blocks|key|1412146|text|另一种选择是使用NumPy+1.8%2B中提供的numpy.full选项|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1412147|a+=+np.full([height,+width,+9],+np.nan)|code-block|syntax|javascript|1412148|这是非常灵活的，您可以填充任何其他您想要的数字。|1412149|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.full.html^0|M|A|M|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Another option is to use <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.full.html" rel="noreferrer"><code>numpy.full</code></a>, an option available in NumPy 1.8+

<pre><code>a = np.full([height, width, 9], np.nan)
</code></pre>

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

blocks|key|1197177|text|我比较了建议的替代方案的速度，发现对于足够大的向量/矩阵填充，除了val+*+ones和array(n+*+[val])之外的所有替代方案都是同样快的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1197178|​|1197179|📷|atomic|1197180|1197181|1197182|重现该图的代码：|1197183|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")|code-block|syntax|javascript|1197184|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/ffd09f4ab6f12a3ebb4fd8df26f0486d.png|imageAlt^0|X|A|18|G|0|0|0|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|K|3|G|5|6|7|1C|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|1E|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1F|8|@]|D|@]|E|$R|S]]|$1|T|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|U|$V|$5|W|X|Y|E|$Z|10|11|-4]]]]

I compared the suggested alternatives for speed and found that, for large enough vectors/matrices to fill, all alternatives except <code>val * ones</code> and <code>array(n * [val])</code> are equally fast.
<a href="https://i.stack.imgur.com/oGR81.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oGR81.png" alt="enter image description here" /></a>
<hr />
Code to reproduce the plot:
<pre class="lang-py prettyprint-override"><code>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=&quot;len(a)&quot;,
)
b.save(&quot;out.png&quot;)
</code></pre>

blocks|key|1196869|text|你熟悉numpy.nan吗？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1196870|您可以创建自己的方法，例如：|1196871|def+nans(shape,+dtype=float):
++++a+=+numpy.empty(shape,+dtype)
++++a.fill(numpy.nan)
++++return+a|code-block|syntax|javascript|1196872|然后|1196873|nans([3,4])|1196874|将输出|1196875|array([[+NaN,++NaN,++NaN,++NaN],
+++++++[+NaN,++NaN,++NaN,++NaN],
+++++++[+NaN,++NaN,++NaN,++NaN]])|1196876|我在mailing+list+thread中找到了这段代码。|1196877|entityMap|0|LINK|mutability|MUTABLE|url|http://www.mail-archive.com/numpy-discussion@scipy.org/msg08790.html^0|3|9|0|0|0|0|0|0|0|2|J|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|19|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1B|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1D|8|@]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|W|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]]]

Are you familiar with <code>numpy.nan</code>?

You can create your own method such as:

<pre><code>def nans(shape, dtype=float):
 a = numpy.empty(shape, dtype)
 a.fill(numpy.nan)
 return a
</code></pre>

Then

<pre><code>nans([3,4])
</code></pre>

would output

<pre><code>array([[ NaN, NaN, NaN, NaN],
 [ NaN, NaN, NaN, NaN],
 [ NaN, NaN, NaN, NaN]])
</code></pre>

I found this code in a <a href="http://www.mail-archive.com/numpy-discussion@scipy.org/msg08790.html" rel="noreferrer">mailing list thread</a>.

blocks|key|1196989|text|如果您没有立即回想起.empty或.full方法，则始终可以使用乘法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1196990|>>>+np.nan+*+np.ones(shape=(3,2))
array([[+nan,++nan],
+++++++[+nan,++nan],
+++++++[+nan,++nan]])|code-block|syntax|javascript|1196991|当然，它也适用于任何其他数值：|1196992|>>>+42+*+np.ones(shape=(3,2))
array([[+42,++42],
+++++++[+42,++42],
+++++++[+42,+42]])|1196993|但是@u0b34a0f6ae的accepted+answer要快3倍(记住numpy语法的是CPU周期，而不是大脑周期;)：|1196994|$+python+-mtimeit+"import+numpy+as+np;+X+=+np.empty((100,100));"+"X[:]+=+np.nan;"
100000+loops,+best+of+3:+8.9+usec+per+loop
(predict)laneh@predict:~/src/predict/predict/webapp$+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|1196995|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/1704853/623735^0|A|6|H|5|0|0|0|0|F|F|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|Q|3|R|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

You can always use multiplication if you don't immediately recall the <code>.empty</code> or <code>.full</code> methods:

<pre><code>&gt;&gt;&gt; np.nan * np.ones(shape=(3,2))
array([[ nan, nan],
 [ nan, nan],
 [ nan, nan]])
</code></pre>

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

<pre><code>&gt;&gt;&gt; 42 * np.ones(shape=(3,2))
array([[ 42, 42],
 [ 42, 42],
 [ 42, 42]])
</code></pre>

But the @u0b34a0f6ae's <a href="https://stackoverflow.com/a/1704853/623735">accepted answer</a> is 3x faster (CPU cycles, not brain cycles to remember numpy syntax ;):

<pre><code>$ 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
</code></pre>

blocks|key|1197079|text|如上所述，numpy.empty()是可行的方法。但是，对于对象，fill()可能不会执行您认为它所执行的操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1197080|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)|code-block|syntax|javascript|1197081|一种方法可以是：|1197082|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)|1197083|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

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

<pre><code>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)
</code></pre>

One way around can be e.g.:

<pre><code>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)
</code></pre>

blocks|key|1197330|text|另一种选择是numpy.broadcast_to(val,n)，无论大小如何，它都会在固定的时间内返回，而且内存效率也是最高的(它返回重复元素的视图)。需要注意的是返回值是只读的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1197331|下面是使用与Nico+Schlömer's+answer中相同的基准测试提出的所有其他方法的性能比较。|1197332|​|1197333|📷|atomic|1197334|1197335|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/45006691/3163607|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/567d765bf67b9b813f6a22bc164cb8f3.png|imageAlt^0|6|P|0|6|M|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|H|3|I|5|6|7|18|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|19|8|@]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|M|3|I|5|6|7|1D|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]|V|$5|W|R|X|E|$Y|Z|10|-4]]]]

Another alternative is <code>numpy.broadcast_to(val,n)</code> 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 <a href="https://stackoverflow.com/a/45006691/3163607">Nico Schlömer's answer</a>.

<a href="https://i.stack.imgur.com/RvrdV.png" rel="noreferrer"><img src="https://i.stack.imgur.com/RvrdV.png" alt="enter image description here"></a>

blocks|key|1197221|text|这里还没有提到的另一种可能性是使用NumPy磁贴：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1197222|a+=+numpy.tile(numpy.nan,+(3,+3))|code-block|syntax|javascript|1197223|还给出了|1197224|array([[+NaN,++NaN,++NaN],
+++++++[+NaN,++NaN,++NaN],
+++++++[+NaN,++NaN,++NaN]])|1197225|我不知道速度对比。|1197226|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

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

<pre><code>a = numpy.tile(numpy.nan, (3, 3))
</code></pre>

Also gives

<pre><code>array([[ NaN, NaN, NaN],
 [ NaN, NaN, NaN],
 [ NaN, NaN, NaN]])
</code></pre>

I don't know about speed comparison.

blocks|key|1412598|text|只是一个警告，使用np.empty()初始化而不随后编辑值可能会导致(内存分配？)问题：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1412599|arr1+=+np.empty(96)
arr2+=+np.empty(96)
print(arr1)
print(arr2)

#+[nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan++1.++1.
#+++1.++1.++2.++2.++2.++2.+nan+nan+nan+nan+nan+nan+nan+nan++0.++0.++0.++0.
#+++0.++0.++0.++0.+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+nan+nan]
#
#+[nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan++1.++1.
#+++1.++1.++2.++2.++2.++2.+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+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan+nan
#++nan+nan+nan+nan+nan+nan]|code-block|syntax|javascript|1412600|在数组中初始化的浮点数在我的脚本中的其他地方使用，但根本不与变量arr1或arr2相关联。怪异的。|1412601|用户@JHBonarius的回答解决了这个问题：|1412602|arr+=+np.tile(np.nan,+96)
print(arr)

#+[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
#++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]|1412603|entityMap^0|9|A|0|0|W|4|11|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

Just a warning that initializing with <code>np.empty()</code> without subsequently editing the values can lead to (memory allocation?) problems:
<pre><code>arr1 = np.empty(96)
arr2 = np.empty(96)
print(arr1)
print(arr2)

# [nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan 1. 1.
# 1. 1. 2. 2. 2. 2. nan nan nan nan nan nan nan nan 0. 0. 0. 0.
# 0. 0. 0. 0. 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 nan nan]
#
# [nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan nan 1. 1.
# 1. 1. 2. 2. 2. 2. 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 nan nan nan nan nan nan nan nan nan nan nan nan
# nan nan nan nan nan nan]
</code></pre>
The floats initialized in the array are used somewhere else in my script but are not associated with variables <code>arr1</code> or <code>arr2</code> at all. Spooky.
Answer from user @JHBonarius solved this problem:
<pre><code>arr = np.tile(np.nan, 96)
print(arr)

# [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
# 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]
</code></pre>

I have the following code:

<pre><code>r = numpy.zeros(shape = (width, height, 9))
</code></pre>

It creates a <code>width x height x 9</code> matrix filled with zeros. Instead, I'd like to know if there's a function or way to initialize them instead to <code>NaN</code>s in an easy way.

Create numpy matrix filled with NaNs

我有以下代码：r = numpy.zeros(shape = (width, height, 9))它创建一个用零填充的width x height x 9矩阵。相反，我想知道是否有一个函数或方法可以以一种简单的方式将它们初始化为NaN。

问创建填充了NaNs的numpy矩阵
EN

回答 8

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问创建填充了NaNs的numpy矩阵EN

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问创建填充了NaNs的numpy矩阵
EN