在Scipy中，为什么idct（dct（a））不等于a？ [英] In scipy why doesn't idct(dct(a)) equal to a?

问题描述

我正在尝试使用python实现JPEG压缩。当我尝试对tiff图像应用DCT，量化，IDCT处理时，我发现scipy.fftpack.dct / idct很奇怪。

I am trying to implement JPEG compression using python. When I tried to apply the DCT, quantization, IDCT process for a tiff image, I found something strange for scipy.fftpack.dct/idct.

由于只有一维scty包中的dct / idct，我是用二维dct

Since there is only 1D dct/idct within scipy package, I was doing this for a 2D dct

import numpy as np
from scipy.fftpack import dct, idct

def dct2(block):
    return dct(dct(block.T).T)

def idct2(block):
    return idct(idct(block.T).T)

我测试了2D dct / idct使用简单的3x3矩阵。我期望在这个测试用例中得到一个True矩阵。

I tested the 2D dct/idct using a simple 3x3 matrix. I was expecting to get a True matrix with this test case.

a = np.random.randint(0,255,9).reshape(3,3)
print a == idct2(dct2(a))

但是事实证明，在idct2（dct2（a））之后，与原始矩阵相比，结果以恒定因子进行缩放。

However it turned out that after idct2(dct2(a)) the result was scaled by a constant factor compared with the original a matrix.

我想问一下是否存在是一种实现一组二维dct / idct的方法，以便在进行idct（dct（a））操作后，我可以获得与输入相同的输出。

I would like to ask if there is a way to implement a set of 2D dct/idct such that after a idct(dct(a)) operation I can get the same output as the input.

推荐答案

对于 dct2 和<$ c $，都需要将缩放比例设置为 ortho c> idct2 ：

You need to set scaling to ortho for both dct2 and idct2:

def dct2 (block):
  return dct(dct(block.T, norm = 'ortho').T, norm = 'ortho')

您不能期望值完全相同，但是在一定误差范围内几乎相同：

also, you cannot expect the values to be exactly the same, but almost the same within some margin of error:

np.allclose (a, idct2(dct2(a)))

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