Python check if a list is nested or not

Question

I have a list, sometimes it is nested, sometimes it is not. Based whether it is nested, the continuation is different. How do I check if this list is nested? True or False should be output.

example:

[1,2,3] --> False

[[1],[2],[3]] --> True

Answer 1

You can use isinstance and a generator expression combined with any. This will check for instances of a list object within your original, outer list.

In [11]: a = [1, 2, 3]  In [12]: b = [[1], [2], [3]]  In [13]: any(isinstance(i, list) for i in a) Out[13]: False  In [14]: any(isinstance(i, list) for i in b) Out[14]: True 

Note that any will return True as soon as it reaches an element that is valid (in this case if the element is a list) so you don't end up iterating over the whole outer list unnecessarily.

Answer 2

We want to check if elements inside outer-list is an instance of list or not, like @Ffisegydd said we can use a generator expression to build a generator and iterate over it using next(), If any element inside the outer-loop is an instance of list then calling next() on the generator will work otherwise if none of the element inside outer-loops belongs to an instance of list then calling next will raise StopIteration

Best case: If it's a nested loop (we can stop iterating as soon as we see the first instance of list)

Worst case: If it's not a nested loop (We need to iterate over all the elements inside the outerlist)

def is_nested_list(l):
    
    try:
          next(x for x in l if isinstance(x,list))
    
    except StopIteration:
        return False
    
    return True