Sort NumPy float array column by column

Question

Following this trick to grab unique entries for a NumPy array, I now have a two-column array, basically of pairs with first element in the range [0.9:0.02:1.1] and the second element in the range [1.5:0.1:2.0]. Let's call this A. Currently, it's completely unsorted, i.e.

In [111]: A
Out[111]: 
array([[ 1.1 ,  1.9 ],
       [ 1.06,  1.9 ],
       [ 1.08,  1.9 ],
       [ 1.08,  1.6 ],
       [ 0.9 ,  1.8 ],
       ...
       [ 1.04,  1.6 ],
       [ 0.96,  2.  ],
       [ 0.94,  2.  ],
       [ 0.98,  1.9 ]])

I'd like to sort it so that each row first increases in the second column, then the first. i.e.

array([[ 0.9 ,  1.5 ],
       [ 0.9 ,  1.6 ],
       [ 0.9 ,  1.7 ],
       [ 0.9 ,  1.9 ],
       [ 0.9 ,  1.9 ],
       [ 0.9 ,  2.  ],
       [ 0.92,  1.5 ],
       ...
       [ 1.08,  2.  ],
       [ 1.1 ,  1.5 ],
       [ 1.1 ,  1.6 ],
       [ 1.1 ,  1.7 ],
       [ 1.1 ,  1.8 ],
       [ 1.1 ,  1.9 ],
       [ 1.1 ,  2.  ]])

but I can't find a sort algorithm that gives both. As suggested here, I've tried A[A[:,0].argsort()] and A[A[:,1].argsort()], but they only sort one column each. I've also tried applying both but the same thing happens.

I apologize if I've missed something simple but I've been looking for this for a while now...

Answer 1

numpy.lexsort will work here:

A[np.lexsort(A.T)]

You need to transpose A before passing it to lexsort because when passed a 2d array it expects to sort by rows (last row, second last row, etc).

The alternative possibly slightly clearer way is to pass the columns explicitly:

A[np.lexsort((A[:, 0], A[:, 1]))]

You still need to remember that lexsort sorts by the last key first (there's probably some good reason for this; it's the same as performing a stable sort on successive keys).

Answer 2

the following will work, but there might be a faster way:

A = np.array(sorted(A,key=tuple))