Python: What does the use of [] mean here?

Question

What is the difference in these two statements in python?

var = foo.bar

and

var = [foo.bar]

I think it is making var into a list containing foo.bar but I am unsure. Also if this is the behavior and foo.bar is already a list what do you get in each case?

For example: if foo.bar = [1, 2] would I get this?

var = foo.bar #[1, 2]

and

var = [foo.bar] #[[1,2]] where [1,2] is the first element in a multidimensional list

Answer 1

[] is an empty list.

[foo.bar] is creating a new list ([]) with foo.bar as the first item in the list, which can then be referenced by its index:

var = [foo.bar]
var[0] == foo.bar # returns True 

So your guess that your assignment of foo.bar = [1,2] is exactly right.

If you haven't already, I recommend playing around with this kind of thing in the Python interactive interpreter. It makes it pretty easy:

>>> []
[]
>>> foobar = [1,2]
>>> foobar
[1, 2]
>>> [foobar]
[[1, 2]]

Answer 2

I think it is making var into a list containing foo.bar but I am unsure. Also if this is the behavior and foo.bar is already a list what do you get in each case?

• Yes, it creates a new list.

• If foo.bar is already a list, it will simply become a list, containing one list.

  h[1] >>> l = [1, 2]
  h[1] >>> [l]
  [[1, 2]]
  h[3] >>> l[l][0]
  [1, 2]
  

Answer 3

Yes, it's making a list containing one element, foo.bar.

If foo.bar is [1,2], you indeed get [[1,2]].

For instance,

>> a=[]
>> a.append([1,2])
>> a[0] 
[1,2]
>> b=[[1,2]]
>> b[0]
[1,2]

To elaborate a bit more on that exact example,

>> class Foos:
>>   bar=[1,2]
>> foo=Foos()
>> foo.bar
[1,2]
>> a=[foo.bar]
>> a
[[1,2]]
>> a[0]
[1,2]

Answer 4

That pretty much means it's an array/list of stuff with foo.bar being the first item in the list/array.