Day 2 and Day 3 changes

python_concepts/dictionary.ipynb

@@ -94,6 +94,35 @@
     "# convert dictionary keys into a list using keys() method\n",
     "list(dict1.keys())"
    ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a 1\n",
+      "{'A': 100, 'B': 200, 'C': 300}\n"
+     ]
+    }
+   ],
+   "source": [
+    "### Dictionary comprehension works similar to list comprehension. Below we will use dict comprehension to convert\n",
+    "### dictionary from one form to another.\n",
+    "### dict.items() return key value pair\n",
+    "\n",
+    "dict1 = {'a': 1}\n",
+    "type(dict1.items())\n",
+    "for k, v in dict1.items():\n",
+    "    print(k, v)\n",
+    "    \n",
+    "dict1 = {'a': 1, 'b': 2, 'c': 3}\n",
+    "dict2 = {k.upper():v*100 for (k,v) in dict1.items()}\n",
+    "print(dict2)\n"
+   ]
   }
  ],
  "metadata": {

python_concepts/itertools.ipynb

@@ -0,0 +1,68 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Itertools"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#### In this notebook we learn all the concepts and use cases of itertools library"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "atif\n",
+      "amit\n",
+      "Naoto\n",
+      "Arti\n",
+      "zoheb\n",
+      "kasif\n"
+     ]
+    }
+   ],
+   "source": [
+    "## Itertools.chain can be used to combine elements of different data structure\n",
+    "\n",
+    "import itertools\n",
+    "group_list = ['atif', 'amit']\n",
+    "group_set = {'Naoto', 'Arti', 'Arti'}\n",
+    "group_tuple = ('zoheb', 'kasif')\n",
+    "for colleagues in itertools.chain(group_list, group_set, group_tuple):\n",
+    "    print(colleagues)\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "dashboard",
+   "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": "3.10.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

python_daily_exercies/day1_25_12_2023.ipynb

@@ -117,6 +117,25 @@
     "for i , j in zip(numbers[::2], numbers[1::2]):\n",
     "    print(i, j)"
    ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "dict1 = {'a':2, 'b':4  , 'c':6, 'd':8 , 'e':10}\n",
+    "\n",
+    "print('g' in dict1)"
+   ]
   }
  ],
  "metadata": {

python_daily_exercies/day2_30_12_2023.ipynb

@@ -0,0 +1,65 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Day 2\n",
+    "#### We will practice list slicing and take up the question from Coding Computing Coach email\n",
+    "#### Question: Suppose you have a list of numbers from 1 to 31, representing the dates in a month.\n",
+    "#### Given that the first Sunday occurs on the 3rd, can you slice out all dates that would be Sundays?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31]\n"
+     ]
+    },
+    {
+     "data": {
+      "text/plain": [
+       "[3, 10, 17, 24, 31]"
+      ]
+     },
+     "execution_count": 3,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "calender = [i for i in range(1,32)]\n",
+    "print(calender)\n",
+    "sundays = calender[2::7]\n",
+    "sundays"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "dashboard",
+   "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": "3.10.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

python_daily_exercies/day3_31_12_2023.ipynb

@@ -0,0 +1,41 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Day 3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##### We have dictionary composed of {'a':1, 'b':2, 'c':3} and we have to convert into {'A':100, 'B':200, 'C':300}"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "vscode": {
+     "languageId": "plaintext"
+    }
+   },
+   "outputs": [],
+   "source": [
+    "dict1 = {'a': 1, 'b': 2, 'c': 3}\n",
+    "dict2 = {k.upper():v*100 for (k,v) in dict1.items()}\n",
+    "# Here dictionary.items returns key value pair\n",
+    "print(dict2)"
+   ]
+  }
+ ],
+ "metadata": {
+  "language_info": {
+   "name": "python"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}