Convert Image to text
#code
🆔 @Python4all_pro
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 -
#code
🆔 @Python4all_pro
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
Python Logo Source Code
#code
🆔 @Python4all_pro
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
؛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 :
🖥 GitHub
🟡 Docks
#library
#python
🆔 @Python4all_pro
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
Convert Animated Gif to video
#code
🆔 @Python4all_pro
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:
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
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
#code
🆔 @Python4all_pro
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:
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
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
• Github
🆔 @Python4all_pro
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
🟡 In general, the PIL library has a lot of features and excellent documentation
#library
#python
🆔 @Python4all_pro
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
🖥 GitHub
#library
#python
🆔 @Python4all_pro
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:
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
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
⭐️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
learneverythingai .pdf
3.8 MB
🖥 zarr - Python library for implementing compressed N-dimensional arrays
- pip install zarr
Zarr provides classes and functions for working with N-dimensional arrays, which behave like #NumPy arrays, but the data is divided into chunks and each chunk is compressed. If anyone is familiar with HDF5, Zarr arrays provide similar functionality, but they are more convenient.
Also, unlike HDF5, Zarr has better multithreading support.
🟡 Docks
🖥 GitHub
#library
#python
🆔 @Python4all_pro
- pip install zarr
Zarr provides classes and functions for working with N-dimensional arrays, which behave like #NumPy arrays, but the data is divided into chunks and each chunk is compressed. If anyone is familiar with HDF5, Zarr arrays provide similar functionality, but they are more convenient.
Also, unlike HDF5, Zarr has better multithreading support.
🟡 Docks
🖥 GitHub
#library
#python
🆔 @Python4all_pro