This is ttv1 code (Timm tools, version 1).
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Returns True if two lists of numbers are different by only a trivially small amount.
This is a non-parametric effect size test. Computes the probability that list2 has smaller
or larger numbers than list1. Two lists are the same if that probability is less
tha some trivial
amount.
The size of trivial
can be adjusted by an optional third argument.
The following values for this argument are recommented:
Romano, Jeanine, et al. Exploring methods for evaluating group differences on the NSSE and other surveys: Are the t-test and Cohen's d indices the most appropriate choices, Annual meeting of the Southern Association for Institutional Research. 2006.
This example compare two lists that differ by an ever growing amont. Note that, until we starting adding more than about 1.5, the two lists are deemed to be the same.
from cliffsDelta import cliffsDelta
from random import random as r
#
base = [r()*10 for _ in range(30)]
print([round(x,4) for x in sorted(base)])
for n in range(1,8,1):
other = [x+ (r()*n/2) for x in base]
print(n/2,
cliffsDelta(base, other))
# output
[0.0211, 0.2545, 0.2835, 0.2904, 0.3059, 0.9386,
1.3436, 2.166, 2.2169, 2.2876, 2.5507, 3.812,
4.2212, 4.3277, 4.3789, 4.4539, 4.4949, 4.9544,
4.9581, 5.4141, 6.5159, 7.2154, 7.6228, 7.6377,
7.8872, 8.3577, 8.4743, 9.0143, 9.3915, 9.4527]
0.5 True
1.0 True
1.5 False # <=== first time we see a difference.
2.0 False
2.5 False
3.0 False
3.5 False
Returns True
if two lists of numbers are similar.
def cd(lst1,lst2, trivial=0.147, fast=True):
This file includes a simple (and slow) version of cliffsDelta plus a trickier (and faster) one that takes advantage of sorted lists and repeated values.
As shown by the _fastWorks function (in the test file):
f = optimized if fast else basic
lt,gt,n = f(lst1,lst2)
print("cd",lt,gt, abs(gt-lt) / (n + 1e-32))
return (abs(gt-lt) / (n + 1e-32)) < trivial
Method #1: simple, slow. Shows the basic idea.
def basic(lst1,lst2):
lt=gt=n=0
for x in lst1:
for y in lst2:
n += 1
if x > y: gt +=1
if x < y: lt +=1
return lt,gt,n
Method #2: trickier, much faster method. Not novice friendly. Exploits sorted lists, repeated values.
def optimized(lst1,lst2):
m, n = len(lst1), len(lst2)
lst2 = sorted(lst2)
j = gt = lt = 0
for repeats,x in runs(sorted(lst1)):
while j <= (n - 1) and lst2[j] < x:
j += 1
gt += j*repeats
while j <= (n - 1) and lst2[j] == x:
j += 1
lt += (n - j)*repeats
return lt,gt,m*n
def runs(lst):
for j,two in enumerate(lst):
if j == 0:
one,i = two,0
if one!=two:
yield j - i,one
i = j
one = two
yield j - i + 1,two
Copyright © 2016,2017 Tim Menzies tim@menzies.us, MIT license v2.
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