Question

我正在尝试调整给定因子的2D numpy数组，在输出中获得一个更小的数组。

从图像文件中读取数组，并且一些值应为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，但输出数组中包含NaN的每个像素都设置为NaN，即使原始像素中有一些有效数据一起插入。

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
创建一个索引数组，以解决原始数组中要为新
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。

我打算在不同输出之间添加关键字（平均值，中位数，输入像素的标准差等）。

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。

有人建议如何更好，更有效地做到这一点吗？

有人知道一个已经实现了所有这些东西的库吗？

感谢。

这里有一个使用以下代码生成的随机像素输入的示例输出：

Example output for random pixel input (see code)

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]))

修改

我对地址ChrisProsser comment添加了一些修改。

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像素改变了他们的价值观。

我的目标只是跳过所有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.        ]]

enter image description here

**第二次编辑**

为了解决Hooked的答案，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），而对于我的小例子，它产生的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]])

enter image description here

Answer 1

您正在对阵列的 windows 进行操作。不是循环遍历数组来构建窗口，而是可以通过操纵其步幅来有效地重构阵列。 numpy库提供了as_strided()函数来帮助解决这个问题。 SciPy CookBook Stride tricks for the Game of Life中提供了一个示例。

以下将使用Efficient Overlapping Windows with Numpy处的广义滑动窗函数 - 我将在最后包含它。

确定新阵列的形状：

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

将数组重组为您需要的窗口，并创建一个标识NaNs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))

可以使用列表推导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。如果内存有问题，使用itertools.izip()代替`zip也会有所帮助。我只是将列表理解用于您的解决方案。

您的职能：

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

我没有与原始功能进行任何时间比较，但它应该更快。

我在这里看到的许多解决方案都是在矢量化操作以提高速度/效率 - 我没有完全掌握它并且不知道它是否可以应用于你的问题。在SO中搜索窗口，数组，移动平均线，矢量化和numpy应该产生类似的问题和答案以供参考。

来自Efficient Overlapping Windows with Numpy 的

sliding_window()：

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)

Answer 2

使用scipy.interpolate在不同的网格上插入点。下面我展示了一个cubic interpolator，它更慢但可能更准确。您会注意到此功能缺少角点像素，然后您可以使用linear或nearest neighbor插值来处理这些最后的值。

enter image description here

import numpy as np
import pylab as plt

# Test data
row = np.linspace(-3,3,50)
X,Y = np.meshgrid(row,row)
Z = np.sqrt(X**2+Y**2) + np.cos(Y) 

# Make some dead pixels, favor an edge
dead = np.random.random(Z.shape)
dead = (dead*X>.7)
Z[dead] =np.nan

from scipy.interpolate import CloughTocher2DInterpolator as intp
C = intp((X[~dead],Y[~dead]),Z[~dead])

new_row = np.linspace(-3,3,25)
xi,yi   = np.meshgrid(new_row,new_row)
zi = C(xi,yi)

plt.subplot(121)
plt.title("Original signal 50x50")
plt.imshow(Z,interpolation='nearest')

plt.subplot(122)
plt.title("Interpolated signal 25x25")
plt.imshow(zi,interpolation='nearest')

plt.show()

调整除NaN之外的2D numpy数组的大小

2 个答案: