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}

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}

您可能会对SciPy Stats包感兴趣。它有你所追求的百分位数函数和许多其他统计上的好处。

Percentile()在numpy中也可用。

import numpy as np
a = np.array([1,2,3,4,5])
p = np.percentile(a, 50) # return 50th percentile, e.g median.
print p
3.0

这张票让我相信他们不会很快将percentile()集成到numpy中。

从Python 3.8开始，标准库附带了quantiles函数，作为统计模块的一部分:

from statistics import quantiles

quantiles([1, 2, 3, 4, 5], n=100)
# [0.06, 0.12, 0.18, 0.24, 0.3, 0.36, 0.42, 0.48, 0.54, 0.6, 0.66, 0.72, 0.78, 0.84, 0.9, 0.96, 1.02, 1.08, 1.14, 1.2, 1.26, 1.32, 1.38, 1.44, 1.5, 1.56, 1.62, 1.68, 1.74, 1.8, 1.86, 1.92, 1.98, 2.04, 2.1, 2.16, 2.22, 2.28, 2.34, 2.4, 2.46, 2.52, 2.58, 2.64, 2.7, 2.76, 2.82, 2.88, 2.94, 3.0, 3.06, 3.12, 3.18, 3.24, 3.3, 3.36, 3.42, 3.48, 3.54, 3.6, 3.66, 3.72, 3.78, 3.84, 3.9, 3.96, 4.02, 4.08, 4.14, 4.2, 4.26, 4.32, 4.38, 4.44, 4.5, 4.56, 4.62, 4.68, 4.74, 4.8, 4.86, 4.92, 4.98, 5.04, 5.1, 5.16, 5.22, 5.28, 5.34, 5.4, 5.46, 5.52, 5.58, 5.64, 5.7, 5.76, 5.82, 5.88, 5.94]
quantiles([1, 2, 3, 4, 5], n=100)[49] # 50th percentile (e.g median)
# 3.0

Quantiles为给定的分布区域返回n - 1个切割点的列表，分隔n个分位数区间(以等概率将dist划分为n个连续区间):

统计数据。分位数(dist， *， n=4, method='exclusive')

在我们的例子中，n(百分位数)是100。

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]

对于系列:用于描述函数

假设df具有以下列sales和id。你想计算销售额的百分比，它是这样工作的，

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