How can I get the second minimum value from each column? I have this array:
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]]
I wish to have output like:
A = [54 53 59 36 40 44]
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 BenchmarkBuilder
from heapq import nsmallest
b = 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
I hope I understood your question correctly but either way here's my solution, im sure there is a more elegent way of doing this but it works
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]
,[35,77,50,76,57,44]]
#rotate the array 90deg
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])
print(result)
Assuming that A is numpy.array (if this holds true please consider adding numpy tag to your question) then you might use apply_along_axis for that following way:
import heap
import numpy as np
A = np.array([[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]])
second_mins = np.apply_along_axis(lambda x:heapq.nsmallest(2,x)[-1], 0, A)
print(second_mins) # [54 53 59 36 40 44]
Note that I used heapq.nsmallest as it does as much sorting as required to get 2 smallest elements, unlike sorted which does complete sort.
>>> A = np.arange(30).reshape(5,6).tolist()
>>> A
[[0, 1, 2, 3, 4, 5],
[6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17],
[18, 19, 20, 21, 22, 23],
[24, 25, 26, 27, 28, 29]]
Updated:
Use set to prevent from duplicate and transpose list using zip(*A)
>>> [sorted(set(items))[1] for items in zip(*A)]
[6, 7, 8, 9, 10, 11]
old: second minimum item in each row
>>> [sorted(set(items))[1] for items in A]
[1, 7, 13, 19, 25]
