Get second minimum values per column in 2D array

Try this, in just one line:

[sorted(i)[1] for i in zip(*A)]

in action:

In [12]: A = [[72, 76, 44, 62, 81, 31],     ...:      [54 ,36 ,82 ,71 ,40, 45],     ...:      [63 ,59, 84, 36, 34 ,51],     ...:      [58, 53, 59, 22, 77 ,64],     ...:      [35 ,77, 60, 76, 57, 44]] In [18]: [sorted(i)[1] for i in zip(*A)]                                                                                                                                                                           Out[18]: [54, 53, 59, 36, 40, 44]

zip(*A) will transpose your list of list so the columns become rows.

and if you have duplicate value, for example:

In [19]: A = [[72, 76, 44, 62, 81, 31],     ...:  [54 ,36 ,82 ,71 ,40, 45],     ...:  [63 ,59, 84, 36, 34 ,51],     ...:  [35, 53, 59, 22, 77 ,64],   # 35    ...:  [35 ,77, 50, 76, 57, 44],]  # 35

If you need to skip both 35s, you can use set():

In [29]: [sorted(list(set(i)))[1] for i in zip(*A)]                                                                                                                                                                Out[29]: [54, 53, 50, 36, 40, 44]

Operations on numpy arrays should be done with numpy functions, so look at this one:

np.sort(A, axis=0)[1, :]

Out[61]: array([54, 53, 59, 36, 40, 44])

you can use heapq.nsmallest

from heapq import nsmallest[nsmallest(2, e)[-1] for e in zip(*A)]

output:

[54, 53, 50, 36, 40, 44]

I added a simple benchmark to compare the performance of the different solutions already posted:

from simple_benchmark import BenchmarkBuilderfrom heapq import nsmallestb = BenchmarkBuilder()@b.add_function()def MehrdadPedramfar(A):    return [sorted(i)[1] for i in zip(*A)]@b.add_function()def NicolasGervais(A):    return np.sort(A, axis=0)[1, :]@b.add_function()def imcrazeegamerr(A):    rotated = zip(*A[::-1])    result = []    for arr in rotated:        # sort each 1d array from min to max        arr = sorted(list(arr))        # add the second minimum value to result array        result.append(arr[1])    return result@b.add_function()def Daweo(A):    return np.apply_along_axis(lambda x:heapq.nsmallest(2,x)[-1], 0, A)@b.add_function()       def kederrac(A):    return [nsmallest(2, e)[-1] for e in zip(*A)]@b.add_arguments('Number of row/cols (A is  square matrix)')def argument_provider():    for exp in range(2, 18):        size = 2**exp        yield size, [[randint(0, 1000) for _ in range(size)] for _ in range(size)]r = b.run()r.plot()

Using zip with sorted function is the fastest solution for small 2d lists while using zip with heapq.nsmallest shows to be the best on big 2d lists