python - skcuda.fft与numpy.fft.rfft不同吗？

我试图将ftf的输出与numpy的ftf进行测试以进行单元测试，但我意识到在失败后不久，这并不是因为我做错了什么，但是skcuda实际上并不会产生相同的答案。我知道它们会有所不同，但是至少其中一个数字与numpy产生的结果相差几个数量级，并且allclose和almost_equal都返回大量错误（对于，对于rtol=1e-6为16％）。我在这里做错了什么？我可以解决这个问题吗？测试文件：import pycuda.autoinitfrom skcuda import fftfrom pycuda import gpuarrayimport numpy as npdef test_skcuda(): array_0 = np.array([[1, 2, 3, 4, 5, 4, 3, 2, 1, 0]], dtype=np.float32) array_1 = array_0 * 10 time_domain_signal = np.array([array_0[0], array_1[0]], dtype=np.float32) fft_point_count = 10 fft_plan = fft.Plan(fft_point_count, np.float32, np.complex64, batch=2) fft_reserved = gpuarray.empty((2, fft_point_count // 2 + 1), dtype=np.complex64) fft.fft(gpuarray.to_gpu(time_domain_signal), fft_reserved, fft_plan) np.testing.assert_array_almost_equal( np.fft.rfft(time_domain_signal, fft_point_count), fft_reserved.get())test_skcuda()断言失败：AssertionError:Arrays are not almost equal to 6 decimals(mismatch 25.0%) x: array([[ 2.500000e+01+0.000000e+00j, -8.472136e+00-6.155367e+00j, -1.193490e-15+2.331468e-15j, 4.721360e-01-1.453085e+00j, 2.664535e-15+0.000000e+00j, 1.000000e+00+0.000000e+00j],... y: array([[ 2.500000e+01+0.000000e+00j, -8.472136e+00-6.155367e+00j, 8.940697e-08+5.960464e-08j, 4.721359e-01-1.453085e+00j, 0.000000e+00+0.000000e+00j, 1.000000e+00+0.000000e+00j],...打印输出：#numpy[[ 2.50000000e+01+0.00000000e+00j -8.47213595e+00-6.15536707e+00j -1.19348975e-15+2.33146835e-15j 4.72135955e-01-1.45308506e+00j 2.66453526e-15+0.00000000e+00j 1.00000000e+00+0.00000000e+00j] [ 2.50000000e+02+0.00000000e+00j -8.47213595e+01-6.15536707e+01j -1.11022302e-14+2.39808173e-14j 4.72135955e+00-1.45308506e+01j 3.55271368e-14+7.10542736e-15j 1.00000000e+01+0.00000000e+00j]]#skcuda[[ 2.5000000e+01+0.0000000e+00j -8.4721355e+00-6.1553669e+00j 8.9406967e-08+5.9604645e-08j 4.7213593e-01-1.4530852e+00j 0.0000000e+00+0.0000000e+00j 1.0000000e+00+0.0000000e+00j] [ 2.5000000e+02+0.0000000e+00j -8.4721359e+01-6.1553673e+01j 1.4305115e-06-4.7683716e-07j 4.7213597e+00-1.4530851e+01j 0.0000000e+00+1.9073486e-06j 1.0000000e+01+0.0000000e+00j]] (adsbygoogle = window.adsbygoogle || []).push({}); 最佳答案 FFT的输出具有与输入值的大小有关的误差。每个输出元素都是通过组合所有输入元素来计算的，因此，它们的大小决定了结果的精度。您正在同一阵列中计算两个1D FFT。它们各自具有不同的幅度输入，因此应具有不同的幅度容差。以下快速代码演示了如何实现此目的。我不知道如何调整numpy.testing中的任何功能来执行此操作。import numpy as nparray_0 = np.array([[1, 2, 3, 4, 5, 4, 3, 2, 1, 0]], dtype=np.float32)array_1 = array_0 * 10time_domain_signal = np.array([array_0[0], array_1[0]], dtype=np.float32)# numpy resulta=np.array([[ 2.50000000e+01+0.00000000e+00j, -8.47213595e+00-6.15536707e+00j, -1.19348975e-15+2.33146835e-15j, 4.72135955e-01-1.45308506e+00j, 2.66453526e-15+0.00000000e+00j, 1.00000000e+00+0.00000000e+00j], [ 2.50000000e+02+0.00000000e+00j, -8.47213595e+01-6.15536707e+01j, -1.11022302e-14+2.39808173e-14j, 4.72135955e+00-1.45308506e+01j, 3.55271368e-14+7.10542736e-15j, 1.00000000e+01+0.00000000e+00j]])# skcuda resultb=np.array([[ 2.5000000e+01+0.0000000e+00j, -8.4721355e+00-6.1553669e+00j, 8.9406967e-08+5.9604645e-08j, 4.7213593e-01-1.4530852e+00j, 0.0000000e+00+0.0000000e+00j, 1.0000000e+00+0.0000000e+00j], [ 2.5000000e+02+0.0000000e+00j, -8.4721359e+01-6.1553673e+01j, 1.4305115e-06-4.7683716e-07j, 4.7213597e+00-1.4530851e+01j, 0.0000000e+00+1.9073486e-06j, 1.0000000e+01+0.0000000e+00j]])# Tolerance for result array row relative to the mean absolute input values# 1e-6 because we're using single-precision floatstol = np.mean(np.abs(time_domain_signal), axis=1) * 1e-6# Compute absolute difference and compare that to our tolearancesdiff = np.abs(a-b)if np.any(diff > tol[:,None]): print('ERROR!!!') (adsbygoogle = window.adsbygoogle || []).push({});