Take a 2D numpy.array
, let's say:
mat = numpy.random.rand(3,3)
In [153]: mat
Out[153]:
array([[ 0.16716156, 0.90822617, 0.83888038],
[ 0.89771815, 0.62627978, 0.34992542],
[ 0.11097042, 0.80858005, 0.0437299 ]])
将索引更改为numpy.nan
非常简单。
以下其中一种方法效果很好:
In [154]: diag = numpy.diag_indices(mat.shape[0], ndim = 2)
In [155]: mat[diag] = numpy.nan
或者
In [156]: numpy.fill_diagonal(mat, numpy.nan)
假设我有一个三维数组,在第三个维度上我希望进行完全相同的处理。
mat = numpy.random.rand(3, 5, 5)
In [158]: mat
Out[158]:
array([[[ 0.65000325, 0.71059547, 0.31880388, 0.24818623, 0.57722849],
[ 0.26908326, 0.41962004, 0.78642476, 0.25711662, 0.8662998 ],
[ 0.15332566, 0.12633147, 0.54032977, 0.17322095, 0.17210078],
[ 0.81952873, 0.20751669, 0.73514815, 0.00884358, 0.89222687],
[ 0.62775839, 0.53657471, 0.99611842, 0.75051645, 0.59328044]],
[[ 0.28718216, 0.84982865, 0.27830082, 0.90604492, 0.43119512],
[ 0.43039373, 0.76557782, 0.58089787, 0.81135684, 0.39151152],
[ 0.70592711, 0.30625204, 0.9753166 , 0.32806864, 0.21947731],
[ 0.74600317, 0.33711673, 0.16203076, 0.6002213 , 0.74996638],
[ 0.63555715, 0.71719058, 0.81420001, 0.28968442, 0.01368163]],
[[ 0.06474027, 0.51966572, 0.006429 , 0.98590784, 0.35708074],
[ 0.44977222, 0.63719921, 0.88325451, 0.53820139, 0.51526687],
[ 0.98529117, 0.46219441, 0.09349748, 0.11406291, 0.47697128],
[ 0.77446136, 0.87423445, 0.71810465, 0.39019846, 0.94070077],
[ 0.09154989, 0.36295161, 0.19740833, 0.17803146, 0.6498038 ]]])
我认为一个合理的做法是:
mat[:, diag] = numpy.nan # doesn't do it
实际上,为了做到这一点,我需要:
In [190]: rng = numpy.arange(5)
In [191]: for i in numpy.arange(mat.shape[0]):
.....: mat[i, rng, rng] = numpy.nan
.....:
In [192]: mat
Out[192]:
array([[[ nan, 0.4040426 , 0.89449522, 0.63593736, 0.94922036],
[ 0.40682651, nan, 0.30812181, 0.01726625, 0.75655994],
[ 0.23925763, 0.41476223, nan, 0.91590111, 0.18391644],
[ 0.99784977, 0.71636554, 0.21252766, nan, 0.24195636],
[ 0.41137357, 0.84705055, 0.60086461, 0.16403918, nan]],
[[ nan, 0.26183712, 0.77621913, 0.5479058 , 0.17142263],
[ 0.17969373, nan, 0.89742863, 0.65698339, 0.95817106],
[ 0.79048886, 0.16365168, nan, 0.97394435, 0.80612441],
[ 0.94169129, 0.10895737, 0.92614597, nan, 0.08689534],
[ 0.20324943, 0.91402716, 0.23112819, 0.2556875 , nan]],
[[ nan, 0.43177039, 0.76901587, 0.82069345, 0.64351534],
[ 0.14148584, nan, 0.35820379, 0.17434688, 0.78884305],
[ 0.85232784, 0.93526843, nan, 0.80981366, 0.57326785],
[ 0.82104636, 0.63453196, 0.5872653 , nan, 0.96214559],
[ 0.69959383, 0.70257404, 0.92471502, 0.50077728, nan]]])
这是一个关于应用程序速度至关重要的问题。如果没有以下基于数组的实现,我将在Cython中使用for循环/赋值。
mat.shape
的最小值而不是mat.shape[0]
来计算diag。 - innoSPG