blocks|key|1354307|text|下面是一个例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1354308|>>>+from+scipy.stats+import+norm
>>>+norm.cdf(1.96)
0.9750021048517795
>>>+norm.cdf(-1.96)
0.024997895148220435|code-block|syntax|javascript|1354309|换句话说，大约95%25的标准正常区间位于两个标准偏差内，以零的标准均值为中心。|1354310|如果您需要反向CDF：|1354311|>>>+norm.ppf(norm.cdf(1.96))
array(1.9599999999999991)|1354312|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here's an example:

<pre><code>&gt;&gt;&gt; from scipy.stats import norm
&gt;&gt;&gt; norm.cdf(1.96)
0.9750021048517795
&gt;&gt;&gt; norm.cdf(-1.96)
0.024997895148220435
</code></pre>

In other words, approximately 95% of the standard normal interval lies within two standard deviations, centered on a standard mean of zero.

If you need the inverse CDF:

<pre><code>&gt;&gt;&gt; norm.ppf(norm.cdf(1.96))
array(1.9599999999999991)
</code></pre>

blocks|key|1354418|text|现在回答这个问题可能为时已晚，但由于谷歌仍在引领人们，我决定在这里写下我的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1354419|也就是说，从Python2.7开始，math库集成了错误函数math.erf(x)|offset|length|style|CODE|1354420|erf()函数可用于计算传统的统计函数，如累积标准正态分布：|1354421|from+math+import+*
def+phi(x):
++++#'Cumulative+distribution+function+for+the+standard+normal+distribution'
++++return+(1.0+%2B+erf(x+/+sqrt(2.0)))+/+2.0|code-block|syntax|javascript|1354422|参考：|1354423|https://docs.python.org/2/library/math.html|1354424|https://docs.python.org/3/library/math.html|1354425|How+are+the+Error+Function+and+Standard+Normal+distribution+function+related?|1354426|entityMap|0|LINK|mutability|MUTABLE|url|1|2|https://stats.stackexchange.com/a/187909^0|0|I|4|U|B|0|0|5|0|0|0|0|17|0|0|0|17|1|0|0|25|2|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@$D|18|E|19|F|G]|$D|1A|E|1B|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1C|8|@$D|1D|E|1E|F|G]]|9|@]|A|$]]|$1|J|3|K|5|L|7|1F|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1H|8|@]|9|@$D|1I|E|1J|1|1K]]|A|$]]|$1|S|3|T|5|6|7|1L|8|@]|9|@$D|1M|E|1N|1|1O]]|A|$]]|$1|U|3|V|5|6|7|1P|8|@]|9|@$D|1Q|E|1R|1|1S]]|A|$]]|$1|W|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|R]]|13|$5|Z|10|11|A|$12|T]]|14|$5|Z|10|11|A|$12|15]]]]

It may be too late to answer the question but since Google still leads people here, I decide to write my solution here.

That is, since Python 2.7, the <code>math</code> library has integrated the error function <code>math.erf(x)</code>

The <code>erf()</code> function can be used to compute traditional statistical functions such as the cumulative standard normal distribution:

<pre><code>from math import *
def phi(x):
 #'Cumulative distribution function for the standard normal distribution'
 return (1.0 + erf(x / sqrt(2.0))) / 2.0
</code></pre>

Ref:

<a href="https://docs.python.org/2/library/math.html" rel="noreferrer">https://docs.python.org/2/library/math.html</a>

<a href="https://docs.python.org/3/library/math.html" rel="noreferrer">https://docs.python.org/3/library/math.html</a>

<a href="https://stats.stackexchange.com/a/187909">How are the Error Function and Standard Normal distribution function related?</a>

blocks|key|1569471|text|改编自这里的http://mail.python.org/pipermail/python-list/2000-June/039873.html|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1569472|from+math+import+*
def+erfcc(x):
++++"""Complementary+error+function."""
++++z+=+abs(x)
++++t+=+1.+/+(1.+%2B+0.5*z)
++++r+=+t+*+exp(-z*z-1.26551223%2Bt*(1.00002368%2Bt*(.37409196%2B
++++++++t*(.09678418%2Bt*(-.18628806%2Bt*(.27886807%2B
++++++++t*(-1.13520398%2Bt*(1.48851587%2Bt*(-.82215223%2B
++++++++t*.17087277)))))))))
++++if+(x+>=+0.):
++++++++return+r
++++else:
++++++++return+2.+-+r

def+ncdf(x):
++++return+1.+-+0.5*erfcc(x/(2**0.5))|code-block|syntax|javascript|1569473|entityMap|0|LINK|mutability|MUTABLE|url|http://mail.python.org/pipermail/python-list/2000-June/039873.html^0|6|1U|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Adapted from here <a href="http://mail.python.org/pipermail/python-list/2000-June/039873.html" rel="noreferrer">http://mail.python.org/pipermail/python-list/2000-June/039873.html</a>

<pre><code>from math import *
def erfcc(x):
 """Complementary error function."""
 z = abs(x)
 t = 1. / (1. + 0.5*z)
 r = t * exp(-z*z-1.26551223+t*(1.00002368+t*(.37409196+
 t*(.09678418+t*(-.18628806+t*(.27886807+
 t*(-1.13520398+t*(1.48851587+t*(-.82215223+
 t*.17087277)))))))))
 if (x &gt;= 0.):
 return r
 else:
 return 2. - r

def ncdf(x):
 return 1. - 0.5*erfcc(x/(2**0.5))
</code></pre>

blocks|key|1569547|text|在未知示例的基础上，在很多库中实现的normdist()函数的Python等效项是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1569548|def+normcdf(x,+mu,+sigma):
++++t+=+x-mu;
++++y+=+0.5*erfcc(-t/(sigma*sqrt(2.0)));
++++if+y>1.0:
++++++++y+=+1.0;
++++return+y

def+normpdf(x,+mu,+sigma):
++++u+=+(x-mu)/abs(sigma)
++++y+=+(1/(sqrt(2*pi)*abs(sigma)))*exp(-u*u/2)
++++return+y

def+normdist(x,+mu,+sigma,+f):
++++if+f:
++++++++y+=+normcdf(x,mu,sigma)
++++else:
++++++++y+=+normpdf(x,mu,sigma)
++++return+y|code-block|syntax|javascript|1569549|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|$]]

To build upon Unknown's example, the Python equivalent of the function normdist() implemented in a lot of libraries would be:

<pre><code>def normcdf(x, mu, sigma):
 t = x-mu;
 y = 0.5*erfcc(-t/(sigma*sqrt(2.0)));
 if y&gt;1.0:
 y = 1.0;
 return y

def normpdf(x, mu, sigma):
 u = (x-mu)/abs(sigma)
 y = (1/(sqrt(2*pi)*abs(sigma)))*exp(-u*u/2)
 return y

def normdist(x, mu, sigma, f):
 if f:
 y = normcdf(x,mu,sigma)
 else:
 y = normpdf(x,mu,sigma)
 return y
</code></pre>

blocks|key|1354493|text|Alex的答案向您展示了标准正态分布(均值=+0，标准差=+1)的解决方案。如果您有正态分布的mean和std+(即sqr(var))，并且您想要计算：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1354494|from+scipy.stats+import+norm

#+cdf(x+<+val)
print+norm.cdf(val,+m,+s)

#+cdf(x+>+val)
print+1+-+norm.cdf(val,+m,+s)

#+cdf(v1+<+x+<+v2)
print+norm.cdf(v2,+m,+s)+-+norm.cdf(v1,+m,+s)|code-block|syntax|javascript|1354495|阅读更多关于cdf+here和scipy使用许多公式here实现正态分布的信息。|1354496|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Cumulative_distribution_function|1|http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.stats.norm.html^0|1B|4|1G|3|1M|8|0|0|6|8|0|Q|4|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]|$9|18|A|19|1|1A]]|E|$]]|$1|M|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

Alex's answer shows you a solution for standard normal distribution (mean = 0, standard deviation = 1). If you have normal distribution with <code>mean</code> and <code>std</code> (which is <code>sqr(var)</code>) and you want to calculate:

<pre><code>from scipy.stats import norm

# cdf(x &lt; val)
print norm.cdf(val, m, s)

# cdf(x &gt; val)
print 1 - norm.cdf(val, m, s)

# cdf(v1 &lt; x &lt; v2)
print norm.cdf(v2, m, s) - norm.cdf(v1, m, s)
</code></pre>

Read more about <a href="https://en.wikipedia.org/wiki/Cumulative_distribution_function" rel="noreferrer">cdf here</a> and scipy implementation of normal distribution with many formulas <a href="http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.stats.norm.html" rel="noreferrer">here</a>.

blocks|key|1354539|text|取自上图：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1354540|from+scipy.stats+import+norm
>>>+norm.cdf(1.96)
0.9750021048517795
>>>+norm.cdf(-1.96)
0.024997895148220435|code-block|syntax|javascript|1354541|对于双尾测试：|1354542|Import+numpy+as+np
z+=+1.96
p_value+=+2+*+norm.cdf(-np.abs(z))
0.04999579029644087|1354543|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|$]]

Taken from above: 

<pre><code>from scipy.stats import norm
&gt;&gt;&gt; norm.cdf(1.96)
0.9750021048517795
&gt;&gt;&gt; norm.cdf(-1.96)
0.024997895148220435
</code></pre>

For a two-tailed test:

<pre><code>Import numpy as np
z = 1.96
p_value = 2 * norm.cdf(-np.abs(z))
0.04999579029644087
</code></pre>

blocks|key|1354584|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1354585|import+math
def+my_cdf(x):
++++return+0.5*(1%2Bmath.erf(x/math.sqrt(2)))|code-block|syntax|javascript|1354586|我在这个页面上找到了这个公式https://www.danielsoper.com/statcalc/formulas.aspx?id=55|offset|length|1354587|entityMap|0|LINK|mutability|MUTABLE|url|https://www.danielsoper.com/statcalc/formulas.aspx?id=55^0|0|0|E|1K|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Simple like this:

<pre><code>import math
def my_cdf(x):
 return 0.5*(1+math.erf(x/math.sqrt(2)))
</code></pre>

I found the formula in this page <a href="https://www.danielsoper.com/statcalc/formulas.aspx?id=55" rel="nofollow noreferrer">https://www.danielsoper.com/statcalc/formulas.aspx?id=55</a>

I am looking for a function in Numpy or Scipy (or any rigorous Python library) that will give me the cumulative normal distribution function in Python.

How to calculate cumulative normal distribution?

我正在寻找Numpy或Scipy (或任何严格的Python库)中的函数，它将给出Python中的累积正态分布函数。

问如何计算累积正态分布？
EN

回答 7

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问如何计算累积正态分布？EN

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问如何计算累积正态分布？
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