بايثون العربي | Python Arabe 🇵🇸

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قناة عربية لتعلّم Python من الصفر بأسلوب مبسّط وعملي مع أمثلة وتمارين ومشاريع تساعدك تبني مهاراتك البرمجية خطوة بخطوة.

Saudi Arabia6 142 Technologies & Applications9 923...

📈 Analytical overview of Telegram channel بايثون العربي | Python Arabe 🇵🇸

Channel بايثون العربي | Python Arabe 🇵🇸 (@pythonarabe) in the Arabic language segment is an active participant. Currently, the community unites 12 278 subscribers, ranking 9 923 in the Technologies & Applications category and 6 142 in the Saudi Arabia region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 12 278 subscribers.

According to the latest data from 28 August, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -170 over the last 30 days and by -6 over the last 24 hours, overall reach remains high.

Verification status: Not verified
Engagement rate (ER): The average audience engagement rate is 3.54%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
Post reach: On average, each post receives 435 views. Within the first day, a publication typically gains 0 views.
Reactions and interaction: The audience actively supports content: the average number of reactions per post is 4.
Thematic interests: Content is focused on key topics such as ذَات, قَنَاة, إِنتِرنِت, بُرمُجَة, تَجمِيعَة.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“قناة عربية لتعلّم Python من الصفر بأسلوب مبسّط وعملي مع أمثلة وتمارين ومشاريع تساعدك تبني مهاراتك البرمجية خطوة بخطوة.”

Thanks to the high frequency of updates (latest data received on 29 August, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Technologies & Applications category.

12 278

Subscribers

-624 hours

-657 days

-17030 days

435

Post views

No data24 hours

No data48 hours

3.54%

Engagement rate

No data

Posts per day

Ads index

beta

Posts Archive

12 278

#Python tuple = (1, 2 ,3) print(2*tuple)

Anonymous voting

12 278

#Python print (True ** False / True)

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i = 0
while i < 3:
 print (i)
 i = i + 1
 print (i+1)

+ ملاحظه لاتغش وتروح تنفذه بمجرر بايثون اريدك تعرف الحل براسك 👀👍 الحل احد هاذة الامثله : 0 2 1 3 2 4 1 0 2 4 3 5 0 1 2 3 4 5 Error

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#Python a = '1 2' print(a * 2) print(a * 0) print(a * -2) ماهو الناتج ؟ 👀 Error 1 2 1 2 1 1 1 1 2 2 2 2

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استعدو بكون معاكم بتمارين ونساطات برمجيه طيلة رمضان انشالله 👍♥️🌺

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for i in range(3): if i == 2: break else : print(i) else : print("Python")

Anonymous voting

12 278

ساشارككم بفكرة بسيطة ستوضح لك الفرق بينpython. و C++ وغيرها من لغات البرمجه امر طباعة كلمة Sorry مئة مرره C++ : #include <iostream> using namespace std; int main() { int i = 1; while (i <= 100) { cout << "sorry" << endl; i++; } } Python : print( "\nSorry" *100 ) اظن الرساله وصلت 👀

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def check(a): print("Even" if a % 2 == 0 else "Odd") check(20)

Anonymous voting

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اخيرا اكتشفنا سرر شعار بايثون 😂😂😂😂😂😂🤣🤣👍

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جربت اكتب خوارزمية تقوم بطباعة مربع او مستطيل من النجوم بحسب ادخال المستخدم المهم هو الفرق بين الكود ب python و الكود ب C++ الكود python : x= int(input("Enter Your Number : ")) for i in range(x): for j in range(4): print("*" , end=" ") print() الكود C++ : #include <iostream> using namespace std; int main() { int x; cout << "Enter Your Number : "; cin >> x; for (int i = 1; i <= x; i++) { for (int j = 1; j <= 4; j++) { cout << "* "; } cout << endl; } return 0; }

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وردة ببايثون 🌺 الكود : import turtle #DrData t = turtle.Turtle() s = turtle.Screen() s.bgcolor('#262626') t.pencolor('#7C909C') t.speed(100) col =( '#ED7864' , '#6E544F' , '#592F2F' , '#6E383E' ) for n in range(5): for x in range(8): t.speed(x+10) for i in range(2): t.pensize(2) t.circle(80+n*20,90) t.lt(90) t.lt(45) t.pencolor(col[n%4]) s.exitonclick()

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برمجت فيروس كورونا 😂😂😂😂👍 الكود : from turtle import * color('cyan') bgcolor('black') speed(10) hideturtle() b = 0 while b< 200: right(b) forward(b * 3) b = b + 1

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for i in 'abcd' [::-1]: print(i)

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var1 = 1 var2 = 2 var3 = "3" print(var1 + var2 + var3)

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x = 1,2_4,3.2 print(sum(x))

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x = "12.0" print(int(x)*2)

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x = 1#2#3 print(x)

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def fun(): return 55 + int(55.55) print(fun())

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def fun(): return "str "+ 'char' print(fun())

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Repost from اسئله برمجيه Programming Questions

for i in range(5): i = i + 1 print(i)

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