为什么vgg.prepare()方法创建给定图像的9个副本? [英] Why does vgg.prepare() method create 9 copies of the given image?

问题描述

将vgg.prepare()应用于以下图像时，会得到以下结果:

I get this result when I apply vgg.prepare() to the following image:

我使用以下代码行:

Image.fromarray(np.uint8(vgg.prepare(pep).reshape(224,224,3)))

并获得包含给定图像的9个副本的图像:

And get an image which is combined of 9 copies of the given image:

推荐答案

我终于明白了你的所作所为... 唯一的错误是.reshape.

I finally got what you did... the only mistake is .reshape.

由于图像是转置的，而不是重塑的，因此必须重新转置才能恢复原始图像.

Because the image is transposed, not reshaped, you have to re-transpose to restore the original image.

pep = pep.transpose((1, 2, 0))  # transpose
pep += [103.939, 116.779, 123.68]  # un-normalize
pep = pep.astype(np.uint8)  # revert dtype
pep = np.flip(pep, axis=2)  # BGR -> RGB
PIL_image = Image.fromarray(pep)  # finally got the original!

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