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python-3.x - Python 中的独立分量分析 (ICA)

转载 作者:行者123 更新时间:2023-12-05 00:56:57 24 4
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python中是否有任何可用的包来执行独立分量分析(ICA)?

请提供一些指针和链接,以便我可以从 python 开始。

最佳答案

Scikit learn 提供了执行独立分量分析的方法。

scikit learn - ICA

print(__doc__)

import numpy as np
import matplotlib.pyplot as plt
from scipy import signal

from sklearn.decomposition import FastICA, PCA

###############################################################################
# Generate sample data
np.random.seed(0)
n_samples = 2000
time = np.linspace(0, 8, n_samples)

s1 = np.sin(2 * time) # Signal 1 : sinusoidal signal
s2 = np.sign(np.sin(3 * time)) # Signal 2 : square signal
s3 = signal.sawtooth(2 * np.pi * time) # Signal 3: saw tooth signal

S = np.c_[s1, s2, s3]
S += 0.2 * np.random.normal(size=S.shape) # Add noise

S /= S.std(axis=0) # Standardize data
# Mix data
A = np.array([[1, 1, 1], [0.5, 2, 1.0], [1.5, 1.0, 2.0]]) # Mixing matrix
X = np.dot(S, A.T) # Generate observations

# Compute ICA
ica = FastICA(n_components=3)
S_ = ica.fit_transform(X) # Reconstruct signals
A_ = ica.mixing_ # Get estimated mixing matrix

# We can `prove` that the ICA model applies by reverting the unmixing.
assert np.allclose(X, np.dot(S_, A_.T) + ica.mean_)

# For comparison, compute PCA
pca = PCA(n_components=3)
H = pca.fit_transform(X) # Reconstruct signals based on orthogonal components

###############################################################################
# Plot results

plt.figure()

models = [X, S, S_, H]
names = ['Observations (mixed signal)',
'True Sources',
'ICA recovered signals',
'PCA recovered signals']
colors = ['red', 'steelblue', 'orange']

for ii, (model, name) in enumerate(zip(models, names), 1):
plt.subplot(4, 1, ii)
plt.title(name)
for sig, color in zip(model.T, colors):
plt.plot(sig, color=color)

plt.subplots_adjust(0.09, 0.04, 0.94, 0.94, 0.26, 0.46)
plt.show()

关于python-3.x - Python 中的独立分量分析 (ICA),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35407669/

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