python converting set to list strange list behavior

Question

I must be missing something because this is strange...

a = ['a', 'b', 'c']
a1 = ['b', 'a']
foo = list( set(a) - set(a1))

** returning **

foo == ['c']
type(foo) == <type 'list'>
foo[0] == 'c'

** now the strange part **

foo = foo.insert(0, 'z')
foo == None

why do list operations like insert, and append cause foo to be None??

the following accomplishes what my top example attempts to but seems ridiculous.

import itertools

a = ['a', 'b', 'c']
a1 = ['b', 'a']

foo = list(set(a) - set(a1))
q = [['z']]
q.append(foo)
q = [i for i in itertools.chain.from_iterable(q)]
q == ['z', 'c']

any insight would be appreciated. thank you.

did you type foo = foo.insert(0, 'z') by accident? that would cause it to be None. — Corley Brigman
– Corley Brigman, Commented Feb 11, 2014 at 20:16
Corley Brigman is correct. I was essentially assigning my list variable the return value of .insert(), which as NPE pointed out is in fact None.. — Ryder Brooks
– Ryder Brooks, Commented Feb 11, 2014 at 20:58
I edited my question to reflect Corley Brigman's comment. Changing foo.insert(0, 'z') to foo = foo.insert(0, 'z'). Thanks everyone — Ryder Brooks
– Ryder Brooks, Commented Feb 11, 2014 at 21:00

NPE · Accepted Answer · 2014-02-11 19:51:33Z

10

foo.insert() returns None, but does change foo in the way you'd expect:

>>> foo = ['c']
>>> foo.insert(0, 'z')
>>> foo
['z', 'c']

If you wish to assign the result to a different variable, here is one way to do it:

>>> foo = ['c']
>>> bar = ['z'] + foo
>>> bar
['z', 'c']

answered Feb 11, 2014 at 19:51

NPE

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2 Comments

salezica Over a year ago

+1, are you perhaps doing print foo.insert(0, 'z')?

Ryder Brooks Over a year ago

the foo.inster(0, 'z') was being used as the return statement in a function. So it was literally ... return foo.insert(0, 'z'). Which explains why I was getting None.. And proves how painfully oblivious I can be sometimes.. Thank you for pointing this out to me.

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