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Day 2 and Day 3 changes
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‎python_concepts/dictionary.ipynb

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

‎python_concepts/itertools.ipynb

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{
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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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"### Itertools"
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]
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},
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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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"#### In this notebook we learn all the concepts and use cases of itertools library"
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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": 1,
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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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"atif\n",
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"amit\n",
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"Naoto\n",
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"Arti\n",
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"zoheb\n",
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"kasif\n"
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]
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}
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],
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"source": [
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"## Itertools.chain can be used to combine elements of different data structure\n",
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"\n",
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"import itertools\n",
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"group_list = ['atif', 'amit']\n",
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"group_set = {'Naoto', 'Arti', 'Arti'}\n",
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"group_tuple = ('zoheb', 'kasif')\n",
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"for colleagues in itertools.chain(group_list, group_set, group_tuple):\n",
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" print(colleagues)\n"
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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": "dashboard",
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"language": "python",
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"name": "python3"
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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": "3.10.6"
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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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}

‎python_daily_exercies/day1_25_12_2023.ipynb

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"for i , j in zip(numbers[::2], numbers[1::2]):\n",
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" print(i, j)"
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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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"False\n"
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]
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}
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],
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"source": [
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"dict1 = {'a':2, 'b':4 , 'c':6, 'd':8 , 'e':10}\n",
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"\n",
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"print('g' in dict1)"
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]
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}
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],
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"metadata": {

‎python_daily_exercies/day2_30_12_2023.ipynb

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{
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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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"### Day 2\n",
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"#### We will practice list slicing and take up the question from Coding Computing Coach email\n",
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"#### Question: Suppose you have a list of numbers from 1 to 31, representing the dates in a month.\n",
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"#### Given that the first Sunday occurs on the 3rd, can you slice out all dates that would be Sundays?"
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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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"[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"
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]
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},
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{
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"data": {
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"text/plain": [
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"[3, 10, 17, 24, 31]"
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]
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"calender = [i for i in range(1,32)]\n",
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"print(calender)\n",
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"sundays = calender[2::7]\n",
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"sundays"
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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": "dashboard",
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"language": "python",
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"name": "python3"
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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": "3.10.6"
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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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}

‎python_daily_exercies/day3_31_12_2023.ipynb

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