Skip to content

M
mooc-rr
Commit d84068d1 authored by adf52b78a1975b5ac63bbd1f3ffc1b59's avatar adf52b78a1975b5ac63bbd1f3ffc1b59
Browse files
Options

Jupyter test. Help.

parent bc064b91
Hide whitespace changes
Inline Side-by-side
Showing with 236 additions and 3 deletions
module2/exo1/Untitled.ipynb 0 → 100644
View file @ d84068d1
{
"cells": [
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"%load_ext rpy2.ipython"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.4"
}
},
"nbformat": 4,
"nbformat_minor": 4
}
module2/exo1/Untitled1.ipynb 0 → 100644
View file @ d84068d1
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# À propos du calcul de $\\pi$\n",
"## En demandant à la lib maths\n",
"Mon ordinateur m’indique que $\\pi$ vaut *approximativement* "
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"3.141592653589793\n"
]
}
],
"source": [
"from math import *\n",
"print (pi)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## En utilisant la méthode des aiguilles de Buffon \n",
"Mais calculé avec la **méthode** des [https://fr.wikipedia.org/wiki/Aiguille_de_Buffon], on obtiendrait comme **approximation** :"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"3.128911138923655"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"import numpy as np\n",
"np.random.seed(seed=42)\n",
"N = 10000\n",
"x = np.random.uniform(size=N, low=0, high=1) \n",
"theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
"2/(sum((x+np.sin(theta))>1)/N)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Avec un argument \"fréquentiel\" de surface\n",
"Sinon, une méthode plus simple à comprendre et ne faisant pas intervenir d’appel à la fonction sinus se base sur le fait que si $X\\sim U(0,1)$ et $Y\\sim U(0,1)$ alors $P[X^2+Y^2\\leq1]=\\pi/4$ (voir [https://fr.wikipedia.org/wiki/M%C3%A9thode_de_Monte-Carlo#D%C3%A9termination_de_la_valeur_de_%CF%80]). Le code suivant illustre ce fait :"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"ename": "SyntaxError",
"evalue": "EOL while scanning string literal (<ipython-input-4-c8a26f7cb22f>, line 13)",
"output_type": "error",
"traceback": [
"\u001b[0;36m File \u001b[0;32m\"<ipython-input-4-c8a26f7cb22f>\"\u001b[0;36m, line \u001b[0;32m13\u001b[0m\n\u001b[0;31m ax.scatter(x[reject],y[reject], c='r’, alpha=0.2, edgecolor=None)\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m EOL while scanning string literal\n"
]
}
],
"source": [
"%matplotlib inline\n",
"import matplotlib.pyplot as plt\n",
"\n",
"np.random.seed(seed=42)\n",
"N = 1000\n",
"x = np.random.uniform(size=N, low=0, high=1)\n",
"y = np.random.uniform(size=N, low=0, high=1)\n",
"\n",
"accept=(x*x+y*y)<=1 \n",
"reject=np.logical_not(accept)\n",
"\n",
"fig, ax=plt.subplots(1) \n",
"ax.scatter(x[reject],y[reject], c='r’, alpha=0.2, edgecolor=None) \n",
"ax.scatter(x[reject],y[reject], c='r’, alpha=0.2, edgecolor=None) \n",
"ax.set_aspect('equal')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Il est alor saisé d’obtenir une approximation (pas terrible) de $\\pi$ en comptant combien de fois, en moyenne, $X^2 + Y^2$ est inferoeir à 1 :"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"ename": "SyntaxError",
"evalue": "invalid syntax (<ipython-input-7-e627d624d667>, line 1)",
"output_type": "error",
"traceback": [
"\u001b[0;36m File \u001b[0;32m\"<ipython-input-7-e627d624d667>\"\u001b[0;36m, line \u001b[0;32m1\u001b[0m\n\u001b[0;31m 4.np.mean(accept)\u001b[0m\n\u001b[0m ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
]
}
],
"source": [
"4.np.mean(accept)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.4"
}
},
"nbformat": 4,
"nbformat_minor": 4
}
module2/exo1/toy_notebook_fr.ipynb
View file @ d84068d1
{ {
"cells": [], "cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# À propos du calcul de $\\pi$\n",
"## En demandant à la lib maths\n",
"Mon ordinateur m’indique que $\\pi$ vaut *approximativement*"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"3.141592653589793\n"
]
}
],
"source": [
"from math import *\n",
"print(pi)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## En utilisant la méthode des aiguilles de Buffon\n",
"Mais calculé avec la **méthode** des [https://fr.wikipedia.org/wiki/Aiguille_de_Buffon], on obtiendrait comme **approximation** :"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": { "metadata": {
"kernelspec": { "kernelspec": {
"display_name": "Python 3", "display_name": "Python 3",
...@@ -16,10 +59,9 @@ ...@@ -16,10 +59,9 @@
"name": "python", "name": "python",
"nbconvert_exporter": "python", "nbconvert_exporter": "python",
"pygments_lexer": "ipython3", "pygments_lexer": "ipython3",
"version": "3.6.3" "version": "3.6.4"
} }
}, },
"nbformat": 4, "nbformat": 4,
"nbformat_minor": 2 "nbformat_minor": 2
} }
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment