Python numpy解包函数 [英] Python numpy unwrap function

问题描述

我希望将弧度数组转换为范围[0, 2*pi)和

I am hoping to convert a array of radians into range [0, 2*pi) and numpy unwrap function is exactly what I need

但是，当我运行以下代码以输入a = [pi, 2*pi, 3*pi]时:

However, when I run the following code to input a = [pi, 2*pi, 3*pi]:

import numpy as np

a = np.array([np.pi, 2*np.pi, 3*np.pi])
np.unwrap(a)

我希望结果接近[pi, 0, pi].但是，输出仍然是:

I expect the results to be close to [pi, 0, pi]. However, the output is still:

array([ 3.14159265,  6.28318531,  9.42477796])

它没有展开.但是，如果我改为运行以下命令而不使用numpy.pi

It is not unwrapped. However, if I instead run the following without using the numpy.pi

a = np.array([3.14159265,  6.28318531,  9.42477796])
np.unwrap(a)

输出正确:

array([  3.14159265e+00,   2.82041412e-09,   3.14159265e+00])

怎么回事?

推荐答案

尽管已接受的答案可以为您提供所需的结果，但我认为这并不是问题的核心，如果我正在解释您的问题正确，是您实际上是要包装您的阶段，而不是包装它.

Although the accepted answer gives you the result you want, I don't think it gets to the heart of the problem which, if I'm interpreting your question correctly, is that you actually want wrap your phase, not unwrap it.

在这种情况下np.unwrap起作用的原因是对您的数据进行了很小的更改，实际上是np.unwrap计算其结果的幼稚方式的结果.它只是在数据中查找局部不连续性并进行相应调整.以这种方式获得所需的结果是采样错误的结果.换句话说，如果通过内插得到a = np.array([np.pi, 3*np.pi/2, 2*np.pi, 5*np.pi/2, 3*np.pi])来改善采样，则调整数据将不再起作用.

The reason np.unwrap works in this case, with small changes to your data, is actually a consequence of the naive way that np.unwrap computes its result; it simply looks for local discontinuities in your data and adjusts accordingly. Getting the result you're looking for in this way is a result of sampling errors. In other words, if you improve your sampling by interpolating to get a = np.array([np.pi, 3*np.pi/2, 2*np.pi, 5*np.pi/2, 3*np.pi]), adjusting the data won't work any more.

更复杂的相位展开方法，例如傅立叶变换方法，即使采样很差，也会使数据无法展开.

A more sophisticated method of phase unwrapping, such as a Fourier transform method, will leave your data unwrapped, even if the sampling is poor.

如果您确实想将数据限制为[0, 2*pi)，则np.unwrap是所需内容的逆.我可以想到的最简单的包装相位的方法是使用模运算符:

If you really want to constrain your data to [0, 2*pi), np.unwrap is the inverse of what you want. The simplest way I can think of to wrap your phase is with the modulo operator:

import numpy as np

a = np.array([np.pi, 2 * np.pi, 3 * np.pi])
a_wrapped = a % (2 * np.pi)
print (a_wrapped)

当然，由于采样错误，np.unwrap(a_wrapped)不会返回原始的a，因此可能不清楚这是相反的.但是，如果您改进采样，则确实会返回原始的a:

Of course, because of the sampling errors, np.unwrap(a_wrapped) does not return your original a, so it may not be clear that this is the inverse. However, if you improve your sampling, it does indeed return the original a:

import numpy as np

a = np.arange(0, 4 * np.pi, np.pi/10)
print (a)
a_wrapped = a % (2 * np.pi)
print (a_wrapped)
a = np.unwrap(a_wrapped)
print (a)

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