144 lines
3.8 KiB
Plaintext
144 lines
3.8 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Rekapitulation Lösung Dictionaries"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"# Ausgangslage\n",
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"\n",
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"salaries_1 = {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500}\n",
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"salaries_2 = {'Barbara': 3000, 'Elke': 3300, 'Michael': 2800, 'Johann': 2000}"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"{'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500}\n",
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"salaries_1: {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500}\n",
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"salaries_all: {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500, 'Barbara': 3000, 'Elke': 3300, 'Michael': 2800, 'Johann': 2000}\n",
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"Summe: 22100\n"
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]
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}
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],
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"source": [
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"salaries_all = {}\n",
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"\n",
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"for key in salaries_1:\n",
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" salaries_all[key] = salaries_1[key]\n",
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"\n",
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"print(salaries_all)\n",
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"\n",
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"for key in salaries_2:\n",
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" salaries_all[key] = salaries_2[key]\n",
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"\n",
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"print(\"salaries_1:\", salaries_1)\n",
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"print(\"salaries_all:\", salaries_all)\n",
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"\n",
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"\n",
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"print(\"Summe:\", sum(salaries_all.values()))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 11,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"salaries_1: {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500}\n",
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"salaries_all: {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500}\n",
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"salaries_1: {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500}\n",
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"salaries_2: {'Barbara': 3000, 'Elke': 3300, 'Michael': 2800, 'Johann': 2000}\n",
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"salaries_all: {'Anna': 2000, 'Mark': 3000, 'Judith': 3500, 'Thomas': 2500, 'Barbara': 3000, 'Elke': 3300, 'Michael': 2800, 'Johann': 2000}\n",
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"8\n"
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]
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}
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],
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"source": [
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"salaries_all = salaries_1.copy()\n",
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"\n",
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"print(\"salaries_1:\", salaries_1)\n",
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"print(\"salaries_all:\", salaries_all)\n",
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"\n",
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"salaries_all.update(salaries_2)\n",
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"\n",
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"print(\"salaries_1:\", salaries_1)\n",
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"print(\"salaries_2:\", salaries_2)\n",
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"print(\"salaries_all:\", salaries_all)\n",
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"\n",
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"print(len(salaries_all))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"{'Anna': 2200.0, 'Mark': 3300.0000000000005, 'Judith': 3850.0000000000005, 'Thomas': 2750.0, 'Barbara': 3300.0000000000005, 'Elke': 3630.0000000000005, 'Michael': 3080.0000000000005, 'Johann': 2200.0}\n"
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]
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}
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],
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"source": [
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"# Alternative 1\n",
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"salaries_new = salaries_all.copy()\n",
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"salaries_new.update((key, value * 1.1) for key, value in salaries_new.items())\n",
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"\n",
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"# Alternative 2\n",
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"results = []\n",
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"for key, value in salaries_all.items():\n",
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" results.append((key, value * 1.1))\n",
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"salaries_new = dict(results)\n",
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"\n",
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"# Alternative 3\n",
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"salaries_new = {}\n",
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"for key, value in salaries_all.items():\n",
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" salaries_new[key] = value * 1.1\n",
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"\n",
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"print(salaries_new)"
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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": "softed",
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"language": "python",
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"name": "softed"
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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": "3.11.8"
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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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