blocks|key|1781|text|您不能将已知的均值传递给np.std或np.var，您必须等待new+standard+library+statistics+module，但在此期间，您可以使用以下公式来节省一点时间：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1782|In+[329]:+a+=+np.random.rand(1000)

In+[330]:+%25%25timeit
+++.....:+a.mean()
+++.....:+a.var()
+++.....:+
10000+loops,+best+of+3:+80.6+µs+per+loop

In+[331]:+%25%25timeit
+++.....:+m+=+a.mean()
+++.....:+np.mean((a-m)**2)
+++.....:+
10000+loops,+best+of+3:+60.9+µs+per+loop

In+[332]:+m+=+a.mean()

In+[333]:+a.var()
Out[333]:+0.078365856465916137

In+[334]:+np.mean((a-m)**2)
Out[334]:+0.078365856465916137|code-block|syntax|javascript|1783|如果你真的想加快速度，试试np.dot来做平方和求和(因为这就是点积)：|1784|In+[335]:+np.dot(a-m,a-m)/a.size
Out[335]:+0.078365856465916137

In+[336]:+%25%25timeit
+++.....:+m+=+a.mean()
+++.....:+c+=+a-m
+++.....:+np.dot(c,c)/a.size
+++.....:+
10000+loops,+best+of+3:+38.2+µs+per+loop|1785|entityMap|0|LINK|mutability|MUTABLE|url|http://www.python.org/dev/peps/pep-0450/^0|C|6|J|6|1G|A|V|12|0|0|0|D|6|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$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|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

You can't pass a known mean to <code>np.std</code> or <code>np.var</code>, you'll have to wait for the <a href="http://www.python.org/dev/peps/pep-0450/">new standard library <code>statistics</code> module</a>, but in the meantime you can save a little time by using the formula:

<pre><code>In [329]: a = np.random.rand(1000)

In [330]: %%timeit
 .....: a.mean()
 .....: a.var()
 .....: 
10000 loops, best of 3: 80.6 µs per loop

In [331]: %%timeit
 .....: m = a.mean()
 .....: np.mean((a-m)**2)
 .....: 
10000 loops, best of 3: 60.9 µs per loop

In [332]: m = a.mean()

In [333]: a.var()
Out[333]: 0.078365856465916137

In [334]: np.mean((a-m)**2)
Out[334]: 0.078365856465916137
</code></pre>

If you really are trying to speed things up, try <code>np.dot</code> to do the squaring and summing (since that's what a dot-product is):

<pre><code>In [335]: np.dot(a-m,a-m)/a.size
Out[335]: 0.078365856465916137

In [336]: %%timeit
 .....: m = a.mean()
 .....: c = a-m
 .....: np.dot(c,c)/a.size
 .....: 
10000 loops, best of 3: 38.2 µs per loop
</code></pre>

blocks|key|1817|text|你也可以通过利用信号的均值、方差和幂之间的关系来避免减法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1818|In+[7]:+import+numpy+as+np

In+[8]:+a+=+np.random.rand(1000)

In+[9]:+%25%25timeit
+++...:+a.mean()
+++...:+a.var()
+++...:+
10000+loops,+best+of+3:+24.7+us+per+loop

In+[10]:+%25%25timeit
++++...:+m+=+a.mean()
++++...:+np.mean((a-m)**2)
++++...:+
100000+loops,+best+of+3:+18.5+us+per+loop

In+[11]:+%25%25timeit
++++...:+m+=+a.mean()
++++...:+power+=+np.mean(a+**+2)
++++...:+power+-+m+**+2
++++...:+
100000+loops,+best+of+3:+17.3+us+per+loop

In+[12]:+%25%25timeit
++++...:+m+=+a.mean()
++++...:+power+=+np.dot(a,+a)+/+a.size
++++...:+power+-+m+**+2
++++...:+
100000+loops,+best+of+3:+9.16+us+per+loop|code-block|syntax|javascript|1819|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can also avoid the subtraction by making use of the relation between mean, variance and power of a signal:

<pre><code>In [7]: import numpy as np

In [8]: a = np.random.rand(1000)

In [9]: %%timeit
 ...: a.mean()
 ...: a.var()
 ...: 
10000 loops, best of 3: 24.7 us per loop

In [10]: %%timeit
 ...: m = a.mean()
 ...: np.mean((a-m)**2)
 ...: 
100000 loops, best of 3: 18.5 us per loop

In [11]: %%timeit
 ...: m = a.mean()
 ...: power = np.mean(a ** 2)
 ...: power - m ** 2
 ...: 
100000 loops, best of 3: 17.3 us per loop

In [12]: %%timeit
 ...: m = a.mean()
 ...: power = np.dot(a, a) / a.size
 ...: power - m ** 2
 ...: 
100000 loops, best of 3: 9.16 us per loop
</code></pre>

blocks|key|1855|text|我不认为NumPy提供了同时返回均值和方差的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1856|但是，SciPy提供了函数scipy.stats.norm.fit()，该函数返回样本的平均值和标准差。该函数以其更具体的目的命名，即将正态分布拟合到样本。|offset|length|style|CODE|1857|示例：|1858|>>>+import+scipy.stats
>>>+scipy.stats.norm.fit([1,2,3])
(2.0,+0.81649658092772603)|code-block|syntax|javascript|1859|请注意，fit()不会将Bessel's+correction应用于标准差，因此，如果您希望进行校正，则必须乘以适当的因子。|1860|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.norm.html|1|https://en.wikipedia.org/wiki/Bessel's_correction^0|0|D|M|D|M|0|0|0|0|4|5|C|J|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@$D|14|E|15|1|16]]|A|$]]|$1|H|3|I|5|6|7|17|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|18|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|19|8|@$D|1A|E|1B|F|G]]|9|@$D|1C|E|1D|1|1E]]|A|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

I don't think NumPy provides a function that returns both the mean and the variance.

However, SciPy provides the function <a href="https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.norm.html" rel="nofollow noreferrer"><code>scipy.stats.norm.fit()</code></a> which returns the mean and standard deviation of a sample. The function is named after its more specific purpose of fitting a normal distribution to a sample.

Example:

<pre><code>&gt;&gt;&gt; import scipy.stats
&gt;&gt;&gt; scipy.stats.norm.fit([1,2,3])
(2.0, 0.81649658092772603)
</code></pre>

Note that <code>fit()</code> does not apply <a href="https://en.wikipedia.org/wiki/Bessel&#39;s_correction" rel="nofollow noreferrer">Bessel's correction</a> to the standard deviation, so if you want that correction, you have to multiply by the appropriate factor.

Using Numpy/Python, is it possible to return the mean AND variance from a single function call?

I know that I can do them separately, but the mean is required to calculate the sample standard deviation. So if I use separate functions to get the mean and variance I am adding unnecesary overhead.

I have tried looking at the numpy docs here (<a href="http://docs.scipy.org/doc/numpy/reference/routines.statistics.html">http://docs.scipy.org/doc/numpy/reference/routines.statistics.html</a>), but with no success.

Numpy mean AND variance from single function?

使用Numpy/Python，可以从单个函数调用中返回均值和方差吗？我知道我可以分别计算它们，但需要平均值来计算样本标准偏差。因此，如果我使用单独的函数来获得均值和方差，则会增加不必要的开销。我试着查看了这里的numpy文档()，但没有成功。

问单个函数的Numpy均值和方差？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问单个函数的Numpy均值和方差？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问单个函数的Numpy均值和方差？
EN