81 lines
2.8 KiB
Plaintext
81 lines
2.8 KiB
Plaintext
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{
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"['dimensions: 1, cases when observation is the nearest: 0.0%',\n",
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" 'dimensions: 2, cases when observation is the nearest: 0.0%',\n",
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" 'dimensions: 3, cases when observation is the nearest: 0.0%',\n",
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" 'dimensions: 10, cases when observation is the nearest: 10.0%',\n",
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" 'dimensions: 20, cases when observation is the nearest: 61.0%',\n",
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" 'dimensions: 30, cases when observation is the nearest: 96.0%',\n",
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" 'dimensions: 40, cases when observation is the nearest: 98.0%',\n",
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" 'dimensions: 50, cases when observation is the nearest: 100.0%',\n",
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" 'dimensions: 60, cases when observation is the nearest: 100.0%',\n",
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" 'dimensions: 70, cases when observation is the nearest: 100.0%',\n",
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" 'dimensions: 80, cases when observation is the nearest: 100.0%',\n",
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" 'dimensions: 90, cases when observation is the nearest: 100.0%']"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"import numpy as np\n",
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"import pandas as pd\n",
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"import random\n",
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"from numpy.linalg import norm\n",
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"\n",
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"dimensions=[1,2,3]+[10*i for i in range(1,10)]\n",
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"nb_vectors=10000\n",
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"trials=100\n",
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"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\n",
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"\n",
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"result=[]\n",
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"for dimension in dimensions:\n",
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" vectors=np.random.normal(0,1,size=(nb_vectors, dimension))\n",
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" successes=0\n",
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" for i in range(trials):\n",
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" i1,i2=random.sample(range(nb_vectors),2)\n",
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" target=(vectors[i1]+vectors[i2])/2\n",
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"\n",
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" distances=pd.DataFrame(enumerate(np.dot(target, vectors.transpose())/norm(target)/norm(vectors.transpose(), axis=0)))\n",
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" distances=distances.sort_values(by=[1], ascending=False)\n",
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" if (i1 in (list(distances[0][:k]))) | (i2 in (list(distances[0][:k]))):\n",
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" successes+=1\n",
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" result.append(successes/trials)\n",
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" \n",
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"[f'dimensions: {i}, cases when observation is the nearest: {100*round(j,3)}%' for i,j in zip(dimensions, result)]"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.9"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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