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SE_BMMF
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Remove all exercises as they were moved to the exercise notebook.

master
paulusja 1 year ago
parent
d20038520c
commit
02018561b0
10 changed files with 0 additions and 1351 deletions
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{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Schreiben Sie ein Python-Programm, das drei Zahlen von der Tastatur einliest und anschließend die\n",
"größte von ihnen ausgibt."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "-"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"is_executing": false,
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"num1 = float(input(\"Erste Zahl:\"))\n",
"num2 = float(input(\"Zweite Zahl:\"))\n",
"num3 = float(input(\"Dritte Zahl:\"))\n",
"\n",
"maximum = num1 if num1 > num2 else num2\n",
"maximum = maximum if maximum > num3 else num3\n",
"print(maximum) \n",
"\n",
"\n",
"if num1 > num2:\n",
" if num1 > num3:\n",
" print(num1)\n",
" else:\n",
" print(num3)\n",
"else:\n",
" if num2 > num3:\n",
" print(num2)\n",
" else:\n",
" print(num3)"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 2_\n",
"\n",
"Schreiben Sie ein Python-Programm, das zwei Zahlen von der Tastatur einliest. Anschließend soll\n",
"nachgefragt werden, ob die beiden Zahlen addiert, subtrahiert, mulitpliziert oder dividiert werden sollen. \n",
"Führen Sie die gewünschte Operation aus und geben Sie das Ergebnis aus. "
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "-"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"num1 = float(input(\"Erste Zahl:\"))\n",
"num2 = float(input(\"Zweite Zahl:\"))\n",
"\n",
"op = input(\"Was soll gemacht werden (+,-,*,/):\")\n",
"if op == \"+\":\n",
" erg = num1 + num2\n",
"elif op == \"-\":\n",
" erg = num1 - num2\n",
"elif op == \"*\":\n",
" erg = num1 * num2\n",
"elif op == \"/\":\n",
" erg = num1 / num2\n",
"else:\n",
" erg = None\n",
" \n",
"print(f\"Unbekannter Operator <{op}>.\") if erg is None else print(f\"{num1} {op} {num2} = {erg}\")"
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
},
"pycharm": {
"stem_cell": {
"cell_type": "raw",
"metadata": {
"collapsed": false
},
"source": []
}
}
},
"nbformat": 4,
"nbformat_minor": 1
}

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@@ -1,146 +0,0 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Schreiben Sie ein Python-Programm, das zunächst eine Zahl von der Tastatur einliest,\n",
"anschließend mittels einer <code>while</code>-Schleife die Binärdarstellung\n",
"der Zahl ermittelt und diese ausgibt."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "-"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"is_executing": false,
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"num = int(input(\"Zahl:\"))\n",
"\n",
"ergebnis = \"\"\n",
"while num > 0:\n",
" ergebnis = str(num % 2) + ergebnis\n",
" num = num // 2\n",
"\n",
"print(ergebnis)"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 2_\n",
"\n",
"Schreiben Sie ein Python-Programm, das einen String bestehend aus <code>0</code> und <code>1</code> \n",
"von der Tastatur einliest, diesen als Binärzahl betrachtet und den dezimalen Wert ausgibt."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "-"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"binaer = input(\"Binärzahl:\")\n",
"\n",
"# Lösung 1\n",
"ergebnis = 0\n",
"index = 0\n",
"while index < len(binaer):\n",
" ergebnis = ergebnis * 2\n",
" if binaer[index] == \"1\":\n",
" ergebnis = ergebnis + 1\n",
" index = index + 1\n",
"\n",
"# Lösung 2 \n",
"index = len(binaer)-1\n",
"ergebnis2 = 0\n",
"while index >=0:\n",
" stellenwert = len(binaer) - 1 - index\n",
" ergebnis2 = ergebnis2 + int(binaer[index])* (2 ** stellenwert ) \n",
" index = index - 1\n",
" \n",
"print(ergebnis)\n",
"print(ergebnis2)"
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
},
"pycharm": {
"stem_cell": {
"cell_type": "raw",
"metadata": {
"collapsed": false
},
"source": []
}
}
},
"nbformat": 4,
"nbformat_minor": 1
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true,
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Schreiben Sie ein Python-Programm, das einen Text von der Tastatur einliest und anschließend\n",
"die Anfangsbuchstaben aller Wörter im Text ausgibt. Dabei soll jeder Buchstabe nur einmal vorkommen."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"text = input(\"Text:\")\n",
"\n",
"worte = text.split()\n",
"startchars = [wort[0] for wort in worte ]\n",
"\n",
"for index, current in enumerate(startchars):\n",
" if current not in startchars[:index]:\n",
" print(current)"
]
},
{
"cell_type": "markdown",
"metadata": {
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 2_\n",
"\n",
"Gegeben sind zwei Listen gleicher Länge mit Artikeln und zugehörgen Preisen, z.B.:\n",
"\n",
"<code>artikel = [ \"Klopapier\", \"Maske\", \"Desinfektionsmittel\" ]</code>\n",
"\n",
"<code>preis = [ 3.5, 2.25, 7.4 ]</code>\n",
"\n",
"Schreiben Sie ein Python-Programm, das eine Ergebnisliste berechnet, in der Artikel und Preis zu Tupeln \n",
"zusammengefasst sind:\n",
"\n",
"<code>ergebnis = [('Klopapier', 3.5), ('Maske', 2.25), ('Desinfektionsmittel', 7.4)] </code>"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "subslide"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"is_executing": false,
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"artikel = [ \"Klopapier\", \"Maske\", \"Desinfektionsmittel\" ]\n",
"preis = [ 3.50, 2.25, 7.4 ]\n",
"\n",
"ergebnis = [(a, preis[index]) for index, a in enumerate(artikel)]\n",
"print(ergebnis)\n",
"\n",
"########\n",
"ergebnis2 = list(zip(artikel, preis))\n",
"print(ergebnis2)\n",
"\n",
"########\n",
"index = 0\n",
"ergebnis3 = []\n",
"while index < len(artikel):\n",
" ergebnis3.append((artikel[index], preis[index]))\n",
" index = index + 1\n",
"print(ergebnis3)"
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
},
"pycharm": {
"stem_cell": {
"cell_type": "raw",
"metadata": {
"collapsed": false
},
"source": []
}
}
},
"nbformat": 4,
"nbformat_minor": 1
}

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@@ -1,153 +0,0 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true,
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Schreiben Sie eine Funktion <code>schnittmenge</code>, die zwei Listen als Parameter erwartet\n",
"und eine Liste zurückgibt, die nur die Elemente beinhaltet, die in beiden übergebenen Listen enthalten waren."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"# Todo: Hier die Funktion anlegen\n",
"print(schnittmenge([1, 2, 3, 10], [3, 4, 1]))"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"is_executing": false,
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"def schnittmenge(liste1, liste2):\n",
" ergebnis = []\n",
" for element in liste2:\n",
" if element in liste1:\n",
" ergebnis.append(element)\n",
" return ergebnis\n",
"\n",
"print(schnittmenge([1, 2, 3, 10], [3, 4, 1]))"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 2_\n",
"\n",
"Schreiben Sie eine Funktion <code>pascal</code>, der eine Liste <code>z</code> von Integer-Zahlen \n",
"übergeben wird. Sei <code>n</code> die Länge dieser Liste. \n",
"Das Ergebnis soll eine neue Liste sein mit folgenden Einträgen\n",
"\n",
"Falls <code>n == 0</code>:\n",
"\n",
"&nbsp;<code>[ 1 ]</code>\n",
"\n",
"Falls <code>n == 1</code>:\n",
"\n",
"&nbsp;<code>[ 1 , 1 ]</code>\n",
"\n",
"Sonst:\n",
"\n",
"&nbsp;<code>[ 1 , z[0]+z[1] , z[1]+z[2] , ... , z[n-2]+z[n-1] , 1 ]</code>"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"# Todo: Hier die Funktion anlegen\n",
"\n",
"print(pascal([]))\n",
"print(pascal([5]))\n",
"print(pascal([5, 2, 7, 3]))"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "subslide"
}
},
"outputs": [],
"source": [
"def pascal(z):\n",
" erg = [1]\n",
" erg.extend([a+b for a, b in zip(z[:-1], z[1:])])\n",
" if len(z) > 0:\n",
" erg.append(1)\n",
" return erg\n",
"\n",
"print(pascal([]))\n",
"print(pascal([5]))\n",
"print(pascal([5, 2, 7, 3]))"
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
},
"pycharm": {
"stem_cell": {
"cell_type": "raw",
"metadata": {
"collapsed": false
},
"source": []
}
}
},
"nbformat": 4,
"nbformat_minor": 1
}

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@@ -1,98 +0,0 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true,
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Schreiben Sie ein Python-Skript, das Studierende in einem Dictionary verwaltet. Schlüssel soll\n",
"die ganzzahlige Matrikelnummer sein, der Wert der Name des Studierenden.\n",
"\n",
"Ihr Skript soll dabei zunächst eine Matrikelnummer einlesen. Ist sie im Dictionary vorhanden,\n",
"soll der gespeicherte Name ausgegeben werden. Anschließend soll das Programm mit dem\n",
"Einlesen der nächsten Matrikelnummer fortgesetzt werden.\n",
"\n",
"Ist die Matrikelnummer nicht vorhanden, soll der Name eingelesen und im Dictionary gespeichert\n",
"werden. Anschließend geht es weiter mit der nächsten Matrikelnummer.\n",
"\n",
"Bei der Eingabe einer negativen Matrikelnummer soll der Studierende aus dem Dictionary\n",
"entfernt werden.\n",
"\n",
"Wird 0 eingegeben, endet das Programm."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"is_executing": false,
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"dict = {}\n",
"matrikel = None\n",
"\n",
"while matrikel != 0:\n",
" matrikel = int(input(\"Matrikelnummer: \"))\n",
" \n",
" if matrikel < 0:\n",
" del(dict[-matrikel])\n",
" elif matrikel in dict:\n",
" print(dict[matrikel])\n",
" elif matrikel > 0:\n",
" name = input(\"Name: \")\n",
" dict[matrikel] = name\n",
" \n",
"print(dict)"
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
}
},
"nbformat": 4,
"nbformat_minor": 1
}

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@@ -1,234 +0,0 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true,
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Schreiben Sie eine Funktion <code>get_it_or_none1</code>, die ein Dictionary und einen\n",
"Schlüsselwert als Parameter erhält. Sie soll den im Dictionary gespeicherten Wert\n",
"zurückgeben, wenn er enthalten ist, andernfalls <code>None</code>.\n",
"\n",
"Vermeiden Sie bei der Implementierung das Auftreten einer Exception, indem Sie vor\n",
"dem Zugriff prüfen, ob der Schlüssel enthalten ist."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"def get_it_or_none1(dict_arg, key):\n",
" if key in dict_arg:\n",
" return dict_arg[key]\n",
" else:\n",
" return None\n",
" \n",
"my_dict = {'a': 1, 2: 'b'}\n",
"print(get_it_or_none1(my_dict, 'a'))\n",
"print(get_it_or_none1(my_dict, 'b'))"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 2_\n",
"\n",
"Implementieren Sie eine Funktion <code>get_it_or_none2</code> mit der gleichen\n",
"Funktionalität. Allerdings verzichten Sie dieses Mal auf die Prüfung und Fangen\n",
"eine evtl. auftretende Exception, um dann <code>None</code> zurückzugeben."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"def get_it_or_none2(dict_arg, key):\n",
" try:\n",
" return dict_arg[key]\n",
" except:\n",
" return None\n",
" \n",
"my_dict = {'a': 1, 2: 'b'}\n",
"print(get_it_or_none2(my_dict, 'a'))\n",
"print(get_it_or_none2(my_dict, 'b'))"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 3_\n",
"\n",
"Überprüfen Sie die Laufzeit der beiden Varianten, indem Sie jede Funktion\n",
"10000-mal so aufrufen, dass <code>None</code> zurückgegeben wird. Was ist schneller?"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"import timeit\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"import timeit\n",
"\n",
"my_dict = {}\n",
"print(f'get_it_or_none1:: {timeit.timeit(lambda: get_it_or_none1(my_dict, 0), number=10000)}')\n",
"print(f'get_it_or_none2:: {timeit.timeit(lambda: get_it_or_none2(my_dict, 0), number=10000)}')\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 4_\n",
"\n",
"Tatsächlich besitzt ein Python-Dictionary eine bisher nicht eingeführte Methode\n",
"<code>get</code>, der neben dem Schlüssel auch ein Default-Wert übergeben werden kann,\n",
"der zurückgegeben wird, wenn der Schlüssel nicht enthalten ist.\n",
"\n",
"<code>dict.get( \\< key \\>, \\< default \\> ) </code>\n",
"\n",
"Überprüfen Sie auch die Laufzeit dieses Ansatzes."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"print(f'get_it_or_none2:: {timeit.timeit(lambda: my_dict.get(0), number=10000)}')\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"Wie erklären Sie sich die Ergebnisse?"
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
}
},
"nbformat": 4,
"nbformat_minor": 1
}

+ 0
- 171
07kapitel/072a Uebung.ipynb View File

@@ -1,171 +0,0 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true,
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Schreiben Sie eine Funktion *prim_sum_up_to*, die die Summe aller Primzahlen von 2 bis *n* berechnet, wobei *n* als\n",
"Parameter übergeben wird."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"def prim_sum_up_to(end):\n",
" prims = []"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"def prim_sum_up_to(n):\n",
" result = 0\n",
" for m in range(2, n+1):\n",
" if is_prim(m):\n",
" result += m\n",
" return result\n",
"\n",
"def is_prim(n):\n",
" divisor = 2\n",
" while divisor * divisor <= n:\n",
" if n % divisor == 0:\n",
" return False\n",
" divisor += 1\n",
" return True\n",
"\n",
"print(prim_sum_up_to(100))\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 2_\n",
"\n",
"Sobald Sie die Funktion mit größeren *n* aufrufen, steigt die Laufzeit des Programms an. Um zu verhindern,\n",
"dass der Anwender fälschlicherweise annimmt, dass das Programm abgestürzt ist, soll jede Sekunde ein\n",
"Stern ausgegeben werden, bis das Ergebnis vorliegt.\n",
"\n",
"Verwenden Sie hierfür das in der Standardbibliothek enthaltene Modul <code>datetime</code> und\n",
"importieren Sie daraus die Objekte <code>datetime</code> und <code>timedelta</code>.\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"from datetime import datetime, timedelta\n",
"\n",
"last_run = datetime.now()\n",
"intervall = timedelta(seconds=1)\n",
"\n",
"def do_only_every_second():\n",
" global last_run\n",
" if last_run + intervall > datetime.now():\n",
" return\n",
" print(\"*\", end='')\n",
" last_run = datetime.now()\n",
"\n",
"\n",
"def is_prim(n):\n",
" divisor = 2\n",
" while divisor * divisor <= n:\n",
" if n % divisor == 0:\n",
" return False\n",
" divisor += 1\n",
" do_only_every_second()\n",
" return True\n",
"\n",
"print(f'\\n{prim_sum_up_to(1000000)}')\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"Vielleicht können Sie das Prinzip in einem zukünftigen Praktikum nutzen, wenn nicht bei jedem Schleifendurchlauf\n",
"etwas passieren soll...."
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
}
},
"nbformat": 4,
"nbformat_minor": 1
}

+ 0
- 137
08kapitel/083a Uebung.ipynb View File

@@ -1,137 +0,0 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"collapsed": true,
"pycharm": {
"name": "#%% md\n"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Sie entwickeln im Team ein CAD-Programm,\n",
"das das Zusammenstellen komplexer Zeichnungen aus einfachen 2D-Elementen erlaubt.\n",
"Gegeben ist bereits eine Klasse *Element*."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
}
},
"outputs": [],
"source": [
"class Element():\n",
"\n",
" def get_bezeichnung(self):\n",
" return \"N.N.\"\n",
"\n",
" def umfang(self):\n",
" raise NotImplementedError()\n",
"\n",
" def __str__(self):\n",
" return self.get_bezeichnung() + f\" Umfang: {self.umfang()}\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Ihr Teamleiter beauftragt Sie, für die Elemente *Kreis* und *Rechteck* Unterklassen\n",
"zur Oberklasse *Element* so zu implementieren,\n",
"dass anschließend die Aufrufe\n",
"\n",
"<code>k = Kreis(radius=3)</code>\n",
"\n",
"<code>r = Rechteck(breite=10, hoehe=5)</code>\n",
"\n",
"möglich sind.\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
}
},
"outputs": [],
"source": [
"import math\n",
"\n",
"\n",
"k = Kreis(radius=3)\n",
"print(k)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
}
},
"outputs": [],
"source": [
"\n",
"r = Rechteck(breite=10, hoehe=5)\n",
"print(r)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"_Aufgabe 2_\n",
"\n",
"Fügen Sie jetzt noch eine weitere Klasse *Quadrat* hinzu. Welche Oberklasse besitzt diese neue Klasse?"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
}
},
"outputs": [],
"source": [
"\n",
"q = Quadrat(seite=4)\n",
"print(q)\n",
"\n",
"\n",
"\n"
]
}
],
"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": "3.8.2"
}
},
"nbformat": 4,
"nbformat_minor": 1
}

+ 0
- 108
09kapitel/094 Uebung.ipynb View File

@@ -1,108 +0,0 @@
{
"cells": [
{
"cell_type": "markdown",
"metadata": {
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"\n",
"Gegeben ist eine binäre Grafikdatei <code>ohm.png</code>."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
}
},
"outputs": [],
"source": [
"from IPython.display import Image\n",
"Image(filename=\"ohm.png\")"
]
},
{
"cell_type": "markdown",
"metadata": {
"pycharm": {
"name": "#%% md\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"source": [
"_Aufgabe 1_\n",
"\n",
"Erstellen Sie ein Python-Programm, dass diese Datei in eine zweite\n",
"Datei <code>logo.png</code> kopiert. Übertragen Sie die Daten bitte\n",
"in 32 Byte Paketen."
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"pycharm": {
"name": "#%%\n"
},
"slideshow": {
"slide_type": "slide"
}
},
"outputs": [],
"source": [
"from IPython.display import Image\n",
"\n",
"\n",
"with open(\"ohm.png\", \"rb\") as source_file:\n",
" with open(\"logo.png\", \"wb\") as dest_file:\n",
" buffer_size = 32\n",
" buffer = source_file.read(buffer_size)\n",
" while len(buffer)>0:\n",
" dest_file.write(buffer)\n",
" buffer = source_file.read(buffer_size)\n",
"\n",
"\n",
"Image(filename=\"logo.png\")"
]
}
],
"metadata": {
"celltoolbar": "Slideshow",
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"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": "3.9.9"
}
},
"nbformat": 4,
"nbformat_minor": 1
}

BIN
09kapitel/ohm.png View File


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