Project Euler #1: Sum of Multiples of 3 and 5 below 1000 Announcing the arrival of Valued...
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Project Euler #1: Sum of Multiples of 3 and 5 below 1000
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Project Euler Problem 12 in PythonFunction to return the sum of multiples within a rangeProject Euler 78: Coin Partitions in PythonProject Euler #14: Longest Collatz sequenceProject Euler #15 in PythonProject Euler #4 - Largest Palindrome Project - PythonProject Euler problem #1 in Python 3Sum of multiples of 3 or 5 using functional programmingProject Euler 001 in C++Project Euler Problem 37 in Python
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
edited 4 mins ago
AJNeufeld
6,9641723
asked 52 mins ago
DjVasuDjVasu
11
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
edited 4 mins ago
AJNeufeld
6,9641723
asked 52 mins ago
DjVasuDjVasu
11
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
edited 4 mins ago
AJNeufeld
6,9641723
asked 52 mins ago
DjVasuDjVasu
11
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
def prob_1():
sum_mult=[] #Create an empty list which will take sum of multiples of 3 and 5
check_sum=0
for i in range(1,1000): #Take numbers till 1000
#if(i)
if( (i%3)==0 or (i%35)==0 ): #divisor condition
sum_mult.append(i)
return sum(sum_mult) #return sum of list
I am just starting out my journey as a programmer, here is my code and I would love to see any critical feedback and other alternative solutions maybe using some clever hack of using lambda function's etc.
python performance python-3.x programming-challenge
python performance python-3.x programming-challenge
edited 4 mins ago
AJNeufeld
6,9641723
asked 52 mins ago
DjVasuDjVasu
11
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 mins ago
AJNeufeld
6,9641723
asked 52 mins ago
DjVasuDjVasu
11
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 mins ago
AJNeufeld
6,9641723
edited 4 mins ago
AJNeufeld
6,9641723
edited 4 mins ago
AJNeufeld
6,9641723
6,9641723
asked 52 mins ago
DjVasuDjVasu
11
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 52 mins ago
DjVasuDjVasu
11
asked 52 mins ago
DjVasuDjVasu
11
11
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
DjVasu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
DjVasu 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
$begingroup$
I expect you made a typo. You don't want (i%35)==0, you want (i%5)==0.
The PEP-8 style guide for Python requires 1 space before and after operators, and after commas. Use PyLint or equivalent tool to ensure you follow all of the PEP-8 guidelines.
check_sum is unused, and can be omitted.
The brackets around the if( ... ): condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings""" to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...) function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I expect you made a typo. You don't want (i%35)==0, you want (i%5)==0.
The PEP-8 style guide for Python requires 1 space before and after operators, and after commas. Use PyLint or equivalent tool to ensure you follow all of the PEP-8 guidelines.
check_sum is unused, and can be omitted.
The brackets around the if( ... ): condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings""" to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...) function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
$endgroup$
add a comment |
$begingroup$
I expect you made a typo. You don't want (i%35)==0, you want (i%5)==0.
The PEP-8 style guide for Python requires 1 space before and after operators, and after commas. Use PyLint or equivalent tool to ensure you follow all of the PEP-8 guidelines.
check_sum is unused, and can be omitted.
The brackets around the if( ... ): condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings""" to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...) function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
$endgroup$
add a comment |
$begingroup$
I expect you made a typo. You don't want (i%35)==0, you want (i%5)==0.
The PEP-8 style guide for Python requires 1 space before and after operators, and after commas. Use PyLint or equivalent tool to ensure you follow all of the PEP-8 guidelines.
check_sum is unused, and can be omitted.
The brackets around the if( ... ): condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings""" to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...) function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
$endgroup$
I expect you made a typo. You don't want (i%35)==0, you want (i%5)==0.
The PEP-8 style guide for Python requires 1 space before and after operators, and after commas. Use PyLint or equivalent tool to ensure you follow all of the PEP-8 guidelines.
check_sum is unused, and can be omitted.
The brackets around the if( ... ): condition are unnecessary. This is Python, not C, C++ or Java:
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
There is no need to create a list just to add up all the numbers after the fact. You are only using each value once, so you could simply add the numbers up as you find them:
def prob_1():
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You should add """doc_strings""" to your functions:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
sum_of_multiples = 0
for i in range(1, 1000): # Take numbers up to but not including 1000
if (i % 3) == 0 or (i % 5) == 0: #divisor condition
sum_of_multiples += i
return sum_of_multiples
You can use list comprehension and the sum(...) function to compute the result, without ever creating the list in memory:
def prob_1():
"""
Compute the sum of all the multiples of 3 or 5 below 1000.
Returns:
The sum of the multiples of 3 or 5, below 1000.
"""
return sum(i for i in range(1000) if i % 3 == 0 or i % 5 == 0)
You can also solve this problem by hand with a pen, a sheet of paper, a calculator and about 1 minute of your time. A program is entirely unnecessary.
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
edited 1 min ago
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
answered 7 mins ago
AJNeufeldAJNeufeld
6,9641723
6,9641723
add a comment |
add a comment |
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
DjVasu is a new contributor. Be nice, and check out our Code of Conduct.
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