调整不包含NaN的二维numpy阵列的大小
我正在尝试调整给定因子的2D numpy数组的大小，以便在输出中获得更小的数组。
该阵列从图像文件中读取，其中一些值应该是NaN（不是来自numpy的数字，np.nan）：它是来自卫星的遥感测量结果，并且仅测量了一些像素。
我发现的适合的包是scypy.misc.imresize，但是包含NaN的输出数组中的每个像素都被设置为NaN，即使原始像素中有一些有效数据被内插在一起。
我的解决方案附在这里，我所做的基本上是：
我正计划在不同的输出（平均值，中值，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〜1Mpx的图像上大约需要3秒。 由于我缺乏Python的经验，我正在寻求改进。
有没有人有建议如何更好，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。
这里有一个用下面代码生成的随机像素输入示例输出：
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。
如果我用其他值代替NaN，假设说非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重新采样的数组之间的差异代替了NaN，表明2个像素的值改变了它们的值。
我的目标是简单地跳过所有的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，遗憾的是它插值了应该是“空”（NaN）的像素的新值，并且对于我的小例子来说，生成的NaN比好的值要多。
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]])
您正在操作阵列的小窗口。 除了循环访问数组以创建窗口外，还可以通过操纵其大步骤来高效地重组数组。 numpy库提供as_strided()函数来帮助解决这个问题。 SciPy CookBook Stride技巧为生命游戏提供了一个例子。
下面将使用在Numpy的Efficient Overlapping Windows中找到的广义滑动窗口函数 - 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。 如果内存有问题，使用itertools.izip()而不是`zip也应该有帮助。 我只是使用列表理解你的解决方案。
你的功能：
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
我没有和原来的功能做过任何时间比较，但它应该更快。
我在这里看到的很多解决方案都是通过向量化操作来提高速度/效率 - 我并不完全掌握它，也不知道它是否可以应用于您的问题。 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)
在不同的网格上使用scipy.interpolate对点进行插值。 下面我已经展示了一个立方体插入器，它比较慢但可能更准确。 你会注意到这个函数没有角落像素，那么你可以使用线性或最近邻居插值来处理这些最后的值。
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()