擅长:python、mysql、java
<p>多亏了罗伯托·费勒!我编写了一个基于<a href="http://www.stata.com/manuals13/rcentile.pdf" rel="nofollow">http://www.stata.com/manuals13/rcentile.pdf</a>的Python函数,它产生的结果与Stata相同:</p>
<pre><code>def centile(arr, percentiles=[50]):
result = {}
s = np.sort(arr)
n = len(s)
for percent in percentiles:
R = float(n + 1) * percent / 100
r, f = int(R), R - int(R)
result['{0}%'.format(percent)] = float(s[r - 1]) + f * (s[r] - s[r - 1])
return result
</code></pre>