如何沿 x 轴以 50% 的几率翻转 3D 数据？

Question

我读了一篇论文，他们提到了

flipped data with 50% chance along x-axis.

给定输入数据为 40x40x24。我怎样才能执行上述要求？我正在尝试使用 python 2.7 的波纹管代码，但我不确定“50% 机会”的含义

data_flip = np.flipud(data)
data_flip = data[:, ::-1, :]

Answer 1

首先，要以 p 的概率从 n 个元素中选出，您可以简单地使用：np.random.rand(n) < p。 r = np.random.rand() generates a number from a uniform distribution over [0, 1), so the probability that r is smaller than some constant p (where p is in [0,1]) is exactly p. This probability is actually the CDF of the distribution，在这种情况下，a=0 和 b=1 是：

F(p) = 0, p<0
       p, 0<=p<=1
       1, p>1

其次，要沿 x 轴翻转数据，请使用 np.fliplr 而不是 np.flipud（沿 y 轴翻转）：

# generate a 3D array size 3x3x5
A = np.array([[1,2,3],[4,5,6],[7,8,9]])
A = np.tile( np.expand_dims(A, axis=2), (1,1,5) )
# index the 3rd axis with probability 0.5
p = 0.5
idxs = np.random.rand(A.shape[2]) < p
# flip left-right the chosen arrays in the 3rd dimension
A[:,:,idxs] = np.fliplr(A[:,:,idxs])

如何沿 x 轴以 50% 的几率翻转 3D 数据？

How to flipped 3D data with 50% chance along x axis?

python

image-processing

computer-vision

python-2.7