我正在寻找类似于Excel的百分位函数的方法。
np.percentile()
函数。import numpy as np
a = np.array([1,2,3,4,5])
p = np.percentile(a, 50) # return 50th percentile, i.e. median.
>>> print(p)
3.0
SciPy有scipy.stats.scoreatpercentile()
,还有许多其他的统计好东西。
df.groupby('key')[['value']].agg(lambda g: np.percentile(g, 10))
。 - patricksurry顺便提一下,如果不想依赖scipy,可以使用纯Python实现的百分位函数。以下是该函数的代码:
## {{{ 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/ }}}
N
中的数据。假设您实际上有一个元组列表N = [(1, 2), (3, 1), ..., (5, 1)]
,并且您想要获取元组的第一个元素的百分位数,那么您可以选择key=lambda x: x[0]
。在计算百分位数之前,您还可以对列表元素应用一些(改变顺序的)转换。 - Elias Strehleimport numpy as np
a = [154, 400, 1124, 82, 94, 108]
print np.percentile(a,95) # gives the 95th percentile
从Python 3.8开始,标准库中的quantiles
函数成为了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
函数接收一个分布 dist
,返回一个长度为 n-1
的列表,表示将分布 dist
分成 n
个连续区间(每个区间的概率相等)所需的 n-1
个切点:
statistics.quantiles(dist, *, n=4, method='exclusive')
在我们的情况下 (percentiles
),n
等于 100
。
以下是无需使用numpy,只使用Python计算百分位数的方法。
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
我通常看到的百分位数定义期望结果是从提供的列表中找到P%的值下面的值...这意味着结果必须来自集合,而不是集合元素之间的插值。 为了达到这个目的,您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
如果你希望获取所提供列表中 P% 的值及以下的值,则可以使用这个简单的修改:
def percentile(N, P):
n = int(round(P * len(N) + 0.5))
if n > 1:
return N[n-2]
else:
return N[0]
或者按照@ijustlovemath建议的简化方式:
def percentile(N, P):
n = max(int(round(P * len(N) + 0.5)), 2)
return N[n-2]
PERCENTILE
函数对您的上述示例返回以下百分位数:3.7 = percentile(A, P=0.3)
,0.82 = percentile(A, P=0.8)
, 20 = percentile(B, P=0.3)
, 42 = percentile(B, P=0.8)
。 - marcon = int(...)
包装在max(int(...),1)
函数中。 - ijustlovemath检查是否安装了scipy.stats模块:
scipy.stats.scoreatpercentile
使用numpy.percentile <https://docs.scipy.org/doc/numpy/reference/generated/numpy.percentile.html>是计算一维 numpy 序列或矩阵百分位数的方便方式。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
然而,如果您的数据中存在任何NaN值,则上述函数将无法使用。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>。
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
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,则可能对整数值的百分位回答感兴趣。如果是这样,请选择插值模式,如“lower”、“higher”或“nearest”。
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}
如果您需要答案是输入numpy数组的成员:
需要补充的是,numpy中的百分位函数默认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开始,您可以使用“interpolation”选项,其中包括“lower”,“higher”或“nearest”。
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
pandas
数据框相关的问题:python - 查找给定列的百分位统计信息 - undefined