# percentile 如何用python / numpy計算百分位數？

``````## {{{ 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/ }}}
``````
python numpy percentile

``````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
``````

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

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

``````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)
``````

``````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
``````

### Tags

python   numpy   percentile