multiples sequence using range

Question

This is basic again and I've searched throughout the website but I can't find anything that is specifically about this.

The problem is to write a function that takes an integer and returns a certain number of multiples for that integer. Then, return a list containing those multiples. For example, if you were to input 2 as your number parameter and 4 as your multiples parameter, your function should return a list containing 2, 4, 6, and 8.

What I have so far:

def multipleList(number, multiples):
  mult = range(number, multiples, number)
  print mult

A test case would be:

print multipleList(2, 9):
>>> [2, 4, 6, 8, 10, 12, 14, 16, 18]

My answer I get:

>>> [2, 4, 6, 8]
None

I know that range has the format (start, stop, step). But how do you tell it to stop after a certain number of times instead of at a certain number?

aldeb · Accepted Answer · 2013-05-30 01:55:44Z

2

try range(number, number*multiples+1, number) instead:

def multipleList(number, multiples):
    return list(range(number, number*multiples+1, number))

print(multipleList(2, 9))

Output:

[2, 4, 6, 8, 10, 12, 14, 16, 18]

edited May 30, 2013 at 1:55

answered May 30, 2013 at 1:48

aldeb

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2 Comments

user2434517 Over a year ago

It still prints "none" at the bottom. Is there a way to take that off?

aldeb Over a year ago

That's because your function returns None. See my example

Ashwini Chaudhary · Accepted Answer · 2013-05-30 01:54:45Z

0

use (number*multiples)+1 as end value in range:

>>> def multipleList(number, multiples):
      mult = range(number, (number*multiples)+1 , number)
      return mult
...     
>>> print multipleList(2,9)
[2, 4, 6, 8, 10, 12, 14, 16, 18]
>>> print multipleList(3, 7)
[3, 6, 9, 12, 15, 18, 21]

The default return value of a function is None, as you're not returning anything in your function so it is going to return None. Instead of printing mult you should return it.

>>> def f():pass
>>> print f()      
None            #default return value

edited May 30, 2013 at 1:54

answered May 30, 2013 at 1:49

Ashwini Chaudhary

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Comments

Mariano Anaya · Accepted Answer · 2013-05-30 02:55:44Z

0

Another approach could be the following one

def mult_list(number, limit):
    multiples = []
    for i in xrange(1, limit+1):
        multiples.append( number * i)
    return multiples

For example: In [8]: mult_list(2,4) Out[8]: [2, 4, 6, 8]

In addition if I do not want parameters an empty list seems consistent with the regular data type:

In [10]: mult_list(2,0)
Out[10]: []

answered May 30, 2013 at 2:55

Mariano Anaya

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