>>>a = [1,2,3,4,5,6]
>>>print (len(a))
6

For one dimensional lists, the above method can be used. len(list_name) returns number of elements in the list.

>>>a = [[1,2,3],[4,5,6]]
>>>nrow = len(a)
>>>ncol = len(a[0])
>>>nrow
2
>>>ncol
3

The above gives the dimension of the list. len(a) returns number of rows. len(a[0]) returns number of rows in a[0] which is the number of columns.

Here's a link to original answer.

By selecting the 0th column in particular, as you've noticed, you reduce the dimensionality:

>>> m = np.random.normal(0, 1, size=(5, 2))
>>> m[:,0].shape
(5,)

You have a lot of options to get a 5x1 object back out. You can index using a list, rather than an integer:

>>> m[:, [0]].shape
(5, 1)

You can ask for "all the columns up to but not including 1":

>>> m[:,:1].shape
(5, 1)

Or you can use None (or np.newaxis), which is a general trick to extend the dimensions:

>>> m[:,0,None].shape
(5, 1)
>>> m[:,0][:,None].shape
(5, 1)
>>> m[:,0, None, None].shape
(5, 1, 1)

Finally, you can reshape:

>>> m[:,0].reshape(5,1).shape
(5, 1)

but I'd use one of the other methods for a case like this.

If I want the shape of each array in the list, do I have to resort to list comprehension or is there a faster way to do this?

List comprehension.

Does numpy simply iterate through the list, checking to make sure that each element is an array of the same size as the previous one, and decide based on that whether to return a tuple or throw an error?

NumPy calls asarray on the list, building an entire array just to get the shape. (This is not something that anyone has bothered to optimize.)