如何绘制一个频谱同样的方式，pylab的specgram（）呢？ [英] How do I plot a spectrogram the same way that pylab's specgram() does?

问题描述

在Pylab中， specgram（）函数创建的幅度给定列表中的频谱，并自动为频谱创建一个窗口。

In Pylab, the specgram() function creates a spectrogram for a given list of amplitudes and automatically creates a window for the spectrogram.

我想生成频谱（瞬时功率由 PXX 给出），通过在其上运行的边缘检测器进行修改，然后绘制的结果。

I would like to generate the spectrogram (instantaneous power is given by Pxx), modify it by running an edge detector on it, and then plot the result.

(Pxx, freqs, bins, im) = pylab.specgram( self.data, Fs=self.rate, ...... )

问题是，每当我试图绘制修改 PXX 使用 imshow 甚至 NonUniformImage ，我碰到下面的错误消息。

The problem is that whenever I try to plot the modified Pxx using imshow or even NonUniformImage, I run into the error message below.

/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/image.py:336: UserWarning：图片不支持非线性轴。
    warnings.warn（图像不支持非线性轴。）

/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/image.py:336: UserWarning: Images are not supported on non-linear axes. warnings.warn("Images are not supported on non-linear axes.")

例如，code我的工作权利的一部分如下。

For example, a part of the code I'm working on right is below.

    # how many instantaneous spectra did we calculate
    (numBins, numSpectra) = Pxx.shape

    # how many seconds in entire audio recording
    numSeconds = float(self.data.size) / self.rate


    ax = fig.add_subplot(212)
    im = NonUniformImage(ax, interpolation='bilinear')

    x = np.arange(0, numSpectra)
    y = np.arange(0, numBins)
    z = Pxx
    im.set_data(x, y, z)
    ax.images.append(im) 
    ax.set_xlim(0, numSpectra)
    ax.set_ylim(0, numBins)
    ax.set_yscale('symlog') # see http://matplotlib.org/api/axes_api.html#matplotlib.axes.Axes.set_yscale
    ax.set_title('Spectrogram 2')

实际问题

你怎么用matplotlib / pylab对数Y轴情节图像样的数据？

Actual Question

How do you plot image-like data with a logarithmic y axis with matplotlib/pylab?

推荐答案

使用令pColor 或 pcolormesh 。 pcolormesh 要快得多，但仅限于直线网格，其中为令pColor能够处理任意形状的细胞。 <罢工> specgram 使用 pcolormesh ，如果我没有记错。（它使用 imshow ）

Use pcolor or pcolormesh. pcolormesh is much faster, but is limited to rectilinear grids, where as pcolor can handle arbitrary shaped cells. specgram uses pcolormesh, if I recall correctly. (It uses imshow.)

作为一个简单的例子：

import numpy as np
import matplotlib.pyplot as plt

z = np.random.random((11,11))
x, y = np.mgrid[:11, :11]

fig, ax = plt.subplots()
ax.set_yscale('symlog')
ax.pcolormesh(x, y, z)
plt.show()

您所看到的差别是由于密谋原始值的 specgram 的回报。什么 specgram 实际上是绘制一个缩放版本。

The differences you're seeing are due to plotting the "raw" values that specgram returns. What specgram actually plots is a scaled version.

import matplotlib.pyplot as plt
import numpy as np

x = np.cumsum(np.random.random(1000) - 0.5)

fig, (ax1, ax2) = plt.subplots(nrows=2)
data, freqs, bins, im = ax1.specgram(x)
ax1.axis('tight')

# "specgram" actually plots 10 * log10(data)...
ax2.pcolormesh(bins, freqs, 10 * np.log10(data))
ax2.axis('tight')

plt.show()

注意，当我们使用绘制的事情 pcolormesh ，没有插值。 （这是 pcolormesh 的点的一部分 - 它只是载体，而不是矩形图像）

Notice that when we plot things using pcolormesh, there's no interpolation. (That's part of the point of pcolormesh--it's just vector rectangles instead of an image.)

如果您想对数尺度的东西，你可以使用 pcolormesh 吧：

If you want things on a log scale, you can use pcolormesh with it:

import matplotlib.pyplot as plt
import numpy as np

x = np.cumsum(np.random.random(1000) - 0.5)

fig, (ax1, ax2) = plt.subplots(nrows=2)
data, freqs, bins, im = ax1.specgram(x)
ax1.axis('tight')

# We need to explictly set the linear threshold in this case...
# Ideally you should calculate this from your bin size...
ax2.set_yscale('symlog', linthreshy=0.01)

ax2.pcolormesh(bins, freqs, 10 * np.log10(data))
ax2.axis('tight')

plt.show()

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