It represents an infinite loop within the structure. An example:

In [1]: l = [1, 2]In [2]: l[0] = lIn [3]: lOut[3]: [[...], 2]

l's first item is itself. It's a recursive reference, and so python can't reasonably display its contents. Instead it shows [...]

Depending on the context here it could different things:

indexing/slicing with Ellipsis

I think it's not implemented for any python class but it should represent an arbitary number of data structure nestings (as much needed). So for example: a[..., 1] should return all the second elements of the innermost nested structure:

>>> import numpy as np>>> a = np.arange(27).reshape(3,3,3)  # 3dimensional array>>> a[..., 1]  # this returns a slice through the array in the third dimensionarray([[ 1,  4,  7],       [10, 13, 16],       [19, 22, 25]])>>> a[0, ...]  # This returns a slice through the first dimensionarray([[0, 1, 2],       [3, 4, 5],       [6, 7, 8]])

and to check for this ... you compare it to an Ellipsis (this is a singleton so recommended is using is:

>>> ... is EllipsisTrue>>> Ellipsis in [...]True# Another (more or less) equivalent alternative to the previous line:>>> any(i is Ellipsis for i in [1, ..., 2]) True

Recursive Datastructures

The other case in which you see an [...] in your output is if you have the sequence inside the sequence itself. Here it stands for an infinite deeply nested sequence (that's not printable). For example:

>>> alist = ['a', 'b', 'c']>>> alist[0] = alist>>> alist[[...], 'b', 'c']# Infinite deeply nested so you can use as many leading [0] as you want>>> alist[0][1] 'b'>>> alist[0][0][0][0][0][1] 'b'>>> alist[0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][1] 'b'

You can even replace it several times:

>>> alist[2] = alist>>> alist[[...], 'b', [...]]>>> alist[1] = alist>>> alist[[...], [...], [...]]

To test if you have any such recursion in your output you need to check if the data-structure itself is also one of the elements:

>>> alist in alistTrue>>> any(i is alist for i in alist)True

Another way to get a more meaningful output is using pprint.pprint:

>>> import pprint>>> pprint.pprint(alist)  # Assuming you only replaced the first element:[<Recursion on list with id=1628861250120>, 'b', 'c']

If your list contains self references Python will display that as [...] rather than trying to recursively print it out, which would lead to an infinte loop:

>>> l = [1, 2, 3]>>> print(l)[1, 2, 3]>>> l.append(l)>>> print(l)[1, 2, 3, [...]]>>> print(l[-1])        # print the last item of list l[1, 2, 3, [...]]>>> print(l[-1][-1])    # print the last item of the last item of list l[1, 2, 3, [...]]

ad infinitum.

A similar situation arises with dictionaries:

>>> d = {}>>> d['key'] = d>>> print(d){'key': {...}}>>> d['key']{'key': {...}}>>> d['key']['key']{'key': {...}}