Automate the Boring Stuff with Python - The Collatz sequence project

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I am new to programming and started learning with Automate the Boring Stuff with Python. I just completed chapter 3's "Collatz sequence" project.

I was wondering what would you do differently and why? Trying to learn how to think like a programmer and become more efficient.

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = (number//2)
 print(nextNumber)
 else:
 nextNumber = (3*number+1)
 print(nextNumber)

print('Type in an integer.')
integer=input()

if integer.isdigit() ==True:
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

while integer.isdigit() != True :
 print('Please try again.')
 integer=input()
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

print('Thank you!')

python collatz-sequence

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edited 49 mins ago

Mast

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asked 6 hours ago

Alex

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Check out our Code of Conduct.

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up vote
4
down vote

favorite

I am new to programming and started learning with Automate the Boring Stuff with Python. I just completed chapter 3's "Collatz sequence" project.

I was wondering what would you do differently and why? Trying to learn how to think like a programmer and become more efficient.

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = (number//2)
 print(nextNumber)
 else:
 nextNumber = (3*number+1)
 print(nextNumber)

print('Type in an integer.')
integer=input()

if integer.isdigit() ==True:
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

while integer.isdigit() != True :
 print('Please try again.')
 integer=input()
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

print('Thank you!')

python collatz-sequence

share|improve this question

edited 49 mins ago

Mast

7,43063584

asked 6 hours ago

Alex

211

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

I am new to programming and started learning with Automate the Boring Stuff with Python. I just completed chapter 3's "Collatz sequence" project.

I was wondering what would you do differently and why? Trying to learn how to think like a programmer and become more efficient.

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = (number//2)
 print(nextNumber)
 else:
 nextNumber = (3*number+1)
 print(nextNumber)

print('Type in an integer.')
integer=input()

if integer.isdigit() ==True:
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

while integer.isdigit() != True :
 print('Please try again.')
 integer=input()
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

print('Thank you!')

python collatz-sequence

share|improve this question

edited 49 mins ago

Mast

7,43063584

asked 6 hours ago

Alex

211

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I am new to programming and started learning with Automate the Boring Stuff with Python. I just completed chapter 3's "Collatz sequence" project.

I was wondering what would you do differently and why? Trying to learn how to think like a programmer and become more efficient.

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = (number//2)
 print(nextNumber)
 else:
 nextNumber = (3*number+1)
 print(nextNumber)

print('Type in an integer.')
integer=input()

if integer.isdigit() ==True:
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

while integer.isdigit() != True :
 print('Please try again.')
 integer=input()
 collatz(int(integer))
 while nextNumber !=1:
 collatz(nextNumber)

print('Thank you!')

python collatz-sequence

python collatz-sequence

share|improve this question

edited 49 mins ago

Mast

7,43063584

asked 6 hours ago

Alex

211

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 49 mins ago

Mast

7,43063584

asked 6 hours ago

Alex

211

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

share|improve this question

edited 49 mins ago

Mast

7,43063584

edited 49 mins ago

Mast

7,43063584

edited 49 mins ago

Mast

7,43063584

7,43063584

asked 6 hours ago

Alex

211

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 6 hours ago

Alex

211

asked 6 hours ago

Alex

211

211

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Alex is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote

One thing you’ll want to learn is not to repeat yourself, when coding.

Consider:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 print(nextNumber)
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You have two identical print statements at the end of both the if and the else clauses. These can be combined and moved out of the if-else:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You are printing in your collatz generator function. If you want the longest sequence starting under 10000000, you’re going to do a lot of printing! Instead, move the printing responsibility to the caller. They know whether they want to print it, or just find the length.

Don’t use global to return one value from a function. Just return the value,

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

and let the caller assign it to whatever variable they want.

nextNumber = collatz( int(number) )
print(nextNumber)
while nextNumber != 1:
 nextNumber = collatz(nextNumber)
 print(nextNumber)

Reordering, to remove one of the calls to collatz:

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

Python’s exception handling is exceptional! Learn it. Rely on it:

number = None
while number is None:
 try:
 number = int( input("Enter a number") )
 except ValueError:
 print("Try again")

 # ... print collatz sequence here

share|improve this answer

edited 46 mins ago

Mast

7,43063584

answered 3 hours ago

AJNeufeld

2,764313

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

One thing you’ll want to learn is not to repeat yourself, when coding.

Consider:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 print(nextNumber)
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You have two identical print statements at the end of both the if and the else clauses. These can be combined and moved out of the if-else:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You are printing in your collatz generator function. If you want the longest sequence starting under 10000000, you’re going to do a lot of printing! Instead, move the printing responsibility to the caller. They know whether they want to print it, or just find the length.

Don’t use global to return one value from a function. Just return the value,

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

and let the caller assign it to whatever variable they want.

nextNumber = collatz( int(number) )
print(nextNumber)
while nextNumber != 1:
 nextNumber = collatz(nextNumber)
 print(nextNumber)

Reordering, to remove one of the calls to collatz:

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

Python’s exception handling is exceptional! Learn it. Rely on it:

number = None
while number is None:
 try:
 number = int( input("Enter a number") )
 except ValueError:
 print("Try again")

 # ... print collatz sequence here

share|improve this answer

edited 46 mins ago

Mast

7,43063584

answered 3 hours ago

AJNeufeld

2,764313

add a comment | 

up vote
3
down vote

One thing you’ll want to learn is not to repeat yourself, when coding.

Consider:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 print(nextNumber)
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You have two identical print statements at the end of both the if and the else clauses. These can be combined and moved out of the if-else:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You are printing in your collatz generator function. If you want the longest sequence starting under 10000000, you’re going to do a lot of printing! Instead, move the printing responsibility to the caller. They know whether they want to print it, or just find the length.

Don’t use global to return one value from a function. Just return the value,

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

and let the caller assign it to whatever variable they want.

nextNumber = collatz( int(number) )
print(nextNumber)
while nextNumber != 1:
 nextNumber = collatz(nextNumber)
 print(nextNumber)

Reordering, to remove one of the calls to collatz:

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

Python’s exception handling is exceptional! Learn it. Rely on it:

number = None
while number is None:
 try:
 number = int( input("Enter a number") )
 except ValueError:
 print("Try again")

 # ... print collatz sequence here

share|improve this answer

edited 46 mins ago

Mast

7,43063584

answered 3 hours ago

AJNeufeld

2,764313

add a comment | 

up vote
3
down vote

up vote
3
down vote

One thing you’ll want to learn is not to repeat yourself, when coding.

Consider:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 print(nextNumber)
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You have two identical print statements at the end of both the if and the else clauses. These can be combined and moved out of the if-else:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You are printing in your collatz generator function. If you want the longest sequence starting under 10000000, you’re going to do a lot of printing! Instead, move the printing responsibility to the caller. They know whether they want to print it, or just find the length.

Don’t use global to return one value from a function. Just return the value,

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

and let the caller assign it to whatever variable they want.

nextNumber = collatz( int(number) )
print(nextNumber)
while nextNumber != 1:
 nextNumber = collatz(nextNumber)
 print(nextNumber)

Reordering, to remove one of the calls to collatz:

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

Python’s exception handling is exceptional! Learn it. Rely on it:

number = None
while number is None:
 try:
 number = int( input("Enter a number") )
 except ValueError:
 print("Try again")

 # ... print collatz sequence here

share|improve this answer

edited 46 mins ago

Mast

7,43063584

answered 3 hours ago

AJNeufeld

2,764313

One thing you’ll want to learn is not to repeat yourself, when coding.

Consider:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 print(nextNumber)
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You have two identical print statements at the end of both the if and the else clauses. These can be combined and moved out of the if-else:

def collatz(number):
 global nextNumber
 if number % 2 == 0:
 nextNumber = number//2
 else:
 nextNumber = 3*number+1
 print(nextNumber)

You are printing in your collatz generator function. If you want the longest sequence starting under 10000000, you’re going to do a lot of printing! Instead, move the printing responsibility to the caller. They know whether they want to print it, or just find the length.

Don’t use global to return one value from a function. Just return the value,

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

and let the caller assign it to whatever variable they want.

nextNumber = collatz( int(number) )
print(nextNumber)
while nextNumber != 1:
 nextNumber = collatz(nextNumber)
 print(nextNumber)

Reordering, to remove one of the calls to collatz:

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

Python’s exception handling is exceptional! Learn it. Rely on it:

number = None
while number is None:
 try:
 number = int( input("Enter a number") )
 except ValueError:
 print("Try again")

 # ... print collatz sequence here

share|improve this answer

edited 46 mins ago

Mast

7,43063584

answered 3 hours ago

AJNeufeld

2,764313

share|improve this answer

share|improve this answer

edited 46 mins ago

Mast

7,43063584

edited 46 mins ago

Mast

7,43063584

edited 46 mins ago

Mast

7,43063584

7,43063584

answered 3 hours ago

AJNeufeld

2,764313

answered 3 hours ago

AJNeufeld

2,764313

answered 3 hours ago

AJNeufeld

2,764313

2,764313

add a comment | 

add a comment | 

Alex is a new contributor. Be nice, and check out our Code of Conduct.

 

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