این سایت با انیمیشن و به صورت بصری به شما الگوریتم هارو یاد میده و میتونید ازش برای یادگیری الگوریتم استفاده کنید.

http://www.algoanim.ide.sk/



🆔 @Python4all_pro

Downloading a YouTube Playlist using #Python


#code

🆔 @Python4all_pro

Tornado: a web framework for asynchronous network operations

Tornado is a web framework and Python library designed to handle network operations asynchronously.

Features of Tornado:

* Non-blocking I/O processing that scales to tens of thousands of active connections.
* Independence from the WSGI standard, which makes it different from most Python web frameworks.
* Integration with the asyncio module from the Python standard library, providing the same event loop.
* Support for long polling and web sockets.

Link: https://www.tornadoweb.org/en/stable/


#library
#python

🆔 @Python4all_pro

Create email and text message SMS alerts with #Python!



🆔 @Python4all_pro

Run #Javascript code using #python subprocess #Library

import subprocess
js_code = """
console.log("Hello, World!");
"""

with open("script.js", "w") as file:
    file.write(js_code)

result = subprocess.run(["node", "script.js"], capture_output=True, text=True)

print(result.stdout)

#code

🆔 @Python4all_pro

Find system name , version processor using #python.

import platform

system_name = platform.system()

system_version = platform.version()

architecture = platform.architecture()[0]

processor = platform.processor()

print("OS:", system_name)
print("OS Version:", system_version)
print("32 or 64:", architecture)
print("Processor :", processor)

#code

🆔 @Python4all_pro

Downloading a Audio file from YouTube Video

Free Code: https://www.clcoding.com/2024/06/downloading-audio-file-from-youtube.html




#code

🆔 @Python4all_pro

Convert Image to text

import pytesseract as t
from PIL import Image
img = Image.open("photo.jpg")
text = t.image_to_string(img)
print(text)

#code

🆔 @Python4all_pro

-Location of Mobile Number Code -

import phonenumbers
from phonenumbers import (timezone,geocoder,carrier)
 
number = input("Enter the phone number with country code : ")
 
# Parsing String to the Phone number
phoneNumber = phonenumbers.parse(number)
 
# printing the timezone using the timezone module
timeZone = timezone.time_zones_for_number(phoneNumber)
print(f"timezone : {timeZone}")
 
# printing the geolocation of the given number using the geocoder module
geolocation = geocoder.description_for_number(phoneNumber,"en")
print(f"location : {geolocation}")
 
# printing the service provider name using the carrier module
service = carrier.name_for_number(phoneNumber,"en")
print(f"service provider : {service}")

#code

🆔 @Python4all_pro

Transform your #Python code in an image



#code

🆔 @Python4all_pro

Python Logo Source Code

import turtle

t = turtle.Turtle()
s = turtle.Screen()
s.bgcolor("black")
t.speed(10)
t.pensize(2)
t.pencolor("white")



def s_curve():
    for i in range(90):
        t.left(1)
        t.forward(1)

def r_curve():
    for i in range(90):
        t.right(1)
        t.forward(1)

def l_curve():
    s_curve()
    t.forward(80)
    s_curve()

def l_curve1():
    s_curve()
    t.forward(90)
    s_curve()

def half():
    t.forward(50)
    s_curve()
    t.forward(90)
    l_curve()
    t.forward(40)
    t.left(90)
    t.forward(80)
    t.right(90)
    t.forward(10)
    t.right(90)
    t.forward(120) #on test
    l_curve1()
    t.forward(30)
    t.left(90)
    t.forward(50)
    r_curve()
    t.forward(40)
    t.end_fill()

def get_pos():
    t.penup()
    t.forward(20)
    t.right(90)
    t.forward(10)
    t.right(90)
    t.pendown()

def eye():
    t.penup()
    t.right(90)
    t.forward(160)
    t.left(90)
    t.forward(70)
    t.pencolor("black")
    t.dot(35)

def sec_dot():
    t.left(90)
    t.penup()
    t.forward(310)
    t.left(90)
    t.forward(120)
    t.pendown()

    t.dot(35)




t.fillcolor("#306998")
t.begin_fill()
half()
t.end_fill()
get_pos()
t.fillcolor("#FFD43B")
t.begin_fill()
half()
t.end_fill()

eye()
sec_dot()



def pause():
    t.speed(2)
    for i in range(100):
        t.left(90)
pause()

#code

🆔 @Python4all_pro

🖥 ؛SearXNG — یک موتور فراجستجوی رایگان است که در پایتون پیاده سازی شده است

git clone https://github.com/searxng/searxng.git searxng
cd searxng
sudo -H ./utils/searxng.sh install all

؛SearXNG — نتایج جستجو و پایگاه داده های مختلف را ترکیب می کند و داده های حساس کاربر را جمع آوری یا ردیابی نمی کند

شروع سریع با Docker :

docker run --rm \
           -d -p 8080:8080 \
           -v "${PWD}/searxng:/etc/searxng" \
           -e "BASE_URL=http://localhost:8080/" \
           -e "INSTANCE_NAME=my-instance" \
           searxng/searxng

🖥 GitHub
🟡 Docks



#library
#python

🆔 @Python4all_pro

Python Projects: Create an Interactive Quiz Application


#پروژه
🆔 @Python4all_pro

Convert Animated Gif to video

from os import remove,popen
import subprocess

def sprocess(a, b="utf-8"):
    p = subprocess.Popen(a,shell=True,stdout=subprocess.PIPE, stderr=subprocess.PIPE).communicate()
    return str(p[0].decode(b)+p[1].decode(b))

vid = await message.reply_to_message.download()

print(sprocess("ffmpeg -y -i '" + vid + "' -map_metadata -1 s.mp4"))

durasi = popen("ffprobe -i '" + vid + "' -show_entries format=duration -v quiet -of csv='p=0'").read()

await message.reply_video(video="s.mp4")

remove("s.mp4")
remove(vid)

#code

🆔 @Python4all_pro

حل سوالات استخدامی سایت leetcode.com

Task: No. 16. 3Sum Closest #medium

Condition:
Given an integer array nums of length n and an integer target, find the three integers in nums whose sum is closest to the target. Returns the sum of three integers. You can assume that each input will have exactly one solution.

Solution:

    def threeSumClosest(self, num, target):
        num.sort()
        result = num[0] + num[1] + num[2]
        for i in range(len(num) - 2):
            j, k = i+1, len(num) - 1
            while j < k:
                sum = num[i] + num[j] + num[k]
                if sum == target:
                    return sum
                
                if abs(sum - target) < abs(result - target):
                    result = sum
                
                if sum < target:
                    j += 1
                elif sum > target:
                    k -= 1
                else:
                    return result
            
        return result

Explanation:
Sort an array:

First we sort the num array. This will allow us to use two pointers to find the closest sum.
Initializing the result:

We initialize the result variable with the sum of the first three elements of the sorted array. This will be our starting closest amount.
Traversing the array:

We use a for loop to iterate through the array. For each element we use two pointers j and k:
j starts immediately after the current element i.
k starts from the end of the array.
Two pointers:

Inside the while loop, while j is less than k, we calculate the sum of the elements num[i], num[j] and num[k].
If sum equals target, then we return sum since we found an exact match.
Result update:

If the current sum sum is closer to target than the previous closest sum result, update result.
Pointer shift:

If sum is less than target, move pointer j to the right to increase the sum.
If sum is greater than target, shift pointer k to the left to decrease the sum.
Return result:

After completing all iterations, we return result, which will contain the sum of three numbers closest to target.
Time and space complexity:
Time complexity: O(n^2), where n is the length of the array. Sorting takes O(n log n) and the basic algorithm with two pointers runs in O(n^2).
Space complexity: O(1) since we only use a few additional variables, and do not use additional memory depending on the size of the input data.

#interview #LeetCode

🆔 @Python4all_pro

Generate AI image with the input text

pip install --upgrade MukeshAPI

from MukeshAPI import api
response = api.ai_image("cute boy pic")
#print(response)
with open("mukesh.jpg", 'wb') as f:
    f.write(response)
print("image generated successfully")

#code

🆔 @Python4all_pro

حل سوالات استخدامی سایت leetcode.com

Task: No. 17. Letter Combinations of a Phone Number #medium

Condition:
Given a string containing the numbers 2 to 9 inclusive, return all possible combinations of letters that the number can represent. Return the answer in any order. The correspondence between numbers and letters (as on telephone buttons) is given below. Note that 1 does not match any letters.

Solution:

class Solution(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        res = []
        nums.sort()
        self.dfs(nums, [], res)
        return res
    
    def dfs(self, nums, path, res):
        res.append(path)
        for i in range(len(nums)):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            self.dfs(nums[i+1:], path + [nums[i]], res)

Explanation:
Sorting:

First we sort the nums array. Sorting makes it easy to handle duplicates because similar items will be placed next to each other.
Recursive dfs method:

The dfs method is used to recursively construct all possible subsets. In dfs, path represents the current subset, and res is a list of all subsets.
Adding a subset to the result:

At each level of recursion, we add the current subset of path to the result of res.
Handling duplicates:

Before continuing the recursion, we check whether the current element nums[i] is a duplicate of the previous element. If so, skip it to avoid adding identical subsets to res.
Recursive construction of subsets:

For each element in nums, starting at the current index, we call dfs on the next elements of the array (nums[i+1:]). This means that we consider subsets that include the current element, and continue to build subsets without the current element.
Path (path) and compartment (nums):

Each time dfs is called, a new subset is created by adding the current element to path. This new list is then passed to the next level of recursion, allowing all possible subsets to be constructed.
Time and space complexity:
Time complexity: O(2^n), where n is the number of elements in nums. This is because there are 2^n subsets for an array of n elements.
Space complexity: O(2^n * n), since each of the 2^n possible subsets may require up to n elements to store.


#interview #LeetCode

🆔 @Python4all_pro