₦airaland Forum

gbolly1151:

from kivy.app import App
from kivy.uix.button import Button

class TestApp(App):
    def build(self):
        # display a button with the text : Hello QPython 
        return Button(text='Hello QPython')

TestApp().run()

Note that there is a documentation on how you can allow it to work on Andriod device

Go to google and search for kviy tutorial pdf or video to see how to compile your kivy python file to work with android,there are 2 or 3 ways to do that,you can pick the anyone you can easily grab

Bro, can you recommend any good kivy book?

Brukx:
Bro, can you recommend any good kivy book?

I haven't try building mobile app before so don't know the best book out there but you can search for kivy tutorial on Google the first index still look ok

HOW TO USE all() KEYWORD

all() in python is a keyword that take iterable object
as aguement and return true only when all the elements in
this iterable are true '''

#Example 1       
string=[True,False,True]
print(all(string))

#output: False
string=[True,True,True]
print(all(string))

#output:True

'''
Example 2
let say you want to validate user input from unknown source
'''
required_field=['name','age','score']
user_value={'name':'peter','age':20,'score':300}

#native aproach
#note that this only check for key and not value itself
#but you can cook it to your taste

def is_valid(require,value):
    for field in required_field:
        if field not in user_value.keys():
              return False
    return True
     
print(is_valid(required_field,user_value))

#output: True
#if you change the key,it will be false on run

#pythonic approch

print(all([field in user_value.keys() for field in required_field]))

#output:True
#if you change the key in user_field it will turn False on run

WHAT IS if __name__=='__main__'?

#short explanation

if __name__=='__main__':
    print('hello')

then, you are telling the computer that if this python file is run by user or robot then
print('hello')


#full explanation
Anytime you open python file,you are actually loading the
file into the memory of the computer (i.e RAM). The name
that will be given to that file at that point wont be the
name of the file but a new name that is called
'__main__' (i.e __name__='__main__') and it is stored together with other data in
memory.

To check that, open any python file and print global data
using global keyword

print(globals())

output
...
{'__builtins__': <module 'builtins' (built-in)>, '__file__': '/storage/sdcard0/qpython/scripts3/.last_tmp.py',
 '__package__': None, '__cached__': None, 
 '__name__': '__main__', '__doc__': None}

from that output above we can see that __name__= '__main__'

so at runtime the name of any python file is '__main__'.

This name is often use proffessionally in python for testing or running the
functionality of a script at runtime.

you might be wondering why we have underscore,This is because
we dont want it to be in conflict when we use 'name' as variable name in our program

now,if you use

if __name__=='__main__':
    print('hello')

then, you are telling the computer that if this python file is run by user or robot then
print('hello')

you can acqually change the variable of __name__

e.g

__name__='changed'

#now let see the changed name of the file at runtime
print(globals())

#output
{'__builtins__': <module 'builtins' (built-in)>, '__file__': '/storage/sdcard0/qpython/scripts3/.last_tmp.py',
 '__package__': None, '__cached__': None,
  '__name__': 'changed', '__doc__': None}

we can see that our __name__ has been assign another variable

gbolly1151:
WHAT IS if __name__=='__main__'?

#short explanation

if __name__=='__main__':
    print('hello')

then, you are telling the computer that if this python file is run by user or robot then
print('hello')


#full explanation
Anytime you open python file,you are actually loading the
file into the memory of the computer (i.e RAM). The name
that will be given to that file at that point wont be the
name of the file but a new name that is called
'__main__' (i.e __name__='__main__') and it is stored together with other data in
memory.

To check that, open any python file and print global data
using global keyword

print(globals())

output
...
{'__builtins__': <module 'builtins' (built-in)>, '__file__': '/storage/sdcard0/qpython/scripts3/.last_tmp.py',
 '__package__': None, '__cached__': None, 
 '__name__': '__main__', '__doc__': None}

from that output above we can see that __name__= '__main__'

so at runtime the name of any python file is '__main__'.

This name is often use proffessionally in python for testing or running the
functionality of a script at runtime.

you might be wondering why we have underscore,This is because
we dont want it to be in conflict when we use 'name' as variable name in our program

now,if you use

if __name__=='__main__':
    print('hello')

then, you are telling the computer that if this python file is run by user or robot then
print('hello')

you can acqually change the variable of __name__

e.g

__name__='changed'

#now let see the changed name of the file at runtime
print(globals())

#output
{'__builtins__': <module 'builtins' (built-in)>, '__file__': '/storage/sdcard0/qpython/scripts3/.last_tmp.py',
 '__package__': None, '__cached__': None,
  '__name__': 'changed', '__doc__': None}

we can see that our __name__ has been assign another variable

I finally learnt OOP. I have completed a project. Its coordinate of a line with slope, distance and to determine if a line is vertical or horizontal and if it is perpendicular or parallel to another.

I also completed a Time project to display the julian day, the day of the week, of that date and gregorian date. Its interesting so far. I couldn't display the calendar without the normal calendar module. I am trying to create my own calendar module but I didnt get that.

I'm also currently working on Polygon.

Predstan:
I finally learnt OOP. I have completed a project. Its coordinate of a line with slope, distance and to determine if a line is vertical or horizontal and if it is perpendicular or parallel to another.

I also completed a Time project to display the julian day, the day of the week, of that date and gregorian date. Its interesting so far. I couldn't display the calendar without the normal calendar module. I am trying to create my own calendar module but I didnt get that.

I'm also currently working on Polygon.

Nice one bro...keep it up

Hey guys, so, I'm trying my hands on this exercise:

The Collatz Sequence Write a function named collatz() that has one parameter named number. If number is even, then collatz() should print number // 2 and return this value. If number is odd, then collatz() should print and return 3 * number + 1. Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1. (Amazingly enough, this sequence actually works for any integer—sooner or later, using this sequence, you’ll arrive at 1! Even mathematicians aren’t sure why. Your program is exploring what’s called the Collatz sequence, sometimes called “the simplest impossible math problem.”) Remember to convert the return value from input() to an integer with the int() function; otherwise, it will be a string value. Hint: An integer number is even if number % 2 == 0, and it’s odd if number % 2 == 1.

Here's my code:

def collatz(number):
	if number % 2 == 0:
		return print(number // 2)
	elif number % 2 == 1:
		return print(3 * number + 1)

number =int(input())
while number != 1:
	collatz(number)

What am I not doing right?

iCode2:
Hey guys, so, I'm trying my hands on this exercise:

The Collatz Sequence Write a function named collatz() that has one parameter named number. If number is even, then collatz() should print number // 2 and return this value. If number is odd, then collatz() should print and return 3 * number + 1. Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1. (Amazingly enough, this sequence actually works for any integer—sooner or later, using this sequence, you’ll arrive at 1! Even mathematicians aren’t sure why. Your program is exploring what’s called the Collatz sequence, sometimes called “the simplest impossible math problem.”) Remember to convert the return value from input() to an integer with the int() function; otherwise, it will be a string value. Hint: An integer number is even if number % 2 == 0, and it’s odd if number % 2 == 1.

Here's my code:

def collatz(number):
	if number % 2 == 0:
		return print(number // 2)
	elif number % 2 == 1:
		return print(3 * number + 1)

number =int(input())
while number != 1:
	collatz(number)

What am I not doing right?

First you need to return the number from the Collatz function as an integer not a print function... use: return number//2.

Then you have to check that the return value from the Collatz is not equal to 1 if not, user keeps inputting number and the Collatz function is called on the number from user.

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1

number =int(input("Enter Number" ))
Num = collatz(number)
while Num != 1:
	continue 

Using Recursion:

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1
def main():
    
    number =int(input("Enter Number" ))
    Num = collatz(number)
    while Num != 1:
        main()
    return print ("Callatz on", number, "is", Num)
    
main()

Predstan:
First you need to return the number from the Collatz function as an integer not a print function... use: return number//2.

Then you have to check that the return value from the Collatz is not equal to 1 if not, user keeps inputting number and the Collatz function is called on the number from user.

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1

number =int(input("Enter Number" ))
Num = collatz(number)
while Num != 1:
	continue 

Using Recursion:

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1
def main():
    
    number =int(input("Enter Number" ))
    Num = collatz(number)
    while Num != 1:
        main()
    return print ("Callatz on", number, "is", Num)
    
main()

That code for recursion doesn't look like one and even the first method will still end up looping and wont end.
Am thinking should look like this

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1

number =int(input("Enter Number" ))
Num = number
while Num != 1:
     Num=collatz(Num)
     print(Num)

Or there is something i don't get?

gbolly1151:
That code for recursion doesn't look like one and even the first method will still end up looping and wont end.
Am thinking should look like this

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1

number =int(input("Enter Number" ))
Num = number
while Num != 1:
     Num=collatz(Num)
     print(Num)

Ok, But on the other hand, look at this bolded part of the question. I was thinking, we should be checking that return value is not equal to 1 and not our input value

iCode2:
Hey guys, so, I'm trying my hands on this exercise:

The Collatz Sequence Write a function named collatz() that has one parameter named number. If number is even, then collatz() should print number // 2 and return this value. If number is odd, then collatz() should print and return 3 * number + 1. Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1. (Amazingly enough, this sequence actually works for any integer—sooner or later, using this sequence, you’ll arrive at 1! Even mathematicians aren’t sure why. Your program is exploring what’s called the Collatz sequence, sometimes called “the simplest impossible math problem.”) Remember to convert the return value from input() to an integer with the int() function; otherwise, it will be a string value. Hint: An integer number is even if number % 2 == 0, and it’s odd if number % 2 == 1.

Predstan:
Ok, But on the other hand, look at this bolded part of the question. I was thinking, we should be checking that return value is not equal to 1 and not our input value

Have you tried to run the code and it works as you expected?

Predstan:
First you need to return the number from the Collatz function as an integer not a print function... use: return number//2.

Then you have to check that the return value from the Collatz is not equal to 1 if not, user keeps inputting number and the Collatz function is called on the number from user.

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1

number =int(input("Enter Number" ))
Num = collatz(number)
while Num != 1:
	continue 

Using Recursion:

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1
def main():
    
    number =int(input("Enter Number" ))
    Num = collatz(number)
    while Num != 1:
        main()
    return print ("Callatz on", number, "is", Num)
    
main()

Thanks but it didn't give the right output.

gbolly1151:
That code for recursion doesn't look like one and even the first method will still end up looping and wont end.
Am thinking should look like this

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1

number =int(input("Enter Number" ))
Num = number
while Num != 1:
     Num=collatz(Num)
     print(Num)

I got the correct output from this. Thanks

iCode2:
Thanks but it didn't give the right output.

Oh Sorry, I hope I haven't confused you

Predstan:
Oh Sorry, I hope I haven't confused you

Nope..

Can anyone recommend good python books with a lot of exercises? I just started out with automate the boring stuff.

iCode2:
Can anyone recommend good python books with a lot of exercises? I just started out with automate the boring stuff.

Download think python

gbolly1151:
Download think python

Alright, thanks.

gbolly1151:
That code for recursion doesn't look like one and even the first method will still end up looping and wont end.
Am thinking should look like this

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1

number =int(input("Enter Number" ))
Num = number
while Num != 1:
     Num=collatz(Num)
     print(Num)

How does this code repeatedly prompt the user for input when the input line isn't in the while loop block? Or is it in the collatz function? Because it doesn't look like you put it there. If it's in the function, then indent it a bit to make it readable and clearer.

DanRay1:
How does this code repeatedly prompt the user for input when the input line isn't in the while loop block? Or is it in the collatz function? Because it doesn't look like you put it there. If it's in the function, then indent it a bit to make it readable and clearer.

We are only required to input a number then it will print out all the collatz number still 1

DanRay1:
How does this code repeatedly prompt the user for input when the input line isn't in the while loop block? Or is it in the collatz function? Because it doesn't look like you put it there. If it's in the function, then indent it a bit to make it readable and clearer.

def collatz(number):
	if number % 2 == 0:
		return number // 2
	elif number % 2 == 1:
		return 3 * number + 1


If __name__=='__main__':
    number =int(input("Enter Number" ))
    Num = number
    while Num != 1:
        Num=collatz(Num)
        print(Num)

Hope it is readable?

gbolly1151:
We are only required to input a number then it will print out all the collatz number still 1

Okay I get it now.

How to create a simple singleton without decorator

Decorator is one of the popular way of creating singleton in python but we wont be using decorator here. before i move to code let me define singleton,it is a way of creating a single object of a class or function. In a program where you only need a single instance of a class you can create a Singleton.

class Test:
    def __init__(self):
        self.num=5
    def fmethod (self,mynum):
        mynum *=2
        return 'fmethod mynum is {0}'.format(mynum)

def singleton(func,*args,**kwargs):
   obj=globals()['__cached__']
   if obj is None:
       globals()['__cached__']=func(*args,**kwargs)
       obj=globals()['__cached__']
   return obj
   
Testobj1=singleton(Test)
Testobj2=singleton(Test)

print(Testobj1 == Testobj2) 
# output : True

Testobj1.num=10
print(Testobj2.num) 
#output : 10

gbolly1151:

#How to format your class output using __str__ method
class myclass:
    def __init__(self):
        pass
    
    def __str__(self):
        return 'my name is class'
        
print(myclass())

Output:
my name is class


#without __str__ method

class myclass:
    def __init__(self):
        pass
        
print(myclass())

Output:
<__main__.myclass object at oxbchjlidk>

I'd prefer the dunder method "repr"....(__repr__)

without explicitly calling the print function on the object, repr returns the object description....

I.e
instead of print(obj) running "obj" alone works better

REMINDER: Range has step argument

range(start,stop,step)

Note that start and step are optional

How to use step in range to print odd and even

for even in range(2,10,2):
    print(even)

#output : 2,4,6,8

for odd in range(3,10,2):
    print(odd)

#output: 3,5,7,9

HOW TO SWAP AN OBJECT

a,b = 4,7
print(a,b) #output : 4,7
a,b=b,a
print(a,b) #output: 7,4

Hi gbolly1151, you seem to be very good with Python. When did you start learning? And are you self-taught?

iCode2:
Hi gbolly1151, you seem to be very good with Python. When did you start learning? And are you self-taught?

I started learning python few years ago and am a self-taught

HOW TO USE LAMBDA

The format of lambda is as below

[lambda keyword] [arguments[seperated by comma]] [semicolon] [logic]

e.g
# with single argument
lambda x: x*x

#multiple argument
lambda x,y: x*y

now assign that with variable name

mul=lambda x,y:x*y

x=5
y=6
print(mul(x,y))

#output 30

KEY DIFFERENCE BETWEEN is AND ==

is operator is use to check if two object have the same id number

== is use to check if two objects have the same value

#e.g
a=[1,2,3]
b=[1,2,3]
print(a is b) #output false
'''
The above print expression will return false because a and b are two objects with
 different id number,let check their id
'''

print(id(a))
print(id(b))

'''
id output is different on different devices,my device print

3066478144
3065959488


'''Since the value are the same,when you use == the output will be True'''

print(a == b) #output True

PLEASE AVIOD THIS MISTAKE IN YOUR CODE

a=1
if a ==3 or 4 or 6:
    print('yes')
else:
   print('no')

#The answer will always be Yes why?


Let me break it down here

a=1
print(a == 3) # False
print(False or 4) # 4 (i.e True)
print(4 or 6) # 4 (i.e True)

now 

if a == 3 or 4 or 6: # this is equivalent to

 if False or True or True: 
    print('yes')
else:
    print('No')

The condition is True so print('yes') will be executed

#solution to the problem
a=1
if a in [3,4,6]:
    print('yes')
else:
   print('no')