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TA贡献2039条经验 获得超7个赞
我注意到两个问题:
您可能想要具有最大绝对值的分量,因此而
np.argsort(-np.abs(D_f))
不是np.argsort(-D_f)
.更巧妙的是:
bigcofs = np.zeros(len(D_f))
是类型float64
并且丢弃了行 处的虚部bigcofs[I] = D_f[I]
。你可以用以下方法解决这个问题bigcofs = np.zeros(len(D_f), dtype=complex)
我在下面稍微改进了您的代码以获得所需的结果:
import numpy as np
from scipy import fft, ifft
import matplotlib.pyplot as plt
n = 1000
limit_low = 0
limit_high = 0.48
N_THRESH = 10
D = 0.5*np.random.normal(0, 0.5, n) + 0.5*np.abs(np.random.normal(0, 2, n) * np.sin(np.linspace(0, 3*np.pi, n))) + np.sin(np.linspace(0, 5*np.pi, n))**2 + np.sin(np.linspace(1, 6*np.pi, n))**2
scaling = (limit_high - limit_low) / (max(D) - min(D))
D = D * scaling
D = D + (limit_low - min(D)) # given data
t = np.linspace(0,D.size-1,D.size) # times
# transformed data
D_fft = fft.fft(D)
# Create boolean mask for N largest indices
idx_sorted = np.argsort(-np.abs(D_fft))
idx = idx_sorted[0:N_THRESH]
mask = np.zeros(D_fft.shape).astype(bool)
mask[idx] = True
# Split fft above, below N_THRESH points:
D_below = D_fft.copy()
D_below[mask] = 0
D_above = D_fft.copy()
D_above[~mask] = 0
#inverse separated functions
D_above = fft.ifft(D_above)
D_below = fft.ifft(D_below)
# plot
plt.ion()
f, (ax1, ax2, ax3) = plt.subplots(3,1)
l1, = ax1.plot(t, D, c="r", label="original")
l2, = ax2.plot(t, D_above, c="g", label="top {} coeff. signal".format(N_THRESH))
l3, = ax3.plot(t, D_below, c="b", label="remaining signal")
f.legend(handles=[l1,l2,l3])
plt.show()
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