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Lösung Blatt 1

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Michael Pochtar 2026-10-05 23:28:07 +02:00
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{
"cells": [
{
"cell_type": "markdown",
"id": "641931b5",
"metadata": {},
"source": [
"# Aufgabe 1\n",
"Arbeiten mit NumPy"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "064ce97d",
"metadata": {},
"outputs": [],
"source": [
"import numpy as np"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "64c14dbd",
"metadata": {},
"outputs": [],
"source": [
"W = np.array([[1., 2, 3], [3, 4, 5]])"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "bf259722",
"metadata": {},
"outputs": [],
"source": [
"W.shape"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "311adc5f",
"metadata": {},
"outputs": [],
"source": [
"W.shape[0]"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "d41d05fe",
"metadata": {},
"outputs": [],
"source": [
"W.shape[1]"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "911571e0",
"metadata": {},
"outputs": [],
"source": [
"W.T"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "aa7d64ee",
"metadata": {},
"outputs": [],
"source": [
"W.T.shape"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "6fee5b46",
"metadata": {},
"outputs": [],
"source": [
"type(W)"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "b064a293",
"metadata": {},
"outputs": [],
"source": [
"type(W.shape)"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "915e6f9e",
"metadata": {},
"outputs": [],
"source": [
"r = np.arange(10)\n",
"print(r)\n",
"r.shape"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "716e7295",
"metadata": {},
"outputs": [],
"source": [
"M = np.arange(12).reshape(3,4)\n",
"print(M)\n",
"print(M.shape)"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "92baea0a",
"metadata": {},
"outputs": [],
"source": [
"M[2,0]\n",
"M[1,2]\n",
"M[1,:]\n",
"M[1]\n",
"M[:,1]\n",
"M[:,[1]]\n",
"M[:,[3, 0, 1, 1]]\n",
"M[:,2:4]\n",
"M[-2,:]\n",
"M[-2:,:]\n",
"M[:,2] = 2"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "6d78e777",
"metadata": {},
"outputs": [],
"source": [
"M > 4\n",
"M[M > 4]\n",
"M[M > 4] = -17\n",
"M"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "27fefbbb",
"metadata": {
"lines_to_next_cell": 1
},
"outputs": [],
"source": [
"s = 0\n",
"for i in range(M.shape[0]):\n",
" for j in range(M.shape[1]):\n",
" s += M[i,j]\n",
" print(s,',',sep='',end='')\n",
" print(s)"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "a00b5af1",
"metadata": {
"lines_to_next_cell": 1
},
"outputs": [],
"source": [
"def func(x):\n",
" print(x * x)\n",
" return x"
]
},
{
"cell_type": "markdown",
"id": "99670ae0",
"metadata": {},
"source": [
"# Aufgabe 2\n",
"Laden und Plotten des Iris-Blüten Datensatzes"
]
},
{
"cell_type": "markdown",
"id": "941aabcc",
"metadata": {},
"source": [
"a) Je Iris-Art gibt es 50 Datenpunkte."
]
},
{
"cell_type": "markdown",
"id": "4e91507b",
"metadata": {},
"source": [
"b) Laden des Datensatzes mit `loadtxt`."
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "530bf005",
"metadata": {},
"outputs": [],
"source": [
"iris_data = np.loadtxt(\"../supplement_files/iris.csv\", delimiter=\",\")"
]
},
{
"cell_type": "markdown",
"id": "25b3a3c1",
"metadata": {},
"source": [
"c + d) Zerlegen der Daten in Matrizen X und T und den Konventionen anpassen.\n",
"\n",
"X = iris_data.T[0:4]\n",
"T = iris_data.T[4:]"
]
},
{
"cell_type": "markdown",
"id": "561b5bc1",
"metadata": {},
"source": [
"e) Scatterplot"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "da07641f",
"metadata": {},
"outputs": [],
"source": [
"import matplotlib.pyplot as plt\n",
"\n",
"# plt.scatter(X[0:1], X[1:2])\n",
"# plt.show()"
]
},
{
"cell_type": "markdown",
"id": "e673709f",
"metadata": {},
"source": [
"f) Scatterplot mit Farben"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "a7d4ead0",
"metadata": {},
"outputs": [],
"source": [
"from matplotlib import colors\n",
"\n",
"cmap = colors.ListedColormap([\"red\", \"green\", \"blue\"])\n",
"\n",
"plt.scatter(X[0:1], X[1:2], c=T, cmap=cmap)\n",
"plt.show()"
]
},
{
"cell_type": "markdown",
"id": "1a5307aa",
"metadata": {},
"source": [
"# Aufgabe 3\n",
"Interaktive 3D Scatter Plots in Python"
]
},
{
"cell_type": "markdown",
"id": "094b1e57",
"metadata": {},
"source": [
"a + b) Iris-Blüten Datensatz mit plotly."
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "da3b79d5",
"metadata": {},
"outputs": [],
"source": [
"import plotly.express as px\n",
"\n",
"df = px.data.iris()\n",
"fig = px.scatter_3d(df, x='sepal_length', y='sepal_width', z='petal_width', color='species')\n",
"fig.show()"
]
},
{
"cell_type": "markdown",
"id": "1466c1b4",
"metadata": {},
"source": [
"c) Plot as html speichern."
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "4c20a687",
"metadata": {},
"outputs": [],
"source": [
"df = px.data.iris()\n",
"fig = px.scatter_3d(df, x='sepal_length', y='sepal_width', z='petal_width', color='species')\n",
"fig.write_html(\"iris.html\")"
]
},
{
"cell_type": "markdown",
"id": "a71792ac",
"metadata": {},
"source": [
"d) Nur Merkmale zwei bis vier."
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "5659a0a1",
"metadata": {},
"outputs": [],
"source": [
"df = px.data.iris()\n",
"fig = px.scatter_3d(df, x=\"sepal_width\", y=\"petal_length\", z=\"petal_width\", color=\"species\")\n",
"fig.show()"
]
},
{
"cell_type": "markdown",
"id": "e5d1f634",
"metadata": {},
"source": [
"# Aufgabe 4\n",
"Der Thyroid Disease Datensatz -- ein etwas gröÿerer Datensatz"
]
},
{
"cell_type": "markdown",
"id": "295ec0d0",
"metadata": {},
"source": [
"a) Der Datensatz hat 22 Merkmale."
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "0d36e655",
"metadata": {},
"outputs": [],
"source": [
"patient_data = np.loadtxt(\"../supplement_files/thyroid-ann-train.data\")\n",
"print(len(patient_data[0]))"
]
},
{
"cell_type": "markdown",
"id": "a313d84d",
"metadata": {},
"source": [
"b) 15 Merkmale sind binär und 6 sind kontinuiertlich. Die Patientengruppe ist nominales Merkmal."
]
},
{
"cell_type": "markdown",
"id": "f6c34fe8",
"metadata": {},
"source": [
"c) Es gibt drei verschiedene Patientengruppen."
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "91a74614",
"metadata": {},
"outputs": [],
"source": [
"X = patient_data.T[0:21]\n",
"T = patient_data.T[21:]\n",
"\n",
"print(np.unique(T).size)"
]
},
{
"cell_type": "markdown",
"id": "8d6abcd4",
"metadata": {},
"source": [
"d) Anzahl der Patienten pro Gruppe."
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "8f075e3d",
"metadata": {},
"outputs": [],
"source": [
"print(np.unique(T, return_counts=True)[1])"
]
},
{
"cell_type": "markdown",
"id": "e11160cb",
"metadata": {},
"source": [
"e) Plotten verschiedener Merkmalskombinationen, eingefärbt nach Patientengruppe."
]
},
{
"cell_type": "markdown",
"id": "a46c1fc9",
"metadata": {},
"source": [
"i) Merkmal 0, 16, 17"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "60a641ca",
"metadata": {},
"outputs": [],
"source": [
"fig = px.scatter_3d(x=X[0], y=X[16], z=X[17], color=T[0].astype(int).astype(str),\n",
" labels={'x': 'Merkmal 0', 'y': 'Merkmal 16', 'z': 'Merkmal 17', 'color': 'Gruppe'})\n",
"fig.update_traces(marker={'size': 3})\n",
"fig.show()"
]
},
{
"cell_type": "markdown",
"id": "65a3fe5e",
"metadata": {},
"source": [
"ii) Merkmal 0, 17, 18"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "2c37db7b",
"metadata": {},
"outputs": [],
"source": [
"fig = px.scatter_3d(x=X[0], y=X[17], z=X[18], color=T[0].astype(int).astype(str),\n",
" labels={'x': 'Merkmal 0', 'y': 'Merkmal 17', 'z': 'Merkmal 18', 'color': 'Gruppe'})\n",
"fig.update_traces(marker={'size': 3})\n",
"fig.show()"
]
},
{
"cell_type": "markdown",
"id": "c1ff9466",
"metadata": {},
"source": [
"iii) Merkmal 18, 19, 20"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "72bfc023",
"metadata": {},
"outputs": [],
"source": [
"fig = px.scatter_3d(x=X[18], y=X[19], z=X[20], color=T[0].astype(int).astype(str),\n",
" labels={'x': 'Merkmal 18', 'y': 'Merkmal 19', 'z': 'Merkmal 20', 'color': 'Gruppe'})\n",
"fig.update_traces(marker={'size': 3})\n",
"fig.show()"
]
},
{
"cell_type": "markdown",
"id": "ed9f06a1",
"metadata": {},
"source": [
"iv) Merkmal 16, 18, 20"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "e90baab5",
"metadata": {},
"outputs": [],
"source": [
"fig = px.scatter_3d(x=X[16], y=X[18], z=X[20], color=T[0].astype(int).astype(str),\n",
" labels={'x': 'Merkmal 16', 'y': 'Merkmal 18', 'z': 'Merkmal 20', 'color': 'Gruppe'})\n",
"fig.update_traces(marker={'size': 3})\n",
"fig.show()"
]
},
{
"cell_type": "markdown",
"id": "7b3af1bc",
"metadata": {},
"source": [
"f) Am besten geeignet ist vermutlich Kombination iv) (16, 18, 20), danach i) (0, 16, 17).\n",
"Entscheidend ist Merkmal 16: Gruppe 1 hat hier deutlich höhere Werte als Gruppe 2 und diese\n",
"wiederum höhere als Gruppe 3. Zusätzlich trennen 18 und 20 Gruppe 1 gut von den anderen.\n",
"\n",
"Am schlechtesten geeignet ist ii) (0, 17, 18), da Merkmal 16 fehlt und sich Gruppe 2 und 3\n",
"dort stark überlappen. iii) (18, 19, 20) trennt Gruppe 1 noch ganz gut ab, Gruppe 2 und 3\n",
"überlappen aber ebenfalls stark."
]
},
{
"cell_type": "markdown",
"id": "40ac7c0e",
"metadata": {},
"source": [
"g) Ja:\n",
"- Merkmal 0 (Alter) ist in i) und ii) vermutlich überflüssig, die Werte unterscheiden sich\n",
" zwischen den Gruppen kaum.\n",
"- Merkmal 19 in iii) unterscheidet sich zwischen den Gruppen ebenfalls kaum.\n",
"- In iv) sind Merkmal 18 und 20 stark korreliert (r ≈ 0.77), eines davon bringt vermutlich\n",
" kaum zusätzliche Information."
]
}
],
"metadata": {
"jupytext": {
"cell_metadata_filter": "-all",
"main_language": "python",
"notebook_metadata_filter": "-all"
}
},
"nbformat": 4,
"nbformat_minor": 5
}