Lambda expression with lists as arguments

Question

I am trying to write an if statement in Python 3:

if n % 2 == 0:
    list.append(2)
elif n % 3 == 0:
    list.append(3)
elif n % 5 == 0:
    list.append(5)

Is there a more space conservative way of doing this? I thought about using lambda expressions, but these bits of code don't seem to work.

if (lambda i: n % i == 0)(i in [2,3,5])
    list.append(i)

and

if (lambda i: n % i == 0)([2,3,5])
    list.append(i)

Is there a way to do this using lambda expressions? Or do I need a different approach? Also, if I use lambda expressions, will I be able to use the value of i for which the condition matches (like appending it to a list)?

I recommend that you first try it with traditional functions and then, once you're comfortable with that, transforming them into lambda one-liners (which isn't actually necessary anyway). — TigerhawkT3
– TigerhawkT3, Commented Apr 30, 2016 at 10:17

Ignacio Vazquez-Abrams · Accepted Answer · 2016-04-30 09:56:47Z

2

>>> n = 33
>>> [i for i in (2, 3, 5) if n % i == 0]
[3]
>>> n = 10
>>> [i for i in (2, 3, 5) if n % i == 0]
[2, 5]

answered Apr 30, 2016 at 9:56

Ignacio Vazquez-Abrams

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Community · Accepted Answer · 2017-05-23 11:44:08Z

1

To get the same result as if/elif/elif statements in the question:

your_list += next(([d] for d in [2, 3, 5] if n % d == 0), [])

Note: only the first divisor is appended to the list. See find first element in a sequence that matches a predicate.

edited May 23, 2017 at 11:44

CommunityBot

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answered Apr 30, 2016 at 10:40

jfs

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Comments

niemmi · Accepted Answer · 2016-04-30 10:26:34Z

0

I can't do it with lambda functions but I think this approach with help u

list = [1,2,3] 
n = 9
[list.append(i) for i in (2,3,5) if n % i == 0]
print list
[1, 2, 3, 3]

and so on ..

edited Apr 30, 2016 at 10:26

niemmi

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answered Apr 30, 2016 at 10:19

Bishoy Atif

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Comments

Mike Müller · Accepted Answer · 2016-04-30 11:01:55Z

0

Like J.F. Sebastian`s answer this gives only the first divisor:

res_list.extend([i for i in (2, 3, 5) if n % i == 0][:1])

The [:1] takes only the first element of the list and gives an empty list if the list is empty (because a slice of an empty list is always an empty list).

Now, extend() adds a one-element list as a single element to your result list or nothing for a zero-element list.

answered Apr 30, 2016 at 11:01

Mike Müller

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