I have this 2d numpy array: [[0,2],[0,3],[0,1],[1,0]]. These elements represent points in x and y axes. So, for example, in [0,2] 0 is the x and 2 is the y.
I would like to get the point with the min x and min y. In this case, it would be [0,1]. I know I can easily do this with a for loop, but it takes too many resources. Is there a numpy function or something to calculate this?
You can first find the min of x, then find the min of y for pair where x equals the min of x:
np.min(xy[xy[:, 0] == np.min(xy[:, 0])], 0)
In [1]: xy
Out[1]:
array([[0, 2],
[0, 3],
[0, 1],
[1, 0]])
In [2]: np.min(xy[xy[:, 0] == np.min(xy[:, 0])], 0)
Out[2]: array([0, 1])
It is relatively fast, even for large array:
In [4]: timeit.timeit('np.min(xy[xy[:, 0] == np.min(xy[:, 0])], 0)',
'import numpy as np; xy = np.random.rand(100000, 2)',
number = 100)
Out[4]: 0.06858562525758316
Old version which was a bit slower (before @OliverW. comment):
In [3]: timeit.timeit('np.min(xy[xy[:, 0] == np.min(xy, 0)[0]], 0)',
'import numpy as np; xy = np.random.rand(100000, 2)',
number = 100)
Out[3]: 0.20433111706540785
np.min(xy, 0)[0] into np.min(xy[:,0]). That way you won't have to find the minimum of the "x-column" too, which is wasteful.
You can use the NumPy built-in np.lexsort to do such a sorting, like so -
xy[np.lexsort(xy.T[::-1])[0]]
lexsort gets the sorted indices that respects such a precedence, but does so in the opposite sense, i.e. the last element, then the second last element and so on until the first element. So, we need to reverse the order of elements, that's why the appended [::-1]. Also, lexsort works along each column instead of each row as needed to solve our case, so we needed that transpose before reversing.
Sample run -
In [235]: xy
Out[235]:
array([[0, 2],
[0, 3],
[0, 1],
[1, 0]])
In [236]: xy[np.lexsort(xy.T[::-1])[0]]
Out[236]: array([0, 1])
