请原谅我的迟到，但这里是大型数组的最快解决方案，iff单精度(f4 float32)是你所需要的。是的，np。南也在工作。

def full_single_prec(n):
    return numpy.full(n, val, dtype='f4')

如果你没有立即回忆起.empty或.full方法，你总是可以使用乘法:

>>> np.nan * np.ones(shape=(3,2))
array([[ nan,  nan],
       [ nan,  nan],
       [ nan,  nan]])

当然，它也适用于任何其他数值:

>>> 42 * np.ones(shape=(3,2))
array([[ 42,  42],
       [ 42,  42],
       [ 42, 42]])

但是@u0b34a0f6ae接受的答案是快3倍(CPU周期，而不是大脑周期来记住numpy语法;):

$ python -mtimeit "import numpy as np; X = np.empty((100,100));" "X[:] = np.nan;"
100000 loops, best of 3: 8.9 usec per loop
(predict)laneh@predict:~/src/predict/predict/webapp$ master
$ python -mtimeit "import numpy as np; X = np.ones((100,100));" "X *= np.nan;"
10000 loops, best of 3: 24.9 usec per loop

如前所述，numpy.empty()是可行的方法。然而，对于对象，fill()可能并不完全像你想象的那样:

In[36]: a = numpy.empty(5,dtype=object)
In[37]: a.fill([])
In[38]: a
Out[38]: array([[], [], [], [], []], dtype=object)
In[39]: a[0].append(4)
In[40]: a
Out[40]: array([[4], [4], [4], [4], [4]], dtype=object)

一种方法可以是:

In[41]: a = numpy.empty(5,dtype=object)
In[42]: a[:]= [ [] for x in range(5)]
In[43]: a[0].append(4)
In[44]: a
Out[44]: array([[4], [], [], [], []], dtype=object)

我比较了建议的替代方案的速度，发现对于足够大的向量/矩阵来填充，除了val * ones和array(n * [val])之外的所有替代方案都同样快。

代码重现情节:

import numpy
import perfplot

val = 42.0


def fill(n):
    a = numpy.empty(n)
    a.fill(val)
    return a


def colon(n):
    a = numpy.empty(n)
    a[:] = val
    return a


def full(n):
    return numpy.full(n, val)


def ones_times(n):
    return val * numpy.ones(n)


def list(n):
    return numpy.array(n * [val])


b = perfplot.bench(
    setup=lambda n: n,
    kernels=[fill, colon, full, ones_times, list],
    n_range=[2 ** k for k in range(20)],
    xlabel="len(a)",
)
b.save("out.png")

>>> width = 2
>>> height = 3

>>> r = np.full((width, height, 9), np.nan)

>>> print(r)

array([[[nan, nan, nan, nan, nan, nan, nan, nan, nan],
        [nan, nan, nan, nan, nan, nan, nan, nan, nan],
        [nan, nan, nan, nan, nan, nan, nan, nan, nan]],

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

>>> r.shape
(2, 3, 9)

另一种选择是使用numpy。full, NumPy 1.8+中可用的选项

a = np.full([height, width, 9], np.nan)

这是非常灵活的，你可以用任何你想要的数字来填充它。