blocks|key|1098177|text|顺便说一句，这里有a+pure-Python+implementation+of+percentile+function，以防有人不想依赖scipy。该函数复制如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1098178|##+{{{+http://code.activestate.com/recipes/511478/+(r1)
import+math
import+functools

def+percentile(N,+percent,+key=lambda+x:x):
++++"""
++++Find+the+percentile+of+a+list+of+values.

++++@parameter+N+-+is+a+list+of+values.+Note+N+MUST+BE+already+sorted.
++++@parameter+percent+-+a+float+value+from+0.0+to+1.0.
++++@parameter+key+-+optional+key+function+to+compute+value+from+each+element+of+N.

++++@return+-+the+percentile+of+the+values
++++"""
++++if+not+N:
++++++++return+None
++++k+=+(len(N)-1)+*+percent
++++f+=+math.floor(k)
++++c+=+math.ceil(k)
++++if+f+==+c:
++++++++return+key(N[int(k)])
++++d0+=+key(N[int(f)])+*+(c-k)
++++d1+=+key(N[int(c)])+*+(k-f)
++++return+d0%2Bd1

#+median+is+50th+percentile.
median+=+functools.partial(percentile,+percent=0.5)
##+end+of+http://code.activestate.com/recipes/511478/+}}}|code-block|syntax|javascript|1098179|entityMap|0|LINK|mutability|MUTABLE|url|http://code.activestate.com/recipes/511478-finding-the-percentile-of-the-values/^0|9|1F|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]]]]

By the way, there is <a href="http://code.activestate.com/recipes/511478-finding-the-percentile-of-the-values/" rel="noreferrer">a pure-Python implementation of percentile function</a>, in case one doesn't want to depend on scipy. The function is copied below:

<pre><code>## {{{ http://code.activestate.com/recipes/511478/ (r1)
import math
import functools

def percentile(N, percent, key=lambda x:x):
 """
 Find the percentile of a list of values.

 @parameter N - is a list of values. Note N MUST BE already sorted.
 @parameter percent - a float value from 0.0 to 1.0.
 @parameter key - optional key function to compute value from each element of N.

 @return - the percentile of the values
 """
 if not N:
 return None
 k = (len(N)-1) * percent
 f = math.floor(k)
 c = math.ceil(k)
 if f == c:
 return key(N[int(k)])
 d0 = key(N[int(f)]) * (c-k)
 d1 = key(N[int(c)]) * (k-f)
 return d0+d1

# median is 50th percentile.
median = functools.partial(percentile, percent=0.5)
## end of http://code.activestate.com/recipes/511478/ }}}
</code></pre>

blocks|key|1313309|text|import+numpy+as+np
a+=+[154,+400,+1124,+82,+94,+108]
print+np.percentile(a,95)+#+gives+the+95th+percentile|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1313310|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import numpy as np
a = [154, 400, 1124, 82, 94, 108]
print np.percentile(a,95) # gives the 95th percentile
</code></pre>

blocks|key|1313295|text|下面是如何在不使用numpy的情况下使用python来计算百分位数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1313296|import+math

def+percentile(data,+perc:+int):
++++size+=+len(data)
++++return+sorted(data)[int(math.ceil((size+*+perc)+/+100))+-+1]

percentile([10.0,+9.0,+8.0,+7.0,+6.0,+5.0,+4.0,+3.0,+2.0,+1.0],+90)
#+9.0
percentile([142,+232,+290,+120,+274,+123,+146,+113,+272,+119,+124,+277,+207],+50)
#+146|code-block|syntax|javascript|1313297|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|$]]

Here's how to do it without numpy, using only python to calculate the percentile.
<pre><code>import math

def percentile(data, perc: int):
 size = len(data)
 return sorted(data)[int(math.ceil((size * perc) / 100)) - 1]

percentile([10.0, 9.0, 8.0, 7.0, 6.0, 5.0, 4.0, 3.0, 2.0, 1.0], 90)
# 9.0
percentile([142, 232, 290, 120, 274, 123, 146, 113, 272, 119, 124, 277, 207], 50)
# 146
</code></pre>

blocks|key|1313273|text|我通常看到的百分位数的定义预期的结果是提供的列表中的值，在该列表下找到P百分比的值……这意味着结果必须来自集合，而不是集合元素之间的插值。要实现这一点，可以使用一个更简单的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1313274|def+percentile(N,+P):
++++"""
++++Find+the+percentile+of+a+list+of+values

++++@parameter+N+-+A+list+of+values.++N+must+be+sorted.
++++@parameter+P+-+A+float+value+from+0.0+to+1.0

++++@return+-+The+percentile+of+the+values.
++++"""
++++n+=+int(round(P+*+len(N)+%2B+0.5))
++++return+N[n-1]

#+A+=+(1,+2,+3,+4,+5,+6,+7,+8,+9,+10)
#+B+=+(15,+20,+35,+40,+50)
#
#+print+percentile(A,+P=0.3)
#+4
#+print+percentile(A,+P=0.8)
#+9
#+print+percentile(B,+P=0.3)
#+20
#+print+percentile(B,+P=0.8)
#+50|code-block|syntax|javascript|1313275|如果您更愿意从提供的列表中获取P%25或更低的值，则使用以下简单修改：|1313276|def+percentile(N,+P):
++++n+=+int(round(P+*+len(N)+%2B+0.5))
++++if+n+>+1:
++++++++return+N[n-2]
++++else:
++++++++return+N[0]|1313277|或者使用@ijustlovemath建议的简化：|1313278|def+percentile(N,+P):
++++n+=+max(int(round(P+*+len(N)+%2B+0.5)),+2)
++++return+N[n-2]|1313279|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The definition of percentile I usually see expects as a result the value from the supplied list below which P percent of values are found... which means the result must be from the set, not an interpolation between set elements. To get that, you can use a simpler function.

<pre><code>def percentile(N, P):
 """
 Find the percentile of a list of values

 @parameter N - A list of values. N must be sorted.
 @parameter P - A float value from 0.0 to 1.0

 @return - The percentile of the values.
 """
 n = int(round(P * len(N) + 0.5))
 return N[n-1]

# A = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
# B = (15, 20, 35, 40, 50)
#
# print percentile(A, P=0.3)
# 4
# print percentile(A, P=0.8)
# 9
# print percentile(B, P=0.3)
# 20
# print percentile(B, P=0.8)
# 50
</code></pre>

If you would rather get the value from the supplied list at or below which P percent of values are found, then use this simple modification:

<pre><code>def percentile(N, P):
 n = int(round(P * len(N) + 0.5))
 if n &gt; 1:
 return N[n-2]
 else:
 return N[0]
</code></pre>

Or with the simplification suggested by @ijustlovemath:

<pre><code>def percentile(N, P):
 n = max(int(round(P * len(N) + 0.5)), 2)
 return N[n-2]
</code></pre>

blocks|key|1313215|text|检查scipy.stats模块：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1313216|+scipy.stats.scoreatpercentile|code-block|syntax|javascript|1313217|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|$]]

check for scipy.stats module: 

<pre><code> scipy.stats.scoreatpercentile
</code></pre>

blocks|key|1098322|text|要计算序列的百分位数，请运行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1098323|from+scipy.stats+import+rankdata
import+numpy+as+np

def+calc_percentile(a,+method='min'):
++++if+isinstance(a,+list):
++++++++a+=+np.asarray(a)
++++return+rankdata(a,+method=method)+/+float(len(a))|code-block|syntax|javascript|1098324|例如：|1098325|a+=+range(20)
print+{val:+round(percentile,+3)+for+val,+percentile+in+zip(a,+calc_percentile(a))}
>>>+{0:+0.05,+1:+0.1,+2:+0.15,+3:+0.2,+4:+0.25,+5:+0.3,+6:+0.35,+7:+0.4,+8:+0.45,+9:+0.5,+10:+0.55,+11:+0.6,+12:+0.65,+13:+0.7,+14:+0.75,+15:+0.8,+16:+0.85,+17:+0.9,+18:+0.95,+19:+1.0}|1098326|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|$]]

To calculate the percentile of a series, run:

<pre><code>from scipy.stats import rankdata
import numpy as np

def calc_percentile(a, method='min'):
 if isinstance(a, list):
 a = np.asarray(a)
 return rankdata(a, method=method) / float(len(a))
</code></pre>

For example:

<pre><code>a = range(20)
print {val: round(percentile, 3) for val, percentile in zip(a, calc_percentile(a))}
&gt;&gt;&gt; {0: 0.05, 1: 0.1, 2: 0.15, 3: 0.2, 4: 0.25, 5: 0.3, 6: 0.35, 7: 0.4, 8: 0.45, 9: 0.5, 10: 0.55, 11: 0.6, 12: 0.65, 13: 0.7, 14: 0.75, 15: 0.8, 16: 0.85, 17: 0.9, 18: 0.95, 19: 1.0}
</code></pre>

blocks|key|1098352|text|计算一维numpy序列或矩阵的百分位数的一种便捷方法是使用numpy.percentile+[https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html](https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html)。示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1098353|import+numpy+as+np

a+=+np.array([0,1,2,3,4,5,6,7,8,9,10])
p50+=+np.percentile(a,+50)+#+return+50th+percentile,+e.g+median.
p90+=+np.percentile(a,+90)+#+return+90th+percentile.
print('median+=+',p50,'+and+p90+=+',p90)+#+median+=++5.0++and+p90+=++9.0|code-block|syntax|javascript|1098354|但是，如果您的数据中有任何NaN值，则上述函数将不起作用。在这种情况下，推荐使用的函数是numpy.nanpercentile+[https://docs.scipy.org/doc/numpy/reference/generated/numpy.nanpercentile.html](https://docs.scipy.org/doc/numpy/reference/generated/numpy.nanpercentile.html)函数：|1098355|import+numpy+as+np

a_NaN+=+np.array([0.,1.,2.,3.,4.,5.,6.,7.,8.,9.,10.])
a_NaN[0]+=+np.nan
print('a_NaN',a_NaN)
p50+=+np.nanpercentile(a_NaN,+50)+#+return+50th+percentile,+e.g+median.
p90+=+np.nanpercentile(a_NaN,+90)+#+return+90th+percentile.
print('median+=+',p50,'+and+p90+=+',p90)+#+median+=++5.5++and+p90+=++9.1|1098356|在上面显示的两个选项中，您仍然可以选择插值模式。请按照下面的示例进行操作，以便更容易理解。|1098357|import+numpy+as+np

b+=+np.array([1,2,3,4,5,6,7,8,9,10])
print('percentiles+using+default+interpolation')
p10+=+np.percentile(b,+10)+#+return+10th+percentile.
p50+=+np.percentile(b,+50)+#+return+50th+percentile,+e.g+median.
p90+=+np.percentile(b,+90)+#+return+90th+percentile.
print('p10+=+',p10,',+median+=+',p50,'+and+p90+=+',p90)
#p10+=++1.9+,+median+=++5.5++and+p90+=++9.1

print('percentiles+using+interpolation+=+',+"linear")
p10+=+np.percentile(b,+10,interpolation='linear')+#+return+10th+percentile.
p50+=+np.percentile(b,+50,interpolation='linear')+#+return+50th+percentile,+e.g+median.
p90+=+np.percentile(b,+90,interpolation='linear')+#+return+90th+percentile.
print('p10+=+',p10,',+median+=+',p50,'+and+p90+=+',p90)
#p10+=++1.9+,+median+=++5.5++and+p90+=++9.1

print('percentiles+using+interpolation+=+',+"lower")
p10+=+np.percentile(b,+10,interpolation='lower')+#+return+10th+percentile.
p50+=+np.percentile(b,+50,interpolation='lower')+#+return+50th+percentile,+e.g+median.
p90+=+np.percentile(b,+90,interpolation='lower')+#+return+90th+percentile.
print('p10+=+',p10,',+median+=+',p50,'+and+p90+=+',p90)
#p10+=++1+,+median+=++5++and+p90+=++9

print('percentiles+using+interpolation+=+',+"higher")
p10+=+np.percentile(b,+10,interpolation='higher')+#+return+10th+percentile.
p50+=+np.percentile(b,+50,interpolation='higher')+#+return+50th+percentile,+e.g+median.
p90+=+np.percentile(b,+90,interpolation='higher')+#+return+90th+percentile.
print('p10+=+',p10,',+median+=+',p50,'+and+p90+=+',p90)
#p10+=++2+,+median+=++6++and+p90+=++10

print('percentiles+using+interpolation+=+',+"midpoint")
p10+=+np.percentile(b,+10,interpolation='midpoint')+#+return+10th+percentile.
p50+=+np.percentile(b,+50,interpolation='midpoint')+#+return+50th+percentile,+e.g+median.
p90+=+np.percentile(b,+90,interpolation='midpoint')+#+return+90th+percentile.
print('p10+=+',p10,',+median+=+',p50,'+and+p90+=+',p90)
#p10+=++1.5+,+median+=++5.5++and+p90+=++9.5

print('percentiles+using+interpolation+=+',+"nearest")
p10+=+np.percentile(b,+10,interpolation='nearest')+#+return+10th+percentile.
p50+=+np.percentile(b,+50,interpolation='nearest')+#+return+50th+percentile,+e.g+median.
p90+=+np.percentile(b,+90,interpolation='nearest')+#+return+90th+percentile.
print('p10+=+',p10,',+median+=+',p50,'+and+p90+=+',p90)
#p10+=++2+,+median+=++5++and+p90+=++9|1098358|如果您的输入数组只包含整数值，那么您可能会对以整数表示的百分数答案感兴趣。如果是这样，选择插值模式，如“较低”、“较高”或“最近”。|1098359|entityMap|0|LINK|mutability|MUTABLE|url|[https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html](https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html)|1|[https://docs.scipy.org/doc/numpy/reference/generated/numpy.nanpercentile.html](https://docs.scipy.org/doc/numpy/reference/generated/numpy.nanpercentile.html)^0|1A|48|0|0|0|1S|4E|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|K|3|L|5|F|7|1B|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1C|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1D|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1E|8|@]|9|@]|C|$]]|$1|S|3|-4|5|6|7|1F|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]]]

A convenient way to calculate percentiles for a one-dimensional numpy sequence or matrix is by using numpy.percentile &lt;<a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html" rel="nofollow noreferrer">https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html</a>>. Example:

<pre><code>import numpy as np

a = np.array([0,1,2,3,4,5,6,7,8,9,10])
p50 = np.percentile(a, 50) # return 50th percentile, e.g median.
p90 = np.percentile(a, 90) # return 90th percentile.
print('median = ',p50,' and p90 = ',p90) # median = 5.0 and p90 = 9.0
</code></pre>

However, if there is any NaN value in your data, the above function will not be useful. The recommended function to use in that case is the numpy.nanpercentile &lt;<a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.nanpercentile.html" rel="nofollow noreferrer">https://docs.scipy.org/doc/numpy/reference/generated/numpy.nanpercentile.html</a>> function:

<pre><code>import numpy as np

a_NaN = np.array([0.,1.,2.,3.,4.,5.,6.,7.,8.,9.,10.])
a_NaN[0] = np.nan
print('a_NaN',a_NaN)
p50 = np.nanpercentile(a_NaN, 50) # return 50th percentile, e.g median.
p90 = np.nanpercentile(a_NaN, 90) # return 90th percentile.
print('median = ',p50,' and p90 = ',p90) # median = 5.5 and p90 = 9.1
</code></pre>

In the two options presented above, you can still choose the interpolation mode. Follow the examples below for easier understanding.

<pre><code>import numpy as np

b = np.array([1,2,3,4,5,6,7,8,9,10])
print('percentiles using default interpolation')
p10 = np.percentile(b, 10) # return 10th percentile.
p50 = np.percentile(b, 50) # return 50th percentile, e.g median.
p90 = np.percentile(b, 90) # return 90th percentile.
print('p10 = ',p10,', median = ',p50,' and p90 = ',p90)
#p10 = 1.9 , median = 5.5 and p90 = 9.1

print('percentiles using interpolation = ', "linear")
p10 = np.percentile(b, 10,interpolation='linear') # return 10th percentile.
p50 = np.percentile(b, 50,interpolation='linear') # return 50th percentile, e.g median.
p90 = np.percentile(b, 90,interpolation='linear') # return 90th percentile.
print('p10 = ',p10,', median = ',p50,' and p90 = ',p90)
#p10 = 1.9 , median = 5.5 and p90 = 9.1

print('percentiles using interpolation = ', "lower")
p10 = np.percentile(b, 10,interpolation='lower') # return 10th percentile.
p50 = np.percentile(b, 50,interpolation='lower') # return 50th percentile, e.g median.
p90 = np.percentile(b, 90,interpolation='lower') # return 90th percentile.
print('p10 = ',p10,', median = ',p50,' and p90 = ',p90)
#p10 = 1 , median = 5 and p90 = 9

print('percentiles using interpolation = ', "higher")
p10 = np.percentile(b, 10,interpolation='higher') # return 10th percentile.
p50 = np.percentile(b, 50,interpolation='higher') # return 50th percentile, e.g median.
p90 = np.percentile(b, 90,interpolation='higher') # return 90th percentile.
print('p10 = ',p10,', median = ',p50,' and p90 = ',p90)
#p10 = 2 , median = 6 and p90 = 10

print('percentiles using interpolation = ', "midpoint")
p10 = np.percentile(b, 10,interpolation='midpoint') # return 10th percentile.
p50 = np.percentile(b, 50,interpolation='midpoint') # return 50th percentile, e.g median.
p90 = np.percentile(b, 90,interpolation='midpoint') # return 90th percentile.
print('p10 = ',p10,', median = ',p50,' and p90 = ',p90)
#p10 = 1.5 , median = 5.5 and p90 = 9.5

print('percentiles using interpolation = ', "nearest")
p10 = np.percentile(b, 10,interpolation='nearest') # return 10th percentile.
p50 = np.percentile(b, 50,interpolation='nearest') # return 50th percentile, e.g median.
p90 = np.percentile(b, 90,interpolation='nearest') # return 90th percentile.
print('p10 = ',p10,', median = ',p50,' and p90 = ',p90)
#p10 = 2 , median = 5 and p90 = 9
</code></pre>

If your input array only consists of integer values, you might be interested in the percentil answer as an integer. If so, choose interpolation mode such as ‘lower’, ‘higher’, or ‘nearest’.

blocks|key|1313324|text|对于系列:使用describe函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1313325|假设您有如下列的df+:+sales和id。你想要计算销售额的百分位数，那么它的工作原理是这样的，|1313326|df['sales'].describe(percentiles+=+[0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1])

0.0:+.0:+minimum
1:+maximum+
0.1+:+10th+percentile+and+so+on|code-block|syntax|javascript|1313327|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

for a series: used describe functions

suppose you have df with following columns sales and id. you want to calculate percentiles for sales then it works like this,

<pre><code>df['sales'].describe(percentiles = [0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1])

0.0: .0: minimum
1: maximum 
0.1 : 10th percentile and so on
</code></pre>

blocks|key|1098361|text|我对数据进行引导，然后绘制出10个样本的置信区间。置信区间显示了概率在5%25到95%25之间的概率范围。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1098362|+import+pandas+as+pd
+import+matplotlib.pyplot+as+plt
+import+seaborn+as+sns
+import+numpy+as+np
+import+json
+import+dc_stat_think+as+dcst

+data+=+[154,+400,+1124,+82,+94,+108]
+#print+(np.percentile(data,[0.5,95]))+#+gives+the+95th+percentile

+bs_data+=+dcst.draw_bs_reps(data,+np.mean,+size=6*10)

+#print(np.reshape(bs_data,(24,6)))

+x=+np.linspace(1,6,6)
+print(x)
+for+(item1,item2,item3,item4,item5,item6)+in+bs_data.reshape((10,6)):
+++++line_data=[item1,item2,item3,item4,item5,item6]
+++++ci=np.percentile(line_data,[.025,.975])
+++++mean_avg=np.mean(line_data)
+++++fig,+ax+=+plt.subplots()
+++++ax.plot(x,line_data)
+++++ax.fill_between(x,+(line_data-ci[0]),+(line_data%2Bci[1]),+color='b',+alpha=.1)
+++++ax.axhline(mean_avg,color='red')
+++++plt.show()|code-block|syntax|javascript|1098363|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|$]]

I bootstrap the data and then plotted out the confidence interval for 10 samples. The confidence interval shows the range where the probabilities will fall between 5 percent and 95 percent probability.
<pre><code> import pandas as pd
 import matplotlib.pyplot as plt
 import seaborn as sns
 import numpy as np
 import json
 import dc_stat_think as dcst

 data = [154, 400, 1124, 82, 94, 108]
 #print (np.percentile(data,[0.5,95])) # gives the 95th percentile

 bs_data = dcst.draw_bs_reps(data, np.mean, size=6*10)

 #print(np.reshape(bs_data,(24,6)))

 x= np.linspace(1,6,6)
 print(x)
 for (item1,item2,item3,item4,item5,item6) in bs_data.reshape((10,6)):
 line_data=[item1,item2,item3,item4,item5,item6]
 ci=np.percentile(line_data,[.025,.975])
 mean_avg=np.mean(line_data)
 fig, ax = plt.subplots()
 ax.plot(x,line_data)
 ax.fill_between(x, (line_data-ci[0]), (line_data+ci[1]), color='b', alpha=.1)
 ax.axhline(mean_avg,color='red')
 plt.show()
</code></pre>

Is there a convenient way to calculate percentiles for a sequence or single-dimensional numpy array?

I am looking for something similar to Excel's percentile function.

I looked in NumPy's statistics reference, and couldn't find this. All I could find is the median (50th percentile), but not something more specific.

How do I calculate percentiles with python/numpy?

有没有一种方便的方法来计算序列或一维numpy数组的百分位数？我正在寻找类似于Excel的百分位函数的东西。我查看了NumPy的统计资料，但找不到这个。我能找到的是中位数(50%)，但不是更具体的东西。

问如何使用python/numpy计算百分位数？
EN

回答 9

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问如何使用python/numpy计算百分位数？
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