要计算一个系列的百分位数，运行:

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))

例如:

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}

如果你需要答案是输入numpy数组的成员:

再加上numpy中的百分位数函数默认情况下将输出计算为输入向量中两个相邻项的线性加权平均。在某些情况下，人们可能希望返回的百分位数是向量的实际元素，在这种情况下，从v1.9.0开始，您可以使用“插值”选项，使用“低”、“高”或“最近”。

import numpy as np
x=np.random.uniform(10,size=(1000))-5.0

np.percentile(x,70) # 70th percentile

2.075966046220879

np.percentile(x,70,interpolation="nearest")

2.0729677997904314

后者是向量中的一个实际条目，而前者是与百分位数相邻的两个向量条目的线性插值

我通常看到的百分位数的定义期望从所提供的列表中找到P个百分比的值…这意味着结果必须来自集合，而不是集合元素之间的插值。为此，可以使用一个更简单的函数。

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

如果你想从所提供的列表中获得等于或低于P百分比的值，那么使用以下简单的修改:

def percentile(N, P):
    n = int(round(P * len(N) + 0.5))
    if n > 1:
        return N[n-2]
    else:
        return N[0]

或者使用@ijustlovemath建议的简化:

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

计算一维numpy序列或矩阵的百分位数的一种方便方法是使用numpy。百分位< https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html >。例子:

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

但是，如果您的数据中有任何NaN值，则上述函数将没有用处。在这种情况下，推荐使用numpy函数。Nanpercentile <https://docs.scipy.org/doc/numpy/reference/generated/numpy.nanpercentile.html>函数:

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

在上面给出的两个选项中，您仍然可以选择插值模式。为了更容易理解，请参考下面的例子。

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

如果您的输入数组只包含整数值，那么您可能对作为整数的百分比答案感兴趣。如果是这样，选择插值模式，如'低'，'高'，或'最近'。

检查scipy。统计模块:

 scipy.stats.scoreatpercentile

顺便说一下，有一个百分位数函数的纯python实现，以防人们不想依赖scipy。函数复制如下:

## {{{ 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/ }}}