Question

我使用4D numpy数组，其中沿数组的第3维计算统计量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)

因此：

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

代表沿该维度的mean, median和std的值。

但是随后我想添加列的平均绝对偏差mad的值，以便统计量为（mean, median, std, mad），但是看来numpy没有提供功能为了那个原因。如何将mad添加到我的统计信息中？

编辑

第一个答案，使用已定义的函数，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：从scipy.stats中将mad_absolute_deviation导入为疯狂

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

通常，我看到MAD指的是中值绝对偏差。如果您要这样做，可以在SciPy库中以scipy.stats.median_absolute_deviation()的形式获得。

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)

Answer 2

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)))轻松地基于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

与numpy ndarray的平均绝对偏差

2 个答案: