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

🖥 Aurora is a static site generator implemented in Python

Aurora is a static site generator implemented in Python.

Aurora supports:

- Create content and pages using markdown, jinja2 and HTML
- Various types of assembly
- Interactive build with fast reload capability for development (reload time up to < 300ms)
- Built-in support for creating site pages


pip3 install aurora-ssg

• Github



🆔 @Python4all_pro

How to easily blur any image using PIL

from PIL import Image, ImageFilter

def blur_image(image_path, output_path, radius):
  image = Image.open(image_path)
  blurred_image = image.filter(ImageFilter.GaussianBlur(radius))
  blurred_image.save(output_path)

blur_image('cat.jpg', 'cat_out.jpg', 3)  
Image.open('cat_out.jpg')

🟡 In general, the PIL library has a lot of features and excellent documentation




#library
#python

🆔 @Python4all_pro

🖥 چگونه با استفاده از پایتون متن مخفی را در داخل یک عکس پنهان کنیم؟

هیچ چیز پیچیده ای نیست، فقط به کتابخانه Stegano نیاز دارید

# pip install stegano

from stegano import lsb
secret = lsb.hide('image.png', 'очень секретный текст')
secret.save('secret_image.png')

print(lsb.reveal('secret_image.png'))

🖥 GitHub



#library
#python

🆔 @Python4all_pro

حل سوالات استخدامی سایت leetcode.com

Problem: No. 18. 4Sum #medium

Condition:
Given an array nums of n integers, return an array of all unique quadruples [nums[a], nums[b], nums[c], nums[d]] such that:
0 <= a, b, c, d < n
a, b, c and d are different.
nums[a] + nums[b] + nums[c] + nums[d] == target
You can return the response in any order.

Solution:

    nums.sort()
    results = []
    self.findNsum(nums, target, 4, [], results)
    return results

def findNsum(self, nums, target, N, result, results):
    if len(nums) < N or N < 2: return

    # solve 2-sum
    if N == 2:
        l,r = 0,len(nums)-1
        while l < r:
            if nums[l] + nums[r] == target:
                results.append(result + [nums[l], nums[r]])
                l += 1
                r -= 1
                while l < r and nums[l] == nums[l - 1]:
                    l += 1
                while r > l and nums[r] == nums[r + 1]:
                    r -= 1
            elif nums[l] + nums[r] < target:
                l += 1
            else:
                r -= 1
    else:
        for i in range(0, len(nums)-N+1): # careful about range
            if target < nums[i]*N or target > nums[-1]*N: # take advantages of sorted list
                break
            if i == 0 or i > 0 and nums[i-1] != nums[i]: # recursively reduce N
                self.findNsum(nums[i+1:], target-nums[i], N-1, result+[nums[i]], results)
    return

Explanation:
Sorting:

First the nums array is sorted. Sorting makes it easier to handle duplicates and speeds up execution using binary search.
Recursive function findNsum:

The findNsum function recursively finds combinations whose sum is equal to the given target. Depending on the value of N, it handles different cases:
For N = 2: This is the "Two Sum" subtask. We use two pointers (l and r) to find pairs of numbers in the array that add up to target.
If the current pair of numbers nums[l] and nums[r] sums to target, add this pair to the results.
We move the pointers left and right, skipping duplicates to avoid repeated combinations.
For N > 2: The function calls itself recursively, decrementing N by 1 and continuing to search for combinations among the remaining elements of the array. We also check if the current element and its combinations are within a valid range (for optimization).
Conditions for exiting recursion:

If the length of the array is less than N or N is less than 2, the function terminates execution, since there is no point in further searching for combinations.
We use conditions to stop execution if the current element is too large or too small to achieve the target value for a given number of elements (N).
Unique combinations:

To avoid duplicates, we check whether the current element is unique compared to previous ones.
Collection of results:

The results for each call to the findNsum function are added to the results list.



#interview #LeetCode

🆔 @Python4all_pro

ربات پولساز تلگرام : #استارز

⭐️از طریق major که مینی اپ جدید و رسمی تلگرام می‌باشد قادر به جمع کردن ستاره خواهید بود.

✍🏼 در آخرین آپدیت تلگرام ، این مسنجر سیستم پرداخت جدیدش که Starts یا ستاره‌ها نام دارد را راه‌اندازی کرد. با خرید ستاره از طریق اپ استور می‌توان پست‌های پولی کانال‌ها را خرید، در ربات‌ها پرداخت انجام داد و ...

❇️بجای خرید ستاره از اپ‌استور، با استفاده از مینی اپ جدید major قادر به دریافت ستاره خواهید بود. در آینده می توان ستاره‌های جمع آوری شده را به TON Coin تبدیل کرد یا از آن‌ها برای خرید اشتراک پریمیوم تلگرام یا موارد دیگر استفاده کرد. پیشنهاد می‌کنیم که حتما این مینی اپ را استارت کنید و با دعوت دوستان ستاره جمع کنید.
با این لینک دعوت ۱۵ تا ستاره رایگان دریافت می‌کنید

با انجام دادن ۳ تسک ربات جمعا 170 استارز رایگان دریافت خواهید کرد

1. کیف پول تون رو بهش وصل کنید
2. ایکس رو چک کنید
3. عضو کانالش بشید

لینک شروع
👇👇
https://t.me/major/start?startapp=84219249



🆔 @Python4all_pro

آموزش کتابخانه pandas در پایتون با مثال



♨️ روی هشتگ ها بزنید و به مطالب بیشتری دسترسی پیدا کنید👇

#علم_داده #کتابخانه
#library #pandas #pdf

کانال آموزش پایتون برای همه ( از صفر تا  100)
👇
🆔 @Python4all_pro

String Methods in Python



🆔 @Python4all_pro

Generate Barcode using Python

#Code

🆔 @Python4all_pro

حل سوالات استخدامی سایت leetcode.com

Problem: No. 19. Remove Nth Node From End of List #medium

Condition:
Given the header of a linked list, remove the nth node from the end of the list and return its header

Solution:

# class ListNode:
#     def init(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        if head.next == None:
            return None
        tmp = head
        size = 0
        
        # find the size of the linked list
        while tmp:
            size += 1
            tmp = tmp.next
        tmp = head
        
        #if we have to remove the first node:
        if n == size: 
            return head.next
        
        for i in range(size-n-1):
            tmp = tmp.next
        tmp.next = tmp.next.next
        return head

Explanation:
Determining the list size:

First we go through the entire list to find its size. This is necessary to determine which node needs to be removed.
The size variable keeps track of the number of nodes in the list.
Removing the first node:

If n is equal to the size of the list, it means the first node should be removed. In this case, we return the next node from the head of the list.
Search for a node before the one you want to delete:

If n is not equal to the size of the list, we need to find the node that comes before the nth node from the end. We do this by moving size-n-1 nodes forward in the list.
Removing a node:

We set the next node for the found node so that it points to the next node after the next one, effectively bypassing the node to be deleted.
Returning a new header:

Returning the head of the list. If the first node was deleted, the new header will be the starting node of the list after the delete.
Time and space complexity:
Time complexity: O(L), where L is the length of the list. We go through the list twice: once to count nodes and again to remove a node.
Space complexity: O(1) because we use a constant amount of extra space independent of the size of the input data.

#Code
#interview #LeetCode


🆔 @Python4all_pro

Python libraries for ML


🆔 @Python4all_pro

Image to pencil sketch using #Python



🆔 @Python4all_pro