是否有方便的方法来计算序列或单维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的统计参考，但找不到这个。我能找到的只是中位数（第50百分位数），但不是更具体的内容。
答案 0 :(得分：230)
您可能对SciPy Stats包感兴趣。它有the percentile function你所追求的以及许多其他统计数据。
在percentile()中也 numpy is available。
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
This ticket让我相信他们不会很快将 percentile()整合到numpy中。
答案 1 :(得分：62)
顺便说一句，有a pure-Python implementation of percentile function，以防万一不想依赖scipy。该功能复制如下：
## {{{ 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/ }}}
答案 2 :(得分：25)
import numpy as np
a = [154, 400, 1124, 82, 94, 108]
print np.percentile(a,95) # gives the 95th percentile
答案 3 :(得分：11)
以下是如何在没有numpy的情况下执行此操作，仅使用python来计算百分位数。
import math
def percentile(data, percentile):
size = len(data)
return sorted(data)[int(math.ceil((size * percentile) / 100)) - 1]
p5 = percentile(mylist, 5)
p25 = percentile(mylist, 25)
p50 = percentile(mylist, 50)
p75 = percentile(mylist, 75)
p95 = percentile(mylist, 95)
答案 4 :(得分：10)
我经常看到的百分位数的定义结果是所提供的列表中的值，其中P％的值被找到...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，您可以使用更简单的功能。
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
如果您希望从提供的列表中获取或低于P％值的值，请使用以下简单修改：
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@ijustlovemath建议的简化：
def percentile(N, P):
n = max(int(round(P * len(N) + 0.5)), 2)
return N[n-2]
答案 5 :(得分：7)
检查scipy.stats模块：
scipy.stats.scoreatpercentile
答案 6 :(得分：2)
要计算系列的百分位数，请运行：
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}
答案 7 :(得分：1)
如果您需要答案成为输入numpy数组的成员：
添加默认numpy中的百分位函数将输出计算为输入向量中两个相邻条目的线性加权平均值。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起，你可以使用＆＃34;插值＆＃34;选项，使用＆＃34; lower＆＃34;，＆＃34;更高＆＃34;或者＆＃34;最近的＆＃34;。
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
后者是向量中的实际条目，而前者是两个与百分位数相邻的向量条目的线性插值
答案 8 :(得分：0)
系列：用于描述功能
假设您的df包含以下列sales和id。您想要计算销售的百分位数，则它的工作原理如下，
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
答案 9 :(得分：0)
从Python 3.8开始，标准库随quantiles函数一起作为statistics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返回给定分布dist的{{1}}切点列表，将n - 1分位数间隔（n分为{{1} }具有相等概率的连续间隔）：
statistics.quantiles（dist，*，n = 4，method ='exclusive'）
其中dist，在我们的情况下（n）是n。
答案 10 :(得分：0)
一种用于计算一维numpy序列或矩阵的百分位数的便捷方法是使用numpy.percentile <https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html>。示例：
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，则上述功能将无用。在这种情况下，建议使用的函数是numpy.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
在上面显示的两个选项中，您仍然可以选择插值模式。请按照以下示例进行操作，以便于理解。
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
如果输入数组仅包含整数值，那么您可能会对百分位数答案作为整数感兴趣。如果是这样，请选择插值模式，例如“较低”，“较高”或“最近”。
答案 11 :(得分：0)
我引导数据，然后绘制了 10 个样本的置信区间。置信区间显示概率落在 5% 到 95% 之间的范围。
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()