与 numpy ndarray 的平均绝对偏差

Question

我使用一个 4D numpy 数组，在该数组中我沿着数组的第三维计算统计信息 mean, meadin, std，如下所示：

import numpy as np
input_shape = (1, 10, 4)
n_sample =20
X = np.random.uniform(0,1, (n_sample,)+input_shape)
X.shape
(20, 1, 10, 4)

然后我以这种方式计算mean, med, 和std-dev：

sta_fuc = (np.mean, np.median, np.std)
stat = np.concatenate([func(X, axis=2, keepdims=True) for func in sta_fuc], axis=2)

这样：

stat.shape
(20, 1, 3, 4)

表示沿该维度的mean, median 和std 的值。

然后我想添加列的平均绝对偏差mad 的值，以便统计信息为 (mean, median, std, mad)，但似乎numpy 没有为此提供函数。如何将mad添加到我的统计信息中？

编辑

至于第一个答案，使用定义的函数，即：

def mad(arr, axis=None, keepdims=True):
    median = np.median(arr, axis=axis, keepdims=True)
    mad = np.median(np.abs(arr-median, axis=axis, keepdims=keepdims),
                    axis=axis, keepdims=keepdims)
    return mad

然后将mad 添加到统计信息中，这会产生错误，如下所示：

sta_fuc = (np.mean, np.median, np.std, mad)
stat = np.concatenate([func(X, axis=2, keepdims=True) for func in sta_fuc], axis=2)

---------------------------------------------------------------------------

TypeError                                 Traceback (most recent call last)

<ipython-input-22-dab51665f952> in <module>()
      1 sta_fuc = (np.mean, np.median, np.std, mad)
----> 2 stat = np.concatenate([func(X, axis=2, keepdims=True) for func in sta_fuc], axis=2)

1 frames

<ipython-input-21-84d735c8c516> in mad(arr, axis, keepdims)
      1 def mad(arr, axis=None, keepdims=True):
      2     median = np.median(arr, axis=axis, keepdims=True)
----> 3     mad = np.median(np.abs(arr-median, axis=axis, keepdims=keepdims),
      4                     axis=axis, keepdims=keepdims)
      5     return mad

TypeError: 'axis' is an invalid keyword to ufunc 'absolute'

EDIT-2

使用@Jussi 建议的scipy 函数也会产生如下错误： from scipy.stats import median_absolute_deviation as mad

sta_fuc = (np.mean, np.median, np.std, mad)
stat = np.concatenate([func(X, axis=2, keepdims=True) for func in sta_fuc], axis=2)

TypeError: median_absolute_deviation() got an unexpected keyword argument 'keepdims'

Answer 1

我不知道使用 numpy 的内置解决方案。但是您可以使用 mad = median(abs(a - median(a))) 相当容易地基于 numpy 函数实现它。

def mad(arr, axis=None, keepdims=True):
    median = np.median(arr, axis=axis, keepdims=True)
    mad = np.median(np.abs(arr-median),axis=axis, keepdims=keepdims)
    return mad

Answer 2

通常，我看到 MAD 指的是中值绝对偏差。如果这是您想要的，它可以在 SciPy 库中以 scipy.stats.median_absolute_deviation() 的形式提供。

自己编写合适的函数也很容易。

编辑：这是一个带有 keepdims 参数的 MAD 函数：

def mad(data, axis=None, scale=1.4826, keepdims=False):
    """Median absolute deviation (MAD).
    
    Defined as the median absolute deviation from the median of the data. A
    robust alternative to stddev. Results should be identical to
    scipy.stats.median_absolute_deviation(), which does not take a keepdims
    argument.

    Parameters
    ----------
    data : array_like
        The data.
    scale : float, optional
        Scaling of the result. By default, it is scaled to give a consistent
        estimate of the standard deviation of values from a normal
        distribution.
    axis : numpy axis spec, optional
        Axis or axes along which to compute MAD.
    keepdims : bool, optional
        If this is set to True, the axes which are reduced are left in the
        result as dimensions with size one.

    Returns
    -------
    ndarray
        The MAD.
    """
    # keep dims here so that broadcasting works
    med = np.median(data, axis=axis, keepdims=True)
    abs_devs = np.abs(data - med)
    return scale * np.median(abs_devs, axis=axis, keepdims=keepdims)