Function doesn't return all results from 'for' loop

Question

I've made a simple function to print out a times table chart depending on the number you decide to run with. The problem I'm having due to my basic understanding of the language is why it only returns the first loop and nothing else.

def timestables(number):
  for a in range(1, number+1):
    b = a*a
    c = a
    return (str(c) + " * " + str(c) + " = " + str(b))

print(timestables(5))

I get the answer ..

1 * 1 = 1

I've tried to rectify this issue by using print instead of return but this ultimately results with a None appearing as well.

def timestables(number):
  for a in range(1, number+1):
    b = a*a
    c = a
    print (str(c) + " * " + str(c) + " = " + str(b))

print(timestables(5))

I get the answer ..

1 * 1 = 1
2 * 2 = 4
3 * 3 = 9
4 * 4 = 16
5 * 5 = 25
None

How can I return all given results from the for loop to avoid a None error?

indentation is significant in python. please fix the indentation in your post — shx2, Apr 19 '13 at 11:10

shx2 · Answer 1 · 2014-01-16T07:42:15.177

16

yield them.

def timestables(number):
  for a in range(1, number+1):
    yield '%s + %s = %s' % (a, a, a*a )

for x in timestables(5):
  print x

This turns your function into a generator function, and you need to iterate over the results, i.e. the result is not a list, but an iterable.

If you need a list, the simplest is to explicitly create one:

res = list(timestables(5))

But again, if you don't, you don't.

IMHO, this is the most pythonic way.

edited Jan 16 '14 at 07:42

answered Apr 19 '13 at 11:11

shx2

61,779
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1

Was just typing this, you beat me to it. – DuncanACoulter Apr 19 '13 at 11:12
...except that using string formatting would make it *more* pythonic (a lot simpler as well) – Volatility Apr 19 '13 at 11:17
@Volatility, yes I agree. except that this question is about the loop, not the strings... – shx2 Apr 19 '13 at 11:19

Volatility · Accepted Answer · 2013-04-19T13:31:15.477

You're returning inside the for loop - and functions stop execution immediately once they hit a return statement.

To work around this, you can use a list to store those values, and then return that list.

def timestables(number):
    lst = []
    for a in range(1, number+1):
        b = a*a
        c = a
        lst.append(str(c) + " * " + str(c) + " = " + str(b))
    return lst

As a side note, you should use string formatting to build the string, like so.

lst.append('{a} * {a} = {b}'.format(a=a, b=a*a))

Now we can get rid of all those intermediate variables (b and c), and we can use a list comprehension instead.

def timestables(number):
    return ['{a} * {a} = {b}'.format(a=a, b=a*a) for a in range(1, number+1)]

If you don't want the function to return a list, but a multi-line string, you can use str.join:

def timestables(number):
    return '\n'.join('{a} * {a} = {b}'.format(a=a, b=a*a) for a in range(1, number+1))

Now we can test the function:

>>> print(timestables(5))
1 * 1 = 1
2 * 2 = 4
3 * 3 = 9
4 * 4 = 16
5 * 5 = 25

Would be better with `yield` or a generator expression than building a list, but +1 for explaining *why* `return` doesn't do what the OP expects. — lvc, Apr 19 '13 at 11:53
Thanks for your detailed response, it looks much cleaner!! I'm trying to figure out how to get this desired result by only using a `print` function however. — Ergo, Apr 19 '13 at 13:21

score 0 · Answer 3 · answered Apr 19 '13 at 11:10

0

You can return an array:

def timestables(number):
    out = []
    for a in range(1, number+1):
        b = a*a
        c = a
        out.append( str(c) + " * " + str(c) + " = " + str(b))

    return out
print(timestables(5))

answered Apr 19 '13 at 11:10

Mariusz Jamro

30,615
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I wish to retain the current format rather than convert it into a list. Thanks for your response though! – Ergo Apr 19 '13 at 12:59
You can always join the list to convert it to string `'\n'.join(out)` – Mariusz Jamro Apr 19 '13 at 14:06

Function doesn't return all results from 'for' loop

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