Python must know Resource

Hi Guys!!!!

Hope you all doing well!!!!.  

Today   I am going to discuss, Puthon must know resource. First Why python is so important?
Python is one of the world’s most popular, in-demand programming languages. This is for many reasons:

• it’s easy to learn
• it’s super versatile
• it has a huge range of modules and libraries(known as Batteries)

Here, I’ve made an attempt at sharing some of famous and must know resource to you guys

You find all the example from my github page

https://github.com/Saurabh-12/Python_Learning
# all or any

x = [True, True, False]
if any(x):
    print("At least one True")
if all(x):
    print("Not one False")
if any(x) and not all(x):
    print("At least one True and one False")

bashplotlib
You want to plot graphs in the console ?

pip install bashplotlib
collections
Python has some great default datatypes and collection . You can check it from my github page. 
https://github.com/Saurabh-12/Python_Learning/blob/master/Python_Collections.py

from collections import OrderedDict, Counter

# Remembers the order the keys are added!

x = OrderedDict(a=1, b=2, c=3)

# Counts the frequency of each character

y = Counter("Hello World!")

print(y)

print(x)

emoji
Yes, really 

pip install emoji

from emoji import emojize

print(emojize(":thumbs_up:"))

geopy

It works by abstracting the APIs of a range of different geocoding services. It enables you to obtain a place’s full street address, latitude, longitude, and even altitude.

pip install geopy

from geopy import GoogleV3

place = "Manjunatha Layout,Munnekolala,Marathali, Bangalore, India"

location = GoogleV3().geocode(place)

print(location.address)

print(location.location)

howdoi
Need to check StackOverflow, but don’t want to leave the terminal?

pip install howdoi

howdoi for loop in python

**kwargs

The double-asterisk in front of a dictionary object lets you pass the contents of that dictionary as named arguments to a function.

dictionary = {"a": 1, "b": 2}

def someFunction(a, b):

print(a + b)

return

# these do the same thing:

someFunction(**dictionary)

someFunction(a=1, b=2)

List comprehensions

One of the best thing in python. Check my git hub page check dedicatd example

https://github.com/Saurabh-12/Python_Learning/blob/master/ListComprehension.py

map

Python supports functional programming through a number of inbuilt features. 

One of the most useful is the map() function — especially in combination with lambda functions.

https://github.com/Saurabh-12/Python_Learning/blob/master/Lamda_Map_example.py

zip

Ever needed to form a dictionary out of two lists?

keys = ['a', 'b', 'c']

vals = [1, 2, 3]

zipped = dict(zip(keys, vals))

print(zipped)

https://github.com/Saurabh-12/Python_Learning/blob/master/ZipExaple.py

Virtual environments

Let say we are working on multiple Python projects at any one time. 

Unfortunately, sometimes two projects will rely on different versions of the same dependency. 

Which do We install in our system?

To solve this :- venv is solution for this . For more detail see my blogs dedicaed to it

http://saurabhsharma123k.blogspot.com/2019/12/python-package-manager.html

uuid

A quick and easy way to generate Universally Unique IDs (or ‘UUIDs’) is through the Python Standard Library’s uuid module.

import uuid

user_id = uuid.uuid4()

print(user_id)

pprint

Python’s default print function does it work quite good.

But if We try to print out any large, nested object, and the result is rather ugly.

pprint comes to rescue us for printing out complex structured objects in an easy-to-read format.

import requests

import pprint

url = 'https://randomuser.me/api/?results=1'

users = requests.get(url).json()

pprint.pprint(users)

Find memory used by an object

import sys

print(sys.getsizeof(5)) # 28

print(sys.getsizeof("Python")) # 55

Find the most frequent element in a list

def most_frequent(list):

return max(set(list), key = list.count)

numbers = [1, 2, 3, 2, 4, 3, 1, 3]

print(most_frequent(numbers)) # 3

Calculate time taken to execute a piece of code

import time

start_time = time.time()

a,b = 5,10

c = a+b

end_time = time.time()

time_taken = (end_time- start_time)*(10**6)

print("Time taken in micro_seconds:", time_taken)

# Time taken in micro_seconds: 3.89577484130859375

Find unique characters in a string

string = "abcbcabdb"

unique = set(string)

new_string = ''.join(unique)

print(new_string) # abcd

Use chained function call

def add(a, b):

return a + b

def subtract(a, b):

return a - b

a, b = 5, 10

print((subtract if a > b else add)(a, b)) # 15

Thanks
Saurabh Sharma
Happy Coding !!!