How does your "2D array" look like?

For example:

>>> a = [
     [12, 18, 6, 3], 
     [ 4,  3, 1, 2], 
     [15,  8, 9, 6]
]
>>> a.sort(key=lambda x: x[1])
>>> a
[[4,  3,  1, 2], 
 [15, 8,  9, 6], 
 [12, 18, 6, 3]]

But I guess you want something like this:

>>> a = [
     [12, 18, 6, 3], 
     [ 4,  3, 1, 2], 
     [15,  8, 9, 6]
]
>>> a = zip(*a)
>>> a.sort(key=lambda x: x[1])
>>> a
[(6,  1,  9), 
 (3,  2,  6), 
 (18, 3,  8), 
 (12, 4, 15)]
>>> a = zip(*a)
>>> a
[(6, 3, 18, 12), 
 (1, 2,  3,  4), 
 (9, 6,  8, 15)
]

Yes. The sorted built-in accepts a key argument:

sorted(li,key=lambda x: x[1])
Out[31]: [['Jason', 1], ['John', 2], ['Jim', 9]]

note that sorted returns a new list. If you want to sort in-place, use the .sort method of your list (which also, conveniently, accepts a key argument).

or alternatively,

from operator import itemgetter
sorted(li,key=itemgetter(1))
Out[33]: [['Jason', 1], ['John', 2], ['Jim', 9]]

Read more on the python wiki.

Regular sorted does it:

print(sorted([[2,3,1,8], [4,7,5,20], [0,-2,2,0]]))

But if you only want to sort by the first columns, use:

print(sorted([[2,3,1,8], [4,7,5,20], [0,-2,2,0]], key=lambda x: x[0]))

They both output:

[[0, -2, 2, 0], [2, 3, 1, 8], [4, 7, 5, 20]]

Source array:

In [215]: a
Out[215]:
array([[4, 8, 2, 0, 5],
       [3, 1, 6, 8, 1],
       [2, 2, 6, 0, 3],
       [9, 7, 6, 7, 8],
       [5, 8, 1, 1, 4]], dtype=int64)

Using Numpy indexing:

In [218]: a[:, a[0].argsort()]
Out[218]:
array([[0, 2, 4, 5, 8],
       [8, 6, 3, 1, 1],
       [0, 6, 2, 3, 2],
       [7, 6, 9, 8, 7],
       [1, 1, 5, 4, 8]], dtype=int64)

Using Pandas:

In [212]: pd.DataFrame(a).sort_values(0, axis=1).values
Out[212]:
array([[0, 2, 4, 5, 8],
       [8, 6, 3, 1, 1],
       [0, 6, 2, 3, 2],
       [7, 6, 9, 8, 7],
       [1, 1, 5, 4, 8]], dtype=int64)

A first step could be to create a nested list, adding every 2 elements to a new sublist:

from itertools import chain
from operator import itemgetter

i = [([0.2,0.10]),0.69, ([0.3,0.67]),0.70, ([0.5,0.68]),0.70, ([0.3,0.67]),0.65]

l = [i[x:x+2] for x in range(0, len(i),2)]
# [[[0.2, 0.1], 0.69], [[0.3, 0.67], 0.7], [[0.5, 0.68], 0.7], [[0.3, 0.67], 0.65]]

And then sort the nested list by the second element in each sublist with operator.itemgetter, and use itertools.chain to flatten the result:

list(chain(*sorted(l, key = itemgetter(1), reverse=True)))

[[0.3, 0.67], 0.7, [0.5, 0.68], 0.7, [0.2, 0.1], 0.69, [0.3, 0.67], 0.65]

Use .argsort() it returns an numpy.array of indices that sort the given numpy.array. You call it as a function or as a method on your array. For example, suppose you have

import numpy as np

arr = np.array([[-0.30565392, -0.96605562],
                [ 0.85331367, -2.62963495],
                [ 0.87839643, -0.28283675],
                [ 0.72676698,  0.93213482],
                [-0.52007354,  0.27752806],
                [-0.08701666,  0.22764316],
                [-1.78897817,  0.50737573],
                [ 0.62260038, -1.96012161],
                [-1.98231706,  0.36523876],
                [-1.07587382, -2.3022289 ]])

You can now call .argsort() on the column you want to sort, and it will give you an array of row indices that sort that particular column which you can pass as an index to your original array.

>>> arr[arr[:, 1].argsort()]
array([[ 0.85331367, -2.62963495],
       [-1.07587382, -2.3022289 ],
       [ 0.62260038, -1.96012161],
       [-0.30565392, -0.96605562],
       [ 0.87839643, -0.28283675],
       [-0.08701666,  0.22764316],
       [-0.52007354,  0.27752806],
       [-1.98231706,  0.36523876],
       [-1.78897817,  0.50737573],
       [ 0.72676698,  0.93213482]])

You can equivalently use numpy.argsort()

>>> arr[np.argsort(arr[:, 1])]
array([[ 0.85331367, -2.62963495],
       [-1.07587382, -2.3022289 ],
       [ 0.62260038, -1.96012161],
       [-0.30565392, -0.96605562],
       [ 0.87839643, -0.28283675],
       [-0.08701666,  0.22764316],
       [-0.52007354,  0.27752806],
       [-1.98231706,  0.36523876],
       [-1.78897817,  0.50737573],
       [ 0.72676698,  0.93213482]])