将float转换为整数 [英] convert float to integer
问题描述
如何缩放以及通过哪个因子将dctmtx系数从float缩放到以下整数值：
how to scale and by which factor to scale dctmtx coefficients from float to get following integer values:
float dctmtx：
float dctmtx:
( (0.3536 0.3536 0.3536 0.3536 0.3536 0.3536 0.3536 0.3536),
(0.4904 0.4157 0.2778 0.0975 -0.0975 -0.2778 -0.4157 -0.4904),
(0.4619 0.1913 -0.1913 -0.4619 -0.4619 -0.1913 0.1913 0.4619),
(0.4157 -0.0975 -0.4904 -0.2778 0.2778 0.4904 0.0975 -0.4157),
(0.3536 -0.3536 -0.3536 0.3536 0.3536 -0.3536 -0.3536 0.3536),
(0.2778 -0.4904 0.0975 0.4157 -0.4157 -0.0975 0.4904 -0.2778),
(0.1913 -0.4619 0.4619 -0.1913 -0.1913 0.4619 -0.4619 0.1913),
(0.0975 -0.2778 0.4157 -0.4904 0.4904 -0.4157 0.2778 -0.0975)
)
整数dctmtx：
(( 125, 122, 115, 103, 88, 69, 47, 24 ),
( 125, 103, 47, -24, -88, -122, -115, -69 ),
( 125, 69, -47, -122, -88, 24, 115, 103 ),
( 125, 24, -115, -69, 88, 103, -47, -122 ),
( 125, -24, -115, 69, 88, -103, -47, 122 ),
( 125, -69, -47, 122, -88, -24, 115, -103 ),
( 125, -103, 47, 24, -88, 122, -115, 69 ),
( 125, -122, 115, -103, 88, -69, 47, -24 )
);
推荐答案
如果您阅读离散余弦变换，会发现基本系数是
If you read up on the discrete cosine transform, you will find that the basic coefficient is
cos(pi*i*(2*j+1)/16), i,j=0..7
然后第一个表由这些值缩放0.5，除了第一行/列，按0.25 * sqrt（2）= 1 / sqrt（8）缩放。这是获得正交矩阵的正确方法。第一列的平方和为8，其他为4。
Then the first table consists of these values scaled by 0.5, except for the first row/column, which is scaled by 0.25*sqrt(2)=1/sqrt(8). Which is the correct way to obtain an orthogonal matrix. The square sum of the first column is 8, of the others 4.
第二个表是将余弦值与125相乘时的舍入结果。在使用转置矩阵计算逆变换时，必须注意正确地重新缩放矢量。
The second table is the rounded results when multiplying the cosine values with 125, uniformly. Here one has to take care to properly rescale the vector when using the transpose matrix to compute the inverse transform.
第一个表转载，第一栏除外：
First table reproduced, except for the first column:
> [[ Cos(pi*i*(2*j+1)/16)/2 : i in [0..7] ]: j in [0..7] ];
[
[ 0.5, 0.49039264, 0.46193977, 0.41573481, 0.35355339, 0.27778512, 0.19134172, 0.09754516 ],
[ 0.5, 0.41573481, 0.19134172, -0.09754516, -0.35355339, -0.49039264, -0.46193977, -0.27778512 ],
[ 0.5, 0.27778512, -0.19134172, -0.49039264, -0.35355339, 0.09754516, 0.46193977, 0.41573481 ],
[ 0.5, 0.09754516, -0.46193977, -0.27778512, 0.35355339, 0.41573481, -0.19134172, -0.49039264 ],
[ 0.5, -0.09754516, -0.46193977, 0.27778512, 0.35355339, -0.41573481, -0.19134172, 0.49039264 ],
[ 0.5, -0.27778512, -0.19134172, 0.49039264, -0.35355339, -0.09754516, 0.46193977, -0.41573481 ],
[ 0.5, -0.41573481, 0.19134172, 0.09754516, -0.35355339, 0.49039264, -0.46193977, 0.27778512 ],
[ 0.5, -0.49039264, 0.46193977, -0.41573481, 0.35355339, -0.27778512, 0.19134172, -0.09754516 ]
]
第二个表，在整数舍入之前
Second table, before integer rounding
> [[ Cos( pi*i*(2*j+1)/16 ) *125 : i in [0..7] ]: j in [0..7] ];
[
[ 125, 122.5982, 115.4849, 103.9337, 88.3883, 69.4463, 47.8354, 24.3863 ],
[ 125, 103.9337, 47.8354, -24.3863, -88.3883, -122.5982, -115.4849, -69.4463 ],
[ 125, 69.4463, -47.8354, -122.5982, -88.3883, 24.3863, 115.4849, 103.9337 ],
[ 125, 24.3863, -115.4849, -69.4463, 88.3883, 103.9337, -47.8354, -122.5982 ],
[ 125, -24.3863, -115.4849, 69.4463, 88.3883, -103.9337, -47.8354, 122.5982 ],
[ 125, -69.4463, -47.8354, 122.5982, -88.3883, -24.3863, 115.4849, -103.9337 ],
[ 125, -103.9337, 47.8354, 24.3863, -88.3883, 122.5982, -115.4849, 69.4463 ],
[ 125, -122.5982, 115.4849, -103.9337, 88.3883, -69.4463, 47.8354, -24.3863 ]
]
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