blocks|key|359098|text|我会将A_Rank等人放入一个2D+NumPy数组中，然后使用numpy.mean()和numpy.std()计算平均值和标准差：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|359099|In+[17]:+import+numpy

In+[18]:+arr+=+numpy.array([A_rank,+B_rank,+C_rank])

In+[20]:+numpy.mean(arr,+axis=0)
Out[20]:+
array([+0.7+++++++,++2.2+++++++,++1.8+++++++,++2.13333333,++3.36666667,
++++++++5.1+++++++])

In+[21]:+numpy.std(arr,+axis=0)
Out[21]:+
array([+0.45460606,++1.29614814,++1.37355985,++1.50628314,++1.15566239,
++++++++1.2083046+])|code-block|syntax|javascript|359100|entityMap|0|LINK|mutability|MUTABLE|url|http://www.numpy.org/|1|http://docs.scipy.org/doc/numpy/reference/generated/numpy.mean.html|2|http://docs.scipy.org/doc/numpy/reference/generated/numpy.std.html^0|3|6|V|C|18|B|I|5|0|V|C|1|18|B|2|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]|$9|16|A|17|1|18]|$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]|U|$5|N|O|P|E|$Q|V]]]]

I would put <code>A_Rank</code> et al into a 2D <a href="http://www.numpy.org/" rel="noreferrer">NumPy</a> array, and then use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.mean.html" rel="noreferrer"><code>numpy.mean()</code></a> and <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.std.html" rel="noreferrer"><code>numpy.std()</code></a> to compute the means and the standard deviations:

<pre><code>In [17]: import numpy

In [18]: arr = numpy.array([A_rank, B_rank, C_rank])

In [20]: numpy.mean(arr, axis=0)
Out[20]: 
array([ 0.7 , 2.2 , 1.8 , 2.13333333, 3.36666667,
 5.1 ])

In [21]: numpy.std(arr, axis=0)
Out[21]: 
array([ 0.45460606, 1.29614814, 1.37355985, 1.50628314, 1.15566239,
 1.2083046 ])
</code></pre>

blocks|key|359123|text|因为Python3.4/+PEP450在标准库中有一个statistics+module，它有一个用于计算像您这样的迭代器的标准偏差的method+stdev：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|359124|>>>+A_rank+=+[0.8,+0.4,+1.2,+3.7,+2.6,+5.8]
>>>+import+statistics
>>>+statistics.stdev(A_rank)
2.0634114147853952|code-block|syntax|javascript|359125|entityMap|0|LINK|mutability|MUTABLE|url|http://www.python.org/dev/peps/pep-0450/|1|http://docs.python.org/3.4/library/statistics.html|2|http://docs.python.org/3.4/library/statistics.html#statistics.stdev^0|R|H|22|5|D|6|0|R|H|1|1V|C|2|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]|$9|14|A|15|1|16]|$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]|U|$5|N|O|P|E|$Q|V]]]]

Since Python 3.4 / <a href="http://www.python.org/dev/peps/pep-0450/" rel="noreferrer">PEP450</a> there is a <a href="http://docs.python.org/3.4/library/statistics.html" rel="noreferrer"><code>statistics module</code></a> in the standard library, which has a <a href="http://docs.python.org/3.4/library/statistics.html#statistics.stdev" rel="noreferrer">method <code>stdev</code></a> for calculating the standard deviation of iterables like yours:

<pre><code>&gt;&gt;&gt; A_rank = [0.8, 0.4, 1.2, 3.7, 2.6, 5.8]
&gt;&gt;&gt; import statistics
&gt;&gt;&gt; statistics.stdev(A_rank)
2.0634114147853952
</code></pre>

blocks|key|359143|text|在Python2.7中，你可以使用NumPy的numpy.std()来提供population+standard+deviation。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|359144|在Python3.4中，statistics.stdev()返回样本标准差。pstdv()函数与numpy.std()相同。|359145|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.std.html|1|https://math.stackexchange.com/questions/15098/sample-standard-deviation-vs-population-standard-deviation|2|https://docs.python.org/3/library/statistics.html^0|N|B|N|B|0|11|T|1|0|C|I|12|7|1C|B|C|I|2|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@$9|W|A|X|1|Y]|$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|H|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]|R|$5|K|L|M|E|$N|S]]]]

In python 2.7 you can use NumPy's <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.std.html" rel="noreferrer"><code>numpy.std()</code></a> gives the <a href="https://math.stackexchange.com/questions/15098/sample-standard-deviation-vs-population-standard-deviation">population standard deviation</a>.

In Python 3.4 <a href="https://docs.python.org/3/library/statistics.html" rel="noreferrer"><code>statistics.stdev()</code></a> returns the sample standard deviation. The <code>pstdv()</code> function is the same as <code>numpy.std()</code>.

blocks|key|141269|text|下面是一些纯Python代码，可以用来计算平均值和标准差。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141270|下面的所有代码都基于Python+3.4%2B中的statistics模块。|offset|length|style|CODE|141271|def+mean(data):
++++"""Return+the+sample+arithmetic+mean+of+data."""
++++n+=+len(data)
++++if+n+<+1:
++++++++raise+ValueError('mean+requires+at+least+one+data+point')
++++return+sum(data)/n+#+in+Python+2+use+sum(data)/float(n)

def+_ss(data):
++++"""Return+sum+of+square+deviations+of+sequence+data."""
++++c+=+mean(data)
++++ss+=+sum((x-c)**2+for+x+in+data)
++++return+ss

def+stddev(data,+ddof=0):
++++"""Calculates+the+population+standard+deviation
++++by+default;+specify+ddof=1+to+compute+the+sample
++++standard+deviation."""
++++n+=+len(data)
++++if+n+<+2:
++++++++raise+ValueError('variance+requires+at+least+two+data+points')
++++ss+=+_ss(data)
++++pvar+=+ss/(n-ddof)
++++return+pvar**0.5|code-block|syntax|javascript|141272|注意:为了提高浮点数求和的准确性，statistics模块使用了一个自定义函数_sum，而不是我用来代替它的内置sum。|141273|现在我们有了这样的例子：|141274|>>>+mean([1,+2,+3])
2.0
>>>+stddev([1,+2,+3])+#+population+standard+deviation
0.816496580927726
>>>+stddev([1,+2,+3],+ddof=1)+#+sample+standard+deviation
0.1|141275|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/python/cpython/blob/master/Lib/statistics.py^0|0|N|A|N|A|0|0|0|H|A|13|4|1K|3|0|0|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|@$D|14|E|15|1|16]]|A|$]]|$1|H|3|I|5|J|7|17|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|18|8|@$D|19|E|1A|F|G]|$D|1B|E|1C|F|G]|$D|1D|E|1E|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1F|8|@]|9|@]|A|$]]|$1|Q|3|R|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|S|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Here's some pure-Python code you can use to calculate the mean and standard deviation.

All code below is based on the <a href="https://github.com/python/cpython/blob/master/Lib/statistics.py" rel="noreferrer"><code>statistics</code></a> module in Python 3.4+.

<pre><code>def mean(data):
 """Return the sample arithmetic mean of data."""
 n = len(data)
 if n &lt; 1:
 raise ValueError('mean requires at least one data point')
 return sum(data)/n # in Python 2 use sum(data)/float(n)

def _ss(data):
 """Return sum of square deviations of sequence data."""
 c = mean(data)
 ss = sum((x-c)**2 for x in data)
 return ss

def stddev(data, ddof=0):
 """Calculates the population standard deviation
 by default; specify ddof=1 to compute the sample
 standard deviation."""
 n = len(data)
 if n &lt; 2:
 raise ValueError('variance requires at least two data points')
 ss = _ss(data)
 pvar = ss/(n-ddof)
 return pvar**0.5
</code></pre>

Note: for improved accuracy when summing floats, the <code>statistics</code> module uses a custom function <code>_sum</code> rather than the built-in <code>sum</code> which I've used in its place.

Now we have for example:

<pre><code>&gt;&gt;&gt; mean([1, 2, 3])
2.0
&gt;&gt;&gt; stddev([1, 2, 3]) # population standard deviation
0.816496580927726
&gt;&gt;&gt; stddev([1, 2, 3], ddof=1) # sample standard deviation
0.1
</code></pre>

blocks|key|141307|text|其他答案涵盖了如何在python中充分执行std+dev，但没有人解释如何执行您所描述的奇怪的遍历。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141308|我假设A-Z是整个人口。如果没有，请看Ome关于如何从样本中推断的答案。|offset|length|141309|因此，要获得每个列表的第一个数字的标准差/平均值，您需要类似以下内容：|141310|#standard+deviation
numpy.std([A_rank[0],+B_rank[0],+C_rank[0],+...,+Z_rank[0]])

#mean
numpy.mean([A_rank[0],+B_rank[0],+C_rank[0],+...,+Z_rank[0]])|code-block|syntax|javascript|141311|要缩短代码并将其推广到任意n位数字，请使用我为您生成的以下函数：|141312|def+getAllNthRanks(n):
++++return+[A_rank[n],+B_rank[n],+C_rank[n],+D_rank[n],+E_rank[n],+F_rank[n],+G_rank[n],+H_rank[n],+I_rank[n],+J_rank[n],+K_rank[n],+L_rank[n],+M_rank[n],+N_rank[n],+O_rank[n],+P_rank[n],+Q_rank[n],+R_rank[n],+S_rank[n],+T_rank[n],+U_rank[n],+V_rank[n],+W_rank[n],+X_rank[n],+Y_rank[n],+Z_rank[n]]+|141313|现在，您可以简单地从A-Z获得所有第n位的标准差和平均值，如下所示：|141314|#standard+deviation
numpy.std(getAllNthRanks(n))

#mean
numpy.mean(getAllNthRanks(n))|141315|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/31366254/1727470^0|0|J|3|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@$D|14|E|15|1|16]]|A|$]]|$1|F|3|G|5|6|7|17|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|18|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|J|7|1C|8|@]|9|@]|A|$K|L]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

The other answers cover how to do std dev in python sufficiently, but no one explains how to do the bizarre traversal you've described.

I'm going to assume A-Z is the entire population. If not see <a href="https://stackoverflow.com/a/31366254/1727470">Ome</a>'s answer on how to inference from a sample.

So to get the standard deviation/mean of the first digit of every list you would need something like this:

<pre><code>#standard deviation
numpy.std([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])

#mean
numpy.mean([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])
</code></pre>

To shorten the code and generalize this to any nth digit use the following function I generated for you:

<pre><code>def getAllNthRanks(n):
 return [A_rank[n], B_rank[n], C_rank[n], D_rank[n], E_rank[n], F_rank[n], G_rank[n], H_rank[n], I_rank[n], J_rank[n], K_rank[n], L_rank[n], M_rank[n], N_rank[n], O_rank[n], P_rank[n], Q_rank[n], R_rank[n], S_rank[n], T_rank[n], U_rank[n], V_rank[n], W_rank[n], X_rank[n], Y_rank[n], Z_rank[n]] 
</code></pre>

Now you can simply get the stdd and mean of all the nth places from A-Z like this:

<pre><code>#standard deviation
numpy.std(getAllNthRanks(n))

#mean
numpy.mean(getAllNthRanks(n))
</code></pre>

blocks|key|141355|text|纯python代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|141356|from+math+import+sqrt

def+stddev(lst):
++++mean+=+float(sum(lst))+/+len(lst)
++++return+sqrt(float(reduce(lambda+x,+y:+x+%2B+y,+map(lambda+x:+(x+-+mean)+**+2,+lst)))+/+len(lst))|code-block|syntax|javascript|141357|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|$]]

pure python code:

<pre><code>from math import sqrt

def stddev(lst):
 mean = float(sum(lst)) / len(lst)
 return sqrt(float(reduce(lambda x, y: x + y, map(lambda x: (x - mean) ** 2, lst))) / len(lst))
</code></pre>

I want to find mean and standard deviation of 1st, 2nd,... digits of several (Z) lists. For example, I have

<pre><code>A_rank=[0.8,0.4,1.2,3.7,2.6,5.8]
B_rank=[0.1,2.8,3.7,2.6,5,3.4]
C_Rank=[1.2,3.4,0.5,0.1,2.5,6.1]
# etc (up to Z_rank )...
</code></pre>

Now I want to take the mean and std of <code>*_Rank[0]</code>, the mean and std of <code>*_Rank[1]</code>, etc. 
(ie: mean and std of the 1st digit from all the (A..Z)_rank lists; 
the mean and std of the 2nd digit from all the (A..Z)_rank lists; 
the mean and std of the 3rd digit...; etc).

Standard deviation of a list

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想找出第一，第二，…的均值和标准差。多个(Z)列表的位数。例如，我有A_rank=[0.8,0.4,1.2,3.7,2.6,5.8]B_rank=[0.1,2.8,3.7,2.6,5,3.4]C_Rank=[1.2,3.4,0.5,0.1,2.5,6.1]# etc (up to Z_rank )...现在我想取*_...

问列表的标准差
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