sorted(Data, key=lambda row: row[1]) should do it.

@steve's answer is actually the most elegant way of doing it.

For the "correct" way see the order keyword argument of numpy.ndarray.sort

However, you'll need to view your array as an array with fields (a structured array).

The "correct" way is quite ugly if you didn't initially define your array with fields...

As a quick example, to sort it and return a copy:

In [1]: import numpy as np

In [2]: a = np.array([[1,2,3],[4,5,6],[0,0,1]])

In [3]: np.sort(a.view('i8,i8,i8'), order=['f1'], axis=0).view(np.int)
Out[3]: 
array([[0, 0, 1],
       [1, 2, 3],
       [4, 5, 6]])

To sort it in-place:

In [6]: a.view('i8,i8,i8').sort(order=['f1'], axis=0) #<-- returns None

In [7]: a
Out[7]: 
array([[0, 0, 1],
       [1, 2, 3],
       [4, 5, 6]])

@Steve's really is the most elegant way to do it, as far as I know...

The only advantage to this method is that the "order" argument is a list of the fields to order the search by. For example, you can sort by the second column, then the third column, then the first column by supplying order=['f1','f2','f0'].

The title says "sorting 2D arrays". Although the questioner uses an (N,2)-shaped array, it's possible to generalize unutbu's solution to work with any (N,M) array, as that's what people might actually be looking for.

One could transpose the array and use slice notation with negative step to pass all the columns to lexsort in reversed order:

>>> import numpy as np
>>> a = np.random.randint(1, 6, (10, 3))
>>> a
array([[4, 2, 3],
       [4, 2, 5],
       [3, 5, 5],
       [1, 5, 5],
       [3, 2, 1],
       [5, 2, 2],
       [3, 2, 3],
       [4, 3, 4],
       [3, 4, 1],
       [5, 3, 4]])

>>> a[np.lexsort(np.transpose(a)[::-1])]
array([[1, 5, 5],
       [3, 2, 1],
       [3, 2, 3],
       [3, 4, 1],
       [3, 5, 5],
       [4, 2, 3],
       [4, 2, 5],
       [4, 3, 4],
       [5, 2, 2],
       [5, 3, 4]])

Python, per se, has no "2d array" -- it has (1d) lists as built-ins, and (1d) arrays in standard library module array. There are third-party libraries such as numpy which do provide Python-usable multi-dimensional arrays, but of course you'd be mentioning such third party libraries if you were using some of them, rather than just saying "in Python", right?-)

So I'll assume that by "2d array" you mean a list of lists, such as:

lol = [ range(10), range(2, 12), range(5, 15) ]

or the like -- i.e. a list with 3 items, each item being a list with 10 items, and the "second row" would be the sublist item lol[1]. Yeah, lots of assumptions, but your question is so maddeningly vague that there's no way to avoid making assumptions - edit your Q to clarify with more precision, and an example!, if you dislike people trying to read your mind (and probably failing) as you currently make it impossible to avoid.

So under these assumptions you can sort each of the 3 sublists in the order required to sort the second one, for example:

indices = range(10)
indices.sort(key = lol[1].__getitem__)
for i, sublist in enumerate(lol):
  lol[i] = [sublist[j] for j in indices]

The general approach here is to sort the range of indices, then just use that appropriately sorted range to reorder all the sublists in play.

If you actually have a different problem, there will of course be different solutions;-).