调整除NaN之外的2D numpy数组的大小

Question

我正在尝试调整给定因子的2D numpy数组,在输出中获得更小的数组.

从图像文件中读取数组,并且一些值应该是NaN(非数字,来自numpy的np.nan):它是来自卫星的遥感测量的结果,并且仅测量了一些像素.

我找到的合适的包是scypy.misc.imresize,但是包含NaN的输出数组中的每个像素都设置为NaN,即使在原始像素中有一些有效数据被插入在一起.

我的解决方案附在此处,我所做的基本上是:

• 根据原始数组形状和所需的缩减系数创建一个新数组
• 创建一个索引数组,以解决原始数组中要为新像素中的每个像素求平均值的所有像素
• 循环通过新的阵列像素并平均所有非NaN像素以获得新的阵列像素值; 它只有NaN,输出将是NaN.

我打算在不同的输出(平均值,中位数,输入像素的标准偏差等)之间添加关键字.

它按预期工作,但在~110px图像上大约需要3秒钟.由于我缺乏python的经验,我正在寻找改进.

有没有人建议如何更好,更有效地做到这一点？

有谁知道已经实现了所有这些东西的图书馆？

谢谢.

这里有一个输出随机像素输入的示例输出,使用下面的代码生成:

import numpy as np
import pylab as plt
from scipy import misc

def resize_2d_nonan(array,factor):
    """
    Resize a 2D array by different factor on two axis sipping NaN values.
    If a new pixel contains only NaN, it will be set to NaN


    Parameters
    ----------

    array : 2D np array

    factor : int or tuple. If int x and y factor wil be the same

    Returns
    -------
    array : 2D np array scaled by factor

    Created on Mon Jan 27 15:21:25 2014

    @author: damo_ma
    """
    xsize, ysize = array.shape

    if isinstance(factor,int):
        factor_x = factor
        factor_y = factor
    elif isinstance(factor,tuple):
        factor_x , factor_y = factor[0], factor[1]
    else:
        raise NameError('Factor must be a tuple (x,y) or an integer')

    if not (xsize %factor_x == 0 or ysize % factor_y == 0) :
        raise NameError('Factors must be intger multiple of array shape')

    new_xsize, new_ysize = xsize/factor_x, ysize/factor_y

    new_array = np.empty([new_xsize, new_ysize])
    new_array[:] = np.nan # this saves us an assignment in the loop below

    # submatrix indexes : is the average box on the original matrix
    subrow, subcol  = np.indices((factor_x, factor_y))

     # new matrix indexs
    row, col  = np.indices((new_xsize, new_ysize))

    # some output for testing
    #for i, j, ind in zip(row.reshape(-1), col.reshape(-1),range(row.size)) :
    #    print '----------------------------------------------'
    #    print 'i: %i, j: %i, ind: %i ' % (i, j, ind)    
    #    print 'subrow+i*new_ysize, subcol+j*new_xsize :'    
    #    print i,'*',new_xsize,'=',i*factor_x
    #    print j,'*',new_ysize,'=',j*factor_y
    #    print subrow+i*factor_x,subcol+j*factor_y
    #    print '---'
    #    print 'array[subrow+i*factor_x,subcol+j*factor_y] : '    
    #    print array[subrow+i*factor_x,subcol+j*factor_y]

    for i, j, ind in zip(row.reshape(-1), col.reshape(-1),range(row.size)) :
        # define the small sub_matrix as view of input matrix subset
        sub_matrix = array[subrow+i*factor_x,subcol+j*factor_y]
        # modified from any(a) and all(a) to a.any() and a.all()
        # see https://stackoverflow.com/a/10063039/1435167
        if not (np.isnan(sub_matrix)).all(): # if we haven't all NaN
            if (np.isnan(sub_matrix)).any(): # if we haven no NaN at all
                msub_matrix = np.ma.masked_array(sub_matrix,np.isnan(sub_matrix))
                (new_array.reshape(-1))[ind] = np.mean(msub_matrix)
            else: # if we haven some NaN
                (new_array.reshape(-1))[ind] = np.mean(sub_matrix)
        # the case assign NaN if we have all NaN is missing due 
        # to the standard values of new_array

    return new_array


row , cols = 6, 4

a = 10*np.random.random_sample((row , cols))
a[0:3,0:2] = np.nan
a[0,2] = np.nan

factor_x = 2
factor_y = 2
a_misc = misc.imresize(a, .5, interp='nearest', mode='F')
a_2d_nonan = resize_2d_nonan(a,(factor_x,factor_y))

print a
print
print a_misc
print
print a_2d_nonan

plt.subplot(131)
plt.imshow(a,interpolation='nearest')
plt.title('original')
plt.xticks(arange(a.shape[1]))
plt.yticks(arange(a.shape[0]))
plt.subplot(132)
plt.imshow(a_misc,interpolation='nearest')
plt.title('scipy.misc')
plt.xticks(arange(a_misc.shape[1]))
plt.yticks(arange(a_misc.shape[0]))
plt.subplot(133)
plt.imshow(a_2d_nonan,interpolation='nearest')
plt.title('my.func')
plt.xticks(arange(a_2d_nonan.shape[1]))
plt.yticks(arange(a_2d_nonan.shape[0]))

编辑

我添加了一些修改来解决ChrisProsser评论.

如果我用一些其他值替换NaN,假设非NaN像素的平均值,它将影响所有后续计算:重新采样的原始数组和使用NaN替换的重采样数组之间的差异显示2个像素更改了它们的值.

我的目标是简单地跳过所有NaN像素.

# substitute NaN with the average value 

ind_nonan , ind_nan = np.where(np.isnan(a) == False), np.where(np.isnan(a) == True)
a_substitute = np.copy(a)

a_substitute[ind_nan] = np.mean(a_substitute[ind_nonan]) # substitute the NaN with average on the not-Nan

a_substitute_misc = misc.imresize(a_substitute, .5, interp='nearest', mode='F')
a_substitute_2d_nonan = resize_2d_nonan(a_substitute,(factor_x,factor_y))

print a_2d_nonan-a_substitute_2d_nonan

[[        nan -0.02296697]
 [ 0.23143208  0.        ]
 [ 0.          0.        ]]

**第二次编辑**

为了解决Hooked的回答,我添加了一些额外的代码.这是一个迭代的想法,遗憾的是它在像素上插入新的值应该是"空"(NaN),并且对于我的小例子,产生的NaN比良好的值更多.

X , Y  = np.indices((row , cols))
X_new , Y_new  = np.indices((row/factor_x , cols/factor_y))

from scipy.interpolate import CloughTocher2DInterpolator as intp
C = intp((X[ind_nonan],Y[ind_nonan]),a[ind_nonan])

a_interp = C(X_new , Y_new)

print a
print
print a_interp

[[        nan,         nan],
 [        nan,         nan],
 [        nan,  6.32826577]])

Answer 1

您正在阵列的小窗口上进行操作。可以通过操纵数组的步幅来有效地重构数组，而不是循环遍历数组来创建窗口。numpy 库提供了as_strided()帮助实现这一点的函数。SciPy CookBook生命游戏的跨步技巧中提供了一个示例。

下面将使用广义的滑动窗口函数，我将把它放在最后。

确定新数组的形状：

rows, cols = a.shape
new_shape = rows / 2, cols / 2

将数组重构为您需要的窗口，并创建一个标识 NaN 的索引数组：

# 2x2 windows of the original array
windows = sliding_window(a, (2,2))
# make a windowed boolean array for indexing
notNan = sliding_window(np.logical_not(np.isnan(a)), (2,2))

可以使用列表理解或生成器表达式来创建新数组。

# using a list comprehension
# make a list of the means of the windows, disregarding the Nan's
means = [window[index].mean() for window, index in zip(windows, notNan)]
new_array = np.array(means).reshape(new_shape)

# generator expression
# produces the means of the windows, disregarding the Nan's
means = (window[index].mean() for window, index in zip(windows, notNan))
new_array = np.fromiter(means, dtype = np.float32).reshape(new_shape)

生成器表达式应该节省内存。itertools.izip()如果内存有问题，使用而不是“zip”也应该有所帮助。我刚刚使用列表理解来解决您的问题。

你的职能：

def resize_2d_nonan(array,factor):
    """
    Resize a 2D array by different factor on two axis skipping NaN values.
    If a new pixel contains only NaN, it will be set to NaN

    Parameters
    ----------
    array : 2D np array

    factor : int or tuple. If int x and y factor wil be the same

    Returns
    -------
    array : 2D np array scaled by factor

    Created on Mon Jan 27 15:21:25 2014

    @author: damo_ma
    """
    xsize, ysize = array.shape

    if isinstance(factor,int):
        factor_x = factor
        factor_y = factor
        window_size = factor, factor
    elif isinstance(factor,tuple):
        factor_x , factor_y = factor
        window_size = factor
    else:
        raise NameError('Factor must be a tuple (x,y) or an integer')

    if (xsize % factor_x or ysize % factor_y) :
        raise NameError('Factors must be integer multiple of array shape')

    new_shape = xsize / factor_x, ysize / factor_y

    # non-overlapping windows of the original array
    windows = sliding_window(a, window_size)
    # windowed boolean array for indexing
    notNan = sliding_window(np.logical_not(np.isnan(a)), window_size)

    #list of the means of the windows, disregarding the Nan's
    means = [window[index].mean() for window, index in zip(windows, notNan)]
    # new array
    new_array = np.array(means).reshape(new_shape)

    return new_array

我没有与你的原始函数进行任何时间比较，但它应该更快。

我在这里看到的许多解决方案都对操作进行矢量化以提高速度/效率 - 我不太了解这一点，也不知道它是否可以应用于您的问题。在 SO 中搜索窗口、数组、移动平均、向量化和 numpy 应该会产生类似的问题和答案以供参考。

sliding_window()请参阅下面的归因：

import numpy as np
from numpy.lib.stride_tricks import as_strided as ast
from itertools import product

def norm_shape(shape):
    '''
    Normalize numpy array shapes so they're always expressed as a tuple, 
    even for one-dimensional shapes.
     
    Parameters
        shape - an int, or a tuple of ints
     
    Returns
        a shape tuple
    '''
    try:
        i = int(shape)
        return (i,)
    except TypeError:
        # shape was not a number
        pass
 
    try:
        t = tuple(shape)
        return t
    except TypeError:
        # shape was not iterable
        pass
     
    raise TypeError('shape must be an int, or a tuple of ints')
 

def sliding_window(a,ws,ss = None,flatten = True):
    '''
    Return a sliding window over a in any number of dimensions
     
    Parameters:
        a  - an n-dimensional numpy array
        ws - an int (a is 1D) or tuple (a is 2D or greater) representing the size 
             of each dimension of the window
        ss - an int (a is 1D) or tuple (a is 2D or greater) representing the 
             amount to slide the window in each dimension. If not specified, it
             defaults to ws.
        flatten - if True, all slices are flattened, otherwise, there is an 
                  extra dimension for each dimension of the input.
     
    Returns
        an array containing each n-dimensional window from a
    '''
     
    if None is ss:
        # ss was not provided. the windows will not overlap in any direction.
        ss = ws
    ws = norm_shape(ws)
    ss = norm_shape(ss)
     
    # convert ws, ss, and a.shape to numpy arrays so that we can do math in every 
    # dimension at once.
    ws = np.array(ws)
    ss = np.array(ss)
    shape = np.array(a.shape)
     
     
    # ensure that ws, ss, and a.shape all have the same number of dimensions
    ls = [len(shape),len(ws),len(ss)]
    if 1 != len(set(ls)):
        raise ValueError(\
        'a.shape, ws and ss must all have the same length. They were %s' % str(ls))
     
    # ensure that ws is smaller than a in every dimension
    if np.any(ws > shape):
        raise ValueError(\
        'ws cannot be larger than a in any dimension.\
 a.shape was %s and ws was %s' % (str(a.shape),str(ws)))
     
    # how many slices will there be in each dimension?
    newshape = norm_shape(((shape - ws) // ss) + 1)
    # the shape of the strided array will be the number of slices in each dimension
    # plus the shape of the window (tuple addition)
    newshape += norm_shape(ws)
    # the strides tuple will be the array's strides multiplied by step size, plus
    # the array's strides (tuple addition)
    newstrides = norm_shape(np.array(a.strides) * ss) + a.strides
    strided = ast(a,shape = newshape,strides = newstrides)
    if not flatten:
        return strided
     
    # Collapse strided so that it has one more dimension than the window.  I.e.,
    # the new array is a flat list of slices.
    meat = len(ws) if ws.shape else 0
    firstdim = (np.product(newshape[:-meat]),) if ws.shape else ()
    dim = firstdim + (newshape[-meat:])
    # remove any dimensions with size 1
    dim = filter(lambda i : i != 1,dim)
    return strided.reshape(dim)

slip_window() 归属
我最初在博客页面上发现了这个，现在是一个损坏的链接：

使用 Numpy 高效重叠窗口 - http://www.johnvinyard.com/blog/?p=268

经过一番搜索，它看起来现在位于Zounds github 存储库中。谢谢约翰·维尼亚德。

请注意，这篇文章已经很老了，并且有很多关于滑动窗口、滚动窗口和图像补丁提取的 Q&A。使用 numpy 的 as_strided有很多一次性的功能，但该函数似乎仍然是唯一处理 nd 窗口的函数。scikits sklearn.feature_extraction.image 库似乎经常被引用用于提取或查看图像补丁。