创建用NaN填充的Numpy矩阵

问题:创建用NaN填充的Numpy矩阵

我有以下代码:

r = numpy.zeros(shape = (width, height, 9))

它创建一个width x height x 9填充零的矩阵。相反,我想知道是否有一种函数或方法可以将它们初始化为NaNs,而方法很简单。

I have the following code:

r = numpy.zeros(shape = (width, height, 9))

It creates a width x height x 9 matrix filled with zeros. Instead, I’d like to know if there’s a function or way to initialize them instead to NaNs in an easy way.


回答 0

您很少需要在numpy中进行矢量操作循环。您可以创建一个未初始化的数组并立即分配给所有条目:

>>> a = numpy.empty((3,3,))
>>> a[:] = numpy.nan
>>> a
array([[ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN]])

我已经在a[:] = numpy.nan这里和a.fill(numpy.nan)Blaenk发布的时间安排了时间:

$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a.fill(np.nan)"
10000 loops, best of 3: 54.3 usec per loop
$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a[:] = np.nan" 
10000 loops, best of 3: 88.8 usec per loop

时序显示优先选择ndarray.fill(..)更快的替代方法。OTOH,我喜欢numpy的便捷实现,在该实现中您可以同时为整个slice分配值,代码的意图非常明确。

请注意,ndarray.fill它是就地执行其操作,因此numpy.empty((3,3,)).fill(numpy.nan)将改为return None

You rarely need loops for vector operations in numpy. You can create an uninitialized array and assign to all entries at once:

>>> a = numpy.empty((3,3,))
>>> a[:] = numpy.nan
>>> a
array([[ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN]])

I have timed the alternatives a[:] = numpy.nan here and a.fill(numpy.nan) as posted by Blaenk:

$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a.fill(np.nan)"
10000 loops, best of 3: 54.3 usec per loop
$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a[:] = np.nan" 
10000 loops, best of 3: 88.8 usec per loop

The timings show a preference for ndarray.fill(..) as the faster alternative. OTOH, I like numpy’s convenience implementation where you can assign values to whole slices at the time, the code’s intention is very clear.

Note that ndarray.fill performs its operation in-place, so numpy.empty((3,3,)).fill(numpy.nan) will instead return None.


回答 1

另一个选择是使用numpy.full,NumPy 1.8+中可用的一个选项

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

这非常灵活,您可以用任何其他所需的数字填充它。

Another option is to use numpy.full, an option available in NumPy 1.8+

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

This is pretty flexible and you can fill it with any other number that you want.


回答 2

我比较了建议的速度替代方案,发现对于足够大的向量/矩阵填充,除val * ones和以外的所有替代方案array(n * [val])都同样快。


复制剧情的代码:

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])


perfplot.show(
    setup=lambda n: n,
    kernels=[fill, colon, full, ones_times, list],
    n_range=[2 ** k for k in range(20)],
    logx=True,
    logy=True,
    xlabel="len(a)",
)

I compared the suggested alternatives for speed and found that, for large enough vectors/matrices to fill, all alternatives except val * ones and array(n * [val]) are equally fast.


Code to reproduce the plot:

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])


perfplot.show(
    setup=lambda n: n,
    kernels=[fill, colon, full, ones_times, list],
    n_range=[2 ** k for k in range(20)],
    logx=True,
    logy=True,
    xlabel="len(a)",
)

回答 3

你熟悉numpy.nan吗?

您可以创建自己的方法,例如:

def nans(shape, dtype=float):
    a = numpy.empty(shape, dtype)
    a.fill(numpy.nan)
    return a

然后

nans([3,4])

将输出

array([[ NaN,  NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN,  NaN]])

我在邮件列表线程中找到了此代码。

Are you familiar with numpy.nan?

You can create your own method such as:

def nans(shape, dtype=float):
    a = numpy.empty(shape, dtype)
    a.fill(numpy.nan)
    return a

Then

nans([3,4])

would output

array([[ NaN,  NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN,  NaN]])

I found this code in a mailing list thread.


回答 4

如果您不立即调用.empty.full方法,则始终可以使用乘法:

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

当然,它也可以与其他任何数值一起使用:

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

但是@ u0b34a0f6ae 可接受的答案快了3倍(CPU周期,而不是记住numpy语法的大脑周期;):

$ 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

You can always use multiplication if you don’t immediately recall the .empty or .full methods:

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

Of course it works with any other numerical value as well:

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

But the @u0b34a0f6ae’s accepted answer is 3x faster (CPU cycles, not brain cycles to remember numpy syntax ;):

$ 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

回答 5

另一种选择是numpy.broadcast_to(val,n),无论大小如何,它都将在恒定时间内返回,并且也是最有效的内存使用方法(它返回重复元素的视图)。需要注意的是,返回值是只读的。

以下是使用与NicoSchlömer的答案相同的基准所建议的所有其他方法的性能的比较。

Another alternative is numpy.broadcast_to(val,n) which returns in constant time regardless of the size and is also the most memory efficient (it returns a view of the repeated element). The caveat is that the returned value is read-only.

Below is a comparison of the performances of all the other methods that have been proposed using the same benchmark as in Nico Schlömer’s answer.


回答 6

如前所述,numpy.empty()是必经之路。但是,对于对象,fill()可能并不能完全按照您的想象:

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)

一种解决方法可以是例如:

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)

As said, numpy.empty() is the way to go. However, for objects, fill() might not do exactly what you think it does:

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)

One way around can be e.g.:

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)

回答 7

此处尚未提及的另一种可能性是使用NumPy tile:

a = numpy.tile(numpy.nan, (3, 3))

还给

array([[ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN]])

我不知道速度比较。

Yet another possibility not yet mentioned here is to use NumPy tile:

a = numpy.tile(numpy.nan, (3, 3))

Also gives

array([[ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN],
       [ NaN,  NaN,  NaN]])

I don’t know about speed comparison.