[PYTHON] Selection of measurement data

what is this

You are a manufacturer's inspection engineer. I have 100 measurement data obtained from a certain sensor. Due to various circumstances, I would like to show that this measurement data ** "may vary" **. I want to maximize the variance by choosing 10 out of 100. However, I want to say that the sensor is normal, so I want to make it ** "mean value is accurate" **.

Try it with Python

Measurement data creation

Create measurement data with random numbers.

%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np, pandas as pd
np.random.seed(1)
measurement data= np.random.normal(50,1,100)
plt.hist(measurement data)
print('standard deviation',measurement data.std())
>>>
Standard deviation 0.885156213832

Solve with mathematical optimization

Maximize the variance. Normally, it is difficult to solve because it is a quadratic integer optimization. Assuming the mean is accurate, $ (value-mean) ^ 2 $ is a fixed value, so the model is linear or mixed integer optimization.

from pulp import *
from ortoolpy import addbinvars
Number of selections= 10
eps = 0.0001

m = LpProblem(sense=LpMaximize)
x = addbinvars(len(measurement data))
m += lpDot((measurement data-50)**2, x)
m += lpSum(x) ==Number of selections
e = lpDot(measurement data, x) /Number of selections
m += 50-eps <= e
m +=           e <= 50+eps
m.solve()
%time m.solve() #Solution
r = np.vectorize(value)(x).astype(int) #result
print(LpStatus[m.status])
>>>
Wall time: 181 ms
Optimal

print('average',measurement data[r>0].mean())
print('standard deviation',measurement data[r>0].std())
>>>
Average 49.9999119632
Standard deviation 1.82811635001

The mean was accurate and the standard deviation was more than double the original data.

that's all