在 numpy 数组中插入 NaN 值

Question

是否有一种快速方法可以用（比如说）线性插值值替换 numpy 数组中的所有 NaN 值？

例如，

[1 1 1 nan nan 2 2 nan 0]

将转换为

[1 1 1 1.3 1.6 2 2  1  0]

Answer 1

让我们首先定义一个简单的辅助函数，以便更直接地处理 NaNs 的索引和逻辑索引：

import numpy as np

def nan_helper(y):
    """Helper to handle indices and logical indices of NaNs.

    Input:
        - y, 1d numpy array with possible NaNs
    Output:
        - nans, logical indices of NaNs
        - index, a function, with signature indices= index(logical_indices),
          to convert logical indices of NaNs to 'equivalent' indices
    Example:
        >>> # linear interpolation of NaNs
        >>> nans, x= nan_helper(y)
        >>> y[nans]= np.interp(x(nans), x(~nans), y[~nans])
    """

    return np.isnan(y), lambda z: z.nonzero()[0]

现在

nan_helper(.)

现在可以像这样使用：

>>> y= array([1, 1, 1, NaN, NaN, 2, 2, NaN, 0])
>>>
>>> nans, x= nan_helper(y)
>>> y[nans]= np.interp(x(nans), x(~nans), y[~nans])
>>>
>>> print y.round(2)
[ 1.    1.    1.    1.33  1.67  2.    2.    1.    0.  ]

---
尽管指定一个单独的函数来完成这样的事情首先看起来有点矫枉过正：

>>> nans, x= np.isnan(y), lambda z: z.nonzero()[0]

最终会带来红利。

因此，每当您使用 NaN 相关数据时，只需将所需的所有（新的 NaN 相关）功能封装在某些特定的辅助函数下即可。您的代码库将更加连贯和可读，因为它遵循易于理解的习惯用法。

插值确实是了解如何完成 NaN 处理的一个很好的上下文，但类似的技术也可以在其他各种上下文中使用。

Answer 2

我想出了这段代码：

import numpy as np
nan = np.nan

A = np.array([1, nan, nan, 2, 2, nan, 0])

ok = -np.isnan(A)
xp = ok.ravel().nonzero()[0]
fp = A[-np.isnan(A)]
x  = np.isnan(A).ravel().nonzero()[0]

A[np.isnan(A)] = np.interp(x, xp, fp)

print A

打印出来

 [ 1.          1.33333333  1.66666667  2.          2.          1.          0.        ]

Answer 3

只需使用 numpy 逻辑和 where 语句即可应用一维插值。

import numpy as np
from scipy import interpolate

def fill_nan(A):
    '''
    interpolate to fill nan values
    '''
    inds = np.arange(A.shape[0])
    good = np.where(np.isfinite(A))
    f = interpolate.interp1d(inds[good], A[good],bounds_error=False)
    B = np.where(np.isfinite(A),A,f(inds))
    return B

Answer 4

对于二维数据，SciPy 的

griddata

对我来说效果相当好：

>>> import numpy as np
>>> from scipy.interpolate import griddata
>>>
>>> # SETUP
>>> a = np.arange(25).reshape((5, 5)).astype(float)
>>> a
array([[  0.,   1.,   2.,   3.,   4.],
       [  5.,   6.,   7.,   8.,   9.],
       [ 10.,  11.,  12.,  13.,  14.],
       [ 15.,  16.,  17.,  18.,  19.],
       [ 20.,  21.,  22.,  23.,  24.]])
>>> a[np.random.randint(2, size=(5, 5)).astype(bool)] = np.NaN
>>> a
array([[ nan,  nan,  nan,   3.,   4.],
       [ nan,   6.,   7.,  nan,  nan],
       [ 10.,  nan,  nan,  13.,  nan],
       [ 15.,  16.,  17.,  nan,  19.],
       [ nan,  nan,  22.,  23.,  nan]])
>>>
>>> # THE INTERPOLATION
>>> x, y = np.indices(a.shape)
>>> interp = np.array(a)
>>> interp[np.isnan(interp)] = griddata(
...     (x[~np.isnan(a)], y[~np.isnan(a)]), # points we know
...     a[~np.isnan(a)],                    # values we know
...     (x[np.isnan(a)], y[np.isnan(a)]))   # points to interpolate
>>> interp
array([[ nan,  nan,  nan,   3.,   4.],
       [ nan,   6.,   7.,   8.,   9.],
       [ 10.,  11.,  12.,  13.,  14.],
       [ 15.,  16.,  17.,  18.,  19.],
       [ nan,  nan,  22.,  23.,  nan]])

我在 3D 图像上使用它，对 2D 切片（4000 个 350x350 切片）进行操作。整个手术还需要一个小时左右:/

Answer 5

首先更改数据的生成方式可能会更容易，但如果不这样做：

bad_indexes = np.isnan(data)

创建一个布尔数组来指示 nan 的位置

good_indexes = np.logical_not(bad_indexes)

创建一个布尔数组，指示良好值所在的区域

good_data = data[good_indexes]

原始数据的受限版本，不包括nans

interpolated = np.interp(bad_indexes.nonzero(), good_indexes.nonzero(), good_data)

通过插值运行所有坏索引

data[bad_indexes] = interpolated

用插值替换原始数据。

Answer 6

或者以温斯顿的答案为基础

def pad(data):
    bad_indexes = np.isnan(data)
    good_indexes = np.logical_not(bad_indexes)
    good_data = data[good_indexes]
    interpolated = np.interp(bad_indexes.nonzero()[0], good_indexes.nonzero()[0], good_data)
    data[bad_indexes] = interpolated
    return data

A = np.array([[1, 20, 300],
              [nan, nan, nan],
              [3, 40, 500]])

A = np.apply_along_axis(pad, 0, A)
print A

结果

[[   1.   20.  300.]
 [   2.   30.  400.]
 [   3.   40.  500.]]

Answer 7

我使用插值来替换所有 NaN 值。

A = np.array([1, nan, nan, 2, 2, nan, 0])
np.interp(np.arange(len(A)), 
          np.arange(len(A))[np.isnan(A) == False], 
          A[np.isnan(A) == False])

输出：

array([1. , 1.33333333, 1.66666667, 2. , 2. , 1. , 0. ])

Answer 8

根据BRYAN WOODS的响应稍微优化的版本。它正确处理源数据的起始值和结束值，并且比原始版本快 25-30%。您也可以使用不同类型的插值（有关详细信息，请参阅 scipy.interpolate.interp1d 文档）。

import numpy as np
from scipy.interpolate import interp1d

def fill_nans_scipy1(padata, pkind='linear'):
"""
Interpolates data to fill nan values

Parameters:
    padata : nd array 
        source data with np.NaN values
    
Returns:
    nd array 
        resulting data with interpolated values instead of nans
"""
aindexes = np.arange(padata.shape[0])
agood_indexes, = np.where(np.isfinite(padata))
f = interp1d(agood_indexes
           , padata[agood_indexes]
           , bounds_error=False
           , copy=False
           , fill_value="extrapolate"
           , kind=pkind)
return f(aindexes)

In [17]: adata = np.array([1, 2, np.NaN, 4])
Out[18]: array([ 1.,  2., nan,  4.])
In [19]: fill_nans_scipy1(adata)
Out[19]: array([1., 2., 3., 4.])

Answer 9

我需要一种在数据的开头和结尾处填充 NaN 的方法，但主要答案似乎没有这样做。

我想出的函数使用线性回归来填充 NaN。这解决了我的问题：

import numpy as np

def linearly_interpolate_nans(y):
    # Fit a linear regression to the non-nan y values

    # Create X matrix for linreg with an intercept and an index
    X = np.vstack((np.ones(len(y)), np.arange(len(y))))
    
    # Get the non-NaN values of X and y
    X_fit = X[:, ~np.isnan(y)]
    y_fit = y[~np.isnan(y)].reshape(-1, 1)
    
    # Estimate the coefficients of the linear regression
    beta = np.linalg.lstsq(X_fit.T, y_fit)[0]
    
    # Fill in all the nan values using the predicted coefficients
    y.flat[np.isnan(y)] = np.dot(X[:, np.isnan(y)].T, beta)
    return y

这是一个示例用例：

# Make an array according to some linear function
y = np.arange(12) * 1.5 + 10.

# First and last value are NaN
y[0] = np.nan
y[-1] = np.nan

# 30% of other values are NaN
for i in range(len(y)):
    if np.random.rand() > 0.7:
        y[i] = np.nan
        
# NaN's are filled in!
print (y)
print (linearly_interpolate_nans(y))

Answer 10

基于 Bryan Woods 的答案，我修改了他的代码，将仅包含

NaN

的列表转换为零列表：

def fill_nan(A):
    '''
    interpolate to fill nan values
    '''
    inds = np.arange(A.shape[0])
    good = np.where(np.isfinite(A))
    if len(good[0]) == 0:
        return np.nan_to_num(A)
    f = interp1d(inds[good], A[good], bounds_error=False)
    B = np.where(np.isfinite(A), A, f(inds))
    return B

简单补充，希望对大家有用。

Answer 11

正如之前的评论所建议的，最好的方法是使用同行评审的实现。 pandas 库有一个针对 1d 数据的插值方法，它将

np.nan

值插值到

Series

或

DataFrame

:

pandas.Series.interpolate 或 pandas.DataFrame.interpolate

文档非常简洁，推荐仔细阅读！我的实现：

import pandas as pd

magnitudes_series = pd.Series(magnitudes)    # Convert np.array to pd.Series
magnitudes_series.interpolate(
    # I used "akima" because the second derivative of my data has frequent drops to 0
    method=interpolation_method,

    # Interpolate from both sides of the sequence, up to you (made sense for my data)
    limit_direction="both",

    # Interpolate only np.nan sequences that have number sequences at the ends of the respective np.nan sequences
    limit_area="inside",

    inplace=True,
)

# I chose to remove np.nan at the tails of data sequence
magnitudes_series.dropna(inplace=True)

result_in_numpy_array = magnitudes_series.values

Answer 12

导入 scipy 对我来说看起来有点大材小用。这是使用 numpy 并保持与 np.interp

相同约定的简单方法

   def interp_nans(x:[float],left=None, right=None, period=None)->[float]:
    """ 
      e.g. [1 1 1 nan nan 2 2 nan 0] -> [1 1 1 1.3 1.6 2 2  1  0]
    
    """
    xp = [i for i, yi in enumerate(x) if np.isfinite(yi)]
    fp = [yi for i, yi in enumerate(x) if np.isfinite(yi)]
    return list(np.interp(x=list(range(len(x))), xp=xp, fp=fp,left=left,right=right,period=period))

Answer 13

使用填充关键字进行插值和外推

如果两侧都存在有限值，则以下解决方案通过 np.interp

 将 nan 值

插值到数组中。 边界处的 Nan 值由

np.pad

处理，具有

constant

或

reflect

等模式。

    import numpy as np
    import matplotlib.pyplot as plt
    
    
    def extrainterpolate_nans_1d(
            arr, kws_pad=({'mode': 'edge'}, {'mode': 'edge'})
            ):
        """Interpolates and extrapolates nan values.
    
        Interpolation is linear, compare np.interp(..).
        Extrapolation works with pad keywords, compare np.pad(..).
    
        Parameters
        ----------
        arr : np.ndarray, shape (N,)
            Array to replace nans in.
        kws_pad : dict or (dict, dict)
            kwargs for np.pad on left and right side
    
        Returns
        -------
        bool
            Description of return value
    
        See Also
        --------
        https://numpy.org/doc/stable/reference/generated/numpy.interp.html
        https://numpy.org/doc/stable/reference/generated/numpy.pad.html
        https://stackoverflow.com/a/43821453/7128154
        """
        assert arr.ndim == 1
        if isinstance(kws_pad, dict):
            kws_pad_left = kws_pad
            kws_pad_right = kws_pad
        else:
            assert len(kws_pad) == 2
            assert isinstance(kws_pad[0], dict)
            assert isinstance(kws_pad[1], dict)
            kws_pad_left = kws_pad[0]
            kws_pad_right = kws_pad[1]
    
        arr_ip = arr.copy()
    
        # interpolation
        inds = np.arange(len(arr_ip))
        nan_msk = np.isnan(arr_ip)
        arr_ip[nan_msk] = np.interp(inds[nan_msk], inds[~nan_msk], arr[~nan_msk])
    
        # detemine pad range
        i0 = next(
            (ids for ids, val in np.ndenumerate(arr) if not np.isnan(val)), 0)[0]
        i1 = next(
            (ids for ids, val in np.ndenumerate(arr[::-1]) if not np.isnan(val)), 0)[0]
        i1 = len(arr) - i1
        # print('pad in range [0:{:}] and [{:}:{:}]'.format(i0, i1, len(arr)))
    
        # pad
        arr_pad = np.pad(
            arr_ip[i0:], pad_width=[(i0, 0)], **kws_pad_left)
        arr_pad = np.pad(
            arr_pad[:i1], pad_width=[(0, len(arr) - i1)], **kws_pad_right)
    
        return arr_pad
    
    
    # setup data
    ys = np.arange(30, dtype=float)**2/20
    ys[:5] = np.nan
    ys[20:] = 20
    ys[28:] = np.nan
    ys[[7, 13, 14, 18, 22]] = np.nan
    
    
    ys_ie0 = extrainterpolate_nans_1d(ys)
    kws_pad_sym = {'mode': 'symmetric'}
    kws_pad_const7 = {'mode': 'constant', 'constant_values':7.}
    ys_ie1 = extrainterpolate_nans_1d(ys, kws_pad=(kws_pad_sym, kws_pad_const7))
    ys_ie2 = extrainterpolate_nans_1d(ys, kws_pad=(kws_pad_const7, kws_pad_sym))
    
    fig, ax = plt.subplots()
    
    
    ax.scatter(np.arange(len(ys)), ys, s=15**2, label='ys')
    ax.scatter(np.arange(len(ys)), ys_ie0, s=8**2, label='ys_ie0, left_pad edge, right_pad edge')
    ax.scatter(np.arange(len(ys)), ys_ie1, s=6**2, label='ys_ie1, left_pad symmetric, right_pad 7')
    ax.scatter(np.arange(len(ys)), ys_ie2, s=4**2, label='ys_ie2, left_pad 7, right_pad symmetric')
    ax.legend()

Answer 14

为此提供了一种灵活、专用的方法，利用 scipy 的 B-Splines 并允许用户控制一些参数（如平滑、外推等）：

https://github.com/DomTomCat/signalProcessing/blob/main/interpolateNonFiniteValues.py