blocks|key|296768|text|你的方法很好。另一种方法是使用函数numpy.random.uniform()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|296769|>>>+numpy.random.uniform(-1,+1,+size=10)
array([-0.92592953,+-0.6045348+,+-0.52860837,++0.00321798,++0.16050848,
+++++++-0.50421058,++0.06754615,++0.46329675,+-0.40952318,++0.49804386])|code-block|syntax|javascript|296770|关于极值的概率:如果它是理想化的、连续的随机数，那么得到一个极值的概率将为0。由于浮点数是连续实数的离散化，因此在现实中有一些正概率得到一些极值。这是某种形式的离散化错误，几乎可以肯定，在您的模拟中，此错误将被其他错误所掩盖。别担心了！|296771|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.uniform.html^0|H|M|H|M|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]]]]

Your approach is fine. An alternative is to use the function <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.uniform.html"><code>numpy.random.uniform()</code></a>:

<pre><code>&gt;&gt;&gt; numpy.random.uniform(-1, 1, size=10)
array([-0.92592953, -0.6045348 , -0.52860837, 0.00321798, 0.16050848,
 -0.50421058, 0.06754615, 0.46329675, -0.40952318, 0.49804386])
</code></pre>

Regarding the probability for the extremes: If it would be idealised, continuous random numbers, the probability to get one of the extremes would be 0. Since floating point numbers are a discretisation of the continuous real numbers, in realitiy there is some positive probability to get some of the extremes. This is some form of discretisation error, and it is almost certain that this error will be dwarved by other errors in your simulation. Stop worrying!

blocks|key|296797|text|请注意，numpy.random.rand允许在一次调用中从均匀分布生成多个样本：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|296798|>>>+np.random.rand(5)
array([+0.69093485,++0.24590705,++0.02013208,++0.06921124,++0.73329277])|code-block|syntax|javascript|296799|它还允许生成给定形状的样本：|296800|>>>+np.random.rand(3,2)
array([[+0.14022471,++0.96360618],+
+++++++[+0.37601032,++0.25528411],+
+++++++[+0.49313049,++0.94909878]])|296801|如您所说，在[-1，1)之间均匀分布的随机数可以通过以下方式生成：|296802|>>>+2+*+np.random.rand(5)+-+1
array([+0.86704088,+-0.65406928,+-0.02814943,++0.74080741,+-0.14416581])|296803|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.rand.html^0|4|H|4|H|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

Note that <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.rand.html" rel="noreferrer"><code>numpy.random.rand</code></a> allows to generate multiple samples from a uniform distribution at one call:

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; np.random.rand(5)
array([ 0.69093485, 0.24590705, 0.02013208, 0.06921124, 0.73329277])
</code></pre>

It also allows to generate samples in a given shape:

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; np.random.rand(3,2)
array([[ 0.14022471, 0.96360618], 
 [ 0.37601032, 0.25528411], 
 [ 0.49313049, 0.94909878]])
</code></pre>

As You said, uniformly distributed random numbers between [-1, 1) can be generated with:

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; 2 * np.random.rand(5) - 1
array([ 0.86704088, -0.65406928, -0.02814943, 0.74080741, -0.14416581])
</code></pre>

blocks|key|296841|text|为了确保包含范围-1，1的极值，我随机生成一个整数数组，该数组的范围为[0,200000001[.后一个整数的值取决于所需的最终numpy数据类型。在这里，我使用numpy+float64，这是用于numpy数组的默认类型。然后，我将numpy数组除以100000000，以生成浮点数并以单位减去。这样做的代码是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|296842|>>>+import+numpy+as+np
>>>+number+=+((np.random.randint(low=0,+high=200000001,+size=5))+/+100000000)+-+1
>>>+print(number)
[-0.65960772++0.30378946+-0.05171788+-0.40737182++0.12998227]|code-block|syntax|javascript|296843|确保不要将这些numpy浮点数转换为python浮点数，以避免舍入错误。|296844|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

To ensure that the extremes of range [-1, 1] are included, I randomly generate a numpy array of integers in the range [0, 200000001[. The value of the latter integer depends on the final numpy data type that is desired. Here, I take the numpy float64, which is the default type used for numpy arrays. Then, I divide the numpy array with 100000000 to generate floats and subtract with unity. Code for this is:
<pre><code>&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; number = ((np.random.randint(low=0, high=200000001, size=5)) / 100000000) - 1
&gt;&gt;&gt; print(number)
[-0.65960772 0.30378946 -0.05171788 -0.40737182 0.12998227]
</code></pre>
Make sure not to transform these numpy floats to python floats to avoid rounding errors.

I want to generate random numbers in the range <code>-1, 1</code> and want each one to have equal probability of being generated. I.e. I don't want the extremes to be less likely to come up. What is the best way of doing this?

So far, I have used:

<pre><code>2 * numpy.random.rand() - 1
</code></pre>

and also:

<pre><code>2 * numpy.random.random_sample() - 1
</code></pre>

What is the best way of getting random numbers in NumPy?

我希望生成范围为-1, 1的随机数，并希望每个随机数都有相同的生成概率。也就是说，我不希望出现极端的可能性更小。做这件事的最好方法是什么？到目前为止，我已经使用了：2 * numpy.random.rand() - 1还有：2 * numpy.random.random_sample() - 1

问在NumPy中获取随机数的最佳方法是什么？
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