#!/usr/bin/python
# -*- coding: utf-8 -*-
# Les nombre multiples de n compris entre 1 et 60
def multiple(n):
    for i in range(1,61):
        r=i//n
        if i%n==0:
            print(i,r)
    return(r)
print(multiple(15))

(Mar-24-2020, 10:30 PM)pythonuser1 Wrote:

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Hello,

#!/usr/bin/python
# -*- coding: utf-8 -*-
# Les nombre multiples de n compris entre 1 et 60
def multiple(n):
    for i in range(1,61):
        r=i//n
        if i%n==0:
            print(i,r)
    return(r)
print(multiple(15))

The output of the script is like this :
./multiple_nombre_v3.py
(15, 1)
(30, 2)
(45, 3)
(60, 4)
4

My script is giving the multiple of a number (here 15 for example) and the script is working.
I want to optimize it. I have these two lines print(i,r) and return(r ). Is there a way I can get rid of one of these two lines and yet have the correct output giving only the multiple of a number ? I am asking this because the return(r ) sends "4" and in the 'if condition' I did not find an another way besides "print(i,r)" to show the multiples of a number.

def multiple(n):
    retDict={}
    for i in range(1,61):
        r=i//n
        if i%n==0:
            retDict.update({i:r})
            #print({i,r})
    return(retDict)

print(multiple(50))

Output:
{50: 1}

multiple(15)

Output:
{15: 1, 30: 2, 45: 3, 60: 4}

(Mar-24-2020, 10:30 PM)pythonuser1 Wrote: I want to optimize it. I have these two lines print(i,r) and return(r ). Is there a way I can get rid of one of these two lines and yet have the correct output giving only the multiple of a number ? I am asking this because the return(r ) sends "4" and in the 'if condition' I did not find an another way besides "print(i,r)" to show the multiples of a number.

• You should use Python 3. Currently you're using Python 2.7, which is outdated since January 2020.
  
• Encoding comment is not needed for Python 3. All source files are encoded with utf8
  
• Instead of comments, you can use docstrings
  
• You should change the shebang to #!/usr/bin/env python3
  
• Your function has side effects. Side effect means, that your function modifies the environment (print to stdout) and return a value. A Pure function takes arguments, does the work and return the result. Nothing else happens beside. Easy fix: remove the print from this function
  
• Use better names for objects. n, i, r are bad names.
  
• Give the user the flexibility to call your function with different range for start and end.
  
• To get rid of return, you could use a generator.
  

#!/usr/bin/env python3


# value is a positional argument
# start and stop are keyword-only-arguments, the * does this
def multiple_of(value, *, start, stop):
    """
    Les nombre multiples de value compris entre start et stop

    start is inclusive
    stop is inclusive
    """
    for factor in range(start, stop + 1):
        if factor % value == 0:
            yield factor, factor // value  # <- this are the values for each step
                                           #    which goes to the consumer


multi_gen = multiple_of(15, start=1, stop=60)
# multi_gen is a generator, lazy evaluation, no values yet
my_results = list(multi_gen) # list consumes the generator
print(my_results)
# a generator can be consumed only once and then it's exhausted
# if you need the elements in a different datatype as a list:

# multi_gen = multiple_of(15, start=1, stop=60)
# my_results = tuple(multi_gen)

# You can also iterate over the generator:
for result in multiple_of(15, start=1, stop=60):
    print(result)


# Bonus
# You can create also a dict from the results.
# my_results_as_dict = dict(multiple_of(15, start=1, stop=60))

import math

def multiple_of(value, *, start=1, stop=60):
    start = math.ceil(start / value) * value
    while start <= stop:
        yield start, start // value
        start += value

multiples = [v for v, m in multiple_of(1000000000, stop=100000000000)]
print(len(multiples))

Output:
100

(Mar-24-2020, 10:30 PM)pythonuser1 Wrote: I want to optimize it. I have these two lines print(i,r) and return(r ). Is there a way I can get rid of one of these two lines and yet have the correct output giving only the multiple of a number ?

# -*- coding: utf-8 -*-
# Les nombre multiples de n compris entre 1 et 60
def multiple(n):
    return int(60/n)

print(multiple(15))

Output:
4