__main__': assert np.allclose(using_unique(), using_digitize())
我以这种方式对两个版本进行了基准测试：
In [107]: %timeit using_unique()10 loops, best of 3: 35.6 ms per loopIn [112]: %timeit using_digitize()100 loops, best of 3: 5.14 ms per loop
