Python 中的自相关代码产生错误(吉他音高检测)
Autocorrelation code in Python produces errors (guitar pitch detection)
This link 提供基于自相关的音调检测算法的代码。我用它来检测简单吉他旋律中的音高。
总的来说,它产生了很好的效果。例如,对于旋律 C4、C#4、D4、D#4、E4,它输出:
262.743653536
272.144441273
290.826273006
310.431336809
327.094621169
与正确的注释相关联。
但是,在某些情况下,例如 this 音频文件(E4、F4、F#4、G4、G#4、A4、A#4、B4)会产生错误:
325.861452246
13381.6439242
367.518651703
391.479384923
414.604661221
218.345286173
466.503751322
244.994090035
更具体地说,这里有三个错误:13381Hz 被错误检测而不是 F4 (~350Hz)(奇怪的错误),还有 218Hz 而不是 A4 (440Hz) 和244Hz 而不是 B4 (~493Hz),这是八度误差。
我假设这两个错误是由不同的原因引起的?这是代码:
slices = segment_signal(y, sr)
for segment in slices:
pitch = freq_from_autocorr(segment, sr)
print pitch
def segment_signal(y, sr, onset_frames=None, offset=0.1):
if (onset_frames == None):
onset_frames = remove_dense_onsets(librosa.onset.onset_detect(y=y, sr=sr))
offset_samples = int(librosa.time_to_samples(offset, sr))
print onset_frames
slices = np.array([y[i : i + offset_samples] for i
in librosa.frames_to_samples(onset_frames)])
return slices
可以看到上面第一个link中的freq_from_autocorr
函数
唯一认为我已经改变的是这一行:
corr = corr[len(corr)/2:]
我已经替换为:
corr = corr[int(len(corr)/2):]
更新:
我注意到我使用的 offset
最小(我用来检测每个音调的信号段最小),我得到的高频(10000+ Hz)错误越多。
具体来说,我注意到在这些情况下(10000+ Hz)不同的部分是 i_peak
值的计算。在没有错误的情况下它在 50-150 的范围内,在错误的情况下它是 3-5.
自相关法并不总是正确的。您可能想要实现更复杂的方法,例如 YIN:
http://audition.ens.fr/adc/pdf/2002_JASA_YIN.pdf
或 MPM:
http://www.cs.otago.ac.nz/tartini/papers/A_Smarter_Way_to_Find_Pitch.pdf
以上两篇论文都值得一读。
您链接的代码片段中的自相关函数不是特别可靠。为了得到正确的结果,需要将第一个峰定位在自相关曲线的左侧。其他开发人员使用的方法(调用 numpy.argmax()
函数)并不总能找到正确的值。
我使用 peakutils 包实现了一个稍微更健壮的版本。我也不保证它是完全健壮的,但无论如何它比您之前使用的 freq_from_autocorr()
函数的版本获得更好的结果。
下面列出了我的示例解决方案:
import librosa
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import fftconvolve
from pprint import pprint
import peakutils
def freq_from_autocorr(signal, fs):
# Calculate autocorrelation (same thing as convolution, but with one input
# reversed in time), and throw away the negative lags
signal -= np.mean(signal) # Remove DC offset
corr = fftconvolve(signal, signal[::-1], mode='full')
corr = corr[len(corr)//2:]
# Find the first peak on the left
i_peak = peakutils.indexes(corr, thres=0.8, min_dist=5)[0]
i_interp = parabolic(corr, i_peak)[0]
return fs / i_interp, corr, i_interp
def parabolic(f, x):
"""
Quadratic interpolation for estimating the true position of an
inter-sample maximum when nearby samples are known.
f is a vector and x is an index for that vector.
Returns (vx, vy), the coordinates of the vertex of a parabola that goes
through point x and its two neighbors.
Example:
Defining a vector f with a local maximum at index 3 (= 6), find local
maximum if points 2, 3, and 4 actually defined a parabola.
In [3]: f = [2, 3, 1, 6, 4, 2, 3, 1]
In [4]: parabolic(f, argmax(f))
Out[4]: (3.2142857142857144, 6.1607142857142856)
"""
xv = 1/2. * (f[x-1] - f[x+1]) / (f[x-1] - 2 * f[x] + f[x+1]) + x
yv = f[x] - 1/4. * (f[x-1] - f[x+1]) * (xv - x)
return (xv, yv)
# Time window after initial onset (in units of seconds)
window = 0.1
# Open the file and obtain the sampling rate
y, sr = librosa.core.load("./Vocaroo_s1A26VqpKgT0.mp3")
idx = np.arange(len(y))
# Set the window size in terms of number of samples
winsamp = int(window * sr)
# Calcualte the onset frames in the usual way
onset_frames = librosa.onset.onset_detect(y=y, sr=sr)
onstm = librosa.frames_to_time(onset_frames, sr=sr)
fqlist = [] # List of estimated frequencies, one per note
crlist = [] # List of autocorrelation arrays, one array per note
iplist = [] # List of peak interpolated peak indices, one per note
for tm in onstm:
startidx = int(tm * sr)
freq, corr, ip = freq_from_autocorr(y[startidx:startidx+winsamp], sr)
fqlist.append(freq)
crlist.append(corr)
iplist.append(ip)
pprint(fqlist)
# Choose which notes to plot (it's set to show all 8 notes in this case)
plidx = [0, 1, 2, 3, 4, 5, 6, 7]
# Plot amplitude curves of all notes in the plidx list
fgwin = plt.figure(figsize=[8, 10])
fgwin.subplots_adjust(bottom=0.0, top=0.98, hspace=0.3)
axwin = []
ii = 1
for tm in onstm[plidx]:
axwin.append(fgwin.add_subplot(len(plidx)+1, 1, ii))
startidx = int(tm * sr)
axwin[-1].plot(np.arange(startidx, startidx+winsamp), y[startidx:startidx+winsamp])
ii += 1
axwin[-1].set_xlabel('Sample ID Number', fontsize=18)
fgwin.show()
# Plot autocorrelation function of all notes in the plidx list
fgcorr = plt.figure(figsize=[8,10])
fgcorr.subplots_adjust(bottom=0.0, top=0.98, hspace=0.3)
axcorr = []
ii = 1
for cr, ip in zip([crlist[ii] for ii in plidx], [iplist[ij] for ij in plidx]):
if ii == 1:
shax = None
else:
shax = axcorr[0]
axcorr.append(fgcorr.add_subplot(len(plidx)+1, 1, ii, sharex=shax))
axcorr[-1].plot(np.arange(500), cr[0:500])
# Plot the location of the leftmost peak
axcorr[-1].axvline(ip, color='r')
ii += 1
axcorr[-1].set_xlabel('Time Lag Index (Zoomed)', fontsize=18)
fgcorr.show()
打印输出如下:
In [1]: %run autocorr.py
[325.81996740236065,
346.43374761017725,
367.12435233192753,
390.17291696559079,
412.9358117076161,
436.04054933498134,
465.38986619237039,
490.34120132405866]
我的代码示例生成的第一个图描绘了每个检测到的起始时间后接下来 0.1 秒的振幅曲线:
代码生成的第二个图显示了在 freq_from_autocorr()
函数内部计算的自相关曲线。垂直红线描绘了每条曲线左侧第一个峰的位置,由 peakutils 包估计。对于其中一些红线,其他开发人员使用的方法得到的结果不正确;这就是为什么他的那个函数的版本偶尔会返回错误的频率。
我的建议是在其他录音上测试 freq_from_autocorr()
功能的修改版本,看看你是否能找到更具挑战性的例子,即使改进版本仍然给出错误的结果,然后发挥创意并尝试开发一种更强大的峰值查找算法,永远不会误火。
This link 提供基于自相关的音调检测算法的代码。我用它来检测简单吉他旋律中的音高。
总的来说,它产生了很好的效果。例如,对于旋律 C4、C#4、D4、D#4、E4,它输出:
262.743653536
272.144441273
290.826273006
310.431336809
327.094621169
与正确的注释相关联。
但是,在某些情况下,例如 this 音频文件(E4、F4、F#4、G4、G#4、A4、A#4、B4)会产生错误:
325.861452246
13381.6439242
367.518651703
391.479384923
414.604661221
218.345286173
466.503751322
244.994090035
更具体地说,这里有三个错误:13381Hz 被错误检测而不是 F4 (~350Hz)(奇怪的错误),还有 218Hz 而不是 A4 (440Hz) 和244Hz 而不是 B4 (~493Hz),这是八度误差。
我假设这两个错误是由不同的原因引起的?这是代码:
slices = segment_signal(y, sr)
for segment in slices:
pitch = freq_from_autocorr(segment, sr)
print pitch
def segment_signal(y, sr, onset_frames=None, offset=0.1):
if (onset_frames == None):
onset_frames = remove_dense_onsets(librosa.onset.onset_detect(y=y, sr=sr))
offset_samples = int(librosa.time_to_samples(offset, sr))
print onset_frames
slices = np.array([y[i : i + offset_samples] for i
in librosa.frames_to_samples(onset_frames)])
return slices
可以看到上面第一个link中的freq_from_autocorr
函数
唯一认为我已经改变的是这一行:
corr = corr[len(corr)/2:]
我已经替换为:
corr = corr[int(len(corr)/2):]
更新:
我注意到我使用的 offset
最小(我用来检测每个音调的信号段最小),我得到的高频(10000+ Hz)错误越多。
具体来说,我注意到在这些情况下(10000+ Hz)不同的部分是 i_peak
值的计算。在没有错误的情况下它在 50-150 的范围内,在错误的情况下它是 3-5.
自相关法并不总是正确的。您可能想要实现更复杂的方法,例如 YIN:
http://audition.ens.fr/adc/pdf/2002_JASA_YIN.pdf
或 MPM:
http://www.cs.otago.ac.nz/tartini/papers/A_Smarter_Way_to_Find_Pitch.pdf
以上两篇论文都值得一读。
您链接的代码片段中的自相关函数不是特别可靠。为了得到正确的结果,需要将第一个峰定位在自相关曲线的左侧。其他开发人员使用的方法(调用 numpy.argmax()
函数)并不总能找到正确的值。
我使用 peakutils 包实现了一个稍微更健壮的版本。我也不保证它是完全健壮的,但无论如何它比您之前使用的 freq_from_autocorr()
函数的版本获得更好的结果。
下面列出了我的示例解决方案:
import librosa
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import fftconvolve
from pprint import pprint
import peakutils
def freq_from_autocorr(signal, fs):
# Calculate autocorrelation (same thing as convolution, but with one input
# reversed in time), and throw away the negative lags
signal -= np.mean(signal) # Remove DC offset
corr = fftconvolve(signal, signal[::-1], mode='full')
corr = corr[len(corr)//2:]
# Find the first peak on the left
i_peak = peakutils.indexes(corr, thres=0.8, min_dist=5)[0]
i_interp = parabolic(corr, i_peak)[0]
return fs / i_interp, corr, i_interp
def parabolic(f, x):
"""
Quadratic interpolation for estimating the true position of an
inter-sample maximum when nearby samples are known.
f is a vector and x is an index for that vector.
Returns (vx, vy), the coordinates of the vertex of a parabola that goes
through point x and its two neighbors.
Example:
Defining a vector f with a local maximum at index 3 (= 6), find local
maximum if points 2, 3, and 4 actually defined a parabola.
In [3]: f = [2, 3, 1, 6, 4, 2, 3, 1]
In [4]: parabolic(f, argmax(f))
Out[4]: (3.2142857142857144, 6.1607142857142856)
"""
xv = 1/2. * (f[x-1] - f[x+1]) / (f[x-1] - 2 * f[x] + f[x+1]) + x
yv = f[x] - 1/4. * (f[x-1] - f[x+1]) * (xv - x)
return (xv, yv)
# Time window after initial onset (in units of seconds)
window = 0.1
# Open the file and obtain the sampling rate
y, sr = librosa.core.load("./Vocaroo_s1A26VqpKgT0.mp3")
idx = np.arange(len(y))
# Set the window size in terms of number of samples
winsamp = int(window * sr)
# Calcualte the onset frames in the usual way
onset_frames = librosa.onset.onset_detect(y=y, sr=sr)
onstm = librosa.frames_to_time(onset_frames, sr=sr)
fqlist = [] # List of estimated frequencies, one per note
crlist = [] # List of autocorrelation arrays, one array per note
iplist = [] # List of peak interpolated peak indices, one per note
for tm in onstm:
startidx = int(tm * sr)
freq, corr, ip = freq_from_autocorr(y[startidx:startidx+winsamp], sr)
fqlist.append(freq)
crlist.append(corr)
iplist.append(ip)
pprint(fqlist)
# Choose which notes to plot (it's set to show all 8 notes in this case)
plidx = [0, 1, 2, 3, 4, 5, 6, 7]
# Plot amplitude curves of all notes in the plidx list
fgwin = plt.figure(figsize=[8, 10])
fgwin.subplots_adjust(bottom=0.0, top=0.98, hspace=0.3)
axwin = []
ii = 1
for tm in onstm[plidx]:
axwin.append(fgwin.add_subplot(len(plidx)+1, 1, ii))
startidx = int(tm * sr)
axwin[-1].plot(np.arange(startidx, startidx+winsamp), y[startidx:startidx+winsamp])
ii += 1
axwin[-1].set_xlabel('Sample ID Number', fontsize=18)
fgwin.show()
# Plot autocorrelation function of all notes in the plidx list
fgcorr = plt.figure(figsize=[8,10])
fgcorr.subplots_adjust(bottom=0.0, top=0.98, hspace=0.3)
axcorr = []
ii = 1
for cr, ip in zip([crlist[ii] for ii in plidx], [iplist[ij] for ij in plidx]):
if ii == 1:
shax = None
else:
shax = axcorr[0]
axcorr.append(fgcorr.add_subplot(len(plidx)+1, 1, ii, sharex=shax))
axcorr[-1].plot(np.arange(500), cr[0:500])
# Plot the location of the leftmost peak
axcorr[-1].axvline(ip, color='r')
ii += 1
axcorr[-1].set_xlabel('Time Lag Index (Zoomed)', fontsize=18)
fgcorr.show()
打印输出如下:
In [1]: %run autocorr.py
[325.81996740236065,
346.43374761017725,
367.12435233192753,
390.17291696559079,
412.9358117076161,
436.04054933498134,
465.38986619237039,
490.34120132405866]
我的代码示例生成的第一个图描绘了每个检测到的起始时间后接下来 0.1 秒的振幅曲线:
代码生成的第二个图显示了在 freq_from_autocorr()
函数内部计算的自相关曲线。垂直红线描绘了每条曲线左侧第一个峰的位置,由 peakutils 包估计。对于其中一些红线,其他开发人员使用的方法得到的结果不正确;这就是为什么他的那个函数的版本偶尔会返回错误的频率。
我的建议是在其他录音上测试 freq_from_autocorr()
功能的修改版本,看看你是否能找到更具挑战性的例子,即使改进版本仍然给出错误的结果,然后发挥创意并尝试开发一种更强大的峰值查找算法,永远不会误火。