3.1 KiB
3.1 KiB
import numpy as np
import pandas as pd
import random
from numpy.linalg import norm
dimensions = [1, 2, 3] + [10 * i for i in range(1, 10)]
nb_vectors = 10000
trials = 100
k = 1 # by setting k=1 we want to check how often the closest vector to the avarage of 2 random vectors is one of these 2 vectors
result = []
for dimension in dimensions:
vectors = np.random.normal(0, 1, size=(nb_vectors, dimension))
successes = 0
for i in range(trials):
i1, i2 = random.sample(range(nb_vectors), 2)
target = (vectors[i1] + vectors[i2]) / 2
distances = pd.DataFrame(
enumerate(
np.dot(target, vectors.transpose())
/ norm(target)
/ norm(vectors.transpose(), axis=0)
)
)
distances = distances.sort_values(by=[1], ascending=False)
if (i1 in (list(distances[0][:k]))) | (i2 in (list(distances[0][:k]))):
successes += 1
result.append(successes / trials)
[
f"dimensions: {i}, cases when observation is the nearest: {100*round(j,3)}%"
for i, j in zip(dimensions, result)
]
['dimensions: 1, cases when observation is the nearest: 0.0%', 'dimensions: 2, cases when observation is the nearest: 0.0%', 'dimensions: 3, cases when observation is the nearest: 0.0%', 'dimensions: 10, cases when observation is the nearest: 13.0%', 'dimensions: 20, cases when observation is the nearest: 61.0%', 'dimensions: 30, cases when observation is the nearest: 96.0%', 'dimensions: 40, cases when observation is the nearest: 98.0%', 'dimensions: 50, cases when observation is the nearest: 100.0%', 'dimensions: 60, cases when observation is the nearest: 100.0%', 'dimensions: 70, cases when observation is the nearest: 100.0%', 'dimensions: 80, cases when observation is the nearest: 100.0%', 'dimensions: 90, cases when observation is the nearest: 100.0%']