Quick-Fix

03-numpy-und-matplotlib/02-matrix.ipynb

@@ -14,11 +14,14 @@
     "Erstellen Sie ein 2D-NumPy-Array `a` mit `arange` und `reshape` wie\n",
     "dargestellt:\n",
     "\n",
-    "$\\begin{bmatrix} 1 & 2 & 3\\\\ 4 & 5 & 6 \\end{bmatrix}$\n",
+    "$\\begin{bmatrix}\n",
+    "1 & 2 & 3\\\\\n",
+    "4 & 5 & 6\n",
+    "\\end{bmatrix}$\n",
     "\n",
     "Hier Ihr Code:"
    ],
-   "id": "0005-c6178e60b944258ad3f127adea03eb4ee0da7c74b1a39b54c9400760909"
+   "id": "0005-db440e97cb5c7f546520ed47264b7fce12db0ccfc7c8adb03bca9113aee"
   },
   {
    "cell_type": "code",
@@ -67,14 +70,18 @@
     "\n",
     "Erstellen Sie ein 1D-NumPy-Array `v` mit `arange` wie dargestellt:\n",
     "\n",
-    "$\\begin{bmatrix} 10\\\\ 8\\\\ 6 \\end{bmatrix}$\n",
+    "$\\begin{bmatrix}\n",
+    "10\\\\\n",
+    "8\\\\\n",
+    "6\n",
+    "\\end{bmatrix}$\n",
     "\n",
     "Führen Sie die Matrix-Vektor-Multiplikation $a \\, v$ durch und speichern\n",
     "das Ergebnis in `w`!\n",
     "\n",
     "Hier Ihr Code:"
    ],
-   "id": "0014-0328546f1dffddc990c09fb5d601b0a937e5de407867597480a75115e3f"
+   "id": "0014-298594a5981320e04d208dd7d227a173f8a4fa65f04cad0e233cf939a12"
   },
   {
    "cell_type": "code",
@@ -278,6 +285,58 @@
     "v[v_sorted_indices]"
    ],
    "id": "0035-f0e4f4e230e723cda122e72eb4cc32c10ac8c30139419ccd2df6a8f01c6"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## d) Zeile hinzufügen\n",
+    "\n",
+    "Fügen Sie $v$ als Zeile zu $a$ hinzu und speichern es in `av`!\n",
+    "\n",
+    "Hier Ihr Code:"
+   ],
+   "id": "0038-60ff5efbf91d48180e27b1e4e4dd3074404e32192dc1356f79296273c72"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [],
+   "id": "0039-44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Tests zu d)\n",
+    "\n",
+    "Wir testen, ob die Elemente von `av` wie erwartet sind:"
+   ],
+   "id": "0041-6c3e8a5629f3b4c854cdac323a8871364284522e5a15f6974a77f00269f"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "array([[ 1,  2,  3],\n",
+      "       [ 4,  5,  6],\n",
+      "       [10,  8,  6]])"
+     ]
+    }
+   ],
+   "source": [
+    "av"
+   ],
+   "id": "0042-bca625b223971909dd88fc93faeb050dc5b34d91c0871661740dcfb9d18"
   }
  ],
  "nbformat": 4,

03-numpy-und-matplotlib/04-mean-temp.ipynb

+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Mittlere Temperaturen\n",
+    "\n",
+    "Die Funktion `load_temp` lädt die mittlere Monatstemperaturen von Januar\n",
+    "2005 bis August 2023 von der Wetterstation 01303 (Essen-Bredeney) als\n",
+    "NumPy-Array:\n",
+    "\n",
+    "| 0: Jahr | 1: Monat | 2: Temperatur in K |\n",
+    "|--------:|---------:|-------------------:|\n",
+    "|    2005 |        1 |             277.02 |\n",
+    "|    2005 |        2 |             274.15 |\n",
+    "|    2005 |        3 |             279.44 |\n",
+    "|       … |        … |                  … |\n",
+    "|    2023 |        7 |             291.75 |\n",
+    "|    2023 |        8 |             291.18 |\n",
+    "\n",
+    "Mitteln Sie die Temperaturen für…\n",
+    "\n",
+    "1.  …die Monate über alle Jahre und speichern es in `month_mean`. So\n",
+    "    erhalten Sie beispielsweise die mittlere Temperatur im Januar\n",
+    "    (unabhängig vom Jahr). Plotten Sie die Ergebnisse.\n",
+    "2.  …die Jahre über alle Monate und speichern es in `year_mean`. So\n",
+    "    erhalten Sie beispielsweise die mittlere Temperatur im Jahr 2005\n",
+    "    (unabhängig vom Monat). Plotten Sie die Ergebnisse.\n",
+    "\n",
+    "Hier Ihr Startcode:"
+   ],
+   "id": "0005-d2ed2a9b5930bbb53dab5d18ab5a1c4bcaabe44815e106bcb42bb7411c2"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import matplotlib.pyplot as plt\n",
+    "\n",
+    "def load_temp():\n",
+    "    FILE_PATH = 'dwd_data-2023-08-31.parquet.zst'\n",
+    "    dwd_data = pd.read_parquet(FILE_PATH)\n",
+    "\n",
+    "    station_ind = dwd_data['station_id'] == '01303'\n",
+    "    param_ind = dwd_data['parameter'] == 'temperature_air_mean_200'\n",
+    "    dwd_data = dwd_data[station_ind & param_ind]\n",
+    "\n",
+    "    dwd_data['year'] = dwd_data['date'].dt.year\n",
+    "    dwd_data['month'] = dwd_data['date'].dt.month\n",
+    "    return dwd_data[['year', 'month', 'value']].to_numpy()\n",
+    "\n",
+    "temp = load_temp()"
+   ],
+   "id": "0006-e4d3cff1c8267ff2b520c715c7e3fb808d00dc6a5155c21ece66a06e444"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Tests\n",
+    "\n",
+    "Wir schauen mal, ob die `shape` passt:"
+   ],
+   "id": "0008-539fc374ab04b1689d3d56775fe49569971d6ae370acf0fb866780c7864"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "(12,)"
+     ]
+    }
+   ],
+   "source": [
+    "np.shape(month_mean)"
+   ],
+   "id": "0009-bcadab89753cff51d9c3ccf5822c5cd62857e16f60b147f35e7b6f32dfa"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Der Mittelwert für Januar sollte ca. 276.276 K betragen:"
+   ],
+   "id": "0010-df2a515e50faefff59e6b552fa4caa1ab108f7d3c2098a41c24ee3680d6"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "276.276 K"
+     ]
+    }
+   ],
+   "source": [
+    "print(f'{month_mean[0]:.3f} K')"
+   ],
+   "id": "0011-91b4879873e05a15752b382e6910b0f9907260a15a24e461078caa00fac"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Der Mittelwert für 2005 sollte ca. 283.485 K betragen:"
+   ],
+   "id": "0012-1851b721844dfa31100516a6bdf306ffb883139d381e456ac704cab86a3"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "283.485 K"
+     ]
+    }
+   ],
+   "source": [
+    "print(f'{year_mean[0]:.3f} K')"
+   ],
+   "id": "0013-0d97c8b776aa3b2fe55e04bb02e3c8f8b31bc133c4aea242af15ff8d9cd"
+  }
+ ],
+ "nbformat": 4,
+ "nbformat_minor": 5,
+ "metadata": {}
+}

03-numpy-und-matplotlib/solutions/02-matrix-sol.ipynb

@@ -14,11 +14,14 @@
     "Erstellen Sie ein 2D-NumPy-Array `a` mit `arange` und `reshape` wie\n",
     "dargestellt:\n",
     "\n",
-    "$\\begin{bmatrix} 1 & 2 & 3\\\\ 4 & 5 & 6 \\end{bmatrix}$\n",
+    "$\\begin{bmatrix}\n",
+    "1 & 2 & 3\\\\\n",
+    "4 & 5 & 6\n",
+    "\\end{bmatrix}$\n",
     "\n",
     "Hier Ihr Code:"
    ],
-   "id": "0005-c6178e60b944258ad3f127adea03eb4ee0da7c74b1a39b54c9400760909"
+   "id": "0005-db440e97cb5c7f546520ed47264b7fce12db0ccfc7c8adb03bca9113aee"
   },
   {
    "cell_type": "code",
@@ -92,14 +95,18 @@
     "\n",
     "Erstellen Sie ein 1D-NumPy-Array `v` mit `arange` wie dargestellt:\n",
     "\n",
-    "$\\begin{bmatrix} 10\\\\ 8\\\\ 6 \\end{bmatrix}$\n",
+    "$\\begin{bmatrix}\n",
+    "10\\\\\n",
+    "8\\\\\n",
+    "6\n",
+    "\\end{bmatrix}$\n",
     "\n",
     "Führen Sie die Matrix-Vektor-Multiplikation $a \\, v$ durch und speichern\n",
     "das Ergebnis in `w`!\n",
     "\n",
     "Hier Ihr Code:"
    ],
-   "id": "0017-0328546f1dffddc990c09fb5d601b0a937e5de407867597480a75115e3f"
+   "id": "0017-298594a5981320e04d208dd7d227a173f8a4fa65f04cad0e233cf939a12"
   },
   {
    "cell_type": "code",
@@ -444,6 +451,81 @@
     "v[v_sorted_indices]"
    ],
    "id": "0051-f0e4f4e230e723cda122e72eb4cc32c10ac8c30139419ccd2df6a8f01c6"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## d) Zeile hinzufügen\n",
+    "\n",
+    "Fügen Sie $v$ als Zeile zu $a$ hinzu und speichern es in `av`!\n",
+    "\n",
+    "Hier Ihr Code:"
+   ],
+   "id": "0054-60ff5efbf91d48180e27b1e4e4dd3074404e32192dc1356f79296273c72"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [],
+   "id": "0055-44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Lösung zu d)\n",
+    "\n",
+    "Wir verwenden `np.vstack` um die Zeile `v` zu `a` hinzuzufügen. Das\n",
+    "Ergebnis ist ein $3 \\times 3$-Array."
+   ],
+   "id": "0057-97452acebf5b4c664fe1476c7982190acc0b2e6dcb4126eab30867457f1"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [
+    "av = np.vstack((a, v))"
+   ],
+   "id": "0058-7137f073beec5783a38b5737aa69b976bdd4863d0afc94e6cc871586665"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Tests zu d)\n",
+    "\n",
+    "Wir testen, ob die Elemente von `av` wie erwartet sind:"
+   ],
+   "id": "0060-6c3e8a5629f3b4c854cdac323a8871364284522e5a15f6974a77f00269f"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "array([[ 1,  2,  3],\n",
+      "       [ 4,  5,  6],\n",
+      "       [10,  8,  6]])"
+     ]
+    }
+   ],
+   "source": [
+    "av"
+   ],
+   "id": "0061-bca625b223971909dd88fc93faeb050dc5b34d91c0871661740dcfb9d18"
   }
  ],
  "nbformat": 4,

03-numpy-und-matplotlib/solutions/04-mean-temp-sol.ipynb

+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Mittlere Temperaturen\n",
+    "\n",
+    "Die Funktion `load_temp` lädt die mittlere Monatstemperaturen von Januar\n",
+    "2005 bis August 2023 von der Wetterstation 01303 (Essen-Bredeney) als\n",
+    "NumPy-Array:\n",
+    "\n",
+    "| 0: Jahr | 1: Monat | 2: Temperatur in K |\n",
+    "|--------:|---------:|-------------------:|\n",
+    "|    2005 |        1 |             277.02 |\n",
+    "|    2005 |        2 |             274.15 |\n",
+    "|    2005 |        3 |             279.44 |\n",
+    "|       … |        … |                  … |\n",
+    "|    2023 |        7 |             291.75 |\n",
+    "|    2023 |        8 |             291.18 |\n",
+    "\n",
+    "Mitteln Sie die Temperaturen für…\n",
+    "\n",
+    "1.  …die Monate über alle Jahre und speichern es in `month_mean`. So\n",
+    "    erhalten Sie beispielsweise die mittlere Temperatur im Januar\n",
+    "    (unabhängig vom Jahr). Plotten Sie die Ergebnisse.\n",
+    "2.  …die Jahre über alle Monate und speichern es in `year_mean`. So\n",
+    "    erhalten Sie beispielsweise die mittlere Temperatur im Jahr 2005\n",
+    "    (unabhängig vom Monat). Plotten Sie die Ergebnisse.\n",
+    "\n",
+    "Hier Ihr Startcode:"
+   ],
+   "id": "0005-d2ed2a9b5930bbb53dab5d18ab5a1c4bcaabe44815e106bcb42bb7411c2"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [
+    "import numpy as np\n",
+    "import pandas as pd\n",
+    "import matplotlib.pyplot as plt\n",
+    "\n",
+    "def load_temp():\n",
+    "    FILE_PATH = 'dwd_data-2023-08-31.parquet.zst'\n",
+    "    dwd_data = pd.read_parquet(FILE_PATH)\n",
+    "\n",
+    "    station_ind = dwd_data['station_id'] == '01303'\n",
+    "    param_ind = dwd_data['parameter'] == 'temperature_air_mean_200'\n",
+    "    dwd_data = dwd_data[station_ind & param_ind]\n",
+    "\n",
+    "    dwd_data['year'] = dwd_data['date'].dt.year\n",
+    "    dwd_data['month'] = dwd_data['date'].dt.month\n",
+    "    return dwd_data[['year', 'month', 'value']].to_numpy()\n",
+    "\n",
+    "temp = load_temp()"
+   ],
+   "id": "0006-e4d3cff1c8267ff2b520c715c7e3fb808d00dc6a5155c21ece66a06e444"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Lösung\n",
+    "\n",
+    "Um für alle Monate (Spalte 1) die mittlere Temperatur (Spalte 2) zu\n",
+    "berechnen, erstellen wir zunächst eine Sequenz von 1 bis 12. Dann\n",
+    "verwenden wir eine List Comprehension um für jeden Monat die mittlere\n",
+    "Temperatur zu berechnen. Dabei ergibt `temp[:, 1] == month` einen\n",
+    "booleschen Vektor, der angibt, ob die Zeile zu dem Monat gehört. Mit\n",
+    "`temp[temp[:, 1] == month, 2]` erhalten wir dann alle Temperaturen für\n",
+    "den Monat, die wir mit `np.mean(...)` mitteln."
+   ],
+   "id": "0008-2cd25c8798f51f8e5a6c713c29b493ac19b894f0ed5571805f0e364e845"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [
+    "all_months = range(1, 13)\n",
+    "month_mean = [np.mean(temp[temp[:, 1] == month, 2]) for month in all_months]"
+   ],
+   "id": "0009-a1253165c038b338462cc82eb29e2f7c48a4a9fc2d4445491a01a7c03c4"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Das Ergebnis plotten wir ganz einfach mit:"
+   ],
+   "id": "0010-ecadd0c3a7487df23a98b0962f4f82f4be5e1843bc8ae8e7e27ef69f618"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [
+    "plt.figure()\n",
+    "plt.plot(month_mean)"
+   ],
+   "id": "0011-3a213d48af32df5375e47a6ecccc9845858537761ddb421498a332a8fb0"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Um für alle Jahre (Spalte 0) die mittlere Temperatur (Spalte 2) zu\n",
+    "berechnen, verwenden wir eine ähnliche List Comprehension. Wir verwenden\n",
+    "`np.unique` um die Jahre sortiert zu extrahieren. Das ergibt einfach\n",
+    "eine Sequenz von 2005 bis 2023. Der Aufruf von `astype(int)` wandelt die\n",
+    "Werte in Integer um, ist aber nicht wichtig. Vor der Mittelung filtern\n",
+    "wir 2023 weg, weil es unvollständig ist und der Mittelwert zu hoch wäre,\n",
+    "da die kalten Temperaturen von September bis Dezember fehlen würden.\n",
+    "Auch dafür hätte man wieder eine `range(2005, 2023)` verwenden können."
+   ],
+   "id": "0012-3ee0a3d69a51adfd7eaf1f4213775cd4d7d7715664ece17e6e89b6ed8fb"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [
+    "all_years = np.unique(temp[:, 0]).astype(int)\n",
+    "all_years = all_years[all_years < 2023]\n",
+    "year_mean = [np.mean(temp[temp[:, 0] == year, 2]) for year in all_years]"
+   ],
+   "id": "0013-ca8699b005a925df38cf6316af64aaf256f49174a83f449e6b447b1cf19"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Das Ergebnis plotten wir ganz einfach mit:"
+   ],
+   "id": "0014-ecadd0c3a7487df23a98b0962f4f82f4be5e1843bc8ae8e7e27ef69f618"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "style": "python"
+   },
+   "outputs": [],
+   "source": [
+    "plt.figure()\n",
+    "plt.plot(all_years, year_mean)"
+   ],
+   "id": "0015-a42d856cfe30a9bcf7f2c514ee6b65372103e8818cf5eb0a0e17d6e6d69"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Wenn wir nicht wüssten, ob noch weitere Jahre außer 2023 unvollständig\n",
+    "sind, könnten wir bei `np.unique` auch `return_counts=True` verwenden:"
+   ],
+   "id": "0016-ed8aabb9d71daa8b180cb6773528e73ff5928d72d7ea8b8e83f7439fc1a"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "array([12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12,  8])"
+     ]
+    }
+   ],
+   "source": [
+    "all_years, counts = np.unique(temp[:, 0], return_counts=True)\n",
+    "counts"
+   ],
+   "id": "0017-d45e342d4be912931cb0a80ee62eec2a2377fb20ca59fea9ea11428f6dd"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Nun lassen sich die Jahre mit weniger als 12 Monaten einfach\n",
+    "herausfiltern:"
+   ],
+   "id": "0018-59550f7dd84c46441758b975aa0b718731fbdc69f9992e9aff1daa99955"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "all_years = all_years[counts == 12]"
+   ],
+   "id": "0019-c5dafbe42aea1dcd7419a1e90abdede4001c4eb471ff52286729745fd4b"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Tests\n",
+    "\n",
+    "Wir schauen mal, ob die `shape` passt:"
+   ],
+   "id": "0021-539fc374ab04b1689d3d56775fe49569971d6ae370acf0fb866780c7864"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "(12,)"
+     ]
+    }
+   ],
+   "source": [
+    "np.shape(month_mean)"
+   ],
+   "id": "0022-bcadab89753cff51d9c3ccf5822c5cd62857e16f60b147f35e7b6f32dfa"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Der Mittelwert für Januar sollte ca. 276.276 K betragen:"
+   ],
+   "id": "0023-df2a515e50faefff59e6b552fa4caa1ab108f7d3c2098a41c24ee3680d6"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "276.276 K"
+     ]
+    }
+   ],
+   "source": [
+    "print(f'{month_mean[0]:.3f} K')"
+   ],
+   "id": "0024-91b4879873e05a15752b382e6910b0f9907260a15a24e461078caa00fac"
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Der Mittelwert für 2005 sollte ca. 283.485 K betragen:"
+   ],
+   "id": "0025-1851b721844dfa31100516a6bdf306ffb883139d381e456ac704cab86a3"
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "output_type": "stream",
+     "name": "stdout",
+     "text": [
+      "283.485 K"
+     ]
+    }
+   ],
+   "source": [
+    "print(f'{year_mean[0]:.3f} K')"
+   ],
+   "id": "0026-0d97c8b776aa3b2fe55e04bb02e3c8f8b31bc133c4aea242af15ff8d9cd"
+  }
+ ],
+ "nbformat": 4,
+ "nbformat_minor": 5,
+ "metadata": {}
+}